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onent, 10 is base 2 = x Evaluate, remember negative exponent is fraction − log x = 10− 1 100 = x Our Solution This lesson has introduced the idea of logarithms, changing between logs and exponents, evaluating logarithms, and solving basic logarithmic equations. In an advanced algebra course logarithms will be studied in much greater detail. 412 10.5 Practice - Logarithmic Functions Rewrite each equation in exponential form. 1) log9 81 = 2 3) log7 1 49 = 2 − 5) log13 169 = 2 2) logb a = 16 − 4) log16 256 = 2 6) log11 1 = 0 Rewrite each equations in logarithmic form. 7) 80 = 1 9) 152 = 225 1 11) 64 6 = 2 Evaluate each expression. 13) log125 5 15) log343 1 7 17) log4 16 19) log6 36 21) log2 64 Solve each equation. 23) log5 x = 1 25) log2 x = − 27) log11 k = 2 2 29) log9 (n + 9) = 4 31) log5 ( 3m) = 3 − 33) log11 (x + 5) = − 35) log4 (6b + 4) = 0 1 37) log5 ( − 39) log2 (10 10x + 4) = 4 5a) = 3 − 8) 17− 2 = 1 289 1 10) 144 2 = 12 12) 192 = 361 14) log5 125 16) log7 1 18) log4 1 64 20) log36 6 22) log3 243 24) log8 k = 3 26) log n = 3 28) log4 p = 4 − 4) = − 8r = 1 30) log11 (x 32) log2 − 34) log7 − 36) log11 (10v + 1) = 3n = 4 38) log9 (7 − 40) log8 (3k 6x) = − 1) = 1 − 1 1 − 2 413 10.6 Functions - Compound Interest Objective: Calculate ﬁnal account balances using the formulas for compound and continuous interest. An application of exponential functions is compound interest. When money is invested in an account (or given out on loan) a certain amount is added to the balance. This money added to the balance is called interest. Once that interest is added to the balance, it will earn more interest during the next compounding period. This idea of earning interest on interest is called compound interest. For example, if you invest S100 at 10% interest compounded annually, after one year you will earn S10 in interest, giving you a new balance of S110. The next year you will earn another 10% or S11, giving you a new balance of S121. The third year you will earn another 10% or S12.10, giving you a new balance of S133.10. This pattern will continue each year until you close the account. There are several ways interest can be paid. The ﬁrst way, as described above, is compounded annually. In this model the interest is paid once per year. But interest can be compounded more often. Some common compounds include compounded semi-annually (twice per year), quarterly (four times per year, such as quarterly taxes), monthly (12 times per year, such as a savings account), weekly (52 times per year), or even daily (365 times per year, such as some student loans). When interest is compounded in any of these ways we can calculate the balance after any amount of time using the following formula: Compound Interest Formula: A = P 1 + nt r n A = Final Amount P = Principle (starting balance) r = Interest rate (as a decimal) n = number of compounds per year t = time (in years) Example 544. If you take a car loan for S25000 with an interest rate of 6.5% compounded quarterly, no payments required for the ﬁrst ﬁve years, what will your balance be at the end of those ﬁve years? P = 25000, r = 0.065, n = 4, t = 5 5 4 A = 25000 1 + · 0.065 4 A = 25000(1.01625)4 Identify each variable Plug each value into formula, evaluate parenthesis 5 Multiply exponents · 414 A = 25000(1.01625)20 Evaluate exponent A = 25000(1.38041977 ) Multiply A = 34510.49 S34, 510.49 Our Solution We can also ﬁnd a missing part of the equation by using our techniques for solving equations. Example 545. What principle will amount to S3000 if invested at 6.5% compounded weekly for 4 years? A = 3000, r = 0.065, n = 52, t = 4 4 52 3000 = P 1 + 0.065 52 · Identify each variable Evaluate parentheses 3000 = P (1.00125)52 4 Multiply exponent 3000 = P (1.00125)208 Evaluate exponent · 3000 = P (1.296719528 ) Divide each side by 1.296719528 1.296719528 1.296719528 2313.53 = P Solution for P S2313.53 Our Solution It is interesting to compare equal investments that are made at several diﬀerent types of compounds. The next few examples do just that. Example 546. If S4000 is invested in an account paying 3% interest compounded monthly, what is the balance after 7 years? P = 4000, r = 0.03, n = 12, t = 7 7 12 A = 4000 1 + · 0.03 12 Identify each variable Plug each value into formula, evaluate parentheses A = 4000(1.0025)12 7 Multiply exponents A = 4000(1.0025)84 Evaluate exponent · A = 4000(1.2333548) Multiply A = 4933.42 S4933.42 Our Solution 415 To investigate what happens to the balance if the compounds happen more often, we will consider the same problem, this time with interest compounded daily. Example 547. If S4000 is invested in an account paying 3% interest compounded daily, what is the balance after 7 years? P = 4000, r = 0.03, n = 365, t = 7 7 365 A = 4000 1 + · 0.03 365 Identify each variable Plug each value into formula, evaluate parenthesis A = 4000(1.00008219 )365 7 Multiply exponent A = 4000(1.00008219 )2555 Evaluate exponent · A = 4000(1.23366741 .) Multiply A = 4934.67 S4934.67 Our Solution While this diﬀerence is not very large, it is a bit higher. The table below shows the result for the same problem with diﬀerent compounds. Compound Annually Balance S4919.50 Semi-Annually S4927.02 S4930.85 S4933.42 S4934.41 S4934.67 Quarterly Monthly Weekly Daily As the table illustrates, the more often interest is compounded, the higher the ﬁnal balance will be. The reason is, because we are calculating compound interest or interest on interest. So once interest is paid into the account it will start earning interest for the next compound and thus giving a higher ﬁnal balance. The next question one might consider is what is the maximum number of compounds possible? We actually have a way to calculate interest compounded an inﬁnite number of times a year. This is when the interest is compounded continuously. When we see the word “continuously” we will know that we cannot use the ﬁrst formula. Instead we will use the following formula: Interest Compounded Continuously: A = Pert A = Final Amount P = Principle (starting balance) e = a constant approximately 2.71828183 . r = Interest rate (written as a decimal) t = time (years) 416 The variable e is a constant similar in idea to pi (π) in that it goes on forever without repeat or pattern, but just as pi (π) naturally occurs in several geometry applications, so does e appear in many exponential applications, continuous interest being one of them. If you have a scientiﬁc calculator you probably have an e button (often using the 2nd or shift key, then hit ln) that will be useful in calculating interest compounded continuously. World View Note: e ﬁrst appeared in 1618 in Scottish mathematician’s Napier’s work on logarithms. However it was Euler in Switzerland who used the letter e ﬁrst to represent this value. Some say he used e because his name begins with E. Others, say it is because exponent starts with e. Others say it is because Euler’s work already had the letter a in use, so e would be the next value. Whatever the reason, ever since he used it in 1731, e became the natural base. Example 548. If S4000 is invested in an account paying 3% interest compounded continuously, what is the balance after 7 years? P = 4000, r = 0.03, t = 7 Identify each of the variables A = 4000e0.03 · 7 Multiply exponent A = 4000e0.21 Evaluate e0.21 A = 4000(1.23367806 ) Multiply A = 4934.71 S4934.71 Our Solution Albert Einstein once said that the most powerful force in the universe is compound interest. Consider the following example, illustrating how powerful compound interest can be. Example 549. If you invest S6.16 in an account paying 12% interest compounded continuously for 100 years, and that is all you have to leave your children as an inheritance, what will the ﬁnal balance be that they will receive? P = 6.16, r = 0.12, t = 100 Identify each of the variables A = 6.16e0.12 · 100 Multiply exponent A = 6.16e12 Evaluate A = 6.16(162, 544.79) Multiply A = 1, 002, 569.52 S1, 002, 569.52 Our Solution In 100 years that one time investment of S6.16 investment grew to over one million dollars! That’s the power of compound interest! 417 10.6 Practice - Compound Interest Solve 1) Find each of the following: a. S500 invested at 4% compounded annually for 10 years. b. S600 invested at 6% compounded annually for 6 years. c. S750 invested at 3% compounded annually for 8 years. d. S1500 invested at 4% compounded semiannually for 7 years. e. S900 invested at 6% compounded semiannually for 5 years. f. S950 invested at 4% compounded semiannually for 12 years. g. S2000 invested at 5% compounded quarterly for 6 years. h. S2250 invested at 4% compounded quarterly for 9 years. i. S3500 invested at 6% compounded quarterly for 12 years. 418 j. All of the above compounded continuously. 2) What principal will amount to S2000 if invested at 4% interest compounded semiannually for 5 years? 3) What principal will amount to S3500 if invested at 4% interest compounded quarterly for 5 years? 4) What principal will amount to S3000 if invested at 3% interest compounded semiannually for 10 years? 5) What principal will amount to S2500 if invested at 5% interest compounded semiannually for 7.5 years? 6) What principal will amount to S1750 if invested at 3% interest compounded quarterly for 5 years? 7) A thousand dollars is left in a bank savings account drawing 7% interest, compounded quarterly for 10 years. What is the balance at the end of that time? 8) A thousand dollars is left in a credit union drawing 7% compounded monthly. What is the balance at the end of 10 years? 9) S1750 is invested in an account earning 13.5% interest compounded monthly for a 2 year period. What is the balance at the end of 9 years? 10) You lend out S5500 at 10% compounded monthly. If the debt is repaid in 18 months, what is the total owed at the time of repayment? 11) A S10, 000 Treasury Bill earned 16% compounded monthly. If the bill matured in 2 years, what was it worth at maturity? 12) You borrow S25000 at 12.2
5% interest compounded monthly. If you are unable to make any payments the ﬁrst year, how much do you owe, excluding penalties? 13) A savings institution advertises 7% annual interest, compounded daily, How much more interest would you earn over the bank savings account or credit union in problems 7 and 8? 14) An 8.5% account earns continuous interest. If S2500 is deposited for 5 years, what is the total accumulated? 15) You lend S100 at 10% continuous interest. If you are repaid 2 months later, what is owed? 419 10.7 Functions - Trigonometric Functions Objective: Solve for a missing side of a right triangle using trigonometric ratios. There are six special functions that describe the relationship between the sides of a right triangle and the angles of the triangle. We will discuss three of the functions here. The three functions are called the sine, cosine, and tangent (the three others are cosecant, secant, and cotangent, but we will not need to use them here). To the right is a picture of a right triangle. Based on which angle we are interested in on a given problem we will name the three sides in relationship to that angle. In the picture, angle A is the angle we will use to name the other sides. The longest side, the side opposite the right angle is always called the hypotenouse. The side across from the angle A is called the opposite side. Opposite Hypotenuse Adjacent A The third side, the side between our angle and the right angle is called the adjacent side. It is important to remember that the opposite and adjacent sides are named in relationship to the angle A or the angle we are using in a problem. If the angle had been the top angle, the opposite and adjacent sides would have been switched. The three trigonometric funtions are functions taken of angles. When an angle goes into the function, the output is a ratio of two of the triangle sides. The ratios are as describe below: sinθ = opposite hypotenuse cosθ = adjacent hypotenuse tanθ = opposite adjacent The “weird” variable θ is a greek letter, pronounced “theta” and is close in idea to our letter “t”. Often working with triangles, the angles are repesented with Greek letters, in honor of the Ancient Greeks who developed much of Geometry. Some students remember the three ratios by remembering the word “SOH CAH TOA” where each letter is the ﬁrst word of: “Sine: Opposite over Hypotenuse; Cosine: Adjacent over Hypotenuse; and Tangent: Opposite over Adjacent.” Knowing how to use each of these relationships is fundamental to solving problems using trigonometry. World View Note: The word “sine” comes from a mistranslation of the Arab word jayb Example 550. 420 Using the diagram at right, ﬁnd each of the following: sinθ, cosθ, tanθ, sinα, cosα, and tanα. First we will ﬁnd the three ratios of θ. from θ, the The hypotenuse is 10, opposite side is 6 and the adjacent side is 8. So we ﬁll in the following: sinθ = opposite hypotenuse = 6 10 = 3 5 5 adjacent = 6 tanθ = opposite cosθ = adjacent hypotenuse = 8 10 = 4 8 = 3 Now we will ﬁnd the three ratios of α. from α, the The hypotenuse is 10, opposite side is 8 and the adjacent side is 6. So we ﬁll in the following: 4 8 θ 6 α 10 6 Opposite of θ Adjacent of α α Adjacent of θ Opposite of α 8 θ 10 Hypotenuse sinα = opposite cosα = adjacent 5 10 = 4 hypotenuse = 8 hypotenuse = 6 10 = 3 6 = 4 adjacent = 8 5 3 tanα = opposite We can either use a trigonometry table or a calculator to ﬁnd decimal values for sine, cosine, or tangent of any angle. We only put angle values into the trigonometric functions, never values for sides. Using either a table or a calculator, we can solve the next example. Example 551. sin 42◦ Use calculator or table 0.669 Our Solution tan 12◦ Use calculator or table 0.213 Our Solution cos 18◦ Use calculator or table 0.951 Our Solution By combining the ratios together with the decimal approximations the calculator or table gives us, we can solve for missing sides of a triangle. The trick will be to determine which angle we are working with, naming the sides we are working with, and deciding which trig function can be used with the sides we have. 421 Example 552. Find the measure of the missing side. x 25◦ 4 We will be using the angle marked 25◦, from this angle, the side marked 4 is the opposite side and the side marked x is the adjacent side. The trig ratio that uses the opposite and adjacent sides is tangent. So we will take the tangent of our angle. tan25◦ = 0.466 1 = 4 x 4 x Tangent is opposite over adjacent Evaluate tan25◦, put over 1 so we have proportion 0.466x = 4 0.466 0.466 Divide both sides by 0.466 Find cross product x = 8.58 Our Solution Example 553. Find the measure of the missing side. We will be using the angle marked 70◦. From this angle, the x is the adjacent the hypotenuse. and the side is 9 The trig ratio that uses adjacent and hypotenuse is the cosine. So we will take the cosine of our angle. Cosine is adjacent over hypotenuse Evaluate cos70◦, put over 1 so we have a proportion x 70◦ 9 cos70◦ = 0.342 1 = x 9 x 9 3.08 = 1x 3.08 = x Our Solution. Find the cross product. 422 10.7 Practice - Trigonometric Functions Find the value of each. Round your answers to the nearest ten-thousandth. 1) cos 71◦ 3) sin 75◦ 2) cos 23◦ 4) sin 50◦ Find the value of the trig function indicated. 5) sin θ 6) tan θ 7) sin θ 24 θ 25 8) sin θ 7 5 θ 4 3 8 15 θ 17 9) sin θ 8 2√ 8 θ 8 3 23√ 7 θ 16 10) cos θ 25 θ 20 15 Find the measure of each side indicated. Round to the nearest tenth. 11) 12) x A C 13 51◦ B B 5 C 56◦ x A 423 13) 15) 17) 19) B 13 C x 24◦ A A x 71◦ C9 B C x 68◦ B A 6 B C6 71.4◦ x A x 52.3◦ B 5 C A A x B 63◦ 7.6 C C 6 52.2◦ B x A B 11 69◦ C x A 14) 16) 18) 20) 424 21) 23) 25) 27) A 5 38◦x B C B x 67◦ A 4 C B x C 67.2◦ 4 A B x 32◦ A 4 C C B x 28◦ 12 A B 12 48◦ C x A A 7 C 16◦ x B A x 13.1 B 64◦ C 22) 24) 26) 28) 425 29) A 2.4 22◦ 31) B 61◦ x 3 C x B A C B 33) 35) 11 A 30◦ x C x B C 75◦ 11 A x B 29◦ A 3.9 C B 12 53◦ A x C C x 47◦ B 3 A B 43◦ x A 3 C 30) 32) 34) 36) 426 37) 39) A C 11 37.1◦ x A B C 13.1 40◦ x B 38) A 40) C 1.4 65◦ B x C 18.1 A 35.5◦ x B 427 10.8 Functions - Inverse Trigonometric Functions Objective: Solve for missing angles of a right triangle using inverse trigonometry. We used a special function, one of the trig functions, to take an angle of a triangle and ﬁnd the side length. Here we will do the opposite, take the side lengths and ﬁnd the angle. Because this is the opposite operation, we will use the inverse function of each of the trig ratios we saw before. The notation we will use for the inverse trig functions will be similar to the inverse notation we used with functions. sin−1 opposite hypotenuse = θ cos−1 adjacent hypotenuse = θ tan−1 opposite adjacent = θ Just as with inverse functions, the 1 is not an exponent, it is a notation to tell us that these are inverse functions. While the regular trig functions take angles as inputs, these inverse functions will always take a ratio of sides as inputs. We can calculate inverse trig values using a table or a calculator (usually pressing shift or 2nd ﬁrst). − Example 554. sinA = 0.5 We don′t know the angle so we use an inverse trig function 1(0.5) = A Evaluate using table or calculator sin− 30◦ = A Our Solution cosB = 0.667 We don′t know the angle so we use an inverse trig function 1(0.667) = B Evaluate using table or calculator cos− 48◦ = B Our Solution tanC = 1.54 We don′t know the angle so we use an inverse trig function 1(1.54) = C Evaluate using table or calculator tan− 57◦ = C Our Solution If we have two sides of a triangle, we can easily calculate their ratio as a decimal and then use one of the inverse trig functions to ﬁnd a missing angle. Example 555. Find the indicated angle. 428 θ 17 12 From angle θ the given sides are the opposite (12) and the hypotenuse (17). The trig function that uses opposite and hypotenuse is the sine Because we are looking for an angle we use the inverse sine sin− 1 12 17 Sine is opposite over hyptenuse, use inverse to ﬁnd angle sin− 1(0.706) Evaluate fraction, take sine inverse using table or calculator 45◦ Our Solution Example 556. Find the indicated angle 5 3 α From the angle α, the given sides are the opposite (5) and the adjacent (3) The trig function that uses opposite and adjacent is the tangent As we are looking for an angle we will use the inverse tangent. tan− 1 5 3 Tangent is opposite over adjacent. Use inverse to ﬁnd angle tan− 1(1.667) Evaluate fraction, take tangent inverse on table or calculator 59◦ Our Solution Using a combination of trig functions and inverse trig functions, if we are given two parts of a right triangle (two sides or a side and an angle), we can ﬁnd all the other sides and angles of the triangle. This is called solving a triangle. When we are solving a triangle, we can use trig ratios to solve for all the missing parts of it, but there are some properties from geometry that may be helpful along the way. The angles of a triangle always add up to 180◦, because we have a right triangle, 90◦ are used up in the right angle, that means there are another 90◦ left in the two acute angles. In other words, the smaller two angles will always add to 90, if we know one angle, we can quickly ﬁnd the other by subtracting from 90. Another trick is on the sides of the angles. If we know two sides of the right triangle, we can use the Pythagorean Theorem to ﬁnd the third side. The Pythagorean Theorem states that if c is the hypotenuse of the triangle, and a and 429 b are the other two sides (legs), then we can use the following formula, a2 + b2 = c2 to ﬁnd a missing side. Often when solving triangles we use trigonometry to ﬁnd one part, then use the angle sum and/or the Pythagorean Theorem to ﬁnd the other two parts. Example 557. Solve the triangle 5 We have one angle and one side. We can use these to ﬁnd either other side. We will ﬁnd the other leg, the adjacent side to the 35◦ angle. 35◦ The 5 is the opposite side, so we will use the tangent to ﬁnd the leg. Tangent is oppos
ite over adjacent 5 tan35◦ = x 5 0.700 x 1 0.700x = 5 0.700 0.700 Divide both sides by 0.700 Find cross product = Evaluate tangent, put it over one so we have a proportion x = 7.1 The missing leg. a2 + b2 = c2 We can now use pythagorean thorem to ﬁnd hypotenuse, c 52 + 7.12 = c2 Evaluate exponents 25 + 50.41 = c2 Add 75.41 = c2 Square root both sides 8.7 = c The hypotenuse 90◦ − 35◦ To ﬁnd the missing angle we subtract from 90◦ 55◦ The missing angle 55◦ 5 8.7 35◦ 7.1 Our Solution In the previous example, once we found the leg to be 7.1 we could have used the sine function on the 35◦ angle to get the hypotenuse and then any inverse trig 430 function to ﬁnd the missing angle and we would have found the same answers. The angle sum and pythagorean theorem are just nice shortcuts to solve the problem quicker. Example 558. Solve the triangle In this triangle we have two sides. We will ﬁrst ﬁnd the angle on the right side, adjacent to 3 and opposite from the 9. 3 Tangent uses opposite and adjacent To ﬁnd an angle we use the inverse tangent. 9 tan− 1 tan− Evaluate fraction 9 3 1(3) Evaluate tangent 71.6◦ The angle on the right side 90◦ − Subtract angle from 90◦ to get other angle 71.6◦ 18.4◦ The angle on the left side a2 + b2 = c2 Pythagorean theorem to ﬁnd hypotenuse 92 + 32 = c2 Evaluate exponents 81 + 9 = c2 Add 90 = c2 3 10√ or 9.5 = c Square root both sides The hypotenuse 9.5 18.4◦ 9 71.6◦ 3 Our Solution World View Note: Ancient Babylonian astronomers kept detailed records of the starts, planets, and eclipses using trigonometric ratios as early as 1900 BC! 431 10.8 Practice - Inverse Trigonometric Functions Find each angle measure to the nearest degree. 1) sin Z = 0.4848 3) sin Y = 0.6561 2) sin Y = 0.6293 4) cos Y = 0.6157 Find the measure of the indicated angle to the nearest degree. 5) 7) 9) 35 ? 32 30 ? 31 3 ? 6 6) 8) 39 ? 46 24 ? 8 10) 16 23 ? 432 Find the measure of each angle indicated. Round to the nearest tenth. 11) B 8 A θ 8 C 13) 15) A θ 11 C B 11 B 7 θ A 4 C 17) A θ 10 C 16 B 12) 14) 16) 18) θ B 7 C 13 C 7 B θ 9.7 A A B 12 θ A 13 C θ 15.3 A B 5.6 C 433 19) 21) 23) 25) C 9.3 θ A 13.2 B B 5 C θ 4 A C 10 B B θ A 12 C 9 θ 15.7 A 20) A θ 7 B A B A θ 15 C 9 θ 6.8 2 C B 22) 24) 26) B 4 15 A θ 6 C C 434 θ B 14 C 6 A Solve each triangle. Round answers to the nearest tenth. 27) A θ 14 C 15 B 29) C 7 A 31) 33) C 28.4 B 62◦ A C 2.9 7 28) 30) B θ 15 B θ 7 32) 34) B 15 C 4 A A 6.3 C C 51◦ 9.3 B A B A 435 35) 7 B 3 C 37) B 16 39) A C 52◦ A A B 45◦ 8 C 36) A 7 38) C 21◦ B A 48◦ B 10.4 C 40) B 14 C 6.8 A 436 437 0.1 2 1) − 2) 5 3) 2 4) 2 6 5 5) − 6) − 7) 8 8) 0 9) 2 − 10) − 11) 4 12) − 13) 3 14) 15) 16) 17) 18) − − − − − 19) − 20 21) 7 − 0.2 1) 7 2 2) 5 4 3) 7 5 Answers - Chapter 0 Answers - Integers 20 43) − 44) 27 45) 24 − 3 46) − 47) 7 48) 3 49) 2 50) 5 51) 2 52) 9 53) 7 54) 10 − 55) 4 56) 10 8 57) − 58) 6 59) 60) 6 9 − − 22) 0 23) 11 24) 9 25) 26) − − 27) − 28) 4 29) 0 30) 31) 32) − − − 33) − 34) 14 35) 8 36) 6 3 4 3 8 4 35 80 56 6 36 10 37) 38) − − 39) − 40) 63 41) − 42) 4 Answers - Fractions 4) 8 3 5) 3 2 6) 5 4 438 7) 5 4 8) 4 3 9) 3 2 10) 8 3 11) 5 2 12) 8 7 13) 7 2 14) 4 3 15) 4 3 16) 3 2 17) 6 5 18) 7 6 19) 3 2 20) 8 7 21) 8 22) 5 3 4 9 2 3 13 4 23) 24) − − 25) − 26) 3 4 27) 33 20 28) 33 56 29) 4 30) 18 7 31) 1 2 32) 19 20 − 59) 37 20 60) − 61) 33 20 62) 3 7 63) 47 56 64) − 65) 2 3 66) − 67) 1 68) 7 8 69) 19 20 70) 71) − − 72) − 73) 34 7 74) 75) 76) − − − 77) − 78) 39 14 5 3 7 6 4 3 2 5 145 56 29 15 23 3 3 8 2 3 5 24 5 6 79) − 80) 1 10 81) 2 82) 62 21 17 15 7 10 8 7 21 26 3 2 5 27 1 10 45 7 1 10 7 33) 3 34) − 35) − 36) 5 14 37) − 38) 20 21 39) 2 9 40) 4 3 41) − 42) 25 21 43) − 44) − 45) 40 9 46) − 47) − 48) 13 15 49) 4 27 50) 32 65 51) 1 15 52) 1 53) − 54) − 55) 2 7 56) 2 57) 3 58) 31 8 − 439 0.3 1) 24 1 2) − 3) 5 4) 180 5) 4 6) 8 7) 1 8) 8 9) 6 0.4 1) 7 2) 29 3) 1 4) 3 5) 23 6) 14 7) 25 8) 46 9) 7 10) 8 11) 5 12) 10 13) 1 14) 6 15) 1 16) 2 17) 36 18) 54 Answers - Order of Operation 6 10 9 22 10) 11) − − 12) − 13) 20 14) − 15) 2 16) 28 17) 18) 40 15 − − Answers - Properties of Algebra 19) 7 20) 38 21) r + 1 4x 2 − 22) − 23) 2n 24) 11b + 7 25) 15v 26) 7x 27) 9x − 7a 1 − 28) − 29) k + 5 30) 31) 32) 33) − − − − 3p 5x − 9 m 9 − 10n 34) 5 r − 35) 10n + 3 − 19) 3 20) 0 21) 22) 23) 18 3 4 − − − 24) 3 25) 2 37) 8x + 32 − 38) 24v + 27 39) 8n2 + 72n 40) 5 9a − 7k2 + 42k 41) − 42) 10x + 20x2 6 − 2n 43) − 44) − 45) 40m 36x 2 − 8m2 − 46) 47) − − 18p2 + 2p 36x + 9x2 48) 32n 8 − 9b2 + 90b 49) 50) − − 28r 4 − 40n 51) − 52) 16x2 80n2 − 20x − 36) 5b 53) 14b + 90 440 54) 60v 7 − 3x + 8x2 89x + 81 68k2 19 − 8k − 90a 34 49p − 10x + 17 55) 56) 57) 58) 59) − − − − − 60) − 61) 10 4n − 30 + 9m 62) − 63) 12x + 60 64) 30r 16r2 74) 2x2 6x 8n2 75) 4p 48 − − 4b2 45 − − 72n 42b − 79v − 8x + 22 20n2 + 80n 42 − 12 + 57a + 54a2 65) − 66) − 67) 79 68) 69) 70) 71) − − − − 20k 75 − 128x 121 72) − 73) 4n2 − 3n − 5 − 3 − 7 − 76) 3x2 + 7x 77) 78) − − v2 + 2v + 2 7b2 + 3b 8 − 79) 4k2 + 12 − 80) a2 + 3a 81) 3x2 15 − n2 + 6 − 5 82) − − 1.1 1) 7 2) 11 5 3) − 4) 4 5) 10 6) 6 7) 19 − 6 8) − 9) 18 10) 6 11) 12) 20 7 − − 108 13) − 14) 5 1.2 1) 4 − Answers - Chapter 1 Answers to One-Step Equations 8 15) − 16) 4 17) 17 18) 4 19) 20 20) − 21) 3 22) 16 208 23) 13 − 9 24) − 25) 15 26) 8 27) 28) 10 204 − − 29) 5 30) 2 11 14 31) − 32) − 33) 14 34) 1 35) 36) 37) 38) 39) 40) 11 15 240 135 16 380 − − − − − − Answers to Two-Step Equations 2) 7 441 3) 14 − 2 4) − 5) 10 12 6) − 7) 0 8) 12 9) 10 − 16 10) − 11) 14 7 12) − 13) 4 5 14) − 15) 16 1.3 3 1) − 2) 6 3) 7 4) 0 5) 1 6) 3 7) 5 4 8) − 9) 0 10) 3 11) 1 12) All real numbers 13) 8 14) 1 15) 7 − 15 16) − 17) 7 18) 12 19) 9 20) 0 21) 11 22) 6 − 10 23) − 24) 13 25) 1 26) 4 9 27) − 28) 15 6 29) − 30) 6 31) 16 − 4 32) − 33) 8 34) 13 − 2 35) − 36) 10 12 37) − 38) 0 39) 12 40) 9 − Answers to General Linear Equations 16) 0 17) 2 18) − 19) − 20) 3 21) 3 22) 23) − − 24) − 25) 8 26) 0 27) − 28) 5 29) − 30 31) − 32) 0 33) − 34) 0 35) 0 36) 37) − − 38) − 39) 5 40) 6 41) 0 42) 2 − 43) No Solution 44) 0 442 45) 12 46) All real numbers 47) No Solution 48) 1 9 49) − 50) 0 1.4 1) 3 4 2) − 3) 6 5 4) 1 6 5) − 6) 25 8 7) 8) − − 9) − 10) 3 2 1.5 4 3 19 6 7 9 1 3 2 1. b = c a 2. h = gi 3. x = gb f 4. y = pq 3 5. x = a 3b 6. y = cb dm 7. m = E c2 8. D = ds S 9. π = 3V 4r3 10. m = 2E v2 Answers to Solving with Fractions 3 2 4 3 5 3 11) 0 12) 4 3 13) − 14) 1 2 15) − 16) 1 17) 0 18) − 19) 1 20) 1 21) 1 2 Answers - Formulas 11. c = b a − 12. x = g + f 13. y = cm + cn 4 3) 14. r = k(a − 5 15. D = 12V πn 16. k = F R − 17. n = P p − 18. L = S c L 2B − 19. D = TL + d 20. Ea = IR + Eg 21. Lo = L 1 + at 443 22) 23) 24 25) 16 26) 27) − − 28) − 29) 4 3 30) 3 2 22. x = c b − a 23. m = p q − 2 24. L = q + 6p 6 25. k = qr + m 26. T = R b − a 27. v = 16t2 + h t 28. h = s πr2 − πr 29. Q2 = Q1 + PQ1 P 30. r1 = L − 2d π − πr2 31. T1 = Rd kAT2 − kA 32. v2 = Pg + V1 V1 2 33. a = c b − x 34. r = d t 35. w = V ℓh 36. h = 3v πr2 1 1 − a − b 37. a = c 38. b = c 39. t = 5 + bw a s 40. w = at 41. x = c 42. x = 3 43. y = 3 − x − b bx 5y − x − 5 45. y = 7 3x − 2 46. a = 7b + 4 5 47. b = 5a 4 − 7 48. x = 8 + 5y 4 49. y = 4x 8 − 5 44. x = 7 2y − 3 50. f = 9c + 160 5 Answers to Absolute Value Equations 1.6 1) 8, 2) 7, 3) 1, 4) 2, 5) 6, 6) 38 9 7) − 8) − 9) 3 29 4 6 , − 2, − 3, 9 39 7 − , − 6 29 3 − 1 , 3 − 9, 15 10) 16 5 11) 7, 12) 13) 1.7 − − 5 3 − 2, 0 14) 3, 15) − 16) 0, 2 − 6 , 0 7 4, 4 3 , 7 2 17 2 , 6 5 6, 2 8 − − 25 3 − − 13 7 21 − 2, 10 7 5 , 1 21) − 22) 6, 23) 1, 24) 7, 17) 18) 19) 20) − − − − 10 3 1 25) 26) − − 1) c 2) x a = k yz = k 3) wx = k 4) r s2 = k 5) f xy = k Answers - Variation 6) jm3 = k 7) h b = k x a b2√ = k 8) 9) ab = k 10) a b = 3 444 27) 6, 16 3 − 28) 2 5 , 0 13 7 , 1 3, 5 2 7 4 3 , − 6, 2 5 29) 30) 31) − − − 32) − 33) 7, 1 5 34) 35) 22 5 , 19 22 , − − 2 13 11 38 − − 36) 0, 12 5 − 11) P rq = 0.5 12) cd = 28 13) t u2 = 0.67 14) e fg = 4 15) wx3 = 1458 23) 241,920,000 cans 32) 1600 km 16) h j = 1.5 a x y2√ 17) = 0.33 18) mn = 3.78 19) 6 k 20) 5.3 k 21) 33.3 cm 22) 160 kg/cm3 1.8 1) 11 2) 5 4 3) − 4) 32 13 5) − 6) 62 7) 16 8) 17 4 9) 35, 36, 37 10) − 11) 43, 14, − 12) 52, 54 42, 13, 41 12 − − − − 13) 61, 63, 65 14) 83, 85, 87 15) 9, 11, 13 16) 56, 56, 68 1.9 1) 6, 16 2) 10, 40 3) 18, 38 24) 3.5 hours 25) 4.29 dollars 26) 450 m 27) 40 kg 28) 5.7 hr 29) 40 lb 30) 100 N 31) 27 min 33) r = 36 34) 8.2 mph 35) 2.5 m 36) V = 100.5 cm3 37) 6.25 km 38) I = 0.25 Answer Set - Number and Geometry 33) 40, 200 34) 60, 180 35) 20, 200 36) 30, 15 37) 76, 532 38) 110, 880 39) 2500, 5000 40) 4, 8 41) 2, 4 42) 3, 5 43) 14, 16 44) 1644 45) 325, 950 17) 64, 64, 52 18) 36, 36, 108 19) 30, 120, 30 20) 30, 90, 60 21) 40, 80, 60 22) 28, 84, 68 23) 24, 120, 36 24) 32, 96, 52 25) 25, 100, 55 26) 45, 30 27) 96, 56 28) 27, 49 29) 57, 83 30) 17, 31 31) 6000, 24000 32) 1000, 4000 Answers - Age Problems 4) 17, 40 5) 27, 31 6) 12, 48 445 7) 31, 36 8) 16, 32 9) 12, 20 10) 40, 16 11) 10, 6 12) 12, 8 13) 26 14) 8 15) 4 16) 3 17) 10, 20 18) 14 1.10 1) 1 1 3 2) 25 1 2 3) 3 4) 10 , 20 1 2 5) 30, 45 6) 3 7) 300 13 8) 10 9) 7 10) 30 11) 150 12) 360 13) 8 2.1 19) 9, 18 20) 15, 20 21) 50, 22 22) 12 23) 72, 16 24) 6 25) 37, 46 26) 15 27) 45 28) 14, 54 29) 8, 4 30) 16, 32 31) 10, 28 32) 12,20 33) 141, 67 34) 16, 40 35) 84, 52 36) 14, 42 37) 10 38) 10, 6 39) 38, 42 40) 5 Answers - Distance, Rate, and Time Problems 14) 10 15) 2 16) 3 17) 48 18) 600 19) 6 20) 120 21) 36 22) 2 23) 570 24) 24, 18 25) 300 26) 8, 16 27) 56 28) 95, 120 29) 180 30) 105, 130 31) 2:15 PM 32) 200 33) 1 3 34) 15 35) 27 4 36) 1 2 37) 3, 2 38) 90 Answers - Chapter 2 446 1) B(4, E( H( K( − − − 3) 6) 9) 12) Answers - Points and Lines 3) C(1, 2) D( 1, 4) 2) − 5, 0) F(2, 1, − 4, 3) 4) I( − − 3) G(1, 3) 1) J(0, 2) − 2) 8) 11) 14) 4) 7) 10) 13) 447 15) 18) 21) 2.2 1) 3 2 2) 5 3) Undeﬁned 1 2 4) − 5) 5 6 2 3 1 6) − 7) − 8) 5 4 1 9) − 10) 0 11) Undeﬁned 12) 16 7 17 31 3 2 − − 13) 14) 2.3 17) 20) 7 13 5 8 5 4 28) 1 16 29) − 30) 2 7 31) − 32) 2 33) − 34) 3 35) − 36) 6 37) − 38) 1 39) 2 40) 1 16) 19) 22) Answers - Slope 7 17 15) 4 3 16)
− 17) 0 18) 5 11 19) 1 2 20) 1 16 21) 22) 11 2 12 31 − − 23) Undeﬁned 24) 24 11 25) 26) 27) 26 27 19 10 1 3 − − − 448 1) y = 2x + 5 2) y = − 3) y = x 4) y = 5) y = − − 6) y = − 7) y = 1 3 8) y = 2 5 6x + ) y = − 10) y = − 11) y = x 12) y = 13) y = 14 4x 3 − 3 4 x + 2 1 10 x 15) y = − 3 10 − 16) y = 1 10x − 2x 1 − x + 70 11 17) y = − 18) y = 6 11 19 20) y = − 21) x = − 22) y = 1 7 8 x + 6 37 10 x 23) y = − 24) y = 5 x 2 1 − 25) y = 4x x + 1 2 3 4x + 3 26) y = − 27) y = − 28) x = 4 2.4 Answers - Slope-Intercept 1 2 x + 1 37) 29) y = − 30) y = 6 5 x + 4 38) 39) 40) 41) 42) 31) 32) 33) 34) 35) 36) 449 1) x = 2 2) x = 1 2) 2) 4) 3) y − 2 = 1 2 (x 4) y 1 = − − 5) y + 5 = 9(x + 1) − − 2 (x 1 6) y + 2 = 2(x − − 1 = 3 4 (x + 4) 2(x 7) y − 8) y + 3 = − − 9) y + 2 = 3x − 10) y 1 = 4(x + 1) − 11 12) y 2 = − − 13) y + 3 = 1 5 (x + 5) 14) y + 4 = 4 = 15) y − 16) y + 4 = 17) y = 2x 2 3(x + 1) 5 4 (x + 1) − − 3 2 x(x 1) − − 3 − 2x + 2 18) y = − 2.5 2 3 10 3 1) 2 2) − 3) 4 4) − 5) 1 6) 6 5 7) 8) 7 3 4 − − Answers - Point-Slope Form 19) 20) y = − 21 10 3 37) y + 2 = 3 2(x + 4) 38) y − 1 = 3 8 (x + 4) 5 = 1 4 (x 39) y − 40) y + 4 = 3) − (x + 1) 41) y + 3 = 42) y + 5 = − − − 3) 8 7(x − 4(x + 1) 1 3 4 x − 11 4 3 2 1 10 43) y = 44) y = − − 45) y = − 46) y = 1 2 x 47) y = − 48) y = 1 3 x + 1 49) y = x + 2 − 50) y = x + 2 51) y = 4x + 3 52 22) y = 23) y = 24) y = 7 4 3 2 5 2 − − − 25) y = − 26) y = 7 x 3 27 28) y = − 3 − 3 29) x = − 30) y = 2x 31) y = − 32) y = 6 x 5 3 = 33) y − 34(x + 4) − 35) y 36) y − − 1 = 1 8(x 5 = − 5) − 8(x + 4) 1 Answers - Parallel and Perpendicular Lines 9) 0 10) 2 11) 3 12) 13) − − 14) − 15) 2 16) − 5 4 3 1 3 3 8 17) x = 2 18) y 19) y 2 = 7 5 (x 4 = 9 2 (x 5) 3) − − − − 20x 3 4(x − 1) − 2) 3 = − 3(x + 1) 21) y 22) y − − 450 23) x = 4 1) − 4 = 7 24) y − 25) y + 5 = 5 (x (x − 26) y + 2 = 1) 1) − 2(x 27) y 28x (x 3 = − − − 5) 1) − 4 (x 1 29) y 2 = − − − 30) y + 5 = 7 3 (x + 3) 3(x 31) y + 2 = 4) 2) − − 1 2(x + 2) 5 = 32) y − 33) y = − 34) y = 3 5 − 2x + 5 x + 5 x x x 35) y = 36 37) y = − 38) y = 5 x 2 39) y = − 40 41) y = x 1 − 42) y = 2x + 1 43) y = 2 44) y = 45) y = 46) y = − − − 47) y = − 48 2x + 5 x + 4 − Answers - Chapter 3 Answers - Solve and Graph Inequalities 18) x < 6: ( , 6) − ∞ 19) a < 12: ( , 12) 20) v > 1: [1, ∞ 21) x > 11: [11, − ∞ ) ) ∞ 2] 22) x 6 18: ( − 23) k > 19: (19, − ∞ ) ∞ 18] , − 10: ( , 10] 24) n 6 25) p < − − 1: ( 26) x 6 20: ( − ∞ , − 1) − ∞ − , 20] 27) m > 2: [2, 28) n 6 5: ( , 5] 29) r > 8: (8] , − 30) x 6 3: ( − 31) b > 1: (1, 32) n > 0: [0, − ∞ ) ∞ ) ∞ ∞ 26, ) ) ∞ 13) x > 110: [110, 26: [ 14) n > − 15) r < 1: ( − , 1) − ∞ 6: ( 16) m 6 − 17) n > 6: [ − − − ∞ 6, − ) ∞ , 6] 33) v < 0: ( , 0) 34) x > 2: (2, − ∞ ) ∞ 451 3.1 1) ( 2) ( 3) ( 4) ( 5) ( 5, 4] , 5] ) ∞ 2 − ∞ 5, 6) ( − 7) m < − 8) m 6 1 9) x > 5 5 2 10) a 6 − 11) b > − 12) x > 1 ⊘ ) 35) No solution: 36) n > 1: (1, ∞ 3.2 37) {All real numbers.} : R 38) p 6 3: ( , 3] − ∞ Answers - Compound Inequalities 4, ) ∞ − ) 1) n 6 − 9 or n > 2: ( 2) m > 4 or m < − 3) x > 5 or x < − 5: ( − 4) r > 0 or r < − 7 : ( 5) x < 6) n < − − , 7: ( − ∞ − 7 or n > 8 : ( ) ∞ [ 9] , − [2, , 5) S − − ∞ , 5) [5, S ∞ 7), S (0) 7), − ∞ − S S (8, ) ∞ 7) − 8 < v < 3: ( − 8) 7 < x < 4: ( 8, 3) 7, 4) − 9) b < 5: ( − , 5) − ∞ 2 6 n 6 6: [ 10) − 11) 7 6 a 6 6: [ − 12) v > 6: [6, ) ∞ 2, 6] 7, 6] − − 13) − 14) 2: [ 2] − 6, − 9, 0] − − 15) 3 < k 6 4: (3, 4] 16) 17) − − 2 6 n 6 4: [ 2 < x < 2: ( 2, 4] 2, 2) − − 18) No solution : ⊘ 1 6 m < 4: [ 19) − 20) r > 8 or r < − 21) No solution : ⊘ 1, 4) − 6: : ( 6) , − − ∞ (8, ) ∞ S 22) x 6 0 or x > 8 : ( , 0] − ∞ (8, ) ∞ 23) No solution : ⊘ 24) n > 5 or n < 1 : ( S , 1) − ∞ [5, ) ∞ S 452 25) 5 6 x < 19: [5, 19) 26) n < 14 or n > 17 : ( − 27) 1 6 v 6 8: [1, 8] 14) , − − ∞ [17, ) ∞ S 28) a 6 1 or a > 19 : ( , 1] − ∞ [19, ) ∞ 29) k > 2 or k < 20 : ( − ∞ 30) {All real numbers.} : R − , 20) S − [2, ) ∞ S 31) − 1 < x 6 1: ( 32) m > 4 or m 6 − − 1, 1] 1 : ( 1] , − − ∞ (4, ) ∞ S Answers - Absolute Value Inequalities 3.3 1) 2) 3) 4) 5) 6) 7) 8) 9) 3, 3 8, 8 3, 3 7, 1 4, 8 4, 20 − − − − − − 2, 4 7, 1 7 3 , 11 3 − − − 10) 7, 2 − 3, 5 11) − 12) 0, 4 13) 1, 4 14) ( , 5) − ∞ 15) ( 16) ( 175, ) S 5 3] S h [9, 1) S (0, ) ∞ S 1) (54, ∞ ) ∞ [3, ) ∞ S [14 ) ∞ 18) ( 19) ( 20) ( 21) ( 22) [ 23) ( 24) ( , − , 2 3) − ∞ − ∞ − ∞ , 0) S 1] , , − − 4] 5 2] − ∞ − ∞ 25) [1, 3] 26) [ 1 2 , 1] 27) ( 4) , − − ∞ ( 3, ) ∞ − 28) [3, 7] 29) [1, 3 2 ] 30) [ 2, − 31) ( − ∞ 4 3] − , 3 2) S ( 5 2 32) ( 1 S 2) , − − ∞ 33) [2, 4] S 34) [ 3, 2] − − 453 ) , ∞ (1, ∞ ) 4.1 1) ( 2) ( 3) ( 4) ( 1, 2) 4, 3) 1, − 3, 1) 3) − − − − 5) No Solution 2, − 3, 1) 6) ( 2) − 7) ( − 8) (4, 4) 3, 9) ( 1) − − 10) No Solution 11) (3, 4) − 4.2 1) (1, 2) ( − 3) − 3, 2) 2, 3) ( − 4) (0, 3) 5) ( 1, − 7, 6) ( − 7) (1, 5) 4, 8) ( − 9) (3, 3) 5) − 2) 8) − − 1) − 10) (4, 4) 11) (2, 6) 12) ( 13) ( − − 3, 3) 6) 2, − 14) (0, 2) 4.3 Answers - Chapter 4 Answers - Graphing 12) (4, 13) (1, 14) ( − 15) (3, 4) − 3) − 1, 3) 4) − 16) No Solution 17) (2, − 18) (4, 1) 2) 23) ( 1, − − 1) 24) (2, 3) 25) ( 26) ( 1, 4, − − − − 2) 3) 27) No Solution 3, 1) 2) 28) ( − 29) (4, − 30) (1, 4) 29) (4, 30) ( − 3) − 1, 5) 31) (0, 2) 7) 32) (0, − 33) (0, 3) 4) 2) 3) 34) (1, 35) (4, − − 36) (8, − 37) (2, 0) 38) (2, 5) 39) ( 4, 8) − 40) (2, 3) 3, 4) 1) 19) ( − 20) (2, − 21) (3, 2) 4, 22) ( 4) − − Answers - Substitution 15) (1, 16) ( 17) ( − − 5) − 1, 0) 1, 8) 18) (3, 7) 19) (2, 3) 8) 20) (8, − 21) (1, 7) 22) (1, 7) 3, 23) ( − 24) (1, − 25) (1, 3) − 3) 26) (2, 1) 27) ( 28) ( 2, 8) 4, 3) − − 2) 454 2, 4) 1) ( − 2) (2, 4) Answers - Addition/Elimination 2) 12) (1, − 13) (0, 4) 3) No solution 14) ( 4) Inﬁnite number of solutions 5) No solution 6) Inﬁnite number of solutions 7) No solution 8) (2, 2) 9) ( − 10) ( 5) − 3, 6) − 3, − 11) ( 2, − − 9) 4.4 1) (1, − 2) (5, − 3) (2, 3, 4) (3, 2) − 2, 1) − 2, 1, 4) 5) ( 6) ( − − 3, 2, 1) − 7) (1, 2, 3) 8) solutions ∝ 9) (0, 0, 0) 10) solutions ∝ 11) (19, 0, 1, 0) − 15) (8, 2) 16) (0, 3) 17) (4, 6) 18) ( 19) ( − − 8) 6, − 2, 3) 20) (1, 2) 4) 21) (0, − 22) (0, 1) 23) ( − 24) (2, 2, 0) 2) − 14) solutions ∝ 15) (2, 1 2 , 2) − solutions 16) ∝ 17)( − 1, 2, 3) 18)( 1, 2, 2) − 19) (0, 2, 1) − 20) no solution 21) (10, 2, 3) Answers - Three Variables 1, 2) 3, 2) 12) solutions ∝ 13) (0, 0, 0) 2) 3) 25) ( 26) ( 27) ( 28) ( 29) ( 1, − 3, 0) 1, − 3, 0) 8, 9) − − − − − 30) (1, 2) 31) ( 32) ( 2, 1) 1, 1) − − 33) (0, 0) 34) Inﬁnite number of solutions 23) (2, 3, 1) 24) ∝ solutions 25) no solutions 26) (1, 2, 4) 25, 18, 25) − 27) ( − , 3 28) ( 2 7 7 , 29) (1, − − 30) (7, 4, 5, 6) 1) 2, − 2 7) − 3, 31) (1, 2, 4, 1) − 1, 0, 4) − 3, 13) − 22) no solution 32) ( − − 4.5 1) 33Q, 70D Answers - Value Problems 455 2) 26 h, 8 n 19) 13 d, 19 q 3) 236 adult, 342 child 20) 28 q 4) 9d, 12q 5) 9, 18 6) 7q, 4h 7) 9, 18 8) 25, 20 9) 203 adults, 226 child 10) 130 adults, 70 students 11) 128 card, 75 no card 12) 73 hotdogs, 58 hamburgers 13) 135 students, 97 non-students 14) 12d, 15q 15) 13n, 5d 16) 8 20c, 32 25c 17) 6 15c, 9 25c 18) 5 4.6 1) 2666.7 2) 2 3) 30 4) 1, 8 5) 8 6) 10 7) 20 8) 16 9) 17.25 21) 15 n, 20 d 22) 20 S1, 6 S5 23) 8 S20, 4 S10 24) 27 25) S12500 @ 12% S14500 @ 13% 26) S20000 @ 5% S30000 @ 7.5% 27) S2500 @ 10% S6500 @ 12% 28) S12400 @ 6% S5600 @ 9% 29) S4100 @ 9.5% S5900 @ 11% 30) S7000 @ 4.5% S9000 @ 6.5% 31) S1600 @ 4%; S2400 @ 8% 32) S3000 @ 4.6% S4500 @ 6.6% Answers - Mixture Problems 10) 1.5 11) 10 12) 8 13) 9.6 14) 36 15) 40, 60 16) 30, 70 17) 40, 20 18) 40, 110 456 33) S3500 @ 6%; S5000 @ 3.5% 34) S7000 @ 9% S5000 @ 7.5% 35) S6500 @ 8%; S8500 @ 11% 36) S12000 @ 7.25% S5500 @ 6.5% 37) S3000 @ 4.25%; S3000 @ 5.75% 38) S10000 @ 5.5% S4000 @ 9% 39) S7500 @ 6.8%; S3500 @ 8.2% 40) S3000 @ 11%; S24000 @ 7% 41) S5000 @ 12% S11000 @ 8% 42) 26n, 13d, 10q 43) 18, 4, 8 44) 20n, 15d, 10q 19) 20, 30 20) 100, 200 21) 40, 20 22) 10, 5 23) 250, 250 24) 21, 49 25) 20, 40 26) 2, 3 27) 56, 144 28) 1.5, 3.5 29) 30 30) 10 31) 75, 25 32) 55, 20 5.1 1) 49 2) 47 3) 24 4) 36 5) 12m2n 6) 12x3 7) 8m6n3 8) x3y6 9) 312 10) 412 11) 48 12) 36 13) 4u6v4 14) x3y3 15) 16a16 16) 16x4y4 5.2 1) 32x8y10 2) 32b13 a2 33) 440, 160 34) 20 35) 35, 63 36) 3, 2 37) 1.2 38) 150 39) 10 40) 30, 20 41) 75 42) 20, 60 43) 25 Answers - Chapter 5 Answers to Exponent Properties 17) 42 18) 34 19) 3 20) 33 21) m2 22) xy3 4 23) 4x2y 3 24) y2 4 25) 4x10y14 26) 8u18v6 27) 2x17y16 28) 3uv 29) x2y 6 30) 4a2 3 31) 64 32) 2a 33) y3 512x24 34) y5x2 2 35) 64m12n12 36) n10 2m 37) 2x2y 38) 2y2 39) 2q7r8p 40) 4x2y4z2 41) x4y16z4 42) 256q4r8 43) 4y4z Answers to Negative Exponents 3) 2a15 b11 4) 2x3y2 457 5) 16x4y8 6) 1 7) y16 x5 8) 32 m5n15 9) 2 9y 10) y5 2x7 11) 1 y2x3 12) y8x5 4 13) u 4v6 14) x7y2 2 15) u2 12v5 y 2x4 17) 2 y7 16) 5.3 1) 8.85 2) 7.44 102 × 4 10− × 10− 2 3) 8.1 × 4) 1.09 5) 3.9 6) 1.5 100 × 10− 2 × 104 × 7) 870000 8) 256 9) 0.0009 10) 50000 11) 2 12) 0.00006 13) 1.4 14) 1.76 3 10− × 10 10− × 18) a16 2b 19) 16a12b12 20) y8x4 4 21) 1 8m4n7 22) 2x16y2 23) 16n6m4 24) 2x y3 25) 1 x15y 26) 4y4 27) u 2v 28) 4y5 29) 8 30) 1 2u3v5 Answers to Scientiﬁc Notation 15) 1.662 16) 5.018 17) 1.56 18) 4.353 19) 1.815 20) 9.836 21) 5.541 22) 6.375 23) 3.025 24) 1.177 25) 2.887 26) 6.351 27) 2.405 6 10− 106 × × 3 10− × 108 104 1 10− × × × 5 10− 4 10− 9 10− 16 10− 6 10− 21 10− × × × × × × 20 10− × 10− 2 28) 2.91 × 458 31) 2y5x4 32) a3 2b3 33) 1 x2y11z 34) a2 8c10b12 35) 1 h3k j6 36) x30z6 16y4 37) 2b14 a12c7 38) m14q8 4p4 39) x2 y4z4 40) mn7 p5 29) 1.196 30) 1.2 × 31) 2.196 2 2 10− × 107 10− × 103 1014 × 32) 2.52 × 33) 1.715 34) 8.404 35) 1.149 36) 3.939 37) 4.6 × 38) 7.474 39) 3.692 40) 1.372 41) 1.034 101 106 × × 109 × 102 103 7 10− 103 106 × × × × 42) 1.2 106 × 5.4 1) 3 2) 7 3) 4) − − 5) − 6) 8 7) 5 10 6 7 1 8) − 9) 12 10) 1 − 11) 3p4 12) − 3p − m3 + 12m2 n3 + 10n2 13) − 14) 8x3 + 8x2 15) 5n4 + 5n Answers to Introduction to Polynomials 16) 2v4 + 6 17) 13p3 1 1 − − − − 3x 22) 18) − 19) 3n3 + 8 20) x4 + 9x2 5 21) 2b4 + 2b + 10 3r4 + 12r2 5x4 + 14x3 4n + 7 3a2 23) − 24) 5n4 25) 7a4 2a − − 26) 12v3 + 3v + 3 27) p2 + 4p − 28) 3m4 2m + 6 − 29) 5b3 + 12b2 + 5 15n4 + 4n 30) − 6
− 6 − 31) n3 32) 33) − − 5n2 + 3 − 6x4 + 13x3 12n4 + n2 + 7 34) 9x2 + 10x2 3r3 + 7r2 + 1 35) r4 − 36) 10x3 6x2 + 3x 8 − − 37) 9n4 + 2n3 + 6n2 38) 2b4 39) − b3 + 4b2 + 4b − 3b4 + 13b3 11b + 19 7b2 − − 40) 12n4 n3 6n2 + 10 − x3 − 4x + 2 41) 2x4 − 42) 3x4 + 9x2 + 4x − 5.5 Answers to Multiply Polynomials 1) 6p 42 − 2) 32k2 + 16k 3) 12x + 6 4) 18n3 + 21n2 5) 20m5 + 20m4 6) 12r 21 − 7) 32n2 + 80n + 48 8) 2x2 − 9) 56b2 7x 4 − 19b − 10) 4r2 + 40r+64 − 15 13) 15v2 26v + 8 − 44a 14) 6a2 − 15) 24x2 − 16) 20x2 17) 30x2 − − − 22x 32 7 − 29x + 6 14xy 18) 16u2 + 10uv 4y2 21v2 − − 19) 3x2 + 13xy + 12y2 20) 40u2 34uv 48v2 − − 21) 56x2 + 61xy + 15y2 11) 8x2 + 22x + 15 12) 7n2 + 43n 42 − 22) 5a2 23) 6r3 − − 459 7ab 24b2 − 43r2 + 12r 35 − 24) 16x3 + 44x2 + 44x + 40 32) 42u4 + 76u3v + 17u2v2 18v4 − 25) 12n3 20n2 + 38n 20 − − 4b2 4b 12 − − 26) 8b3 − 27) 36x3 24x2y + 3xy2 + 12y3 − 28) 21m3 + 4m2n 8n3 − 29) 48n4 − 16n3 + 64n2 6n + 36 − 30) 14a4 + 30a3 13a2 − 31) 15k4 + 24k3 + 48k2 + 27k + 18 − 12a + 3 33) 18x2 34) 10x2 35) 24x2 36) 16x2 37) 7x2 − 38) 40x2 12 15x − 55x + 60 18x 15 − − − − 44x 12 − − 49x + 70 10x 5 − − 39) 96x2 6 − 40) 36x2 + 108x + 81 5.6 1) x2 2) a2 3) 1 − 4) x2 64 − 16 − 9p2 9 − 49n2 25 64 64 − − 9 5) 1 − 6) 64m2 7) 25n2 8) 4r2 − 9) 16x2 10) b2 − 11) 16y2 12) 49a2 − 13) 16m2 14) 9y2 − 49 − − x2 49b2 64n2 − 9x2 Answers to Multiply Special Products 4y2 15) 36x2 − 16) 1 + 10n + 25n2 17) a2 + 10a + 25 18) v2 + 8v + 16 19) x2 16x + 64 − 12n + 36n2 20) 1 21) p2 + 14p + 49 22) 49k2 − 98k + 49 − 70n + 25n2 40x + 25 23) 49 − 24) 16x2 − 25) 25m2 80m + 64 − 26) 9a2 + 18ab + 9b2 27) 25x2 + 70xy + 49y2 28) 16m2 8mn + n2 − 29) 4x2 + 8xy + 4y2 30) 64x2 + 80xy + 25y2 31) 25 + 20r + 4r2 32) m2 14m + 49 − 33) 4 + 20x + 25x2 49 49 34) 64n2 − 35) 16v2 36) b2 − − 16 37) n2 25 − 38) 49x2 + 98x + 49 39) 16k2 + 16k + 4 40) 9a2 64 − 5.7 1) 5x + 1 4 + 1 2x Answers to Divide Polynomials 3) 2n3 + n2 10 + 4n 2) 5x3 9 + 5x2 + 4x 9 4) 3k2 8 + k 2 + 1 4 460 5) 2x3 + 4x2 + x 2 6) 5p3 4 + 4p2 + 4p 19) x 3 + 3 10x − − 20 33) 3n2 9n 10 − − 8 n + 6 − 5 9) x 7) n2 + 5n + 1 5 8) m2 3 + 2m + 3 10 + 9 x + 8 − 10 11) n + 8 9 − 12 13) v + 8 − 14) x 3 − − v 10 − 5 x + 7 15 16) x 6 x − − − 17) 5p + 4 + 3 4 9p + 4 18) 8k 9 − 1 − 3k 1 − 6.1 1) 9 + 8b2 2) x 5 − 3) 5(9x2 − 4) 1 + 2n2 5) 5) 7(8 − 6) 10(5x 7) 7ab(1 5p) 8y) 5a) − − 8) 9x2y2(3y3 8x) − 21) r 1 + 2 4x + 3 − 22) m + 4 + 1 m − 1 23) n + 2 24) x 4 + 4 2x + 3 − 25) 9b + 5 5 3b + 8 − 26) v + 3 3v − 9 5 − 7 5 4x − 3 4n + 5 7 − 27) x 7 − − 28) n 7 − − 29) a2 + 8a 30) 8k2 2k − − 31) x2 32) x2 − − 10 4x − 8x + − 34) k2 35) x2 − − 3k 9 1 − k − − 7x + 3 + 1 x + 7 36) n2 + 9n 37) p2 + 4p 1 + 3 2n + 3 1 + 4 9p + 9 − − 38) m2 8m + 7 − 7 8m + 7 − 39) r2 + 3r − 4 40) x2 + 3x − 8 4 r − − 7 + 5 2x + 6 41) 6n2 3n − − 3 + 5 2n + 3 42) 6b2 + b + 9 + 3 4b − 6v + 6 + 1 43) v2 7 4v + 3 − Answers - Chapter 6 Answers - Greatest Common Factor 9) 3a2b( 1 + 2ab) − 10) 4x3(2y2 + 1) 11) 5x2(1 + x + 3x2) − 12) 8n5( − 13) 10(2x4 4n4 + 4n + 5) 3x + 3) − 14) 3(7p6 + 10p2 + 9) 15) 4(7m4 + 10m3 + 2) 461 16) 2x( − 5x3 + 10x + 6) 17) 5(6b9 + ab 3a2) − 18) 3y2(9y5 + 4x + 3) 19) − 8a2b (6b + 7a + 7a3) 20) 5(6m6 + 3mn2 5) − 21) 5x3y2z(4x5z + 3x2 + 7y) 22) 3(p + 4q 5q2r2) − 23) 10(5x2y + y2 + 7xz2) 6.2 1) (8r2 2) (5x2 3) (n2 − 4) (2v2 5)(5r − 8)(7x − 3)(3n 1) − 2) − 2) − 1)(7v + 5) − 5) (3b2 7)(5b + 7) − 6) (6x2 + 5)(x 8) − 7) (3x2 + 2)(x + 5) 8) (7p2 + 5)(4p + 3) 9) (7x2 4)(5x 4) − 10)(7n2 5)(n + 3) − − 24) 10y4z3 (3x5 + 5z2 x) − 25) 5q(6pr p + 1) − 26) 7b(4 + 2b + 5b2 + b4) 27) 3( − 6n5 + n3 7n + 1) − 28) 3a2(10a6 + 2a3 + 9a + 7) 29) 10x11( 30) 4x2( − 31) 4mn( 4 − 6x4 − − 2x + 5x2 5x3) − x2 + 3x + 1) 8n7 + m5 + 3n3 + 4) − 5 + 3y3 − 2xy3 4xy) − − 32) 2y7( Answers - Grouping 11) (7x + 5)(y − 12) (7r2 + 3)(6r 7) 21) (4u + 3)(8v 5) − 7) 22) 2(u + 3)(2v + 7u) − 13) (8x + 3)(4y + 5x) 14) (3a + b2)(5b 15) (8x + 1)(2y 2) 7) − − 8) 16) (m + 5)(3n − 17) (2x + 7y2)(y 4x) − 5)(5n + 2) y)(8y + 7) 1)(y + 7) 18) (m 19) (5x 20) (8x − − − 23) (5x + 6)(2y + 5) 24) (4x 25) (3u − − 5y2)(6y 5) − 2u) 7)(v − 26) (7a 2)(8b − 27) (2x + 1)(8y 7) 3x) − − 6.3 1) (p + 9)(p + 8) 2) (x 3) (n − − 8)(x + 9) 8)(n 1) − Answers - Trinomials where a = 1 4) (x 5)(x + 6) − 5) (x + 1)(x 10) − 6) (x + 5)(x + 8) 462 7) (b + 8)(b + 4) 17) (u 5v)(u 3v) 27) 5(a + 10)(a + 2) − − 18) (m + 5n)(m 19) (m + 4n)(m − − 8n) 2n) 10)(b 7) − 7)(x + 10) 8) (b − 9) (x − 10) (x − 3)(x + 6) 20) (x + 8y)(x + 2y) 11) (n 5)(n − 12) (a + 3)(a 3) − 9) − 13) (p + 6)(p + 9) 14) (p + 10)(p 3) − 7) 8)(n − 5n)(m 10n) − 15) (n − 16) (m − 6.4 21) (x 22) (u 23) (x − − − 9y)(x 2y) − 7v)(u 2v) − 3y)(x + 4y) 24) (x + 5y)(x + 9y) 25) (x + 6y)(x 2y) − 26) 4(x + 7)(x + 6) 28) 5(n 29) 6(a 30) 5(v − − − 8)(n 1) − 4)(a + 8) 1)(v + 5) 31) 6(x + 2y)(x + y) 32) 5(m2 + 6mn 18n2) − 33) 6(x + 9y)(x + 7y) 34) 6(m − 9n)(m + 3n) 1) (7x 6)(x 6) − − 2) (7n 2)(n − 3) (7b + 1)(b + 2) − 6) 4) (7v + 4)(v 5) (5a + 7)(a 4) 4) − − 6) Prime 7) (2x 1)(x 2) − − 8) (3r + 2)(r 2) − 9) (2x + 5)(x + 7) 10) (7x 11) (2b − − 6)(x + 5) 3)(b + 1) 12) (5k 6)(k − 13) (5k + 3)(k + 2) − 4) Answers - Trinomials where a 1 15) (3x 16) (3u − − 5)(x 4) − 2v)(u + 5v) 17) (3x + 2y)(x + 5y) 18) (7x + 5y)(x y) − 7y)(x + 7y) 19) (5x − 20) (5u 4v)(u + 7v) − 21) 3(2x + 1)(x 7) 6) − − 22) 2(5a + 3)(a 23) 3(7k + 6)(k 5) 24) 3(7n 25) 2(7x − − 26) (r + 1)(4r − 6)(n + 3) 2)(x 4) − 3) − 27) (x + 4)(6x + 5) 29) (k 30) (r − − 4)(4k 1) − 1)(4r + 7) 31) (x + 2y)(4x + y) 32) 2(2m2 + 3mn + 3n2) 3n)(4m + 3n) 33) (m − 34) 2(2x2 3xy + 15y2) − 35) (x + 3y)(4x + y) 36) 3(3u + 4v)(2u 3v) − 37) 2(2x + 7y)(3x + 5y) 38) 4(x + 3y)(4x + 3y) 39) 4(x 2y)(6x y) − − 40) 2(3x + 2y)(2x + 7y) 14) (3r + 7)(r + 3) 28) (3p + 7)(2p 1) − 6.5 1) (r + 4)(r 2) (x + 3)(x 4) − 3) − Answers - Factoring Special Products 3) (v + 5)(v 4) (x + 1)(x 5) 1) − − 463 27) 2(2x 3y)2 − 28) 5(2x + y)2 29) (2 m)(4 + 2m + m2) − 30) (x + 4)(x2 4x + 16) − 31) (x 4)(x2 + 4x + 16) − 32) (x + 2)(x2 2x + 4) − u)(36 + 6u + u2) 33) (6 − 34) (5x − 6)(25x2 + 30x + 36) 35) (5a − 4)(25a2 + 20a + 16) 36) (4x 3)(16x2 + 12x + 9) − 37) (4x + 3y)(16x2 12xy + 9y2) − 38) 4(2m 3n)(4m2 + 6mn + 9n2) − 39) 2(3x + 5y)(9x2 − 40) 3(5m + 6n)(25m2 15xy + 25y2) 30mn + 36n2) − 41) (a2 + 9)(a + 3)(a − 42) (x2 + 16)(x + 4)(x 3) 4) − z) − 5) (p + 2)(p − 6) (2v + 1)(2v 7) (3k + 2)(3k 8) (3a + 1)(3a 2) − − − 1) 2) 1) 3) 9) 3(x + 3)(x − 10) 5(n + 2)(n − 11) 4(2x + 3)(2x 2) 3) − 12) 5(25x2 + 9y2) 13) 2(3a + 5b)(3a 5b) − 14) 4(m2 + 16n2) 15) (a 1)2 − 16) (k + 2)2 17) (x + 3)2 4)2 3)2 2)2 18) (n 19) (x − − 20) (k − 21) (5p − 22) (x + 1)2 23) (5a + 3b)2 24) (x + 4y)2 25) (2a − 26) 2(3m 5b)2 2n)2 − 6.6 − 3) 1) 3(2a + 5y)(4z 2) (2x − 3) (5u 5)(x − 4v)(u − 4) 4(2x + 3y)2 − 1)2 43) (4 + z2 )(2 + z)(2 1) 44) (n2 + 1)(n + 1)(n − 45) (x2 + y2)(x + y)(x − 46) (4a2 + b2)(2a + b)(2a − 47) (m2 + 9b2)(m + 3b)(m y) b) 3b) − 48) (9c2 + 4d2)(3c + 2d)(3c 2d) − Answers - Factoring Strategy 3h) 6) 5(4u 7) n(5n − − 3u2) x)(v − 3)(n + 2) v) 8) x(2x + 3y)(x + y) 5) 2( − x + 4y)(x2 + 4xy + 16y2) 9) 2(3u 10) 2(3 − − 2)(9u2 + 6u + 4) 4x)(9 + 12x + 16x2) 464 11) n(n 1) − 12) (5x + 3)(x 13) (x − 14) 5(3u 3y)(x 5v)2 − 5) y) − − 15) (3x + 5y)(3x 5y) − 3y)(x2 + 3xy + 9y2) 16) (x − 17) (m + 2n)(m 27) (3x 4)(9x2 + 12x + 16) − 28) (4a + 3b)(4a 29) x(5x + 2) 30) 2(x − 31) 3k(k 2)(x − 5)(k − − 32) 2(4x + 3y)(4x 3b) − 3) 4) 3y) − 4x)(n + 3) 2n) − 33) (m − 18) 3(2a + n)(2b − 19) 4(3b2 + 2x)(3c 3) 2d) − 20) 3m(m + 2n)(m 4n) 21) 2(4 + 3x)(16 − − 12x + 9x2) 22) (4m + 3n)(16m2 12mn + 9n2) 23) 2x(x + 5y)(x − − 2y) 24) (3a + x2)(c + 5d2) 25) n(n + 2)(n + 5) n)(16m2 + 4mn + n2) 26) (4m − 6.7 34) (2k + 5)(k 2) − 35) (4x y)2 − 36) v(v + 1) 37) 3(3m + 4n)(3m 4n) − 38) x2(x + 4) 5y)(x + 4y) 39) 3x(3x − 40) 3n2(3n 41) 2(m − 42) v2(2u 1) − 2n)(m + 5n) 5v)(u 3v) − − 1) 7, 2) − 3) 1, 4) − 5) − 6) 4, 7) 2, 2 − 4, 3 4 , 7 − 5 2 5, 5 − 8 7 − 5, 6 8) 9) 10) 11) 12 Answers - Solve by Factoring 13) 4, 0 14) 8, 0 15) 1, 4 16) 4, 2 17) 3 7 , 18) − 19) 4 , 7 20) 1 4 21) − 22) 8, − − 4 8 3 2 23) 8, − 24) 4, 0 465 , 1 25) 8 3 , 26) 27) 28) − − − 29) − 30) 2, 3 − 7, 7 − 6 8 7 4, 5 2 6 5 , , − − − − − − 31) 32) 33) 34) 35) 4 5 , 6 − 36) 5 3 , 2 − Answers - Chapter 7 7.1 1) 3 2) 1 3 3) 1 5 − 4) undeﬁned 5) 1 2 6) 6 10 7) − 8) 0, 2 5 2 9) − 10) 0, 10 − 11) 0 12) − 13) − 14) 0, 15) − 16) 0, 1 7 10 3 2 1 2 − 8, 4 17) 7x 6 18) 3 n 19) 3 5a 7.2 1) 4x2 2) 14 3 Answers - Reduce Rational Expressions 20) 7 8k 21) 4 x 22) 9x 2 23) 3m − 10 4 24) 10 9n2(9n + 4) 25) 10 2p + 1 26) 1 9 27) 1 x + 7 28) 7m + 3 9 29) 8x 7(x + 1) 30) 7r + 8 8r 31) n + 6 n 5 − 32) b + 6 b + 7 37) 3a 5 − 5a + 2 38) 9 p + 2 39) 2n 1 − 9 40) 3x 5 − 5(x + 2) 41) 2(m + 2) 5m 3 − 42) 9r 5(r + 1) 43) 2(x 3x 4) 4 − − 44) 5b 8 − 5b + 2 45) 7n 4 − 4 46) 5(v + 1) 3v + 1 47) (n 1)2 − 6(n + 1) 48) 49) 7x 6 − (3x + 4)(x + 1) 7a + 9 2(3a 2) − 50) 2(2k + 1) 9(k 1) − 9 33) v 10 − 34) 3(x 3) − 5x + 4 7 7 35) 2x 5x 36) k − − 8 − k + 4 Answers - Multipy and Divide 3) 63 10n 4) 63 10m 466 5) 3x2 2 6) 5p 2 7) 5m 8) 7 10 9) r 6 − r + 10 10) x + 4 11) 2 3 12) 9 − b 5 13) x 14) v 10 10 − 7 1 − 15) x + 1 16) a + 10 a 6 − 17) 5 18) p − p − 10 4 7.3 1) 18 2) a2 3) ay 4) 20xy 5) 6a2c3 6) 12 7) 2x 8) x2 8 − 2x − x 9) x2 − 10) x2 − 3 − 12 − 11x + 30 19) 3 5 20) x + 10 x + 4 21) 4(m − 5m2 5) 22) 7 23) x + 3 4 24) n 9 − n + 7 25) b + 2 8b 26) v 9 − 5 1 − 6 27) n − 28 29) 30) 7 8(k + 3) 31) x 4 − x + 3 32) 9(x + 6) 10 33) 9m2(m + 10) 34) 10 9(n + 6) 35) p + 3 6(p + 8) 36) x 8 − x + 7 37) 5b b + 5 38) n + 3 8 39) r − 40) 18 5 41) 3 2 42) 1 a + 2b 43) 1 x + 2 44) 3(x 2) − 4(x + 2) Answers - Least Common Denominators 467 11) 12a4b5 12) 25x3y5z 13) x (x 3) − 14) 4(x 2) − 15) (x + 2)(x 16) x(x − 4) − 7)(x + 1) 17) (x + 5)(x 5) − 3)2(x + 3) 18) (x − 19) (x + 1)(x + 2)(x + 3) 20) (x 2)(x − 6a4 10a3b2 , 2b 10a3b2 5)(x + 3) − 21) 22) 23) 3x2 + 6x (x 4)(x + 2) − x2 + 4x + 4 3)(x + 2) (x − , (x 2x 8 4)(x + 2) − − , x2 (x 6x + 9 − 3)(x + 2) − 24) x(x 5 , 2x x(x 6) 3x 12 6) , − x(x 6) − − − x2 4x 4)2(x + 4) − , 25) (x − (x 5x + 1 5)(x + 2) − x2 + 7x + 6 26) 27) − 3x2 + 12x 4)2(x + 4) (x − , (x 4x + 8 5) (x +
2) − 6)(x + 6)2, 2x2 (x 9x 18 − − 6)(x + 6)2 − (x − 28) (x 3x2 + 4x + 1 4)(x + 3)(x + 1) − , (x − 2x2 8x 4)(x + 3)(x + 1) − 4x 3)(x + 2) , x2 + 4x + 4 3)(x + 2) (x − 3x2 + 15x 4)(x 2)(x + 5) − 29) 30) (x − (x − 7.4 1) 6 a + 3 4 2) x − 3) t + 7 x2 − 4)(x , (x − 4x + 4 2)(x + 5) − , 5x 4)(x − − (x − 20 2)(x + 5) Answers - Add and Subtract 4) a + 4 a + 6 5) x + 6 x 5 − 6) 3x + 4 x2 468 d2 − 27) − (x + 3)(x + 1) 6x2 3xy − − 2x2y2 28) (x 2x + 3 1)(x + 4) − 40) 2 r + s 5 7) 24r 8) 7x + 3y x2y2 9) 15t + 16 18t3 10) 5x + 9 24 11) a + 8 4 12) 5a2 + 7a 9a2 3 − 13 7x − 4x c2 + cd c2d2 13) − 14) − 15) 3y2 4x 16) x2 17) − 1 − z2 + 5z z2 1 18) 11x + 15 4x(x + 5) − 19) 14 x2 20) x2 x2 3x 4 x 25 − − − − 7.5 1) x 2) 1 x − − y 1 y 3) − a a + 2 4) − 6) b3 + 2b b 2 − − 8b 7) 2 5 8) 4 5 9) 1 2 − 10) − 11) x2 1 2 x 1 x2 + x + 1 − − − − 21) 4t 5 4(t 3) 22) 2x + 10 (x + 3)2 23) 24) 20x 6 − 15x(x + 1) 9a 4(a 5) − 25) t2 + 2ty y2 26) 2x2 − x(x y2 − t2 − 10x + 25 5) − 3 x 34) 2x + 7 x2 + 5x + 6 35) 2x − 5x x2 − 8 − 14 36) − 3x2 + 7x + 4 x) 3(x + 2)(2 − 37) a a2 2 9 − − 38) 2 y2 y − 39) z 2z 3 1 − − 29) 30) 31) 32) x 8 − (x + 8)(x + 6) 2) 2x 5 − 3)(x (x − − 5x + 12 x2 + 5x + 6 41) 42) 5(x 1) − (x + 1)(x + 3) 5x + 5 x2 + 2x 15 − 4x + 1 (x + 1)(x − 2) 43) − (x − (x 29) 3)(x + 5) − 33) 2x + 4 x2 + 4x + 3 44) Answers - Complex Fractions 5x 10 − x2 + 5x + 4 3a + 3 2a − − 4a2 12) 2a2 − 13) x 3 14) 3x + 2 b) 15) 4b(a − a 16) x + 2 x 1 − 17) x 5 − x + 9 3)(x + 5) 5x + 4 − 18) 19) (x − 4x2 − 1 3x + 8 20) 1 x + 4 21) x 2 − x + 2 22) x 7 − x + 5 469 23) x 3 − x + 4 24) − 2a 3a 2 − 4 − b 2 − 2b + 3 25) − 26) x + y y x − 27) a 3b − a + 3b 28) 2x x2 + 1 − 2 y 29) − 30) x2 31) y − xy 1 − x 32) x2 − y xy + y2 x − 33) x2 + y2 xy 34) 2x 1 − 2x + 1 7.6 1) 40 3 = a 2) n = 14 3 3) k = 12 7 4) x = 16 5) x = 3 2 6) n = 34 7) m = 21 8) x = 79 8 9) p = 49 10) n = 25 40 3 11) b = − 12) r = 36 5 13) x = 5 2 14) n = 32 5 7.7 1) − 2) − 3) 3 4) − 5) 2 6) 1 3 1 , 2 2 3 3, 1 1, 4 7) 1 − 35) 3x)2 (1 − x2(x + 3)(x − 3) 36) x + y xy Answers - Proportions 16 7 15) a = 6 7 16) v = − 17) v = 69 5 18) n = 61 3 19) x = 38 3 20) k = 73 3 21) x = 22) x = 8, 5 7, 5 − − 23) m = 7, 8 − 24) x = 25) p = 26) n = 27) n = 28) n = 3, 9 2 7, − 6, 9 1 4, 1 − − − − − − 7, 1 1, 3 29) x = − 30) x = − 31) S9.31 32) 16 33) 2.5 in 34) 12.1 ft 35) 39.4 ft 36) 3.1 in 37) T: 38, V: 57 38) J: 4 hr, S: 14 hr 39) S8 40) C: 36 min, K: 51 min Answers - Solving Rational Equations 8) 9) 10) 1 3 5 − − − 11) − 12) 5, 10 7 15 5, 0 13) 16 3 , 5 14) 2, 13 470 8 15) − 16 17) − 18) − 19) 3 2 20) 10 21) 0, 5 2, 5 3 22) − 23) 4, 7 1 24) − 25) 2 3 7.8 1) 12320 yd 2) 0.0073125 T 3) 0.0112 g 4) 135,000 cm 5) 6.1 mi 6) 0.5 yd2 7) 0.435 km2 8) 86,067,200 ft2 9) 6,500,000 m3 10) 239.58 cm3 11) 0.0072 yd3 12) 5.13 ft/sec 13) 6.31 mph 14) 104.32 mi/hr 15) 111 m/s 8.1 1) 7 5√ 2) 5 5√ 26) 1 2 27) 3 10 28) 1 29) 30) 2 3 1 − − 31) 13 4 32) 1 10 33) − 34) 7 4 Answers - Dimensional Analysis 16) 2,623,269,600 km/yr 17) 11.6 lb/in2 18) 63,219.51 km/hr2 19) 32.5 mph; 447 yd/oz 20) 6.608 mi/hr 21)17280 pages/day; 103.4 reams/month 22) 2,365,200,000 beats/lifetime 23) 1.28 g/L 24) S3040 25) 56 mph; 25 m/s 26) 148.15 yd3; 113 m3 27) 3630 ft2, 522,720 in2 28) 350,000 pages 29) 15,603,840,000 ft3/week 30) 621,200 mg; 1.42 lb Answers - Chapter 8 Answers - Square Roots 3) 6 4) 14 471 5) 2 3√ 6) 6 2√ 7) 6 3√ 8) 20 2√ 9) 48 2√ 10) 56 2√ 11) − 112 2√ 21 7√ 12) − 13) 8 3n√ 14) 7 7b√ 15) 14v 16) 10n n√ 17) 6x 7√ 8.2 1) 5 53√ 2) 5 33√ 3) 5 63√ 4) 5 23√ 5) 5 73√ 6) 2 33√ 8 64√ 7) − 16 34√ 8) − 9) 12 74√ 10) 6 34√ 2 74√ 11) − 12) 15 34√ 4√ 13) 3 8a2 4√ 14) 2 4n3 18) 10a 2a√ 19) 20) 21) 22) 10k2 20p2 7√ 56x2 16 2n√ − − − − 23) 30 m√ − 24) 32p 7√ 25) 3xy 5√ 26) 6b2a 2a√ 27) 4xy xy√ 28) 16a2b 2√ 29) 8x2y2 5√ 30) 16m2n 2n√ Answers - Higher Roots 5√ 15) 2 7n3 5√ 2 3x3 16) − 17) 2p 75√ 18) 2x 46√ 19) − 20) − 21) 4v2 7√ 6 7r 7√ 16b 3b 3√ 6v 22) 20a2 23√ 23) 24) − − 25) 28n2 53√ 8n2 3√ 3xy 5x2 − 26) 4uv u2 3√ 27) − 3√ 2xy 4xy 3√ 28) 10ab ab2 4x3√ 29) 4xy2 472 31) 24y 5x√ 32) 56 2mn√ 33) 35xy 5y√ 34) 12xy 35) 36) 37) − − − 12u 5uv√ 30y2x 2x√ 48x2z2y 5√ 38) 30a2c 2b√ 39) 8j2 5hk√ 40) 41) 42) − − − 4yz 2xz√ 12p 6mn√ 32p2m 2q√ 30) 3xy2 73√ 31) 32) − − 21xy2 3√ 3y 3 8y2 7x2y2 33) 10v2 3u2v2 3√ p 34) − 3√ 40 6xy 3√ 12 3ab2 35) − 36) 9y 5x3√ 37) 38) − − 18m2np2 3 2m2p 12mpq 4 p 5p3 39) 18xy 4 p 8xy3z2 p 18ab2 40) − 41) 14hj2k2 4√ 5ac 4√ 8h2 42) 8.3 − 18xz 4 4x3yz3 p 1) 6 5√ Answers - Adding Radicals 21) − 4 6√ + 4 5√ 3 6√ 5 3√ − 3 2√ + 6 5√ 3√ − 5 6√ 5 6√ 3 3√ 2) 3) 4) 5) 6) − − − − − 7) 3 6√ + 5 5√ 8) 9) 10) 11) 12) 5√ + 3√ 8 2√ − − 6 6√ + 9 3√ 3 6√ + 3√ 2 5√ 6 6√ − − − − 2 2√ 13) − 14) 8 5√ 3√ − 15) 5 2√ 16) 17) 9 3√ 3 6√ − − 3√ − 18) 3 2√ + 3 6√ 12 2√ + 2 5√ 3 2√ 19) 20) 8.4 − − 48 5√ 1) − 2) 25 6√ − 3) 6m 5√ 25r2 2r√ 4) − 5√ 22) − 23) 8 6√ 3 6√ − 9 3√ + 4 2√ − 6√ − 24) − 25) 2 23√ 10 3√ 26) 6 53√ 3 33√ − 34√ 27) − 28) 10 44√ 29) 24√ 3 34√ − 30) 5 64√ + 2 44√ 31) 6 34√ 3 44√ − 32) − 6 34√ + 2 64√ 33) 2 24√ + 34√ + 6 44√ 2 34√ 9 54√ 3 24√ − 34) 35) − 65√ − 6 25√ − 6 67√ + 3 57√ 36) 37√ − 37) 4 55√ 4 65√ − 11 27√ 4 46√ − 2 57√ − 6 56√ 38) 39) − − 4 26√ − Answers - Multiply and Divide Radicals 5) 2x2 x3√ 6) 6a2 5a3√ 7) 2 3√ + 2 6√ 8) 5 2√ + 2 5√ 473 45 5√ 9) − − 10 15√ 10) 45 5√ + 10 15√ 11) 25n 10√ + 10 5√ 12) 5 3√ 9 5v√ − 13) − 14) 16 2 4 2√ − 9 3√ − 15) 15 11 5√ − 16) 30 + 8 3√ + 5 15√ + 4 5√ 17) 6a + a 10√ + 6a 6√ + 2a 15√ 18) − 4p 10√ + 50 p√ 19) 63 + 32 3√ 10 m√ + 25 2√ + 2m√ 5 − 20) 21) − 3√ 25 25) 26) 15√ 3 10√ 15 27) 4 3√ 9 28) 4 5√ 5 29) 5 3xy√ 12y2 30) 4 3x√ 16xy2 31) 6p√ 3 32) 2 5n√ 5 33) 34) 35) 36) 103√ 5 153√ 4 103√ 8 4√ 8 8 4√ 37) 5 10r2 2 38) 4√ 4n2 mn 22) 15√ 4 23) 1 20 24) 2 8.5 1) 4 + 2 3√ 3 2) − 4 + 3√ 12 3) 2 + 3√ 5 4) 1 3√ − 4 5) 2 13√ − 52 5 65√ 6) 85√ + 4 17√ 68 7) 9 6√ − 3 8) 30√ 2 3√ − 18 Answers - Rationalize Denominators 9) 15 5√ − 43 5 2√ 10) − 5 3√ + 20 5√ 77 11) 10 2 2√ − 23 12) 2 3√ + 2√ 2 13) − 9 3√ 12 − 11 2 2√ 4 − 14) − 15) 3 16) 5√ 3√ − 5√ − 2 17) 1 + 2√ 474 29) a2 − 2a b√ + b a2 b − b√ 30) a√ − 31) 3 2√ + 2 3√ 32) a b√ + b a√ a b − 33) a b√ + b a√ ab 34) 24 − 4 6√ + 9 2√ 15 − 3 3√ 35) − 1 + 5√ 4 36) 2 5√ − 5 2√ − 30 10 + 5 10√ 37) − 5 2√ + 10 − 5 3√ + 6√ 38) 8 + 3 6√ 10 18) 16 3√ + 4 5√ 43 19) 2√ 1 − 20) 3 + 2 3√ 21) 2√ 22) 2√ 23) a√ 24) 3 2 2√ − a√ 25) 26) 1 3 27) 4 − 28) 2 5√ 8.6 2 3√ + 2 6√ 3 2√ − − 2 15√ + 3√ + 3 2 − 1) ( m5√ )3 2) 1 4√ )3 ( 10r 3) ( 7x√ )3 4) 1 3√ )4 ( 6b 5) (6x)− 3 2 1 2 6) v 7 4 7) n− 8) (5a) 1 2 Answers - Rational Exponents 15) 1 1 2b 1 2 a 16) 1 17) 1 3a2 3 2 1 4 35 8 7 6 26) a 2b 27) m n 25 12 18) y x 5 6 19) u2v 11 2 20) 1 1 2 21) y 22) v2 7 2 u 28) 1 5 4x 3 2 y 29) 1 n 3 4 1 3 1 4 30) y x 31) xy 4 3 4 3 32) x y 10 3 1 2 3 4 33) u v 475 9) 4 10) 2 11) 8 12) 1 1000 4 3 y 13) x 14) 4 v 1 3 5 2 23) b 3 4 7 4a 3 17 6 7 4 24) 2y x 1 12 25) 3y 2 34) x 15 4 y 17 4 8.7 4 1) 4x2y3 p2) 3xy3 p3) 6 8x2y3z4 p4) 5√ 2x 5) 3√ 36xy 3y 5 6) x3y4z2 p7) 4 x2y3 p8) 5 8x4y2 p9) 4 x3y2z p10) 5y√ 3 11) 2xy2 p12) 4 3x2y3 6√ p13) 5400 14) 12√ 300125 6 15) 49x3y2 p16) 15 27y5z5 p17) 6 x3(x 2)2 − 3x(y + 4)2 p18) 4 p19) 10 x9y7 10√ p20) 4a9b9 21) 12 x11y10 20√ p22) a18b17 23) 20√ a18b17c14 Answers - Radicals of Mixed Index 24) 30 x22y11z27 p25) a a4√ 26) x x√ 27) b b9 10√ 28) a a5 12√ 6 29) xy xy5 10√ 30) a ab7 p 31) 3a2b ab 4√ 32) 2xy2 6 2x5y 33) x 12 p 59049xy11z10 34) a2b2c2 p 6√ a2b c2 6 35) 9a2(b + 1) 243a5(b + 1)5 36) 4x(y + z)3 p 6 2x(y + z) p 37) 12√ a5 38) 15√ x7 39) 12 x2y5 40) p 15√ a7b11 b 41) 10√ ab9c7 42) yz 10 xy8z3 20 43) p44) 12 p (3x 1)3 − (2 + 5x)5 p45) 15 (2x + 1)4 p46) 12 (5 3x) − p 476 8.8 1) 11 2) 3) 4) − − − 5) − 6) 8) 9) − 10) 13 11) 48 12) 24 13) 40 14) 32 15) − 16) 28 4i − 4i 3 + 9i 6i 13i − − 12i 11i 6i 2i 8i − − − − 49 21i − 17) 11 + 60i 18) − 19) 80 20) 36 32 − − 128i − 10i 36i 21) 27 + 38i 9.1 1) 3 2) 3 3) 1, 5 4) no solution 5) 2 ± Answers - Complex Numbers 22) 28 + 76i − 23) 44 + 8i 24) 16 25) − 26) 18i − 3 + 11i 1 + 13i − 27) 9i + 5 28) − 2 3i 3 − 9 29) 10i − 6 30) 4i + 2 3 31) 3i 6 − 4 32) 5i + 9 9 33) 10i + 1 2i 34) − 35) − 40i + 4 101 36) 9i 45 − 26 37) 56 + 48i 85 6i 38) 4 − 13 39) 70 + 49i 149 40) − 36 + 27i 50 41) − 5 30i 37 − 42) 48i − 85 56 43) 9i 44) 3i 5√ 45) 2 5√ − 2 6√ 46) − 47) 1 + i 3√ 2 48) 2 + i 2√ 2 i 49) 2 − 50) 3 + 2i 2√ 2 51) i i 52) − 53) 1 54) 1 55) 1 − 56) i 57) 58) 1 i − − Answers - Chapter 9 Answers - Solving with Radicals 6) 3 7) 1 4 8) no solution 9) 5 10) 7 477 11) 6 12) 46 13) 5 14) 21 15) 3 2 − 16) 7 3 − 9.2 1) 2) ± − 3) ± 4) 3 5) 6) ± − 7) − 8) 1 , 5 9) − 9.3 5 3√ 2 2 2√ 2 6√ 3, 11 5 − 1 3 5 1) 225; (x 2) 144; (a 15)2 12)2 − − − 17)2 15 − 3) 324; (m 18)2 4) 289; (x 5) 225 1 1 − 4 ; (x 324; (r 4; (y − 4 ; (p 2 )2 18)2 − 1 2)2 2 )2 17 − 6) 7) 1 8) 289 9) 11, 5 10) 4 + 2 7√ , 4 11) 4 + i 29√ , 4 2 7√ − 12) − 13) − − 1 + i 42√ , 2 + i 38√ 2 2 , − 1 − − i 38√ − 2 14) 3 + 2i 33√ 3 , 3 − 2i 33√ 3 15) 5 + i 215√ 5 , 5 − i 215√ 5 16) − 4 + 3 2√ 4 4 , − − 4 3 2√ Answers - Solving with Exponents 10) − 3 2√ 1 ± 2 11) 65, 12) 5 63 − 13) − 14) − 15) 11 , 2 16) 17) − − 7 11 2 , 5 2 5 2 − 191 64 3 8 , − 5 8 18) 9 8 19) 5 4 20) No Solution 34 3 , 10 − 21) − 22) 3 23) 17 2 − 24) No Solutoin Answers - Complete the Square 17) − 5 + 86√ , 18) 8 + 2 29√ , 8 − − 19) 9, 7 86√ 5 − 2 29√ − 1 + i 21√ , 20) 9, 1 21) − 22) 1, 23) 3 2 , 24) 3, − − 3 7 2 1 − 5 + 2i, 25) − − 26) 7 + 85√ , 7 5 − 1 − − i 21√ 2i − 85√ i 29√ 27) 7, 3 i 42√ 28) 4, 14 − 29) 1 + i 2√ , 1 i 2√ − 30) 5 + i 105√ 5 , 5 − i 105√ 5 31) 4 + i 110√ 2 , 4 − i 110√ 2 32) 1, 3 − 478 i 39√ − 33) 4 + i 39√ , 4 1. 34) − 35) 7, 1 7 − 36) 2, 37) − 6 − 6 + i 258√ 6 , − 6 − i 258√ 6 38) − 6 + i 111√ 3 , − 6 − i 111√ 3 39) 5 + i 130√ 5 , 5 − i 130√ 5 40) 2, 41) − 4 − 5 + i 87√ 2 , − 5 i 87√ − 2 42) − 7 + 181√ 2 , − 7 181√ − 2 43) 3 + i 271√ 7 , 3 − i 271√ 7 44) − 1 + 2i 6√ 2 , − 1 2i 6√ − 2 9.4 1) i 6√ 2 , i 6√ 2 − 45) 7 + i 139√ 2 , 7 − i 139√ 2 46) 5 + 3i 7√ 2 , 5 − 3i 7√ 2 47) 12 5 , 4 − 48) 1 + i 511√ 4 , 1 − i 511√ 4 49) 9 + 21√ 2 , 9 − 21√ 2 50) 1 + i 163√ 2 , 1 − i163 2 51) − 5 + i 415√ 8 , − 5 − i 415√ 8 52) 11 + i 95√ 6 , 1
1 − i 95√ 6 53) 5 + i 191√ 2 , 5 − i 191√ 2 54) 8, 7 55) 1, 56) 3, 5 2 3 2 − − Answers - Quadratic Formula 3 + 141√ 6 15) − , − 3 141√ − 6 5√ 17) − 16) 3√ , 3√ − 3 + 401√ 14 3 , − 401√ − 14 3 , 2) i 6√ i 6√ 3 3) 2 + 5√ , 2 − − 4) 5) 6√ 6 , − 6√ 6 6√ 2 , 6√ 2 − 1 + i 29√ 5 6) − , − 1 i 29√ − 5 1 3 7) 1, − 8) 1 + 31√ 2 9) 3, 3 − 10) i 2√ , 11) 3, 1 , 1 − 31√ 2 i 2√ − 12) − 13) − 1 + i, 1 i − 3 − , − i 55√ − 4 3 + i 55√ 4 14) − 3 + i 159√ 12 , − 3 i 159√ 12 − 18) − 5 + 137√ 8 , − 5 137√ − 8 19) 2, 20) 5, 21) − 22) 3, 23) 7 2 , 24√ 3 i 3√ 1 , − − 2 i 3√ 3 , − − 3 25) 7 + 3 21√ 10 , 7 − 3 21√ 10 26) − 5 + 337√ 12 5 , − 337√ − 12 27) − 3 + i 247√ 16 , − 3 i 247√ 16 − 479 28) 3 + 33√ 6 , 3 − 33√ 6 29) 1, − − 30) 2 2√ , 3 2 2 2√ 31) 4, 32) 2, − 4 − − 4 33) 4, 9 34) 2 + 3i 5√ − 7 9 2 35) 6, − 36) 5 + i 143√ 14 , 5 − i 143√ 14 37) − 3 + 345√ 14 3 , − 345√ − 14 38) 6√ 2 , 39) 26√ 2 − , 6√ 2 26√ 2 − 1 + 141√ 10 1 , − 141√ − 10 , 2 − 3i 5√ 7 40) − 9.5 Answers - Build Quadratics from Roots NOTE: There are multiple answers for each problem. Try checking your answers because your answer may also be correct. 14x + 13 = 0 22x + 40 = 0 − − − 7x + 10=0 9x + 18 = 0 1) x2 2) x2 3) x2 4) x2 5) x2 6) x2 7) x2 = 0 8) x2 + 7x + 10 = 0 8x + 16 = 0 9x = 0 − − − 7x 44 = 0 − 2x − 9) x2 − 10) x2 − 11) 16x2 12) 56x2 13) 6x2 14) 6x2 − − 3 = 0 − 16x + 3 = 0 75x + 25 = 0 − 5x + 1 = 0 7x + 2 = 0 31x + 12 = 0 20x + − − 9x 25 = 0 15) 7x2 16) 9x2 − 17) 18x2 − 18) 6x2 7x − 19) 9x2 + 53x − 20) 5x2 + 2x = 0 21) x2 22) x2 − 23) 25x2 24) x2 25) x2 26) x2 − 27) 16x2 3 = 0 28) x2 + 121 = 0 − 7 = 0 12 = 0 11 = 0 1 = 0 1 = 0 − − − − 29) x2 + 13 = 0 30) x2 + 50 = 0 31) x2 4x − − 2 = 0 32) x2 + 6x + 7 = 0 33) x2 − 2x + 10 = 0 34) x2 + 4x + 20 = 0 35) x2 − 12x + 39 = 0 36) x2 + 18x + 86 = 0 37) 4x2 + 4x 5 = 0 − 12x + 29 = 0 38) 9x2 − 39) 64x2 96x + 38 = 0 − 40) 4x2 + 8x + 19 = 0 9.6 1) 2) 3) 4) 1, 2, i, 5√ 2 2√ 2 Answers - Quadratic in Form 5) 6) 7) 8) 1, 3, 3, 6, 7 1 4 2 ± ± ± ± ± ± ± ± 480 9) 2, 4 ± ± 10) 2, 3, 1 i 3√ , − 3 3i 3√ ± 2 ± i 3√ , − i 3√ 3 ± 2 − 2, 3, 1 6√ , ± 2i 6√ 2 ± , ± 2i 3√ 3 1 3 − 125, 343 5 4 , 1 5 11) − 12) ± 13) ± 14) 1 4 , 15) − 16) − 17) 1, ± ± ± ± ± 18) 19) 20) 21) 22) 23) , 1 1 2 ± i 3√ 4 1 , − ± 2 i 3√ 3√ ± − 2, 3√ i, ± i 5√ , i 2√ ± 2√ 2 2√ , i, ± 2 ± 6 1, 2 2√ ± 24) 2, ± 23√ , 25) 1√ i 3√ , − 6√ i 108 23√ ± 2 i 3√ 1 , − ± 2 26) 1 2 , − 1 1, − ± 4 i 3√ , 1 ± i 3√ 2 9.7 1) 6 m x 10 m 2) 5 3) 40 yd x 60 yd 4) 10 ft x 18 ft 5) 6 x 10 6) 20 ft x 35 ft 7) 6” x 6” 8) 6 yd x 7 yd 9) 4 ft x 12 ft 27) 1, 2, i, ± ± 2i ± ± 28) 6, 0 (b + 3), 7 b − 4 4, 6 29) 30) − − 31) − 32) 8, 33) − 34) 2, 35) − 36) 5 , 0 2 1 − 2, 10 6 − 1, 11 4 3 − 6√ , 37) 4, 38) ± 1, 39) ± 40) 0, ± 41) 511 , 3 − 1, − 2√ , 5 3 ± 1 3 2 − 1339 24 42) 43) 44) 3, 1, 3, − ± − ± − − 45) 1, − ± 46) 1, 2, 1 , 3 1 2 , − 2, 1 3 1 − Answers - Rectangles 10) 1.54 in 11) 3 in 12) 10 ft 13) 1.5 yd 14) 6 m x 8 m 15) 7 x 9 16) 1 in 17) 10 rods 18) 2 in 481 19) 15 ft 20) 20 ft 21) 1.25 in 22) 23.16 ft 23) 17.5 ft 24) 25 ft 25) 3 ft 26) 1.145 in 9.8 1) 4 and 6 2) 6 hours 3) 2 and 3 4) 2.4 5) C = 4, J = 12 6) 1.28 days 7) 1 1 3 days 8) 12 min 9) 8 days 10) 15 days 9.9 Answers - Teamwork 11) 2 days 12) 4 4 9 days 13) 9 hours 14) 12 hours 15) 16 hours 16) 7 1 2 min 17) 15 hours 18) 18 min 19) 5 1 4 min 20) 3.6 hours 21) 24 min 22) 180 min or 3 hrs 23) Su = 6, Sa = 12 24) 3 hrs and 12 hrs 25) P = 7, S = 17 1 2 26) 15 and 22.5 min 27) A = 21, B = 15 28) 12 and 36 min Answers - Simultaneous Product 1) (2, 36), ( 2) ( 9, − 3) (10, 15), ( − 4) (8, 15), ( 18, − 20), ( 4) − 40, − 5 3) − 12) − 90, − 10, − − 5) (5, 9), (18, 2.5) 6) (13, 5), ( 20, − − 13 4 ) 9.10 1) 12 2) S4 3) 24 4) 55 5) 20 6) 30 7) 25 @ S18 8) 12 @ S6 9.11 9 2 ) 7) (45, 2), ( 8) (16, 3), ( − − 10, 9) − 8) 4) − 6, 3, 9) (1, 12), ( − 10) (20, 3), (5, 12) − 11) (45, 1), ( 12) (8, 10), ( − − , 5 3 − 10, 27) 8) − Answers - Revenue and Distance 9) 60 mph, 80 mph 17) r = 5 10) 60, 80 11) 6 km/hr 12) 200 km/hr 13) 56, 76 14) 3.033 km/hr 15) 12 mph, 24 mph 18) 36 mph 19) 45 mph 20) 40 mph, 60 mph 21) 20 mph 16) 30 mph, 40 mph 22) 4 mph 482 1) (-2,0) (4,0) (0,-8) (1,-9) 2) (-1,0) (3, 0) (0, -3) (1, -4) 3) (0,10) (1,0) (5,0) (3,-8) 4) 5) 6) (0,16) (2,0) (0,-18) (1,0) (-10,0) (4,0) (3, -2) (3,0) (3,8) (5,0) Answers - Graphs of Quadratics 13) (4,8) (2,0) (6,0) (0,-24) 14) (-3,0) (-1,-8) (1,0) (0,-6) (3,0) (4,3) (5,0) (0,-45) (1,0) (2,3) (3,0) 7) 8) 9) (-9,0) (0,5) (-1,0) (2,9) 15) (5,0) (-3,0) (-1,0) (0,9) (-2,-3) (0,45) (-3,0) 16) 17) (0,75) (3,0) (5,0) (4,-5) (0,15) (-1,0) 18) (-3,0) (-2,-5) (1,0) (2,1) (3,0) (0,-3) (3,4) (5,0) (1,0) (0,-5) 10) 11) 12) (3,0) (4,2) (5,0) (0,-30) 483 19) (-6,5) (-7,0) (-5,0) 20) (1,0) (2,5) (3,0) (0,-175) (-15,0) Answers - Chapter 10 10.1 1) a. yes b. yes c. no d. no e. yes f. no g. yes h. no 2) all real numbers 3) x 6 5 4 4) t 0 5) all real numbers 6) all real numbers 1, 4 7) x > 16 8) x − 9) x > 4, x 5 10) x ± 4 11) − 5 3 25 12) − 13) 2 10.2 1) 82 2) 20 3) 46 4) 2 5) 5 Answers - Function Notation 14) 85 7 15) − 16) 7 17 9 6 21 17) − 18) − 19) 13 20) 5 21) 11 22) − 23) 1 24) 4 − 21 25) − 26) 2 27) 28) 60 32 − − 29) 2 30) 31 32 31) 64x3 + 2 − 32) 4n + 10 33) − 1 + 3x 34) − 35) 2 12+a 4 3 2 · 3n2 1 + 2 − − 36) 1 + 1 16 x2 37) 3x + 1 38) t4 + t2 39) 5− x 3 − −2+n 40) 5 2 + 1 Answers - Operations on Functions 6) 30 − 3 7) − 8) 140 9) 1 10) 43 − 484 11) 100 74 9 26 − x3 12) − 13) 1 5 14) 27 15) − 16) n2 17) 18) − − x2 19) − 20) 2t2 − − 8t 2n 4x 2 − − x3 + 2x2 3 − 8x + 2 21) 4x3 + 25x2 + 25x 22) − 23) x2 2t3 15t2 − 4x + 5 25t − − 24) 3x2 + 4x 25) n2 + 5 3n + 5 9 − 2x + 9 26) − 27) − 2a + 5 3a + 5 28) t3 + 3t2 3t + 5 − 29) n3 + 8n + 5 30) 4x + 2 x2 + 2x 31) n6 − 32) 18n2 9n4 + 20n2 15n 25 − − 33) x + 3 34) 2 3 − 35) t4 + 8t2 + 2 10.3 1) Yes 36) 3n n2 − 6 4n − − x3 2x − 3x + 4 37) − − 38) x4 39) − 4x2 3 − − n2 − 3 2n 40) 32 + 23n 8 − n3 155 41) − 42) 5 43) 21 44) 4 45) 103 46) 12 47) 50 − 48) 112 49) 176 50) 147 51) 16x2 + 12x 4 − 8a + 14 8a + 2 52) 53) − − 54) t 55) 4x3 12n 16 − 2n2 − 2x + 8 56) − 57) − 58) 27t3 − 16t 5 59) − − 60) 3x3 + 6x2 4 − 108t2 + 141t 60 − Answers - Inverse Functions 2) No 485 3 − 2)3 + 2 1 − 3) Yes 4) Yes 5) No 6) Yes 7) No 8) Yes 9) Yes 10) No 11) f − 1(x) = x 5√ 12) g− 1(x) = (x 13) g− 1(x) = 4 − 2x − x 14) f − 1(x) = − 3 + 3x x 15) f − 1(x) = − 2x − x + 2 2 16) g− 1(x) = 3x 9 − 10.4 1) 0 1 2) − 3) 0 5) 6) − − 7) − 8) 0 9) − 10) 0 11) 5 6 12) 0 13) 14) 2 5 6 − − 17) f − 18) f − 1(x) = − 1(x) = 15 + 2x 5x + 10 5 19) g− 20) f − 1(x) = − 1(x) = − x3√ + 1 4x + 12 3 21) f − 1(x) = x3√ + 3 2x5 + 2 22) g− 23) g− 24) f − 1(x) = − 1(x) = x x − 3x 1(x − 2x 1 25) f − 1(x) = − x 26) h− 27) g− 28) g− 29) g− 1(x) = − x − 4x + 8 5 3x + 2 1(x) = − 1(x) = − 1(x) = − x + 1 5 30) f − 1(x) = 5 + 4x 5 31) g− 3√ 1(x) = x + 1 32) f − 1(x) = − 5 x + 3 2 33) h− 34) g− q − 1(x) = ( 2x + 4)3 4 2 + 1 3√ 1(x) = x − 2x + 1 1(x) = − x 1 − 36) f − 1(x) = − x 1 − x 37) f − 1(x) = 2x + 7 x + 3 4x 3 x 38) f − 1(x) = − 39) g− 40) g− 1(x) = − 1(x) = − 3x + 1 2 35) f − 3 Answers - Exponential Functions 15) 1 16) 1 − 29) 0 30) No solution 17) No solution 4 3 18) − 19) 1 4 − 20) 3 4 21) No solution − 22) 0 23) − 24) 2 5 25) − 26) 1 4 27) − 28) 1 3 3 2 1 1 2 486 31) 1 32) 3 33) 1 3 34) 2 3 35) 0 36) 0 37) 3 8 38) 39) 1 3 − − 40) No solution 10.5 1) 92 = 81 2) b− 16 = a 2 = 1 49 4) 162 = 256 3) 7− 5) 132 = 169 6) 110 = 1 7) log8 1 = 0 2 1 289 = 8) log17 − 9) log15 225 = 2 10) log144 12 = 1 2 11) log64 2 = 1 6 12) log19 361 = 2 13) 1 3 14) 3 10.6 1) Answers - Logarithmic Functions 1 3 15) − 16) 0 17) 2 3 18) − 19) 2 20) 1 2 21) 6 22) 5 23) 5 24) 512 25) 1 4 26) 1000 27) 121 28) 256 29) 6552 30) 45 11 125 3 1 4 54 11 2401 3 1 2 1 11 621 10 31) 32) 33) 34) 35) 36) − − − − − − 37) − 38) 283 243 39) 2 5 40) 3 Answers - Interest Rate Problems a. 740.12; 745.91 b. 851.11; 859.99 c. 950.08; 953.44 e. 1209.52; 1214.87 f. 1528.02; 1535.27 g. 2694.70; 2699.72 d. 1979.22; 1984.69 h. 3219.23; 3224.99 i. 7152.17; 7190.52 2) 1640.70 3) 2868.41 4) 2227.41 5) 1726.16 6) 1507.08 7) 2001.60 8) 2009.66 9) 2288.98 10) 6386.12 11) 13742.19 487 12) 28240.43 13) 12.02; 3.96 14) 3823.98 15) 101.68 10.7 1) 0.3256 2) 0.9205 3) 0.9659 4) 0.7660 5) 7 25 6) 8 15 7) 7 16 8) 3 5 9) 2√ 2 10) 4 5 11) 16.1 12) 2.8 13) 32 10.8 1) 29◦ 2) 39◦ 3) 41◦ 4) 52◦ 5) 24◦ 6) 32◦ 7) 15◦ 8) 18◦ 9) 27◦ 10) 35◦ Answers - Trigonometric Functions 14) 8.2 15) 26.1 16) 16.8 17) 2.2 18) 9.8 19) 17.8 20) 10.3 21) 3.9 22) 10.6 23) 10.2 24) 8.9 25) 9.5 26) 24.4 27) 4.7 28) 14.6 29) 1 30) 8 31) 1.5 32) 7.2 33) 5.5 34)2 35) 41.1 36) 3.2 37) 18.2 38) 3.3 39) 17.1 40) 22.2 Answers - Inverse Trigonometric Functions 11) 36◦ 12) 61.7◦ 13) 54◦ 14) 46.2◦ 15) 55.2◦ 16) 42.7◦ 17) 58◦ 18) 20.1◦ 19) 45.2◦ 20) 73.4◦ 488 21) 51.3◦ 22) 45◦ 23) 56.4◦ 24) 48.2◦ 25) 55◦ 26) 30.5◦ 27) 47◦ 28) 15.5◦ 29) 30◦ 30) 59◦ 32) m∠B = 22.8◦, m∠A = 67.2◦, c = 16.3 33) m∠B = 22.5◦, m∠A = 67.5◦, c = 7.6 34) m∠A = 39◦, b = 7.2, a = 5.9 35) m∠B = 64.6◦, m∠A = 25.4◦, b = 6.3 36) m∠A = 69◦, b = 2.5, a = 6.5 37) m∠B = 38◦, b = 9.9, a = 12.6 38) m∠B = 42◦, b = 9.4, c = 14 39) m∠A = 45◦, b = 8, c = 11.3 40) m∠B = 29.1◦, m∠A = 60.9◦, 31) m∠B = 28◦, b = 15.1, c = 32.2 a = 12.2 489
20 From this table we can see that the ratio of the column areas, 68 : 92 : 40, is exactly the same as the ratio of the frequencies, 34 : 46 : 20. In a histogram, the area of a column represents the frequency of the corresponding class, so that the area must be proportional to the frequency. We may see this written as ‘area ∝ frequency’. This also means that in every histogram, just as in the example above, the ratio of column areas is the same as the ratio of the frequencies, even if the classes do not have equal widths. Also, there can be no gaps between the columns in a histogram because the upper boundary of one class is equal to the lower boundary of the neighbouring class. A gap can appear only when a class has zero frequency. The axis showing the measurements is labelled as a continuous number line, and the width of each column is equal to the width of the class that it represents. When we construct a histogram, since the classes may not have equal widths, the height of each column is no longer determined by the frequency alone, but must be calculated so that area ∝ frequency. The vertical axis of the histogram is labelled frequency density, which measures frequency per standard interval. The simplest and most commonly used standard interval is 1 unit of measurement. For example, a column representing 85 objects with masses from 50 to 60 kg has a frequency density of per gram and so on. 85objects − (60 50) kg = KEY POINT 1.3 8.5 objects per kilogram or 0.0085 objects For a standard interval of 1 unit of measurement, rearranged to give Frequency density = class frequency class width , which can be Class frequency = class width × frequency density In a histogram, we can see the relative frequencies of classes by comparing column areas, and we can make estimates by assuming that the values in each class are spread evenly over the whole class interval. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 1.2 The masses, m kg, of 100 children are grouped into two classes, as shown in the table. Mass m( kg) No. children f( ) ø <m40 50 40 ø <m50 70 60 a Illustrate the data in a histogram. b Estimate the number of children with masses between 45 and 63kg. Answer a Mass m( kg) ø <m40 50 ø <m50 70 No. children f( ) Class width (kg) 40 10 60 20 Frequency density ÷ 40 10 = 4 ÷ 60 20 = 3 Frequency density is calculated for the unequal-width intervals in the table. The masses are represented in the histogram, where frequency density measures number of children per 1kg or simply children per kg 40 45 50 55 Mass (m kg) 60 63 65 70 b There are children with masses from 45 to 63kg in both classes, so we must split this interval into two parts: 45 –50 and 50 – 63. × 40 = 20 children 1 2 Frequency width = × frequency density = (63 50) kg − × 3 children 1kg = = 13 × 3 children 39 children Our estimate is + 20 39 = 59 children. The class 40−50 kg has a mid-value of 45kg and a frequency of 40. Our estimate for the interval 50–63kg is equal to the area corresponding to this section of the second column. We add together the estimates for the two intervals. Copyright Material - Review Only - Not for Redistribution TIP Column areas are equal to class frequencies. For example, the area of the first column is 4 children 1kg (50 40) kg − × = 40 children. TIP If we drew column heights of 8 and 6 instead of 4 and 3, then frequency density would measure children per 2 kg. The area of the first column would be (50 40) kg − × 8 children 2 kg = 40 children. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data Consider the times taken, to the nearest minute, for 36 athletes to complete a race, as given in the table below. TIP Time taken (min) No. athletes f( ) 13 4 14 15− 14 16 18− 18 Gaps of 1 minute appear between classes because the times are rounded. Frequency densities are calculated in the following table. Time taken (t min) 12.5 ø <t 13.5 13.5 ø <t 15.5 15.5 ø <t 18.5 No. athletes f( ) Class width (min) Frequency density 4 1 14 2 18 3 4 ÷ = 1 4 14 ÷ = 2 7 18 ÷ = 3 6 This histogram represents the race times, where frequency density measures athletes per minute 12.5 13.5 15.5 Time taken (t min) 18.5 WORKED EXAMPLE 1.3 Use the histogram of race times shown previously to estimate: a the number of athletes who took less than 13.0 minutes b the number of athletes who took between 14.5 and 17.5 minutes c the time taken to run the race by the slowest three athletes. Answer Use class boundaries (rather than rounded values) to find class widths, otherwise incorrect frequency densities will be obtained. TIP The class with the highest frequency does not necessarily have the highest frequency density. TIP Do think carefully about the scales you use when constructing a histogram or any other type of diagram. Sensible scales, such as 1cm for 1, 5, 10, 20 or 50 units, allow you to read values with much greater accuracy than scales such as 1cm for 3, 7 or 23 units. For similar reasons, try to use as much of the sheet of graph paper as possible, ensuring that the whole diagram will fit before you start to draw it. 9 We can see that two blocks represent one athlete in the histogram. So, instead of calculating with frequency densities, we can simply count the number of blocks and divide by 2 to estimate the number of athletes involved. a b ÷ = 4 2 2 athletes There are four blocks to the left of 13.0 minutes. 38 2 19 athletes ÷ = There are + 14 24 = 38 blocks between 14.5 and 17.5 minutes. c Between 18.0 and 18.5 minutes. The slowest three athletes are represented by the six blocks to the right of 18.0 minutes. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXPLORE 1.1 Refer back to the table in Section 1.3 that shows the percentage scores of 100 students who took an examination. Discuss what adjustments must be made so that the data can be represented in a histogram. How could we make these adjustments and is there more than one way of doing this? TIP It is not acceptable to draw the axes or the columns of a histogram freehand. Always use a ruler! EXERCISE 1B 1 In a particular city there are 51 buildings of historical interest. The following table presents the ages of these buildings, given to the nearest 50 years. Age (years) No. buildings f( ) 50–150 15 200–300 18 350–450 12 500–600 6 a Write down the lower and upper boundary values of the class containing the greatest number of buildings. b State the widths of the four class intervals. c Illustrate the data in a histogram. d Estimate the number of buildings that are between 250 and 400 years old. 10 2 The masses, m grams, of 690 medical samples are given in the following table. Mass (m grams) No. medical samples f( ) ø <m4 12 224 ø <m12 24 396 ø <m24 28 p a Find the value of p that appears in the table. b On graph paper, draw a histogram to represent the data. c Calculate an estimate of the number of samples with masses between 8 and 18 grams. 3 The table below shows the heights, in metres, of 50 boys and of 50 girls. Height (m) No. boys f( ) No. girls f( ) 1.2– 7 10 1.3– 11 22 1.6– 26 16 1.8–1.9 6 2 a How many children are between 1.3 and 1.6 metres tall? b Draw a histogram to represent the heights of all the boys and girls together. c Estimate the number of children whose heights are 1.7 metres or more. 4 The heights of 600 saplings are shown in the following table. Height (cm) No. saplings f( ) 0– 64 5– 232 15– 240 30– u 64 a Suggest a suitable value for u, the upper boundary of the data. b Illustrate the data in a histogram. c Calculate an estimate of the number of saplings with heights that are: i less than 25cm ii between 7.5 and 19.5cm. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data 5 Each of the 70 trainees at a secretarial college was asked to type a copy of a particular document. The times taken are shown, correct to the nearest 0.1 minutes, in the following table. Time taken (min) No. trainees f( ) 2.6–2.8 15 2.9–3.0 25 3.1–3.2 20 3.3–3.7 10 a Explain why the interval for the first class has a width of 0.3 minutes. b Represent the times taken in a histogram. c Estimate, to the nearest second, the upper boundary of the times taken by the fastest 10 typists. d It is given that 15 trainees took between 3.15 and b minutes. Calculate an estimate for the value of b when: i >b 3.15 ii <b 3.15. 6 A railway line monitored 15% of its August train journeys to find their departure delay times. The results are shown below. It is given that 24 of these journeys were delayed by less than 2 minutes 20 16 12 8 4 0 2 4 12 Time (min) 18 20 11 a How many journeys were monitored? b Calculate an estimate of the number of these journeys that were delayed by: i 1 to 3 minutes ii 10 to 15 minutes. c Show that a total of 2160 j
ourneys were provided in August. d Calculate an estimate of the number of August journeys that were delayed by 3 to 7 minutes. State any assumptions that you make in your calculations. 7 A university investigated how much space on its computers’ hard drives is used for data storage. The results are shown below. It is given that 40 hard drives use less than 20GB for data storage 20 60 120 200 Storage space (GB) a Find the total number of hard drives represented. b Calculate an estimate of the number of hard drives that use less than 50GB. c Estimate the value of k, if 25% of the hard drives use k GB or more. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 8 The lengths of the 575 items in a candle maker’s workshop are represented in the histogram. a What proportion of the items are less than 25cm long? b Estimate the number of items that are between 12.4 and 36.8cm long. c The shortest 20% of the workshop’s items are to be recycled. Calculate an estimate of the length of the shortest item that will not be recycled. 9 The thicknesses, k mm, of some steel sheets are represented in the histogram. It is given that 0.4<k for 180 sheets. a Find the ratio between the frequencies of the three classes. Give your answer in simplified form. b Find the value of n, given that frequency density measures sheets per n mm. c Calculate an estimate of the number of sheets for which: i <k 0.5 ii 0.75 kø < 0.94 . 25 20 15 10 12 10 20 30 40 50 Length (l cm) 0.2 0.4 0.6 0.8 Thickness (k mm) 1.0 1.2 d The sheets are classified as thin, medium or thick in the ratio 1: 3 : 1. Estimate the thickness of a medium sheet, giving your answer in the form < < your values for a and b? a k b. How accurate are PS 10 The masses, in kilograms, of the animals treated at a veterinary clinic in the past year are illustrated in a histogram. The histogram has four columns of equal height. The following table shows the class intervals and the number of animals in two of the classes. Mass (kg) No. animals f( ) 3–5 a 6–12 371 13–32 1060 33–44 b a Find the value of a and of b, and show that a total of 2226 animals were treated at the clinic. b Calculate an estimate of the lower boundary of the masses of the heaviest 50% of these animals. PS 11 The minimum daily temperature at a mountain village was recorded to the nearest days. The results are grouped into a frequency table and a histogram is drawn. °0.5 C on 200 consecutive The temperatures ranged from °0.5 C to °2 C on n days, and this class is represented by a column of height h cm. The temperatures ranged from –2.5 to column that represents these temperatures. ° –0.5 C on d days. Find, in terms of n h, and d , the height of the PS 12 The frequency densities of the four classes in a histogram are in the ratio 4 : 3 : 2 : 1. The frequencies of these classes are in the ratio 10 : 15 : 24 : 8. Find the total width of the histogram, given that the narrowest class interval is represented by a column of width 3cm. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data PS 13 The percentage examination scores of 747 students are given in the following table. Score %%( ) No. students f( ) p –50 165 51–70 240 71–80 195 81–q 147 Given that the frequency densities of the four classes of percentage scores are in the ratio 5 : 8 : 13 : 7, find the value of p and of q. DID YOU KNOW? Bar charts first appeared in a book by the Scottish political economist William Playfair, entitled The Commercial and Political Atlas (London, 1786). His invention was adopted by many in the following years, including Florence Nightingale, who used bar charts in 1859 to compare mortality in the peacetime army with the mortality of civilians. This helped to convince the government to improve army hygiene. In the past few decades histograms have played a very important role in image processing and computer vision. An image histogram acts as a graphical representation of the distribution of colour tones in a digital image. By making adjustments to the histogram, an image can be greatly enhanced. This has had great benefits in medicine, where scanned images are used to diagnose injury and illness. FAST FORWARD In Chapter 2, Section 2.3 and in Chapter 8, Section 8.1, we will see how a histogram or bar chart can be used to show the shape of a set of data, and how that shape provides information on average values. 1.4 Representation of continuous data: cumulative frequency graphs A cumulative frequency graph can be used to represent continuous data. Cumulative frequency is the total frequency of all values less than a given value. If we are given grouped data, we can construct the cumulative frequency diagram by plotting cumulative frequencies (abbreviated to cf ) against upper class boundaries for all intervals. We can join the points consecutively with straight-line segments to give a cumulative frequency polygon or with a smooth curve to give a cumulative frequency curve. For example, a set of data that includes 100 values below 7.5 and 200 values below 9.5 will have two of its points plotted at (7.5, 100) and at (9.5, 200). We plot points at upper boundaries because we know the total frequencies up to these points are precise. From a cumulative frequency graph we can estimate the number or proportion of values that lie above or below a given value, or between two values. There is no rule for deciding whether a polygon or curve is the best type of graph to draw. It is often difficult to fit a smooth curve through a set of plotted points. Also, it is unlikely that any two people will draw exactly the same curve. A polygon, however, is not subject to this uncertainty, as we know exactly where the line segments must be drawn. 13 TIP A common mistake is to plot points at class mid-values but the total frequency up to each mid-value is not precise – it is an estimate. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 By drawing a cumulative frequency polygon, we are making exactly the same assumptions that we made when we used a histogram to calculate estimates. This means that estimates from a polygon should match exactly with estimates from a histogram. WORKED EXAMPLE 1.4 The following table shows the lengths of 80 leaves from a particular tree, given to the nearest centimetre. Lengths (cm) No. leaves f( ) 1–2 8 3–4 20 5–7 38 8–9 10 10–11 4 Draw a cumulative frequency curve and a cumulative frequency polygon. Use each of these to estimate: a the number of leaves that are less than 3.7 cm long b the lower boundary of the lengths of the longest 22 leaves. Answer Lengths l( cm) l < 0.5 l < 2..5 7.5 9.5 11.5 14 Addition of frequencies No. leaves cf( ) 0 0 8+ + + 0 8 20 + + 0 8 20 38 + 0 8 20 38 10 + + + + + 0 8 20 38 10 4 + + + + 0 8 28 66 76 80 Before the diagrams can be drawn, we must organise the given data to show upper class boundaries and cumulative frequencies. 76 can also be calculated as 66 + 10, using the previous cf value 80 60 40 20 0 2 4 6 8 10 12 Length (l cm) a The polygon gives an estimate of 20 leaves. The curve gives an estimate of 18 leaves. b The polygon gives an estimate of 6.9cm. The curve gives an estimate of 6.7 cm. We plot points at (0.5,0), (2.5,8), (4.5,28), (7.5,66), (9.5,76) and (11.5, 80). We then join them in order with ruled lines and also with a smooth curve, to give the two types of graph. Copyright Material - Review Only - Not for Redistribution REWIND In Worked example 1.2, we made estimates from a histogram by assuming that the values in each class are spread evenly over the whole class interval. TIP • Do include the lowest class boundary, which has a cumulative frequency of 0, and plot this point on your graph. • The dotted lines show the workings for parts a and b. • When constructing a cumulative frequency graph, you are advised to use sensible scales that allow you to plot and read values accurately. • Estimates from a polygon and a curve will not be the same, as they coincide only at the plotted points. • Note that all of these answers are only estimates, as we do not know the exact shape of the cumulative frequency graph between the plotted points. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data EXPLORE 1.2 Four histograms (1– 4) that represent four different sets of data with equal-width intervals are shown. 1 3 fd 0 fd 0 2 4 fd 0 fd 0 a Which of the following cumulative frequency graphs (A–D) could represent the same set of data as each of the histograms (1– 4) above? A C cf 0 cf 0 B D cf 0 cf 0 How do column heights affect the shape of a cumulative frequency graph? b Why could a cumulative frequency graph never look like the sketch shown below? cf 0 Copyright Material - Review Only - Not for Redistribution 15 FAST FORWARD We will use cu
mulative frequency graphs to estimate the median, quartiles and percentiles of a set of data in Chapter 2, Section 2.3 and in Chapter 3, Section 3.2. WEB LINK You can investigate the relationship between histograms and cumulative frequency graphs by visiting the Cumulative Frequency Properties resource on the Geogebra website (www.geogebra.org). Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXERCISE 1C 1 The reaction times, t seconds, of 66 participants were measured in an experiment and presented below. Time (t seconds) No. participants cf( ) t < 1.5 t < 3.0 t < t < 4.5 6.5 t < 8.5 t < 11.0 t < 13.0 0 3 8 32 54 62 66 a Draw a cumulative frequency polygon to represent the data. b Use your graph to estimate: i the number of participants with reaction times between 5.5 and 7.5 seconds ii the lower boundary of the slowest 20 reaction times. 2 The following table shows the widths of the 70 books in one section of a library, given to the nearest 16 centimetre. Width (cm) No. books f( ) 10–14 3 15–19 13 20–29 25 30–39 24 40–44 5 a Given that the upper boundary of the first class is 14.5cm, write down the upper boundary of the second class. b Draw up a cumulative frequency table for the data and construct a cumulative frequency graph. c Use your graph to estimate: i the number of books that have widths of less than 27 cm ii the widths of the widest 20 books. 3 Measurements of the distances, x mm, between two moving parts inside car engines were recorded and are summarised in the following table. There were 156 engines of type A and 156 engines of type B. Distance x( mm) x < 0.10 x < 0.35 x < 0.60 x < 0.85 x < 1.20 Engine A cf( ) Engine B cf( ) 0 0 16 8 84 52 134 120 156 156 a Draw and label two cumulative frequency curves on the same axes. b Use your graphs to estimate: i the number of engines of each type with measurements between 0.30 and 0.70 mm ii the total number of engines with measurements that were less than 0.55 mm. c Both types of engine must be repaired if the distance between these moving parts is more than a certain fixed amount. Given that 16 type A engines need repairing, estimate the number of type B engines that need repairing. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data 4 The diameters, d cm, of 60 cylindrical electronic components are represented in the following cumulative frequency graph ( 60 40 20 0 0.1 0.2 0.3 0.4 0.5 0.6 Diameter (d cm) a Find the number of components such that 0.2 ø d < 0.4 , and explain why your answer is not an estimate. b Estimate the number of components that have: i a diameter of less than 0.15cm ii a radius of 0.16cm or more. c Estimate the value of k, given that 20% of the components have diameters of k mm or more. d Give the reason why 0.1–0.2 cm is the modal class. 17 5 The following cumulative frequency graph shows the masses, m grams, of 152 uncut diamonds 160 120 80 40 0 P 4 8 12 16 20 24 Mass (m grams) a Estimate the number of uncut diamonds with masses such that: i ø < 17m 9 ii 7.2 ø < m . 15.6 b The lightest 40 diamonds are classified as small. The heaviest 40 diamonds are classified as large. Estimate the difference between the mass of the heaviest small diamond and the lightest large diamond. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 c The point marked at P(24, 152) on the graph indicates that the 152 uncut diamonds all have masses of less than 24 grams. Each diamond is now cut into two parts of equal mass. Assuming that there is no wastage of material, write down the coordinates of the point corresponding to P on a cumulative frequency graph representing the masses of these cut diamonds. 6 The densities, d g/cm3, of 125 chemical compounds are given in the following table. Density d( g/cm )3 No. compounds cf( ) < d 1.30 32 < d 1.38 77 < d 1.42 92 < d 1.58 107 < d 1.70 125 Find the frequencies a b c , , and d given in the table below. Density d( g/cm )3 No. compounds f( ) 0– a 1.30– b 1.38– c 1.42–1.70 d PS 7 The daily journey times for 80 bank staff to get to work are given in the following table. Time (t min) No. staff cf( ) t 10< 3 t 15< 11 20< t 24 25< t 56 t 30< 68 45< t 76 t 60< 80 18 a How many staff take between 15 and 45 minutes to get to work? b Find the exact number of staff who take + x y 2 minutes or more to get to work, given that 85% of the staff take less than x minutes and that 70% of the staff take y minutes or more. M 8 A fashion company selected 100 12-year-old boys and 100 12-year-old girls to audition as models. The heights, h cm, of the selected children are represented in the following graph 100 75 50 25 0 Boys Girls 140 150 160 170 180 Height (h cm) a What features of the data suggest that the children were not selected at random? b Estimate the number of girls who are taller than the shortest 50 boys. c What is the significance of the value of h where the graphs intersect? d The shortest 75 boys and tallest 75 girls were recalled for a second audition. On a cumulative frequency graph, show the heights of the children who were not recalled. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data 9 The following table shows the ages of the students currently at a university, given by percentage. Ages are rounded down to the number of whole years. Age (years) Students %%( ) 18< 0 18–19 20 – 21 22 – 24 25– 28 29– 35 36– 44 27 51 11 5 4 2 a Represent the data in a percentage cumulative frequency polygon. b The oldest 8% of these students qualify as ‘mature’. Use your polygon to estimate the minimum age requirement for a student to be considered mature. Give your answer to the nearest month. c Of the 324 students who are 18–19 years old, 54 are not expected to find employment within 3 months of finishing their course. i Calculate an estimate of the number of current students who are expected to find employment within 3 months of finishing their course. ii What assumptions must be made to justify your calculations in part c i? Are these assumptions reasonable? Do you expect your estimate to be an overestimate or an underestimate? PS 10 The distances, in km, that 80 new cars can travel on 1 litre of fuel are shown in the table. Distance (km) No. cars f( ) 4.4– 5 6.6– 7 8.8– 52 12.1– 12 15.4–18.7 4 These distances are 10% greater than the distances the cars will be able to travel after they have covered more than 100 000 km. Estimate how many of the cars can travel 10.5km or more on 1 litre of fuel when new, but not after they have covered more than 100 000 km. 19 PSM 11 A small company produces cylindrical wooden pegs for making garden chairs. The lengths and diameters of the 242 pegs produced yesterday have been measured independently by two employees, and their results are given in the following table. Length ( cm) l No. pegs ( )cf 1.0< l 2.0< l 2.5< l 3.0< l 3.5< l 4.0< l 4.5< l 0 0 8 40 110 216 242 Diameter ( cm) d No. pegs ( )cf d 1.0< d 1.5< d 2.0< d 2.5< d 3.0< 0 60 182 222 242 a On the same axes, draw two cumulative frequency graphs: one for lengths and one for diameters. b Correct to the nearest millimetre, the lengths and diameters of n of these pegs are equal. Find the least and greatest possible value of n. c A peg is acceptable for use when it satisfies both l ù and <d 2.8 2.2. Explain why you cannot obtain from your graphs an accurate estimate of the number of these 242 pegs that are acceptable. Suggest what the company could do differently so that an accurate estimate of the proportion of acceptable pegs could be obtained. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXPLORE 1.3 The following table shows data on the masses, m grams, of 150 objects. Mass ( g)m m 0< m 12< m 30< m 53< m 70< m 80< No. objects cf( ) 0 24 60 106 138 150 By drawing a cumulative frequency polygon, the following estimates will be obtained: a Number of objects with masses less than 20 g = 40 objects. TIP The six plotted points, whose coordinates you know, are joined by straight lines. b Arranged in ascending order, the mass of the 100th object 50 g. = However, we can calculate these estimates from the information given in the table without drawing the polygon. Investigate the possible methods that we can use to calculate these two estimates. 1.5 Comparing different data representations Pictograms, vertical line graphs, bar charts and pie charts are useful ways of displaying qualitative data and ungrouped quantita
tive data, and people generally find them easy to understand. Nevertheless, it may be of benefit to group a set of raw data so that we can see how the values are distributed. Knowing the proportion of small, medium and large values, for example, may prove to be useful. For small datasets we can do this by constructing stem-and-leaf diagrams, which have the advantage that raw values can still be seen after grouping. 20 In large datasets individual values lose their significance and a picture of the whole is more informative. We can use frequency tables to make a compact summary by grouping but most people find the information easier to grasp when it is shown in a graphical format, which allows absolute, relative or cumulative frequencies to be seen. Although some data are lost by grouping, histograms and cumulative frequency graphs have the advantage that data can be grouped into classes of any and varied widths. The choice of which representation to use will depend on the type and quantity of data, the audience and the objectives behind making the representation. Most importantly, the representation must show the data clearly and should not be misleading in any way. The following chart is a guide to some of the most commonly used methods of data representation. FAST FORWARD In Chapter 2 and Chapter 3, we will see how grouping effects the methods we use to find measures of central tendency and measures of variation. Qualitative data Discrete quantitative data Continuous data Ungrouped Grouped Pictogram, vertical line graph, bar chart, pie chart, sectional bar chart Small amount Stem-and-leaf diagram Large amount Histogram Cumulative frequency graph Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data EXERCISE 1D 1 Jamila noted each student’s answer when her year group was asked to name their favourite colour. a List the methods of representation that would be suitable for displaying Jamila’s data. b Jamila wishes to emphasise that the favourite colour of exactly three-quarters of the students is blue. Which type of representation from your list do you think would be the most effective for Jamila to use? Explain why you have chosen this particular type of representation. 2 A large number of chickens’ eggs are individually weighed. The masses are grouped into nine classes, each of width 2 grams, from 48 to 66g. Name a type of representation in which the fact could be seen that the majority of the eggs have masses from 54 to 60 g. Explain how the representation would show this. 3 Boxes of floor tiles are to be offered for sale at a special price of $75. The boxes claim to contain at least 100 tiles each. a Why would it be preferable to use a stem-and-leaf diagram rather than a bar chart to represent the numbers of tiles, which are 112, 116, 107, 108, 121, 104, 111, 106, 105 and 110? b How may the seller benefit if the numbers 12, 16, 7, 8, 21, 4, 11, 6, 5 and 10 are used to draw the stem-and- leaf diagram instead of the actual numbers of tiles? 4 A charity group’s target is to raise a certain amount of money in a year. At the end of the first month the group raised 36% of the target amount, and at the end of each subsequent month they manage to raise exactly half of the amount outstanding. 21 a How many months will it take the group to raise 99% of the money? b Name a type of representation that will show that the group fails to reach its target by the end of the year. Explain how this fact would be shown in the representation. 5 University students measured the heights of the 54 trees in the grounds of a primary school. As part of a talk on conservation at a school assembly, the students have decided to present their data using one of the following diagrams 12 Heights of trees (m) Heights of 54 trees tall short medium 2 to 3 m 3 to 5 m 5 to 8 m a Give one disadvantage of using each of the representations shown. b Name and describe a different type of representation that would be appropriate for the audience, and that has none of the disadvantages given in part a. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 6 The percentage scores of 40 candidates who took a Health and Safety test are given: 77 44 65 84 52 60 35 83 68 66 50 68 65 57 60 50 93 38 46 55 45 69 61 64 40 66 91 59 61 74 70 75 42 65 85 63 73 84 68 30 a Construct a frequency table by grouping the data into seven classes with equal-width intervals, where the first class is 30–39. Label this as Table 1. b It is proposed that each of the 40 candidates is awarded one of three grades, A, B or C. Construct a new frequency table that matches with this proposal. Label this as Table 2. c A student plans to display all three versions of the data (i.e. the raw data, the data in Table 1 and the data in Table 2) in separate stem-and-leaf diagrams. For which version(s) of the data would this not be appropriate? Suggest an alternative type of representation in each case. M 7 Last year Tom renovated an old building during which he worked for at least 9 hours each week. By plotting four points in a graph, he has represented the time he spent working 52 51 50 49 22 0 34 36 38 40 42 44 Time spent working (hours) a What can you say about the time that Tom spent working on the basis of this graph? b Explain why Tom’s graph might be considered to be misleading. c Name the different types of representation that are suitable for displaying the amount of time that Tom worked each week throughout the year. Consider the benefits of each type of representation and then fully describe (but do not draw) the one you believe to be the most suitable. PS 8 The following table shows the focal lengths, l mm, of the 84 zoom lenses sold by a shop. For example, there are 18 zoom lenses that can be set to any focal length between 24 and 50 mm. Focal length l( mm) No. lenses f( ) 24–50 18 50–108 100–200 150–300 250–400 30 18 12 6 a What feature of the data does not allow them to be displayed in a histogram? b What type of diagram could you use to illustrate the data? Explain clearly how you would do this. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data M 9 The following table shows estimates, in hundred thousands, of the number of people living in poverty and the populations, in millions, of the countries where they live. (Source: World Bank 2011/12) Country No. living in poverty ××( 10 )5 Population ××( 10 )6 Chile Sri Lanka Malaysia Georgia Mongolia 24.8 17.25 18.4 20.65 10.8 28.33 7.91 4.47 7.50 2.74 Represent, in a single diagram, the actual numbers and the relative poverty that exists in these countries. In what way do the two sets of data in your representation give very different pictures of the poverty levels that exist? Which is the better representation to use and why? EXPLORE 1.4 Past, present and predicted world population figures by age group, sex and other categories can be found on government census websites. You may be interested, for example, in the population changes for your own age group during your lifetime. This is something that can be represented in a diagram, either manually, using spreadsheet software or an application such as GeoGebra, and for which you may like to try making predictions by looking for trends shown in the raw data or in any diagrams you create. Checklist of learning and understanding ● Non-numerical data are called qualitative or categorical data. ● Numerical data are called quantitative data, and are either discrete or continuous. ● Discrete data can take only certain values. ● Continuous data can take any value, possibly within a limited range. ● Data in a stem-and-leaf diagram are ordered in rows with intervals of equal width. ● In a histogram, column area ∝ frequency, and the vertical axis is labelled frequency density. ● Frequency density = class frequency class width and Class frequency = class width frequency density × . ● In a cumulative frequency graph, points are plotted at class upper boundaries. 23 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 END-OF-CHAPTER REVIEW EXERCISE 1 1 The weights of 220 sausages are summarised in the following table. Weight (grams) Cumulative frequency 20< 0 30< 20 40< 50 45< 100 50< 160 60< 210 70< 220 i State how many sausages weighed between 50 g and 60 g. ii On graph paper, draw a histogram to represent the weights of the sausages. [1] [4] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q4 November 2011[Adapted] 2 The lengths of children’s feet, in centimetres, are classified as 14–16, 17–19, 20–22 and so on. State the lower class boundary, the class width and the class mid-value for the lengths given as 17–19. 3 The capacities of ten engines, in litres, are
given rounded to 2 decimal places, as follows: 1.86, 2.07, 1.74, 1.08, 1.99, 1.96, 1.83, 1.45, 1.28 and 2.19. These capacities are to be grouped in three classes as 1.0–1.4, 1.5– 2.0 and 2.1– 2.2. a Find the frequency of the class 1.5–2.0 litres. b Write down two words that describe the type of data given about the engines. 4 Over a 10-day period, Alina recorded the number of text messages she received each day. The following stem-and-leaf diagram shows her results Key: 1 5 represents 15 messages 24 messages? a On how many days did she receive more than 10 but not more than 15 b How many more rows would need to be added to the stem-and-leaf diagram if Alina included data for two extra days on which she received 4 and 36 messages? Explain your answer. 5 The following stem-and-leaf diagram shows eight randomly selected numbers between 2 and 4 represents 2.1 Key: 2 1 Given that a b– find the value of a and of b. = 7 and that the sum of the eight numbers correct to the nearest integer is 24, [2] [1] [2] [1] [2] [3] 6 Eighty people downloaded a particular application and recorded the time taken for the download to complete. The times are given in the following table. Download time (min) No. downloads cf( ) 3< 6 5< 18 6< 66 10< 80 a Find the number of downloads that completed in 5 to 6 minutes. b On a histogram, the download times from 5 to 6 minutes are represented by a column of height 9.6cm. Find the height of the column that represents the download times of 6 to 10 minutes. [1] [2] PS 7 A histogram is drawn with three columns whose widths are in the ratio 1: 2 : 4. The frequency densities of these classes are in the ratio 16 : 12 : 3, respectively. a Given that the total frequency of the data is 390, find the frequency of each class. b The classes with the two highest frequencies are to be merged and a new histogram drawn. Given that the height of the column representing the merged classes is to be 30 cm, find the correct height for the remaining column. c Explain what problems you would encounter if asked to construct a histogram in which the classes with the two lowest frequencies are to be merged. [3] [3] [1] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data P 8 The histograms below illustrate the number of hours of sunshine during August in two regions, A and B. Neither region had more than 8 hours of sunshine per day. Region A Region Hours of sunshine Hours of sunshine a Explain how you know that some information for one of the regions has been omitted. [2] b After studying the histograms, two students make the following statements. ● Bindu: There was more sunshine in region A than in region B during the first 2 weeks of August. ● Janet: In August there was less sunshine in region A than in region B. 25 Discuss these statements and decide whether or not you agree with each of them. In each case, explain your reasoning. [3] 9 A hotel has 90 rooms. The table summarises information about the number of rooms occupied each day for a period of 200 days. Number of rooms occupied 1–20 21– 40 41–50 51–60 61–70 71–90 Frequency 10 32 62 50 28 18 i Draw a cumulative frequency graph on graph paper to illustrate this information. ii Estimate the number of days when more than 30 rooms were occupied. iii On 75% of the days at most n rooms were occupied. Estimate the value of n. [4] [2] [2] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q5 June 2011 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 26 Chapter 2 Measures of central tendency In this chapter you will learn how to: ■ find and use different measures of central tendency ■ calculate and use the mean of a set of data (including grouped data) either from the data itself or − and use such totals in solving problems that may x b ) from a given total xΣ or a coded total ( Σ involve up to two datasets. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency PREREQUISITE KNOWLEDGE Where it comes from What you should be able to do Check your skills IGCSE / O Level Mathematics Calculate the mean, median and mode for individual and discrete data. Use a calculator efficiently and apply appropriate checks of accuracy. 1 Find the mean, median and mode of the numbers 7.3, 3.9, 1.3, 6.6, 9.2, 4.7, 3.9 and 3.1. 2 Use a calculator to evaluate 6 1.7 × 8 1.9 11 2.1 + + × 6 8 11 + + × and then check that your answer is reasonable. 27 Three types of average There are three measures of central tendency that are commonly used to describe the average value of a set of data. These are the mode, the mean and the median. ● The mode is the most commonly occurring value. ● The mean is calculated by dividing the sum of the values by the number of values. ● The median is the value in the middle of an ordered set of data. We use an average to summarise the values in a set of data. As a representative value, it should be fairly central to, and typical of, the values that it represents. If we investigate the annual incomes of all the people in a region, then a single value (i.e. an average income) would be a convenient number to represent our findings. However, choosing which average to use is something that needs to be thought about, as one measure may be more appropriate to use than the others. Deciding which measure to use depends on many factors. Although the mean is the most familiar average, a shoemaker would prefer to know which shoe size is the most popular (i.e. the mode). A farmer may find the median number of eggs laid by their chickens to be the most useful because they could use it to identify which chickens are profitable and which are not. As for the average income in our chosen region, we must also consider whether to calculate an average for the workers and managers together or separately; and, if separately, then we need to decide who fits into which category. EXPLORE 2.1 Various sources tell us that the average person: ● laughs 10 times a day ● falls asleep in 7 minutes ● sheds 0.7 kg of skin each year ● grows 944km of hair in a lifetime ● produces a sneeze that travels at 160 km/h ● has over 97 000 km of blood vessels in their body ● has a vocabulary between 5000 and 6000 words. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 The average adult male is 172.5cm tall and weighs 80 kg. The average adult female is 159cm tall and weighs 68kg. What might each of these statements mean and how might they have been determined? Have you ever met such an average person? How could this information be useful? You can find a variety of continuously updated figures that yield interesting averages at http://www.worldometers.info/. 2.1 The mode and the modal class As you will recall, a set of data may have more than one mode or no mode at all. The following table shows the scores on 25 rolls of a die, where 2 is the mode because it has the highest frequency. Score on die Frequency ( ) In a set of grouped data in which raw values cannot be seen, we can find the modal class, which is the class with the highest frequency density. WORKED EXAMPLE 2.1 28 Find the modal class of the 270 pencil lengths, given to the nearest centimetre in the following table. Length x( cm) No. pencils ( )f 4–7 8–10 11–12 100 90 80 Answer Length ( cm) x No. pencils ( )f 3.5 < <x 7.5 7.5 < <x 10.5 10.5 < <x 12.5 100 90 80 Width (cm) Frequency density 4 3 2 100 ÷ = 4 25 90 3 30 ÷ = 80 2 ÷ = 40 ( 40 30 20 10 0 3.5 7.5 10.5 12.5 Length (cm) The modal class is 11–12 cm (or, more accurately, 10.5 12.5cm). << x Class boundaries, class widths and frequency density calculations are shown in the table. Although the histogram shown is not needed to answer this question, it is useful to see that, in this case, the modal class is the one with the tallest column and the greatest frequency density, even though it has the lowest frequency. Copyright Material - Review Only - Not for Redistribution REWIND We saw how to calculate frequency density in Chapter 1, Section 1.3. KEY POINT 2.1 In histograms, the modal class has the greatest column height. If there is no modal class then all classes have the same frequency density. TIP The modal class does not contain the most pencils but it does contain the greatest number of pencils per centimetre. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency WORKED EXAMPLE 2.2 Two classes of data have interval widths in the ratio 3:2. Given that there is no modal class and that the frequency of the first class is 48, find the frequency of the second class. Answer Let the freq
uency of the second class be x. 48: 3:2 x = x 48 x 2 3 32 = = The second class has a frequency of 32. Or, let the frequency of the second class be x. x 2 x 48 3 32 = = EXERCISE 2A ‘No modal class’ means that the frequency densities of the two classes are equal, so class frequencies are in the same ratio as interval widths. Alternatively, frequencies are proportional to interval widths. TIP In the special case where all classes have equal widths, frequency densities are proportional to frequencies, so the modal class is the class with the highest frequency. 1 Find the mode(s) of the following sets of numbers. 29 a 12, 15, 11, 7, 4, 10, 32, 14, 6, 13, 19, 3 b 19, 21, 23, 16, 35, 8, 21, 16, 13, 17, 12, 19, 14, 9 2 Which of the eleven words in this sentence is the mode? 3 Identify the mode of x and of y in the following tables4 27 –3 28 –2 29 –1 27 0 25 4 Find the modal class for x and for y in the following tables. x f 0– 5 4– 9 14–20 8 y f 3–6 66 7–11 12–20 80 134 5 A small company sells glass, which it cuts to size to fit into window frames. How could the company benefit from knowing the modal size of glass its customers purchase? 6 Four classes of continuous data are recorded as 1–7, 8–16, 17–20 and 21–25. The class 1–7 has a frequency of 84 and there is no modal class. Find the total frequency of the other three classes. PS 7 Data about the times, in seconds, taken to run 100 metres by n adults are given in the following table. Time x( s) 13.6 <<x 15.4 15.4 <<x 17.4 17.4 <<x 19.8 No. adults ( )f a b 27 By first investigating the possible values of a and of b, find the largest possible value of n, given that the modal class contains the slowest runners. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PS 8 Three classes of continuous data are given as 0–4, 4–10 and 10–18. The frequency densities of the classes 0– 4 and 10–18 are in the ratio 4 : 3 and the total frequency of these two classes is 120. Find the least possible frequency of the modal class, given that the modal class is 4–10. 2.2 The mean The mean is referred to more precisely as the arithmetic mean and it is the most commonly known average. The sum of a set of data values can be found from the mean. Suppose, for example, that 12 values have a mean of 7.5 : = Mean sum of values number of values sum of values 12 You will soon be performing calculations involving the mean, so here we introduce notation that is used in place of the word definition used above. and sum of values 7.5 12 × , so 7.5 = = = . 90 We use the upper case Greek letter ‘sigma’, written Σ, to represent ‘sum’ and x to represent the mean, where x represents our data values. The notation used for ungrouped and for grouped data are shown on separate rows in the following table. Sum Data values Frequency of Number of Sum of data Mean data values data values values Ungrouped Grouped Σ Σ x x – f 30 n Σf Σx x = xfΣ or Σfx x = x Σ n xf Σ f Σ WORKED EXAMPLE 2.3 Five labourers, whose mean mass is 70.2 kg, wish to go to the top of a building in a lift with some cement. Find the greatest mass of cement they can take if the lift has a maximum weight allowance of 500 kg. TIP f for grouped = Σn data. TIP fx xf = Σ indicates the Σ sum of the products of each value and its frequency. For example, the sum of five 10s and six 20s is (20 6) (10 5) + × × = (6 20) 170. (5 10) = × + × × = Answer Σ = × x x n = = 70.2 5 × 351kg 351 y + = y = 500 149 We first rearrange the formula x to find the sum of the labourers’ masses. = x Σ n We now form an equation using y to represent the greatest possible mass of cement. KEY POINT 2.2 For ungrouped data, the mean is = x Σ n For grouped data, x . The greatest mass of cement they can take is 149 kg. x = xf Σ f Σ or fx Σ f Σ . Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency WORKED EXAMPLE 2.4 Find the mean of the 40 values of x, given in the following table. x f Answer x f xf 31 5 31 5 32 7 32 7 33 9 33 9 34 8 34 8 35 11 35 11 fΣ = 40 155 224 297 272 385 xfΣ = 1333 The mean, x = xf Σ f Σ = 1333 40 = 33.325. We find the sum of the 40 values by adding together the products of each value of x and its frequency. This is done in the row headed xf in the table. The 40 values of x have a sum of 1333. Combined sets of data There are many different ways to combine sets of data. However, here we do this by simply considering all of their values together. To find the mean of two combined sets, we divide the sum of all their values by the total number of values in the two sets. For example, by combining the dataset 1, 2, 3, 4 with the dataset 4, 5, 6, we obtain a new set of data that has seven values in it: 1, 2, 3, 4, 4, 5, 6. Note that the value 4 appears twice. 1 2 3 4 + + + 4 Individually, the sets have means of 5. The combined sets 4 5 6 + + 3 2.5 and = = have a mean of . WORKED EXAMPLE 2.5 A large bag of sweets claims to contain 72 sweets, having a total mass of 852.4g. A small bag of sweets claims to contain 24 sweets, having a total mass of 282.8g. What is the mean mass of all the sweets together? Answer Total number of sweets = 72 24 + = 96 . Total mass of We find the total number of sweets and their total mass. sweets = 852.4 282.8 1135.2 g + = Mean mass = 1135.2 96 = 11.825g Copyright Material - Review Only - Not for Redistribution TIP 31 Note that the mean of the combined sets is not +2.5 5 2 . TIP Our answer assumes that the masses given are accurate to 1 decimal place; that the numbers of sweets given are accurate; and that the masses of the bags are not included in the given totals. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 2.6 A family has 38 films on DVD with a mean playing time of 1 hour 32 minutes. They also have 26 films on video cassette, with a mean playing time of 2 hours 4 minutes. Find the mean playing time of all the films in their collection. Answer 64films 38 26 + = (1h 32 min 38) × + (2 h 4 min 26) × = = (92 38) × 6720 min + (124 26) × We find the total number of films and their total playing time. Mean playing time = = 6720 64 105 min or 1h 45 min. The 64 films have a total playing time of 6720 minutes. P EXPLORE 2.2 In Worked example 2.6, the mean playing time of 105 minutes is not equal to +92 124 2 . The mean of A and B ≠ but this is not always the case. mean of A mean of B + 2 32 Suppose two sets of data, A and B, have m and n values with means respectively. In what situations will the mean of A and B together be equal to mean of mean of A B + 2 ? A Σ m and B Σ n , TIP The symbol ≠ means ‘is not equal to’. Means from grouped frequency tables When data are presented in a grouped frequency table or illustrated in a histogram or cumulative frequency graph, we lose information about the raw values. For this reason we cannot determine the mean exactly but we can calculate an estimate of the mean. We do this by using mid-values to represent the values in each class. We use the formula x = xf Σ f Σ , given in Key point 2.2, to calculate an estimate of the mean, where x now represents the class mid-values. WORKED EXAMPLE 2.7 Coconuts are packed into 75 crates, with 40 of a similar size in each crate. 46 crates contain coconuts with a total mass from 20 up to but not including 25kg . 22 crates contain coconuts with a total mass from 25 up to but not including 40 kg. 7 crates contain coconuts with a total mass from 40 up to but not including 54kg. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency a Calculate an estimate of the mean mass of a crate of coconuts. b Use your answer to part a to estimate the mean mass of a coconut. Answer a Mass (kg) No. crates ( )f 20– 46 Mid-value ( )x 22.5 xf 1035 25– 22 32.5 715 40–54 7 47.0 329 Σ =f 75 Σ =xf 2079 Estimate for mean, x = xf Σ f Σ = 2079 75 = 27.72 kg b Estimate for the mean mass of a coconut 27.72 40 0.693kg = = We tabulate the data to include class mid-values, x, and the products xf . We estimate that the 75 crates have a total mass of 2079 kg. We divide our answer to part a by 40 because each crate contains 40 coconuts. When gaps appear between classes of grouped data, class boundaries should be used to find class mid-values. The following example shows a situation in which using incorrect boundaries leads to an incorrect estimate of the mean. 33 WORKED EXAMPLE 2.8 Calculate an estimate of the mean age of a group of 50 students, where there are sixteen 18-year-olds, twenty 19-year-olds and fourteen who are either 20 or 21 years old. Answer Age ( years ) A Mid-value No. students xf ( )x ( )f < 18 <A 19 < 19 <A 20 < 20 <A 22 18.5 19.5 21.0 16 20 14 296 390 294 =Σf 50 =Σxf 980 Estimate of the mean age is xf Σ f Σ 980 50 19.6years = = The 18-year-olds are all aged from 18 up to but not including 19 yea
rs. The 19-year-olds are all aged from 19 up to but not including 20 years. The 20- and 21-yearolds are all aged from 20 up to but not including 22 years. The age groups and necessary totals are shown in the table. Copyright Material - Review Only - Not for Redistribution TIP Incorrect mid-values of 18, 19 and 20.5 give an incorrect estimated mean of 19.1 years. FAST FORWARD We will see how the mean is used to calculate the variance and standard deviation of a set of data in Chapter 3, Section 3.3. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXERCISE 2B 1 Calculate the mean of the following sets of numbers: a 28, 16, 83, 72, 105, 55, 6 and 35 b 7.3, 8.6, 11.7, 9.1, 1.7 and 4. and 7 3 8. 2 a The mean of 15, 31, 47, 83, 97, 119 and p2 is 63. Find the possible values of p. b The mean of 6, 29, 3, 14, , ( q q + 8), q 2 and (10 – ) is 20. Find the possible values of q. q 3 Given that: a n = 14 and x =Σ 325.5, find x. b n = 45 and y = 23.6, find the value of yΣ . c d e z =Σ 4598 and z = 52.25, find the number of values in the set of data. Σ xf = 86 and x 7 1 6= , find the value of fΣ . f =Σ 135 and x = 0.842, find the value of xfΣ . 4 Find the mean of x and of y given in the following tables. a x f 18.0 18.5 19.0 19.5 20.0 8 10 17 24 1 34 5 For the data given in the following table, it is given that 3.62 3.65 3.68 3.71 3.74 127 209 322 291 251 13 9 a 10 11 Calculate the value of a. 6 Calculate an estimate of the mean of x and of y given in the following tables. a b x f y f 0 << x 2 2 << x 4 4 << x 8 8 << x 14 8 9 11 2 13 << y 16 16 << y 21 21 << y 28 28 << y 33 33 << y 36 7 17 29 16 11 7 An examination was taken by 50 students. The 22 boys scored a mean of 71% and the girls scored a mean of 76%. Find the mean score of all the students. 8 A company employs 12 drivers. Their mean monthly salary is $1950. A new driver is employed and the mean monthly salary falls by $8. Find the monthly salary of the new driver. MP 9 The mean age of the 16 members of a karate club is 26 years and 3 months. One member leaves the club and the mean age of those remaining is 26 years. Find the age of the member who left the club. Give a reason why your answer might not be very accurate. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency M 10 The following table shows the hourly rates of pay, in dollars, of a company’s employees. Hourly rate ($) No. employees ( )f 6 8 7 11 8 17 109 1 a Is the mean a good average to use here? Give a reason for your answer. b Find the mean rate of pay for the majority of the employees. PS P 11 A train makes a non-stop journey from one city to another and back again each day. Over a period of 30 days, the mean number of passengers per journey is exactly 61.5. Exact one-way ticket prices paid by these passengers are given by percentage in the following table. Price ($) Passengers ( )% 34 30 38 41 45 29 a Calculate the total revenue from ticket sales, and explain why your answer is an approximation. b The minimum and maximum possible revenues differ by k$ . Find the value of k. 12 The heights, in centimetres, of 54 children are represented in the following diagram. The children are split into two equal-sized groups: a ‘tall half’ and a ‘short half’. Calculate an estimate of the difference between the mean heights of these two groups of children. 0 140 144 150 156 159 35 Height (cm) M 13 The following table summarises the number of tomatoes produced by the plants in the plots on a farm. No. tomatoes No. plots ( )f 20–29 329 30– 49 413 50–79 704 80–100 258 a Calculate an estimate of the mean number of tomatoes produced by these plots. b The tomatoes are weighed accurately and their mean mass is found to be 156.50 grams. At market they are sold for $3.20 per kilogram and the total revenue is $50350. Find the actual mean number of tomatoes produced per plot. c Why could your answer to part b be inaccurate? 14 Twenty boys and girls were each asked how many aunts and uncles they have. The entry 4 / 5 in the following table, for example, shows that 4 boys and 5 girls each have 3 aunts and 2 uncles. a Find the mean number of uncles that the boys have. b For the boys and girls together, calculate the mean number of: i aunts ii aunts and uncles/ 0 0 / 0 0 / 0 1/ 0 Aunts 1 0 / 2 3 / 4 1/ 1 0 / 0 7 / 11 Suggest an alternative way of presenting the data so that the calculations in parts a and b would be simpler to make. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PS 15 A calculated estimate of the mean capacity of 120 refrigerators stored at a warehouse is 348 litres. The capacities are given in the following table. Capacity (litres) No. refrigerators ( )f 160– 12 200– 28 320– 48 400– p 32 A delivery of n new refrigerators, all with capacities between 200 and 320 litres arrives at the warehouse. This causes the mean capacity to decrease by 8 litres. Find the value of n and state what assumptions you are making in your calculations. PS 16 A carpet fitter is employed to fit carpet in each of the 72 guest bedrooms at a new hotel. The following table shows how many rooms were completed during the first 10 days of work. No. rooms completed No. days ( )f 5 2 6 or 7 8 Based on these figures, estimate how many more days it will take to finish the job. What assumptions are you making in your calculations? PS 17 In the figure opposite, a square of side 8cm is joined edge-to-edge to a semicircle, with centre O. P is 2 cm from O on the figure’s axis of symmetry. X 8 Y Points X and Y are fixed but the position of Z is variable on the shape’s perimeter. 8 P O 36 a Find the mean distance from P to X Y, and Z when angle POZ is equal to: i 180 ° ii 135 . ° 6 Z b Find obtuse angle POZ, so that the mean distance from P to X Y, and Z is identical to the mean distance from P to X and .Y EXPLORE 2.3 Six cards, numbered 1, 2, 3, 4, 5 and 6, are placed in a bag, as shown. 15 different pairs of cards can be selected without replacement from the bag. Three of these pairs are {2,3}, {6,4} and {5, 1}. Make a list of all 15 unordered pairs and find the mean of each. We will denote these mean values by X2. Choose a suitable method to represent the values of X2 and their frequencies. Find X2, the mean of the values of X2. Repeat the process described above for each of the following: 1 2 5 4 3 6 ● the six possible selections of five cards, denoting their means by X5 ● the 15 possible selections of four cards, denoting their means by X 4 ● the 20 possible selections of three cards, denoting their means by X3 ● the six possible selections of one card, denoting their means by X1. FAST FORWARD We will study the number of ways of selecting objects in Chapter 5, Section 5.3. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency What does the single value of X6 represent? Do the values of and X5 have anything in common? Can you suggest reasons for any of the common features that you observe? Investigate the values of X r when there are a different number of consecutively numbered cards in the bag. Coded data To code a set of data, we can transform all of its values by addition of a positive or negative constant. The result of doing this produces a set of coded data. One reason for coding is to make the numbers easier to handle when performing manual calculations. Also, it is sometimes easier to work with coded data than with the original data (by arranging the mean to be a convenient number, such as zero, for example). To find the mean of 101, 103, 104, 109 and 113, for example, we can use the values 1, 3, 4, 9 and 13. Our x values are 101, 103, 104, 109 and 113, so 1, 3, 4, 9 and 13 are corresponding values of x( – 100). Mean of the coded values is 100) x( Σ − 5 = 1 3 4 9 13 + + + + 5 = . 6 We subtracted 100 from each x value, so we simply add 100 to the mean of the coded values to find the mean of x. FAST FORWARD We will study the standard normal variable, which has a mean of 0, in Chapter 8, Section 8.2. FAST FORWARD ) ( x b We will see how to use coded totals, such as )2 Σ − and x b to find measures of variation in Chapter 3, Section 3.3. ( Σ − , 37 Mean( x ) mean( – 100) 100 106 or x x + = = 100) = x( Σ − 5 + 100 106 = Refer to the following diagram. If we add b– to the set of x values, they are all translated by b– and so is their mean. TIP ( ), b. The If we remove the bracket from Σ −x b we obtain Σ − Σx term Σb means ‘the sum of all the bs’ and there are n of them, so ( Σ − = Σ − . x nb x b ) So, mean translated by –b Values of x − b mean of x − b Values of x mean of x KEY POINT 2.3 For ungrouped data For grouped data These formulae can be summarised by writing = x mean( – ) x b + . b Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge
University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 For two datasets coded as (x − a) and (y − b), we can use the totals Σx and Σy to find the mean of the combined set of values of x and y. WORKED EXAMPLE 2.9 The exact age of an individual boy is denoted by b, and the exact age of an individual girl is denoted by g. Exactly 5 years ago, the sum of the ages of 10 boys was 127.0 years, so b Σ( − 5) = 127.0 . In exactly 5 years’ time, the sum of the ages of 15 girls will be 351.0 years, so g Σ( + 5) = 351.0 . Find the mean age today of a the 10 boys b the 15 girls c the 10 boys and 15 girls combined. Answers a = b 127 10 + = 5 17.7 years ( b Σ − 5) b = Σ − b Σ = 127 50 177 and = + × (10 5) 127, so = 177 10 = b We update the boys’ past mean age by addition. Alternatively, we expand the brackets. = 17.7 years. g 351 b = 15 Σ + ( g − = 5 18.4 years We backdate the girls’ future mean age by subtraction. 5) g = Σ + (15 5) × = 351, so Alternatively, we expand the brackets. 38 g Σ = 351 75 − = 276 and g = 276 15 = 18.4 years. c g b Σ + Σ 10 15 + = 177 276 + 25 = 18.12 years WORKED EXAMPLE 2.10 Forty values of x are coded in the following table. 3x – Frequency 0– 9 18– 13 24–32 18 Calculate an estimate of the mean value of x. Answer 39 9) × x = = = 24.45 + (21 13) × 40 + (28 18) × + 3 We calculate an estimate for the mean of the coded data using class mid-values of 9, 21 and 28, and then add 3 to obtain our estimate for x. TIP It is not necessary to decode the values of x – 3. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency EXERCISE 2C 1 For 10 values denoted x, it is given that x = 7.4. Find: a Σx b x( Σ + 2) c x( Σ − 1) 2 Twenty-five values of z are such that z( Σ − 7) = 275 . Find z. 3 Given q = 22 and ( qΣ − 4) = 3672 , find the number of values of q. 4 The lengths of 2500 bolts, x mm, are summarised by ( xΣ − 40) = 875 . Find the mean length of the bolts. P 5 Six data values are coded by subtracting 13 from each. Five of the coded values are 9.3, 5.4, 3.9, 7.6 and 2.2, and the mean of the six data values is 17.6. Find the sixth coded value. 6 The SD card slots on digital cameras are designed to accommodate a card of up to 24 mm in width. Due to low sales figures, a manufacturer suspects that the machine used to cut the cards needs to be recalibrated. The widths, w mm, of 400 of these cards were measured and are coded in the following table, where x w= – 24. w – 24 (mm) –0.15 << x –0.1 –0.1 << x 0 0 << x 0.1 0.1 << x 0.2 No. cards ( )f 32 360 6 2 a Suggest a reason why the widths have been coded in this way. b What percentage of the SD cards are too wide to fit into the slots? c Use the coded data to estimate the mean width of these 400 cards. 7 Sixteen bank accounts have been accidentally under-credited by the following amounts, denoted by x$ . 917.95 917.98 918.03 917.97 918.01 917.94 918.05 918.07 918.02 917.93 918.01 917.88 918.10 917.85 918.11 917.94 To calculate x manually, Fidel and Ramon code these figures using x( – 917) and x( – 920), respectively. Who has the simpler maths to do? Explain your answer. PS 8 Throughout her career, an athlete has been timed in 120 of her 400-metre races. Her times, denoted by t seconds, were recorded on indoor tracks 45 times and are summarised by Calculate her average 400-metre running time and comment on the accuracy of your answer. , and on outdoor tracks where 60) 83.7 = . 38.7 t( Σ − t( Σ − 65) = − P 9 All the interior angles of n triangular metal plates, denoted by y°, are measured. a State the number of angles measured and write down the value of y. b Hence, or otherwise, find the value of y( Σ − 30). 39 FAST FORWARD You will study the mean of linear combinations of random variables in the Probability & Statistics 2 Coursebook, Chapter 3. 10 A dataset of 20 values is denoted by x where Σ − values is denoted by y where Σ − y( 2) = 1) 58. Another dataset of 30 36. Find the mean of the 50 values of x and y. x( = Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 11 Students investigated the prices in dollars ($) of 1 litre bottles of a certain drink at 24 shops in a town and at 16 shops in surrounding villages. Denoting the town prices by t and the village prices by v, the students’ data are summarised by the 0.56. totals Σ − 1.1) 1.44 and ( v Σ − 1.2) = = ( t Find the mean price of 1 litre of this drink at all the shops at which the students collected their data. A set of data can be coded by multiplication as well as by addition of a constant. Suppose the monthly take-home salaries of four teachers are $3600, $4200, $3700 and $4500, which have mean x = $4000. What happens to the mean if all the teachers receive a 10% increase but must pay an extra $50 in tax each month? To find their new take-home salaries, we multiply the current salaries by 1.1 and then subtract 50. KEY POINT 2.4 For ungrouped data, x = 1 a   Σ ( ) ax b − n + b .   For grouped data, x = 1 a    Σ ( ) ax b f − f Σ + b .    The new take-home salaries are $3910, $4570, $4020 and $4900. + + 3910 4570 4020 4900 + 4 = $4350. The mean is The original data, x, has been coded by multiplication and by addition as x1.1 – 50. The mean of the coded data is 4350, which is equal to (1.1 4000) – 50, where × 4000 . x= Data coded as ax b– has a mean of ax b− . To find x from a total such as ax b ‘ b– ’ and undo ‘ a× ’, in that order. That is: Σ ( − , we can find the mean of the coded data, then undo ) 40 x = (4350 50) 1.1 or + ÷ 1 1.1 × (4350 50) + = 4000. These formulae can be summarised by writing 1 a [mean( × = x ) ax b − + ]. b TIP ) ax b − can be Σ( . rewritten as a x nb Σ − WORKED EXAMPLE 2.11 The total area of cloth produced at a textile factory is denoted by xΣ and is measured in square metres. Find an expression in x for the area of cloth produced in square centimetres. Answer 1m 100 cm = 2 1m 100 cm 10 000 cm = = 2 2 2 Total area, in square centimetres, is Σ 10 000 x or 10 000Σ . x WORKED EXAMPLE 2.12 For the 20 values of x summarised by (2 xΣ − 3) 104, find x. = We convert the measurements of x from m2 to cm2. Answer 104 20 x = = 5.2 5.2 3 + 2 = 4.1 We first find the mean of the coded values. Knowing that x − = then undo the ‘ 2× ’, in that order, to find x. 3 5.2, we undo the ‘ – 3’ and 2 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency Alternatively, we can expand the brackets in − us to find the value of xΣ . 3), which allows (2 xΣ FAST FORWARD We will see how to use coded totals such as ( Σ measures of variation in Chapter 3, Section 3.3. ( ax b Σ − )2 ax b − and to find ) Σ (2 x − 3) 104 = 2 x Σ − (20 3) 104 = × x Σ = 82 x = 82 20 = 4.1 EXERCISE 2D 1 The masses, x kg, of 12 objects are such that x = 0.475. Find the value of Σ 1000 x and state what it represents. 2 The total mass of gold extracted from a mine is denoted by xΣ , which is measured in grams. Find an expression in x for the total mass in: a carats, given that 1 carat is equivalent to 200 milligrams b kilograms. 3 The area of land used for growing wheat in a region is denoted by wΣ hectares. Find an expression in w for the total area in square kilometres, given that 1 hectare is equivalent to 10 000 m2. 4 Speeds, measured in metres per second, are denoted by x. Find the constant k such that kx denotes the speeds in kilometres per hour. 5 The wind speeds, x miles per hour (mph), were measured at a coastal location at midday on 40 consecutive days and are presented in the following table. 41 Speed ( mph) x No. days ( )f 15 < x 17< 17 < x 20< 20 < x 24< 24 < x 25< 9 13 14 4 Abel wishes to calculate an estimate of the mean wind speed in kilometres per hour (km/h). He knows that a distance of 5 miles is approximately equal to 8km. a Explain how Abel can calculate his estimate without converting the given boundary values from miles per hour to kilometres per hour. b Use the wind speeds in mph to estimate the mean wind speed in km/h. 6 Given that 15 values of x are such that (3 xΣ − 2) = 528, find x and find the value of b such that Σ (0.5 x b − ) 138. = 7 For 20 values of y, it is given that Σ ( ax b − ) = 400 and Σ ( bx a − ) = 545. Given also that x = 6.25, find the value of a and of b. P 8 The midpoint of the line segment between A and B is at (5.2,–1.2). Find the coordinates of the midpoint after the following transformations have been applied to A and to B. a T: Translation by the vector   7 − 4 .   Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Math
ematics: Probability & Statistics 1 b E: Enlargement through the origin with scale factor 5. c Transformations T and E are carried out one after the other. Investigate whether the location of the mid- point of AB is independent of the order in which the transformations are carried out. M PS 9 Five investors are repaid, each with their initial investment increased by %p plus a fixed ‘thank you’ bonus of $q. The woman who invested $20 000 is repaid double her investment and the man who invested $7500 is repaid triple his investment. Find the total amount that the five people invested, given that the mean amount repaid to them was $33000. Do you think the method of repayment is fair? Give a reason for your answer. 10 One of the units used to measure pressure is pounds per square inch (psi). The mean pressure in the four tyres of a particular vehicle is denoted by x psi. Given that 1 pound is approximately equal to 0.4536kg and that 1 metre is approximately equal to 39.37 inches, express the sum of the pressures in the four tyres of this vehicle in grams per cm2. 2.3 The median You will recall that the median splits a set of data into two parts with an equal number of values in each part: a bottom half and a top half. In a set of n ordered values, the median is at the value half-way between the 1st and the nth. Consider a DIY store that opens for 12 hours on Monday and for 15 hours on Saturday. The numbers of customers served during each hour on Monday and on Saturday last week are shown in the following back-to-back stem-and-leaf diagram. 42 Monday (12) Saturday (15 Key: 0 2 2 represents 20 customers on Monday and 22 customers on Saturday To find the median number of customers served on each of these days, we need to find their positions in the ordered rows of the back-to-back stem-and-leaf diagram. from left to right. The median is at the For Saturday, there are n = 15 values arranged in ascending order from top to bottom and   n 1 +  2 In the first row, we have the 1st to 4th values, and in the second row we have the 5th to 10th values, so the 8th value is 38. 15 1 + 2 value. 8th th ( ) = = The median number of customers on Saturday was 38. For Monday, there are n = 12 values arranged in ascending order from top to bottom and   from right to left. The median is at the value, so we locate the th 6.5th n 1 +  2 12 1 + 2 th ) ( = = median mid-way between the 6th and 7th values. In the first row, we have the 1st to 6th values and the 6th is 28. The first value in the second row is the 7th value, which is 30. The median number of customers on Monday was 28 30 + 2 = 29. KEY POINT 2.5 For n ordered values, the median is at the n 1 +   th value.  2 For even values of n, the median is the mean of the two middle values. TIP We can find the 8 th value by counting down and left to right from 22 or by counting up and right to left from 49. TIP Take care when locating values at the left side of a back- to-back stem-and-leaf diagram; they ascend from right to left, and descend from left to right, as we move along each row. When data appear in an ordered frequency table of individual values, we can use cumulative frequencies to investigate the positions of the values, knowing that the median TIP is at the n 1 +  2   th value. n is equal to the total frequency fΣ . Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency WORKED EXAMPLE 2.13 The following table shows 65 ungrouped readings of x. Cumulative frequencies and the positions of the readings are also shown. Find the median value of x. x 40 41 42 43 44 f 11 23 19 8 4 cf 11 34 53 61 65 Positions 1st to 11th 12th to 34th 35th to 53rd 54th to 61st 62nd to 65th Answer Median value of x is 41. The total frequency is 65, and n + 2 65 1 + 2 = = 1 33, so the median is at the 33rd value. From the table, we see that the 12th to 34th values are all equal to 41. Estimating the median In large datasets and in sets of continuous data, values are grouped and the actual values cannot be seen. This means that we cannot find the exact value of the median but we can estimate it. The method we use to estimate the median for this type of data is by reading its value from a cumulative frequency graph. We estimate the median to be the value whose cumulative frequency is equal to half of the total frequency. Consider the masses of 300 museum artefacts, which are represented in the following cumulative frequency graph ( 300 250 200 150 100 50 0 1 2 3 4 median 5 Mass (kg) The set of data has a total of n = 300 values. An estimate for the median is the mass of the n 2 = 300 2 = 150th artefact. Copyright Material - Review Only - Not for Redistribution TIP Frequencies must be taken in account here. Although 41 is not the middle of the five values of x, it is the middle of the 65 readings. REWIND We studied cumulative frequency graphs in Chapter 1, Section 1.4. KEY POINT 2.6 43 On a cumulative frequency graph with total frequency n the median is at the n 2 th value. f= Σ , TIP The graph is only an n 2 estimate, so we use rather than to n 1 + 2 estimate the median. This ensures that we arrive at the same position for the median whether we count up from the bottom or down from the top of the cumulative frequency axis. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 We draw a horizontal line from a cumulative frequency value of 150 to the graph. Then, at the point of intersection, we draw a perpendicular, vertical line down to the axis showing the masses. Reading from the graph, we see that the median mass is approximately 2.6kg. DID YOU KNOW? The concept of representing many different measurements with one representative value is quite a recent invention. There are no historical examples of the mean, median or mode being used before the 17th century. In trying to find the longitude of Ghanza in modern-day Afghanistan, and in studying the characteristics of metals, the 11th century Persian Al-Biruni is one of the earliest known users of a method for finding a representative measure. He used the number in the middle of the smallest and largest values (what we would call the mid-range) ignoring all but the minimum and maximum values. The mid-range was used by Isaac Newton and also by explorers in the 17th and 18th centuries to estimate their geographic positions. It is likely that measuring magnetic declination (i.e. the variation in the angle of magnetic north from true north) played a large part in the growth of the mean’s popularity. 44 Choosing an appropriate average Selecting the most appropriate average to represent the values in a set of data is a matter for discussion in most situations. Just as it may be possible to choose an average that represents the data well, so it is often possible to choose an average that badly misrepresents the data. The purpose and motives behind choosing an average value must also be considered as part of the equation. Consider a student whose marks out of 20 in 10 tests are: 3,4,6,7,8,11,12,13,17 and 17. The three averages for this set of data are: mode 17, = mean 9.8 and = median 9.5. = If the student wishes to impress their friends (or parents), they are most likely to use the mode as the average because it is the highest of the three. Using either the mean or median would suggest that, on average, the student scored fewer than half marks on these tests. Some of the features of the measures of central tendency are given in the following table. TIP Do not confuse the median’s position (150th) with its value (2.6 kg). FAST FORWARD We will use cumulative frequency graphs to estimate the quartiles, the interquartile range and percentiles in Chapter 3, Section 3.2. FAST FORWARD The mid-range, as you will discover in Chapter 3, Section 3.2, is not the same as the median. Advantages Unlikely to be affected by extreme values. Useful to manufacturers that need to know the most popular styles and sizes. Can be used for all sets of qualitative data. Disadvantages Ignores most values. Rarely used in further calculations. Takes all values into account. Frequently used in further calculations. The most commonly understood average. Can be used to find the sum of the data values. Cannot be found unless all values are known. Likely to be affected by extreme values. Mode Mean Median Can be found without knowing all of the values. Relatively unaffected by extreme values. Only takes account of the order of the values and so ignores most of them. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency As an example of the effect of an extreme value, consider the dataset 40,40,70,100,130 and 250. If we increase the largest value from 250 to 880, the mode and median are unchanged (i.e. 40 and 85), but the mean increases by 100% from 105 to 210. Although the median is usually unaffected by extreme values, this is not always the case, as the Libor scandal shows. DID YOU KNOW? LIBOR (London Interbank Offered Rates) are average interest rates that the world’s leading banks charge each other for short-term loans. They determine the prices that people and b
usinesses around the world pay for loans or receive for their savings. They underpin over US $450 trillion worth of investments and are used to assess the health of the world’s financial system. A B C D 2.6% 2.8% 3.0% 3.1% LIBOR = 2.9% The highest and lowest 25% of the daily rates submitted by a small group of leading banks are discarded and a LIBOR is then fixed as the mean of the middle 50%. The above diagram shows a simple example. Consider how the LIBOR would be affected if bank D submitted a rate of 2.5% instead of 3.1%. Several leading banks have been found guilty of manipulating the LIBOR by submitting false rates, which has so far resulted in them being fined over US $9 billion. You can find out more about the LIBOR scandal by searching news websites. Consider the number of days taken by a courier company to deliver 100 packages, as given in the following table and represented in the bar chart. No. days No. packages ( )f 1 10 2 40 3 25 4 14 5 8 6 3 45 40 30 20 10 No. days 5 6 A curve has been drawn over the bars to show the shape of the data. The mode is 2 days. The median is between the 50th and 51st values, which is 2.5 days. The mean = (1 10) × + (2 40) × + (3 25) × + 100 (4 14) × + (5 8) × + (6 3) × = 2.79 days Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 The mean is the largest average and is to the side of the curve’s longer tail. The mode is the smallest average and is to the side of the curve’s shorter tail. The median is between the mode and the mean. A set of data that is not symmetrical is said to be skewed. When the curve’s longer tail is to the side of the larger values, as in the previous bar chart, the data are said to be positively skewed. When the longer tail is to the side of the smaller values, the data are said to be negatively skewed. Generally, we find that: Mode < median < mean when the data are positively skewed. Mean < median < mode when the data are negatively skewed. EXERCISE 2E FAST FORWARD In Chapter 3, we will use a measure of central tendency and a measure of variation to better describe the values in a set of data. In Chapter 8, we will study sets of data called normal distributions in which the mode, mean and median are equal. 1 The number of patients treated each day by a dentist during a 20-day period is shown in the following stem-and-leaf diagram Key: 1 5 represents 15 patients a Find the median number of patients. 46 b On eight of these 20 days, the dentist arrived late to collect their son from school. If they decide to use their average number of patients as a reason for arriving late, would they use the median or the mean? Explain your answer. c Describe a situation in which it would be to the dentist’s advantage to use a mode as the average. 2 a Find the median for the values of t given in the following table. t f 7 4 8 7 9 9 10 14 11 16 12 41 13 9 b What feature of the data suggests that t is less than the median? Confirm whether or not this is the case. 3 a Find the median and the mode for the values of x given in the following table. x f 4 14 5 13 6 4 7 12 8 15 b Give one positive and one negative aspect of using each of the median and the mode as the average value for x. c Some values in the table have been incorrectly recorded as 8 instead of 4. Find the number of incorrectly recorded values, given that the true median of x is 5.5. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency 4 The following graph illustrates the times taken by 112 people to complete a puzzle. 120 ( 80 40 0 2 4 6 8 10 Time (min) Estimate the median time taken. The median is used to divide these people into two groups. Find the median time taken by each of the groups. a b 5 The masses, m kilograms, of 148 objects are summarised in the following table. Mass ( kg)m m 0< m 0.2< m 0.3< m 0.5< m 0.7< m 0.8< cf 0 16 28 120 144 148 47 Construct a cumulative frequency polygon on graph paper, and use it to estimate the number of objects with masses that are: a within 0.1kg of the median b more than 200 g from the median. 6 A teacher recorded the quiz marks of eight students as 11, 13, 15, 15, 17, 18, 19 and 20. They later realised that there was a typing error, so they changed the mark of 11 to 1. Investigate what effect this change has on the mode, mean and median of the students’ marks. 7 The following table shows the lifetimes, to the nearest 10 days, of a certain brand of light bulb. Lifetime (days) 90–100 110–120 130–140 150–160 170–190 200–220 230–260 No. light bulbs ( )f 12 28 54 63 41 16 6 a Use upper class boundaries to represent the data in a cumulative frequency graph and estimate the median lifetime of the light bulbs. b How might the manufacturer choose a value to use as the average lifetime of the light bulbs in a publicity campaign? Based on the figures in the table, investigate whether it would be to the manufacturer’s advantage to use the median or the mean. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 8 It is claimed on the packaging of a brand of battery that they can run a standard kitchen clock continuously for ‘at least 150 days on average’. Tests are carried out to find the length of time, t hours, that a standard kitchen clock runs using one of these batteries. The results are shown in the following table. Time (t hours) 3000 < t < 3096 3096 < t < 3576 3576 < t < 3768 3768 < t < 3840 No. batteries ( )f 34 66 117 33 What could the words on the packaging mean? Test the claim by finding the mean, the median and the modal class. What conclusions, if any, can you make about the claim? 9 Homes in a certain neighbourhood have recently sold for $220 000, $242 000, $236 000 and $3500 000. A potential buyer wants to know the average selling price in the neighbourhood. Which of the mean, median or mode would be more helpful? Explain your answer. 10 A study was carried out on 60 electronic items to find the currents, x amperes, that could be safely passed through them at a fixed voltage before they overheat. The results are given in the two tables below. Current (x amperes) No. items that do not overheat Current (x amperes) No. items that overheat 0.5 60 0.5 0 a Find the value of p, of q and of r. 1.5 48 1.5 p 2.0 20 2.0 q 3.5 6 3.5 r 5.0 0 5.0 60 b Cumulative frequency graphs are drawn to illustrate the data in both tables. 48 i Describe the transformation that maps one graph onto the other. ii Explain the significance of the point where the two graphs intersect. PS P 11 The lengths of extra-time, t minutes, played in the first and second halves of 100 football matches are summarised in the following table. Extra-time (t min) First halves ( )cf Second halves ( )cf <t 1 24 6 <t 2 62 17 <t 4 80 35 <t 5 92 82 <t 7 97 93 <t 9 100 100 a Explain how you know that the median extra-time played in the second halves is greater than in the first halves. b The first-half median is exactly 100 seconds. i Find the upper boundary value of k, given that the second-half median is k times longer than the first- half median. ii Explain why the mean must be greater than the median for the extra-time played in the first halves. PS 12 Eighty candidates took an examination in Astronomy, for which no candidate scored more than 80%. The examiners suggest that five grades, A, B, C, D and E, should be awarded to these candidates, using upper grade boundaries 64, 50, 36 and 26 for grades B, C, D and E, respectively. In this case, grades A, B, C, D and E, will be awarded in the ratio 1: 3 : 5 : 4 : 3. a Using the examiners’ suggestion, represent the scores in a cumulative frequency polygon and use it to estimate the median score. b All of the grade boundaries are later reduced by 10%. Estimate how many candidates will be awarded a higher grade because of this. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency 13 The values of x shown in the following table are to be represented in a bar chart. x Frequency 5 2 6 5 7 9 8 10 9 9 10 5 11 2 a i Sketch a curve that shows the shape of the data. ii Find the mode, mean and the median of x. b The two smallest values of x (i.e. 5 and 5) are changed to 21 and 31. Investigate the effect that this has on the mode, the mean, the median and on the shape of the curve. c If, instead, the two largest values of x (i.e. 11 and 11) are changed to –9 and b, so that the mean of x decreases by 1, find the value of b and investigate the effect that this has on the mode, the median and the shape of the curve. 14 A histogram is drawn to illustrate a set of continuous data whose mean and median are equal. Make sketches of the different types of curve that could be drawn to represent the shape of the histogram. 15 Students’ marks in a Biol
ogy examination are shown by percentage in the following table. Marks ( )% Frequency ( )% 20– 5 30– 10 40– 20 50– 30 60– 20 70– 10 80–90 5 a Without drawing an accurate histogram, describe the shape of the set of marks. What does the shape suggest about the values of the mean, the median and the mode? b Information is provided about the marks in examinations in two other subjects: Chemistry: mode > median > mean Physics: mean > median > mode Sketch a curve to show the shape of the distribution of marks in each of these exams. 49 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Chapter 2: Measures of central tendency Checklist of learning and understanding ● Measures of central tendency are the mode, the mean and the median. ● For ungrouped data, the mode is the most frequently occurring value. ● For grouped data, the modal class has the highest frequency density and the greatest height column in a histogram. ● For ungrouped data, x ● For grouped data, x = x n= Σ . xf Σ f Σ or fx Σ f Σ . ● The formulae for ungrouped and grouped coded data can be summarised by: x = mean ( – ) mean ( ax b − ) + b ] ● For ungrouped coded data   Σ ( ) ax b − n + b   ● For grouped coded data 50 x = 1 a    Σ ( ) ax b f − f Σ + b     ● For ungrouped data, the median is at the  n 1 + 2   th value. n 2 ● For grouped data, we estimate the median to be at the th value on a cumulative frequency graph. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency END-OF-CHAPTER REVIEW EXERCISE 2 1 For each of the following sets of data, decide whether you would expect the mean to be less than, equal to or greater than the median and the mode. a The ages of patients receiving long-term care at a hospital. b The numbers of goals scored in football matches. c The heights of adults living in a particular city. 2 The mean mass of 13 textbooks is 875 grams, and n novels have a total mass of 13 706 grams. Find the mean mass of a novel, given that the textbooks and novels together have a mean mass of 716.6 grams. 3 Nine values are 7, 13, 28, 36, 13, 29, 31, 13 and x. a Write down the name and the value of the measure of central tendency that can be found without knowing the value of x. b If it is known that x is greater than 40, which other measure of central tendency can be found and what is its value? c If the remaining measure of central tendency is 25, find the value of x. 4 For the data shown in the following table, x has a mean of 7.15. x Frequency 3 a 6 b 10 15 c d [1] [1] [1] [3] [1] [1] [2] a Find the mean value of y given in the following table. [1] 51 y 11 14 18 Frequency a b c 23 d b Find a calculated estimate of the mean value of z given in the following table. z Frequency 2– a 8– b 14– 24–34 c d 5 The table below shows the number of books read last month by a group of children. No. books No. children 2 3 3 8 4 15 5 q a If the mean number of books read is exactly 3.75, find the value of q. b Find the greatest possible value of q if: i the modal number of books read is 4 ii the median number of books read is 4. 6 The following table gives the heights, to the nearest 5cm, of a group of people. Heights (cm) 120–135 140–150 155–160 165–170 175 –185 No. people 30 p 12 16 21 Given that the modal class is 140–150 cm, find the least possible value of p. Copyright Material - Review Only - Not for Redistribution [2] [2] [1] [1] [3] Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Chapter 2: Measures of central tendency 7 The following histogram illustrates the masses, m kilograms, of the 216 sales of hay that a farmer made to customers last year. a Show that a calculated estimate of the mean is equal to the median. [4] b Estimate the price per kilogram at which the hay was sold, given that these sales generated exactly $1944. Why is it possible that none of the customers actually paid this amount per kilogram for the hay? [4] 8 An internet service provider wants to know how customers rate its services. A questionnaire asks customers to tick one of the following boxes.2 5.4 3.6 1.8 0 30 40 50 60 70 80 Mass (m kg) excellent □ good □ average □ poor □ very poor □ a How might the company benefit from knowing each of the available average responses of its customers? [2] b What additional benefit could the company obtain by using the following set of tick boxes instead? 52 excellent = 5 □ good 4 □ = average = 3 □ poor = 2 □ very poor 1 □ = [2] 9 The numbers of items returned to the electrical department of a store on each of 100 consecutive days are given in the following table. No. items No. days 0 49 1 16 2 10 3 9 4 7 5 5 p6– 4 a Write down the median. b Is the mode a good value to use as the average in this case? Give a reason for your answer. c Find the value of p, given that a calculated estimate of the mean is 1.5. d Sketch a curve that shows the shape of this set of data, and mark onto it the relative positions of the mode, the mean and the median. [1] [1] [3] [2] 10 As part of a data collection exercise, members of a certain school year group were asked how long they spent on their Mathematics homework during one particular week. The times are given to the nearest 0.1 hour. The results are displayed in the following table. Time spent (t hours) 0.1 < < t 0.5 0.6 < < t 1.0 1.1 < < t 2.0 2.1 < <t 3.0 3.1 < < t 4.5 Frequency 11 15 18 30 21 i Draw, on graph paper, a histogram to illustrate this information. [5] ii Calculate an estimate of the mean time spent on their Mathematics homework by members of this year group. [3] Cambridge International AS & A Level Mathematics 9709 Paper 6 Q5 June 2008 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency [3] [3] [3] [2] [3] [3] [5] 53 11 For 150 values of x, it is given that Σ ( x 1) − + Σ ( x − 4) = 4170. Find x. PS 12 On Monday, a teacher asked eight students to write down a number, which is denoted by x. On Tuesday, when one of these students was absent, they asked them to add 1 to yesterday’s number and write it down. Find the number written down on Monday by the student who was absent on Tuesday, given that x = , and that the mean of Monday’s and Tuesday’s numbers combined was 27 1 3. 30 1 4 13 A delivery of 150 boxes, each containing 20 items, is made to a retailer. The numbers of damaged items in the boxes are shown in the following table. No. damaged items No. boxes ( )f 0 100 1 10 2 10 3 10 4 10 5 10 6 or more 0 a Find the mode, the mean and the median number of damaged items. b Which of the three measures of central tendency would be the most appropriate to use as the average in this case? Explain why using the other two measures could be misleading. 14 The monthly salaries, w dollars, of 10 women are such that Σ w( − 3000) = − 200. The monthly salaries, m dollars, of 20 men are such that Σ m( − 4000) 120. = a Find the difference between the mean monthly salary of the women and the mean monthly salary of the men. b Find the mean monthly salary of all the women and men together. 15 For 90 values of x and 64 values of y, it is given that Σ − x ( 1) = 72.9 and ( y Σ + 1) = 201.6. Find the mean value of all the values of x and y combined. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 54 Chapter 3 Measures of variation In this chapter you will learn how to: ■ find and use different measures of variation ■ use a cumulative frequency graph to estimate medians, quartiles and percentiles ■ calculate and use the standard deviation of a set of data (including grouped data) either from the data itself or from given totals xΣ and Σx2, or coded totals ( Σ − totals in solving problems that may involve up to two datasets. x b ) and Σ ( x b − )2 and use such Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation PREREQUISITE KNOWLEDGE Where it comes from What you should be able to do Check your skills IGCSE / O Level Mathematics Accurately label and read from an axis, using a given scale. Substitute into and manipulate algebraic formulae containing squares and square roots. 1 The numbers 2 and 18 are marked on an axis 20 cm apart. How far apart are the numbers 4.5 and 17.3 on this axis? 2 If =   , find the positive value of: a y when =x 1
3, =a 4 and =b 352 b x when =y 12, =a . 5 and b 11= How do we best summarise a set of data? A measure of central tendency alone does not describe or summarise a set of data fully. Although it may tell us the location of the more central values or the most common values, it tells us nothing about how widely spread out the values are. Two sets of data can have the same mean, median or mode, yet they can be completely different. A better description of a set of data is given by a measure of central tendency and a measure of variation. Variation is also known as spread or dispersion. Consider the runs scored by two batters in their past eight cricket matches, which are given in the following table. Batter A Batter B 25 2 30 70 31 1 26 0 31 43 28 29 24 Total: 224 1 104 3 Total: 224 55 The mean number of runs scored by A and by B is the same; namely, the patterns of the number of runs are clearly very different. The numbers for batter A are quite consistent, whereas the numbers for batter B are quite varied. This consistency (or lack of it) can be indicated by a measure of variation, which shows how spread out a set of data values are. 28. However, 224 8 ÷ = Three commonly used measures of variation are the range, interquartile range and standard deviation. 3.1 The range As you will recall, the range is the numerical difference between the largest and smallest values in a set of data. One advantage of using the range is that it is easy to calculate. However, it does not take the more central values into account but uses only the most extreme values. It is often more informative to state the minimum and maximum values rather than the difference between them. For example, in a test for which the lowest mark is 6 and the highest mark is 19, the range is 19 – 6 13. = For grouped data, we can find a minimum and maximum possible range, using the lower and upper boundary values of the data. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 3.1 To the nearest centimetre, the tallest and shortest pupils in a class are 169cm and 150 cm. Find the least and greatest possible range of the students’ heights. Answer Least possible range 168.5 – 150.5 = Greatest possible range 169.5 – 149.5 = 18cm = The intervals in which the given heights, h, lie are 168.5 150.5cm 169.5cm 149.5 and ø < h ø < h . = 20 cm 3.2 The interquartile range and percentiles The lower quartile, median and upper quartile, as you will recall, divide the values in a dataset into four parts, with an equal number of values in each part. These three measures are commonly abbreviated by: ● Q1 for the lower quartile ● Q2 for the median (or middle quartile) ● Q3 for the upper quartile. The interquartile range is the numerical difference between the upper quartile and the lower quartile, and gives the range of the middle half (50%) of the values, as shown in the following diagram. 56 interquartile range smallest value Q1 Q2 Q3 largest value KEY POINT 3.1 Interquartile range upper quartile – lower quartile = or IQR 1. –3 = Q Q The interquartile range is often preferred to the range because it gives a measure of how varied the more central values are. It is relatively unaffected by extreme values, also called outliers, and can be found even when the exact values of these are not known. TIP Ungrouped data The positions of the lower and upper quartiles depend on whether there are an odd or even number of values in the set of data. One method that we can use to find the quartiles is as follows. For an even number of ordered values: we split the data into a lower half and an upper half. Then Q1 and Q3 are the medians of the lower half and upper half, respectively. For an odd number of ordered values: we split the data into a lower half and an upper half at the median, which we then discard. Again, Q1 and Q3 are the medians of the lower half and upper half, respectively. Copyright Material - Review Only - Not for Redistribution th always at the In a set of ungrouped data, the median is n 1 +  2   value. However, it is advisable to find the quartiles by inspection rather than by memorising formulae for their positions. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation WORKED EXAMPLE 3.2 Find the interquartile range of the eight ordered values 2, 5, 9, 13, 29, 33, 49 and 55. Answer 1st 2 2nd 5 3rd 9 4th 13 5th 29 6th 33 7th 49 8th 55 Q1 Q2 Q3 The ordered values are shown, with their positions indicated. IQR = = = = Q Q − 1 3 33 49 + 2 41 7 − 34 − 5 9 + 2 WORKED EXAMPLE 3.3 Find the interquartile range of the seven values 69, 17, 43, 6, 73, 77 and 39. Answer 1st 6 IQR 2nd 17 Q1 3rd 39 4th 43 Q2 5th 69 6th 73 Q3 7th 77 The ordered values and their positions are shown. 57 = = = 3 – Q Q 1 73 – 17 56 REWIND We studied stemand-leaf diagrams in Chapter 1, Section 1.2. WORKED EXAMPLE 3.4 Find the interquartile range of the 13 grouped values shown in the following stem-and-leaf diagram. 14 15 16 Key: 14 2 represents 142 Answer Q2 is at the   13 1 + 2   which is 153 . th 7th value, = Q 1 = Q 3 = 144 148 + 2 157 159 + 2 − IQR 158 146 = = 146 = 158 = 12 We identify the median as 153, which we now discard. This leaves a lower half (142 to 151) and an upper half (155 to 168), with six values in each. The median of a group of six values is the 3.5th value. These are marked by red dots in the stemand-leaf diagram. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXPLORE 3.1 In this activity, you will investigate the value of the median in relation to the smallest and largest values in a set of data, and also in relation to the lower and upper quartiles. For each ordered set of data, A to D, write down these five values: the smallest value; the lower quartile; the median; the upper quartile; and the largest value. Set A: 2, 2, 3, 11, 11, 21, 22. Set B: 6, 6, 6, 11, 13, 17, 19, 20. Set C: 9, 15, 28, 32, 35, 49. Set D: 5, 7, 9, 10, 11, 12, 12, 16, 17. It may be useful to mark the five values for each dataset on a number line. Use your results to decide which of the following statements are always, sometimes or never true. 1 The median is mid-way between the smallest and largest values. 2 The median is mid-way between the lower and upper quartiles. 3 The interquartile range is equal to exactly half of the range. 4 58 REWIND We estimated the median from a cumulative frequency graph in Chapter 1, Section 1.4. Grouped data We can use a cumulative frequency graph to estimate values in any position in a set of data. This includes the lower quartile, the upper quartile and any chosen percentile. KEY POINT 3.2 For grouped data with total frequency n following table. f= Σ , the positions of the quartiles are shown in the Quartile lower ( )1Q median ( )2Q upper ( )3Q Position n 4 or 1 4 f Σ n 2 or 1 2 Σ f n3 4 or 3 4 Σ f The nth percentile is the value that is n% of the way through a set of data. Q Q,1 2 and Q3 are the 25th, 50th and 75th percentiles, respectively. In an ordered dataset with, say, 320 values, Q Q,1 240th values, and the 90th percentile is at the 2 and Q3 are at the 80th, 160th and 288th value. (0.90 320) × = The range of the middle 80% of a dataset is the difference between the 10th and 90th percentiles. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation WORKED EXAMPLE 3.5 The following graph illustrates the times, in minutes, taken by 500 people to complete a task. Use the graph to find an estimate of: a the greatest possible range b the interquartile range c the 95th percentile 500 475 400 375 300 250 200 125 100 0 59 5 10 Q1 15 Q3 Time (min) 20 25 30 95th percentile Answer a 30 – 2 = 28 min The greatest possible range is equal to the width of the polygon. b Q 1 Q 3 IQR ≈ ≈ = ≈ = 8.0 min 14.5 min 3 – Q Q 1 14.5 – 8.0 6.5 min We locate the quartiles, then estimate their values by reading from the graph. Lower quartile: Upper quartile: = n 4 3 n 4 = 500 4 3 500 × 4 = 125 th value = 375 th value c ≈ 24.0 min The 95th percentile is at the (0.95 500) × = 475th value. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Box-and-whisker diagrams A box-and-whisker diagram (or box plot) is a graphical representation of data, showing some of its key features. These features are its smallest and largest values, its lower and upper quartiles, and its median. If drawn by hand, the diagram is best drawn on graph paper and must include a scale. It takes the
form shown in the following diagram, which shows some features of a dataset denoted by x. 0 5 10 15 x smallest value Q1 Q2 box whisker Q3 largest value Key features of the data for x represented in the box-and-whisker diagram are: Median: Range 14 – 1 13 = – IQR Q Q 1 =Q 6 2 = = = 3 11 – 4 = 7 The following box-and-whisker diagram is a representation of a dataset denoted by y, drawn using the same scale as the previous diagram. 0 5 10 15 y 60 The following table shows a measure of central tendency and two measures of variation for each of x and y, and we can use these to make comparisons. Median Range IQR Dataset x Dataset y 6 9 13 7 15 – 0 15 = 12 – 3 9 = By comparing medians, values of x are, on average, less than values of y. By comparing ranges and interquartile ranges, values of y are more varied than values of x. We can assess the skewness of a set of data using the quartiles in a box-and-whisker diagram. In the previous box-and-whisker diagram for x, 1, and the longer tail of the curve drawn over a bar chart would be to the side of the larger values. This means that the data for x is positively skewed In the previous box-and-whisker diagram for y, 2, and the longer tail of the curve would be to the side of the smaller values. This means that the data for y is negatively skewed reasonably symmetrical set of data would have 1 Copyright Material - Review Only - Not for Redistribution TIP The whisker (which shows the range) is not drawn through the box (which shows the interquartile range). Items and, where appropriate, units such as ‘Length (cm)’ and ‘Mass (kg)’ must be indicated on the diagram. REWIND We looked briefly at positively and negatively skewed sets of data in Chapter 2, Section 2.3. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation EXERCISE 3A 1 Find the range and the interquartile range of the following sets of data. a 5, 8, 13, 17, 22, 25, 30 b 7, 13, 21, 2, 37, 28, 17, 11, 2 c 42, 47, 39, 51, 73, 18, 83, 29, 41, 64 d 113, 97, 36, 81, 49, 41, 20, 66, 28, 32, 17, 107 e 4.6, 0, –2.6, 0.8, –1.9, –3.3, 5.2, –3.2 2 a Find the range and the interquartile range of the dataset represented in the following box plot. 0 1 2 3 4 b What type of skewness would you expect this set of data to have? 3 The following stem-and-leaf diagram shows the marks out of 50 obtained by 15 students in a Science test Key: 2 5 represents a mark of 25 out of 50 61 a Find the range and interquartile range of the marks. b Illustrate the data in a box-and-whisker diagram on graph paper and include a scale. c For this set of data, express Q3 in terms of Q1 and Q2. 4 The numbers of fouls made in eight hockey matches and in eight football matches played at the weekend are shown in the following back-to-back stem-and-leaf diagram. Hockey (8) Football (8 Key: 6 1 8 represents 16 fouls in hockey and 18 fouls in football a Is it true to say that the numbers of fouls in the two sports are equally varied? Explain your answer. b Draw two box-and-whisker diagrams using the same scales. Write a sentence to compare the numbers of fouls committed in the two sports. 5 Rishi and Daisy take the same seven tests in Mathematics, and both students’ marks improve on successive tests. Their percentage marks are as follows. Rishi’s marks Daisy’s marks 15 24 28 33 39 42 50 51 65 69 72 78 83 86 a Explain why it would not be useful to use the range or the interquartile range alone as measures for comparing the marks of the two students. b Name two measures that could be used together to give a meaningful comparison of the two students’ marks. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 6 The following table shows the maximum speeds, s km/h, of Maximum speed No. vintage some vintage cars. s( km/ h) cars f( ) a On the same sheet of graph paper, construct a cumulative frequency polygon and a box-and-whisker diagram to illustrate the data. b Use your box-and-whisker diagram to assess what type of skewness the data have. 7 Twenty adults are selected at random, and each is asked to state the number of trips abroad that they have made. The results are shown in the following back-to-back stem-and-leaf diagram. <s 35 s <<< s <<< s <<< s <<< s <<< s <<< 40 45 50 55 70 75 35 40 45 50 55 70 0 20 65 110 27 13 5 a i Draw box-and-whisker diagrams, using the same scales for males and for females. ii Interpret the key features of the data represented in your diagrams and compare the data for the two groups of adults. b In a summary of the data, a student writes, ‘The females have visited more countries than the males.’ Is this statement justified? Give a reason to support your answer. Males (9) Females (11 Key: 3 1 1 represents 13 trips for a male and 11 trips for a female 8 The resistances, in ohms (Ω , of 100 conductors are represented in the following graph. ) 62 ) 100 80 60 40 20 0 0.1 0.2 0.3 0.4 0.5 0.6 Resistance (Ω) Find, to an appropriate degree of accuracy, an estimate of: a c the interquartile range the percentile that is equal to 0.192 Ω b d the 90th percentile the range of the middle 40% of the resistances. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 9 The areas, in cm2, of some circuit boards are represented in the following graph. Chapter 3: Measures of variation ( 200 150 100 50 0 8 16 24 32 40 48 56 Area (cm2) a State the greatest possible range of the data. b Construct a box-and-whisker diagram to illustrate the data. c Find the range of the middle 60% of the areas. d An outlier is an extreme value that is more than 1.5 times the interquartile range above the upper quartile or more than 1.5 times the interquartile range below the lower quartile. Find the areas that define the outliers in this set of data, and estimate how many there are. How accurate is your answer? 63 10 A company manufactures right-angled brackets for use in the construction industry. A sample of brackets are measured, and the number of degrees by which their angles deviate from a right angle are summarised in the following table. a Draw a cumulative frequency polygon to illustrate these deviations. b Estimate the median and the interquartile range of the bracket angles, giving both answers correct to 1 decimal place. c A bracket is considered unsuitable for use if its angle deviates from a right angle by more than 1.2°. Estimate what percentage of this sample is unsuitable for use, giving your answer correct to the nearest integer. Deviation from 90° d( ) No. brackets f( ) <d –1.5 –1.5 ø < d –1.0 –1.0 ø < d –0.5 –0.5 ø < d 0.0 0.0 ø < d 0.5 0.5 ø < d 1.0 1.0 ø < d 1.5 1.5 ø < d 2.5 0 24 46 61 34 34 20 17 11 The following table shows the cumulative frequencies for values of x. x cf < 0 0 < 10 12 < 15 30 < 25 90 < 30 102 < 40 120 Without drawing a cumulative frequency graph, find: a b the interquartile range the 85th percentile. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 M 12 Fifty 10-gram samples of a particular type of mushroom are collected by volunteers at a university and tested. The following table shows the mass of toxins, in hundredths of a gram, in these samples. Mass (/0.01g) No. 10 g samples 0– 2 4– 19 11– 23 17– 3 20–30 3 a Draw a cumulative frequency curve to illustrate the data. b Use your curve to estimate, correct to 2 decimal places: i the interquartile range ii the range of the middle 80%. c It was found that toxins made up between 0.75% and 2.25% of the mass of n of these samples. Use your curve to estimate the value of n. d Make an assessment of the variation in the percentage of toxic material in these samples. Can you suggest any possible reasons for such variation? M 13 A 9-year study was carried out on the pollutants released when biomass fuels are used for cooking. Researchers offered nearly 1000 people living in 12 villages in southern China access to clean biogas and to improved kitchen ventilation. Some people took advantage of neither; some changed to clean fuels; some improved their kitchen ventilation; and some did both. The following diagram shows data on the concentrations of nitrogen dioxide in these people’s homes at the end of the study. 64 Groups: Neither Clean fuels only Ventilation only Both Nitrogen dioxide pollutant concentration (mg/m3) 0 0.25 0.5 0.75 Study the data represented in the diagram and then write a brief analysis that summarises the results of this part of the study. DID YOU KNOW? The study of human physical growth, auxology, is a multidisciplinary science involving genetics, health sciences, sociology and economics, among others. Exceptional height variation in populations that share a genetic background and environmental factors is sometimes due to dwarfism or gigantism, which are medical conditions caused by specific genes or abnormalities in the production of hormones. In
regions of poverty or warfare, environmental factors, such as chronic malnutrition during childhood, may result in delayed growth and/or significant reductions in adult stature even without the presence of these medical conditions. At the time of their meeting in London in 2014, Chandra Bahadur Dangi (at 54.6 cm) and Sultan Kosen (at 254.3 cm) were the shortest and tallest adults in the world. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation EXPLORE 3.2 The interquartile range is based around the median. In this exploration, we investigate a possible way to define variation based on the mean. Choose a set of five numbers with a mean of 10. The deviation of a number tells us how far and to which side of the mean it is. Numbers greater than the mean have a positive deviation, whereas numbers less than the mean have a negative deviation, as indicated in the following diagram. mean negative deviation positive deviation Find the deviation of each of your five numbers and then calculate the mean deviation. Compare and discuss your results, and investigate other sets of numbers. Can you predict what the result will be for any set of five numbers with a mean of 10? Can you justify your prediction? What would you expect to happen if you started with any set of five numbers? 3.3 Variance and standard deviation In the Explore 3.2 activity, you discovered that the mean deviation is not a useful way of measuring the variation of a dataset because the positive and negative deviations cancel each other out. So, if we want a measure of variation around the mean, we need to ensure that each deviation is positive or zero. We can do this by calculating the mean distance of the data values from the mean, which TIP means we x x− calculate −x x and remove the minus sign if the answer is negative. 65 we call the ‘mean absolute deviation from the mean’, x x Σ − n . However, it is hard to calculate this accurately or efficiently for large sets of data and it is difficult to work with algebraically, so this approach is not used in practice. ) 2 for all data values and find Alternatively, we can calculate the squared deviation, their mean. This is the ‘mean squared deviation from the mean’, which we call the variance of the data. ( −x x Σ 2 ) ( Var( x ) = x x − n For measurements and deviations in metres, say, the variance is in m2. So, to get a measure of variation that is also in metres, we take the square root of the variance, which we call the standard deviation. Standard deviation of x = Var( x ) = )2 ( x x Σ − n This looks no easier to calculate than the ‘mean absolute deviation from the mean’, however, the formula for variance can be simplified (see appendix at the end of this chapter) to give: Var  n   We can find the variance and standard deviation from n, Σx and Σx2, which are the number of values, their sum and the sum of their squares, respectively. We often use the abbreviation ) to represent the standard deviation of X . XSD( Copyright Material - Review Only - Not for Redistribution TIP To find Σx2, we add up the squares of the data values. A common error is to add up the data values and then square the answer but this would be written as instead. 2x) (Σ Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 TIP To find each value of x f2 multiply x by xf . If we multiply x by f and then square the answer, we will obtain which is not required. (which is the same as fx2), we can either multiply x2 by f or we can 2 x f xf )2 = ( 2, A low standard deviation indicates that most values are close to the mean, whereas a high standard deviation indicates that the values are widely spread out from the mean. Consider a drinks machine that is supposed to dispense 400 ml of coffee per cup. We would expect some variation in the amount dispensed, yet if the standard deviation is high then some customers are likely to feel cheated and some risk being injured because of their overflowing cups! KEY POINT 3.3 For ungrouped data: Standard deviation = Variance = ) , where x = x Σ n . For grouped data: Standard deviation = Variance = ( , where x = xf Σ f Σ . TIP We can remember the formula for variance as ‘mean of the squares minus square of the mean’. 66 WORKED EXAMPLE 3.6 For the set of five numbers 3, 9, 15, 24 and 29, find: a the standard deviation b which of the five numbers are more than one standard deviation from the mean. Answer a Variance = 2 2 x Σ n − x Σ + 15 + 2 24 2 29 + 5 2   80 5 2 1732 5 −   346.4 16 − = = = = Standard deviation 90.4 = = 90.4 9.51 b 16 9.51 6.49 = − 16 9.51 25.51 = + −   + + 3 9 15 24 29 + 5 + We subtract the square of the mean from the mean of the squares to find the variance. 2   We take the square root of the variance to find the standard deviation, correct to 3 significant figures. We find the values that are 9.51 below and 9.51 above, using the mean of 16. The numbers are 3 and 29. Identify which of the five numbers are outside the range 6.49 number < < . 25.51 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation WORKED EXAMPLE 3.7 Find the standard deviation of the values of x given in the following table, correct to 3 significant figures. x 12 14 16 Answer x 12 14 16 f 13 28 10 f 13 28 10 xf 156 392 160 2 x f ××= x xf 12 156 1872 = × 14 392 5488 × = 16 160 × = 2560 Σ =f 51 Σ =xf 708 2x fΣ = 9920 SD 9920 51 −   708 51   = 1.34 What happens if we use a rounded value for the mean? The table shown opposite is an extended frequency table that is used to find fΣ , Σxf and x f2Σ , which are needed to calculate the standard deviation. We use the totals 51, 708 and 9920 to find the standard deviation. TIP Note how the values of x f2 are calculated. TIP Always use the exact value of the mean to calculate variance and standard deviation. 67 Correct to 1 decimal place, the mean in Worked example 3.7 is 708 ÷ 51 13.9. = If we use x 13.9 = error of 0.2. in our calculation, we obtain SD( x ) = 9920 51 − 13.9 2 = 1.14 . This is an The rounded mean has caused a substantial error (0.2 is about 15% of the correct value 1.34). So, when calculating the variance or standard deviation, always use xf Σ f Σ , rather than a rounded value for the mean. x Σ or n When data are grouped, actual values cannot be seen, but we can calculate estimates of the variance and standard deviation. The formulae in Key point 3.3 are used to do this, where x now represents class mid-values and x = xf Σ f Σ is a calculated estimate of the mean. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 3.8 Calculate an estimate of the standard deviation of the heights of the 20 children given in the following table. Height (metres) No. children f( ) Answer 1.2– 2 1.4– 12 1.5–1.7 6 We extend the frequency table to include class mid-values x( find the totals Σf , Σxf and x f2Σ shown in the table opposite. ), and to , as Height (m) 1.2– 1.4– 1.5–1.7 No. children f( ) Mid-value x( ) xf x f2 2 1.3 2.6 12 1.45 17.4 6 1.6 9.6 fΣ = 20 xfΣ = 29.6 3.38 25.23 15.36 2x fΣ = 43.97 Estimate of standard deviation 43.97 20 −   29.6 20   0.0081 0.09 m = = = = 68 EXPLORE 3.3 Four students analysed data that they had collected. Their findings are given below. 1 Property prices in a certain area of town have a high standard deviation. 2 The variance of the monthly sales of a particular product last year was high. 3 The standard deviation of students’ marks in a particular examination was close to zero. 4 The times taken to perform a new medical procedure have a low variance. Discuss the students’ findings and give a possible description of each of the following. 1 The type of environment and the people purchasing property in this area of town. 2 The type of product being sold. 3 The usefulness of the examination. 4 The efficiency of the teams performing the medical procedures. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation Although standard deviation is far more commonly used as a measure of variation than the interquartile range, it may not always be ideal because it can be significantly affected by extreme values. The interquartile range may be better, and a box-and-whisker diagram is often much more useful as a visual representation of data than the mean and standard deviation. Some features of the standard deviation are compared to the interquartile range in the following table. Advantages of standard deviation Disadvantages of standard deviation Much simpler to
calculate than the IQR. Far more affected by extreme values than the IQR. Data values do not have to be ordered. Gives greater emphasis to large deviations than to small deviations. Easier to work with algebraically when doing more advanced work. Takes account of all data values. EXERCISE 3B 1 Find the mean and the standard deviation for these sets of numbers. a 27, 43, 29, 34, 53, 37, 19 and 58. b 6.2, 8.5, 7.7, 4.3, 13.5 and −11.9. − − 2 Last term Abraham sat three tests in each of his science subjects. His raw percentage marks for the tests, in the order they were completed, are listed. Biology Chemistry Physics 21 33 45 41 53 65 51 63 75 69 a Calculate the variance of Abraham’s marks in each of the three subjects. b Comment on the three values obtained in part a. Do the same comments apply to Abraham’s mean mark for the tests in the three subjects? Justify your answer. 3 The following table shows the number of pets owned by each of 35 families. No. pets No. families f( ) 0 6 1 12 2 9 3 4 4 3 5 1 Find the mean and variance of the number of pets. 4 The numbers of cobs produced by 360 maize plants are shown in the following table. No. cobs No. plants f( ) 0 11 1 75 2 185 3 81 4 8 a Calculate the mean and the standard deviation. b Find the interquartile range and give an example of what it tells us about this dataset that the standard deviation does not tell us. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 5 The times spent, in minutes, by 30 girls and by 40 boys on an assignment are detailed in the following table. Time spent (min) No. girls f( ) No. boys f( ) 20– 6 15 30– 14 11 40– 60–80 7 7 3 7 a For the boys and for the girls, calculate estimates of the mean and standard deviation. b It is required to make a comparison between the times spent by the two groups. i What do the means tell us about the times spent? ii Use the standard deviations to compare the times spent by the two groups. 6 The lengths, correct to the nearest centimetre, of 50 rods are given in the following table. Length (cm) 15–17 18–24 25–29 30–37 No. rods f( ) 13 18 11 8 Calculate an estimate of the standard deviation of the lengths. 7 For the dataset denoted by x in the following table, k is a constant. x f 15 2k 16 17 5+k – 3k 18 10 19 8 20 3 70 Find the value of k and calculate the variance of x, given that 17=x . 8 The following table illustrates the heights, in centimetres, of 150 children. Height (cm) No. children f( ) 140 up to 144 144 up to 150 150 up to 160 160 up to 165 a b 69 28 a Given that a calculated estimate of the mean height is exactly 153.14 cm, show that 142 a + 147 b = 7726 , and evaluate a and b. b Calculate an estimate of the standard deviation of the heights. PS 9 Kristina plans to raise money for charity. Her plan is to walk 217 km in 7 days so that she walks km2 2 – n n +k compare this with the interquartile range. on the nth day. Find the standard deviation of the daily distances she plans to walk, and PS 10 The mass of waste produced by a school during its three 13-week terms is given in tonnes, correct to 2 decimal places, in the following table. Mass of waste (tonnes) 0.15– 0.29 0.30– 0.86 0.87–1.35 1.36–2.00 No. weeks f( ) 5 8 20 6 a Calculate estimates of the mean and standard deviation of the mass of waste produced per week, giving both answers correct to 2 decimal places b No waste is produced in the 13 weeks of the year that the school is closed. If this additional data is included in the calculations, what effect does it have on the mean and on the standard deviation? Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation PS 11 The ages, in whole numbers of years, of a hotel’s 50 staff are given in the following table. Calculated estimates of the mean and variance are 37.32 and 69.1176, respectively. Age (years) No. staff f( ) 23–30 14 31–37 x 38– 45 46–59 y 6 Exactly 1 year after these calculations were made, Gudrun became the 51st staff member and the mean age became exactly 38 years. Find Gudrun’s age on the day of her recruitment, and determine what effect this had on the variance of the staff’s ages. What assumptions must be made to justify your answers? PS P 12 Refer to the following diagram. In position 1, a 10-metre rod is placed 10 metres from a fixed point, P. Six small discs, A to F , are evenly spaced along the length of the rod. The rod is rotated anti-clockwise about its centre by ° to position 2. The distances from P to the discs are denoted by x. 30α = P 10 Position 1 Position 2 A B C D E F 10 P 10 A B 10 C D E F a What effect does the 30° rotation have on values of x? Investigate this by first considering the effect on the average distance from P to the discs. b Find two values that can be used as measures of the change in the variation of x caused by the rotation. 71 c Use the values obtained in parts a and b to summarise the changes in the distances from P to the discs caused by the rotation. 2Σx is constant for all values of α? (Hint: to do this, you need only to show that d Can you prove that constant for 0 < 90α° < °). 2Σx is EXPLORE 3.4 Twenty adults completed as many laps of a running track as they could manage in 30 minutes. The following table shows how many laps they completed. Completed laps No. adults f( ) 4–8 6 9–13 10 14–18 4 Two students, Andrea and Billie, were asked to calculate an estimate of the standard deviation. Their working and answers, which you should check carefully, are shown below. Andrea: ( Billie: 2 6 6 × ( ) + 2 ( 11 10 × 20 ) + 2 ( 16 4 × ) 2 × 2 6.5 6 11.5 10 16.5 4 × 20 ) + ) + × ( ( 2 − ) 2 10.5 3.5 = 2 11 − 3.5 = Compare Andrea and Billie’s approaches. What have they done differently and why do you think they did so? Is one of their answers better than the other? If so, in what way? Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Calculating from totals For ungrouped data, we calculate variance (Var) and standard deviation (SD) from totals n, Σx and 2Σx . For grouped data, we calculate using totals Σf , Σxf and 2Σx f . In both cases, we can rearrange the formula for variance if we wish to evaluate one of the totals. WORKED EXAMPLE 3.9 Given that , 25=n Σ =x 275 and Var( 7=x ) , find 2Σx . Answer 2 2 x Σ 25 −   275 25   = 7 2 Σ x = = × 25   3200 7 + ( 275 25 2 )   Substitute the given values into the formula for variance, then rearrange the terms to make x2Σ the subject. 72 Combined sets of data In Chapter 2, Section 2.2, sets of data were combined by simply considering all of their values together, and we learned how to find the mean. Here, we consider the variation of datasets that have been combined in the same way. The variance and standard deviation of a combined dataset are calculated from its totals, which are the sums of the totals of the two sets from which it has been made. 4 and 2 3 The two sets {1, 2, 3, 4} and {4, 5, 6} individually have variances of 1 1 set {1, 2, 3, 4, 4, 5, 6} has a variance of approximately 2.53. . The combined WORKED EXAMPLE 3.10 The heights, cmx , of 10 boys are summarised by Σ =x 1650 The heights, cmy , of 15 girls are summarised by Σ =y 2370 = 2Σ and x 2Σ =y and 275 490 . . 377835 Calculate, to 3 significant figures, the standard deviation of the heights of all 25 children together. Answer 2 Σ x 2 y + Σ = 275490 377835 + = 653325 Σ + Σ = x y 1650 2370 + = 4020 For the 25 children, we find the sum of the squares of their heights and the sum of their heights. Standard deviation 653 325 25 16.6cm = = 2 −   4020 25   We substitute the three sums into the formula for standard deviation and evaluate this to the required degree of accuracy. TIP The variance of two combined datasets x and y is not (in general) equal to Var( x In Worked example 3.10, Var(boys) 324 = and Var(girls) 225 but Var(boys and girls) = ≠ ) + Var( y ) 2 . 324 225 + 2 . Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation If two sets of data, denoted by x and y, have nx and ny values, respectively, then the mean y) (Σ + Σ and and variance of their combined values are found using the totals n x () . KEY POINT 3.4 The mean of x and y combined is . The variance of x and y combined is   .  We can rearrange the formulae in Key point 3.4 if we wish to find one of the totals involved. WORKED EXAMPLE 3.11 In an examination, the percentage marks of the 120 boys are denoted by x, and the percentage marks of the 80 girls are denoted by y. The marks are summarised by the totals x =Σ , 7020 2xΣ = 424 320 and 2yΣ = . 352 130 Calculate the girls’ mean mark, given that the standard deviation for all these students is 10. Answer 424320 352130 + 120 80 + −   7020 y + Σ 120 80 + 3882.25 − ( 7020 y + Σ 40 000 2   2 ) 2
10 = = 100 We substitute the given values into the formula for variance, knowing that this is equal to102. Then multiply throughout by 40 000. 73 155 290 000 − 7020 7020 ( ( = 4000 000 = = = 151290 000 151290 000 − 7020 5280 Then we rearrange to make yΣ (the total marks for the girls) the subject. We take the positive square root, as we know that all of the y values are non-negative. Girls’ mean mark = 5280 80 = 66 Divide the total marks for the girls by the number of girls. EXERCISE 3C 1 Given that: a b c d e 2Σ = v 5480 , v Σ = 288 and n 64= , find the variance of v. 2Σ w 2Σ x f = 4000 , w 5.2= and n 36= , find the standard deviation of w . = 6120 , f Σ = 40 and the standard deviation of x is 12, find xfΣ . Σ xf = 2800 , f Σ = 50 and the variance of x is 100, find x f2Σ . 193144 2Σ = , t number of data values of t. 2324 t Σ = and that the standard deviation of t is 3, find the Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 2 A building is occupied by n companies. The number of people employed by these companies is denoted by x. Find the mean number of employees, given that 2Σ and that the standard deviation of x is 18. x , x 220 8900 Σ = = 2Σ = 3 Twenty-five values of p are such that p 2Σ q are such that q 387 = of p and q together. and q 6114 Σ = 6006 and p , and 25 values of . Calculate the variance of the 50 values Σ = 388 4 In a class of 30 students, the mean mass of the 14 boys is 63.5kg and the mean mass of the girls is 57.3kg. Calculate the mean and standard deviation of the masses of all the students together, given that the sums of the squares of the masses of the boys and girls are 58444kg2 and 56222 kg2, respectively. 5 The following table shows the front tyre pressure, in psi, of five 4-wheeled vehicles, A to E. Front-left tyre Front-right tyre A 26 24 B 29 27 C 30 31 D 34 30 E 26 28 a Show that the variance of the pressure in all of these front tyres is 7.65psi2. 2Σ b Rear tyre pressures for these five vehicles are denoted by x. Given that x 7946 and that the variance of the pressures in all of the front and rear tyres on these five vehicles together is 31.6275psi2, find the mean pressure in all the rear tyres. = 74 2Σ 6 The totals x = 7931 , x 397 Σ = and y Σ = 499 are given by 29 values of x and n values of y. All the values of x and y together have a variance of 52. a Express y2Σ in terms of n. b Find the value of n for which y 2 Σ − Σ x 2 = . 10 PS 7 The five values in a dataset have a sum of 250 and standard deviation of 15. A sixth value is added to the dataset, such that the mean is now 40. Find the variance of the six values in the dataset. PS 8 A group of 10 friends played a mini-golf competition. Eight of the friends tied for second place, each with a score of 34, and the other two friends tied for first place. Find the winning score, given that the standard deviation of the scores of all 10 friends was 1.2 and that the lowest score in golf wins. PSM 9 An author has written 15 children’s books. The first eight books that she wrote contained between 240 and 250 pages each. The next six books contained between 180 and 190 pages each. Correct to 1 decimal place, the standard deviation of the number of pages in the 15 books together is 31.2. Show that it is not possible to determine a specific calculated estimate of the number of pages in the author’s 15th book. PS 10 A set of n pieces of data has mean x and standard deviation S. Another set of n2 pieces of data has mean x and standard deviation 1 2 S. Find the standard deviation of all these pieces of data together in terms of S. FAST FORWARD We will see how standard deviation and probabilities are linked in the normal distribution in Chapter 8, Section 8.2. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation EXPLORE 3.5 The following table shows three students’ marks out of 20 in the same five tests. Amber Buti Chen 1st 12 11 15 2nd 17 16 20 3rd 11 10 14 4th 9 8 12 5th 16 15 19 x x – 1 x 3+ Note that Buti’s marks are consistently 1 less than Amber’s and that Chen’s marks are consistently 3 more than Amber’s. This is indicated in the last column of the table. For each student, calculate the variance and standard deviation. Can you explain your results, and do they apply equally to the range and interquartile range? Coded data What effect does addition of a constant to all the values in a dataset have on its variation? And how can we find the variance and standard deviation of the original data from the coded data? In the Explore 3.5 activity, you discovered that the datasets x, x – 1 and x 3+ have identical measures of variation. The effect of adding –1 or 3+ is to translate the whole set of values, which has no effect on the pattern of spread, as shown in the following diagram. The marks of the three students have the same variance and the same standard deviation. 8 9 10 11 12 13 14 15 16 17 18 19 20 REWIND We saw in Chapter 2, Section 2.2 that the mean of a set of data can be found from a coded total such as Σ − . ) x b ( 75 Amber Buti Chen KEY POINT 3.5 For ungrouped data −   ) ( x b Σ − n   2 Σ 2 x n For grouped data −   ) ( x b f Σ − f Σ   2 These formulae can be summarised by writing Var( ) Var( – ) x b . x = Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 For two datasets coded as (x − a) and (x − b), we can use the coded totals x a) 2xΣ , yΣ and y b)2 Σ( − and y b) the combined set of values of x and y. , x a)2 Σ( − 2yΣ , from which we can find the variance of to find xΣ , Σ( − Σ( − , WORKED EXAMPLE 3.12 Eight values of x are summarised by the totals ( xΣ − 2 10) = 1490 and ( xΣ − 10) 100 = . Twelve values of y are summarised by the totals ( + 5) y Σ 2 = 5139 and ( + 5) Σ y = 234 . Find the variance of the 20 values of x and y together. Answer 100 8 x = + 10 = 22.5, so Σ = × x 8 22.5 180. = Var(x) = Var(x − 10) = 2 1490 8 −   100 8   = 30 . 2 x Σ 8 − 22.5 2 = 30 , so Σ 2 =x 4290 . y = 234 12 − = 5 14.5, so Σ = y 12 14.5 174. × = We find the totals xΣ and 2xΣ . We find the totals yΣ and 2yΣ . 76 2 5139 12 −   234 12   = 48 . Var(y) = Var(y + 5) = y Σ 12 , so Σ 14.5 48 = − 2 2 2 =y Var(x and y + 12 . 3099 12 +   = 4290 + 3099 20 − ( 180 + 174 20 2 ) = 56.16 WORKED EXAMPLE 3.13 It is known that 20 girls each have at least one brother. The number of brothers that they have is denoted by x. Information about the values of x – 1 is given in the following table. x – 1 No. girls f Use the coded values to calculate the standard deviation of the number of brothers, to 3 decimal places. Answer x – 1 No. girls f( ) ( x – 1) f ( x – 1) 16 32 3 5 15 45 4 1 4 16 Σ =f 20 ( x Σ − 1) f = 39 ( x Σ − 1) 2 f = 97 We extend the frequency table to find the necessary totals. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation We know that the standard deviation of x and the standard deviation of x( – 1) are equal. SD( ) SD( = x x − 1) 1) f   ( x Σ − f Σ 2   − 2 = 2 f ( 1) x Σ − f Σ −   39 20   97 20 1.023 = = EXERCISE 3D 1 Two years ago, the standard deviation of the masses of a group of men and a group of women were 8kg and 6kg, respectively. Today, all the men are 5kg heavier and all the women are 3kg lighter. Find the standard deviation for each group today. 2 Twenty readings of y are summarised by the totals y( Σ − 5) 2 = 890 and y( Σ − 5) 130 = . Find the standard deviation of y. 3 The amounts of rainfall, r mm, at a certain location were recorded on 365 consecutive days and are summarised by Calculate the mean daily rainfall and the value of r2Σ . r( Σ − 3) = 2 9950 and r( Σ − . 3) 1795.8 = 4 Exactly 20 years ago, the mean age of a group of boys was 15.7 years and the sum of the squares of their ages was 16000. If the sum of the squares of their ages has increased by 8224 in this 20-year period, find the number of boys in the group. 77 5 Readings from a device, denoted by y, are such that , y 105 Σ = the standard deviation of y is 13. Find the number of readings that were taken. y( Σ − 2775 3) = 2 and M 6 Mei measured the heights of her classmates and, after correctly analysing her data, she found the mean and standard deviation to be 163.8cm and 7.6cm. Decide whether or not these measures are valid, given the fact that Mei measured all the heights from the end of the tape measure, which is exactly 1.2 cm from the zero mark. Explain your answers. M 7 A transport company runs 21 coaches between two cities every week. In the past, the mean and variance of the journey times were 4 hours 35 minutes and 53.29 minutes2. What would be the mean and standard deviation of the times if all the coaches departed 10 minutes later and arrived 5 minutes earlier than in the past? Are there any situations i
n which achieving this might actually be possible? PS 8 During a sale, a boy bought six pairs of jeans, each with leg length x cm. He also bought four pairs of pants, each with leg length x( – 2) cm. The boy is quite short, so his father removed 4cm from the length of each trouser leg. Find the variance of the leg lengths after his father made the alterations. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 P 9 a Find the mean and standard deviation of the first seven positive even integers. b Without using a calculator, write down the mean and standard deviation of the first seven positive odd integers. c Find an expression in terms of n for the variance of the first n positive even integers. What other measure can be found using this expression? 10 Each year Upchester United plays against Upchester City in a local derby match. The number of goals scored in a match by United is denoted by u and the number of goals scored in a match by City is denoted by c. The number of goals scored in the past 15 matches are summarised by 2Σ = c . and c 19 Σ = u( Σ − u( Σ − 9 = , 25 39 1) 1) = , 2 a How many goals have been scored altogether in these 15 matches? 2Σ b Show that u = . 58 c Find, correct to 3 decimal places, the variance of the number of goals scored by the two teams together in these 15 matches. 11 Twenty values of x are summarised by x( Σ − 1) 2 = 132 and x( Σ − 1) = 44 . Eighty values of y are summarised by y( Σ + 2 1) = 17 704 and y( Σ + 1) 1184 . = a Show that x 64 Σ = 2Σ and that x = 240 . b Calculate the value of yΣ and of 2yΣ . 78 c Find the exact variance of the 100 values of x and y combined. 12 The heights, x cm, of 200 boys and the heights, y cm, of 300 girls are summarised by the following totals: x( Σ − 160) 2 = 18 240 , x( Σ − 160) 1820 = , y( Σ − 150) 2 = 20 100 , y( Σ − 150) = . 2250 a Find the mean height of these 500 children. b By first evaluating 2xΣ and 2yΣ , find the variance of the heights of the 500 children, including appropriate units with your answer. What effect does multiplication of all the values in a dataset have on its variation? And how can we find the variance and standard deviation of the original values from the coded data? Consider the total cost of hiring a taxi for which a customer pays a fixed charge of $3 plus $4 per kilometre travelled. Using y for the total cost and x for the distance travelled in kilometres, the cost can be calculated from the equation y are shown in the following table. + . Some example values x4 = 3 1+  → 1+  → Distance x( km) Cost y($ ) 1 7 2 11 3 15  → 4+  → 4+ Copyright Material - Review Only - Not for Redistribution FAST FORWARD You will study the variance of linear combinations of random variables in the Probability & Statistics 2 Coursebook, Chapter 3. REWIND We saw in Chapter 2, Section 2.2 that the mean of a set of data can be found from a coded total such as − . ) ax b Σ ( Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation When the distance changes or varies by 1+ , the cost changes or varies by 4+ . Variation in cost is affected by multiplication ( 4)× but not by addition ( 3)+ . If we consider the graph of y variation of y. If we increase the x-coordinate of a point on the line by 1, its y-coordinate increases by 4. + , then it is only the gradient of the line that affects the x4 = 3 Multiplying x {1, 2, 3} of spread. = by 4 ‘stretches’ the whole set to {4, 8, 12}, which affects the pattern Adding 3+ to {4, 8, 12} simply translates the whole set to y {7, 11, 15} on the pattern of spread. = , which has no effect Journeys of 1, 2 and 3km, costing $7, $11 and $15 are represented in the following diagram 10 11 12 13 14 15 Distance Cost x y = 4x + 3 In the diagram, we see that the range of y is 4 times the range of x, so the range of x is 1 4 times the range of y. For x {1, 2, 3} the following results: , you should check to confirm 3 {7, 11, 15} and x4 + = = SD( x ) = 1 4 × SD(4 x + and 3) Var( x ) = 1 16 × Var(4 x + 3) . 79 KEY POINT 3.6 x For ungrouped data For grouped data ( ax b − n −   Σ ) ( ax b − n   2     Σ 2 ) f ( ax b − f Σ −   Σ ( ) ax b f − f Σ   2     These formulae can be summarised by writing Var( x ) = 1 2 a × Var( ax b − ) or Var( ax b − ) = 2 a × Var( x ) . WORKED EXAMPLE 3.14 The standard deviation of the prices of a selection of brand-name products is $24. Imitations of these products are all sold at 25% of the brand-name price. Find the variance of the prices of the imitations. Answer Var(0.25 ) x 2 0.25 Var( × x ) 0.25 2 2 24 × 36 = = = Denoting the brand-name prices by x and 0.25 , we use the the imitation prices by 2= ) Var( fact that to find a × Var x x Var( (0.25 ). ax x ) TIP The units for variance in this case are ‘dollars squared’. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 3.15 Given that Σ x(3 − 1) 2 = 9136 , Σ x(3 − 1) 53 = and n 10= , find the value of x2Σ . Answer 3 x 1 − = x = = 53 10 1 3 2.1 ×   53 10 +  1  We first find x, knowing that the mean of the coded values is 1 less than 3 times the mean of x; that is, mean x (3 – 1) 3 x − . 1 = Var( x ) = × Var(3 x − 1) We form and solve an equation knowing that 2 2.1 − = ×   9136 10 2 5.3 −   Var( x ) 1 2= a × Var( ax b − ) and Var(3 – 1) x = 9136 10 5.32 . − 1 2 3 1 9 2 x Σ 10 2 x Σ 10 − 4.41 98.39 = ∴ Σ x 2 = 1028 EXERCISE 3E 80 1 The range of prices of the newspapers sold at a kiosk is $0.80. After 6p.m. all prices are reduced by 20%. Find the range of the prices after 6p.m. 2 Find the standard deviation of x, given that 2Σ x4 = 14600 , Σ x2 = 420 and n . 20= 3 The values of x given in the table on the left have a standard deviation of 0.88. Find the standard deviation of the values of y 13 b 19 c 4 The temperatures, °T Celsius, at seven locations in the Central Kalahari Game Reserve were recorded at 4p.m. one January afternoon. The values of T , correct to 1 decimal place, were: 32.1, 31.7, 31.2, 31.5, 31.9, 32.2 and 32.7. a Evaluate Σ T10( − 30) and Σ 100( T 30)2 . − b Use your answers to part a to calculate the standard deviation of T . c By 5p.m. the temperature at each location had dropped by exactly 0.75 C. Find the variance of the ° temperatures at 5p.m. 5 Building plots are offered for sale at $315 per square metre. The seller has to pay a lawyer’s fee of $500 from the money received. Salome’s plot is 240 square metres larger than Nadia’s plot. How much more did the seller receive from Salome than from Nadia after paying the lawyer’s fees? 6 Temperatures in degrees Celsius ( C)° + . formula F 1.8C 32 = can be converted to temperatures in degrees Fahrenheit ( F)° using the a The temperatures yesterday had a range of °15 C. Express this range in degrees Fahrenheit. b Temperatures elsewhere were recorded at hourly intervals in degrees Fahrenheit and were found to have mean 54.5 and variance 65.61. Find the mean and standard deviation of these temperatures in degrees Celsius. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation M 7 Ten items were selected from each of four sections at a supermarket. Details of the prices of those items, in dollars, on 1st April and on 1st June are shown in the following table. Bakery Household Tinned food Fruit & veg 1st April 1st June Mean 2.50 5.00 4.00 2.00 SD 0.40 1.20 0.60 0.40 Mean 2.25 4.75 4.10 2.00 SD 0.36 1.25 0.60 0.50 For which section’s items could each of the following statements be true? Briefly explain each of your answers. a The total cost of the items did not change. b The price of each item changed by the same amount. c The proportional change in the price of each item was the same. PS 8 The lengths of 45 ropes used at an outdoor recreational centre can be extended by 30% when stretched. The sum of the squares of their stretched lengths is 0.0507 km2 and their natural lengths, x metres, are summarised by x( . Find the mean natural length of these 45 ropes. 1200 20) 2 Σ − = PS 9 Over a short period of time in 2016, the value of the pound sterling (£) fell by 15.25% against the euro (€). Find the percentage change in the value of the euro against the pound over this same period. Appendix to Section 3.3 In this appendix, we show how the two formulae for variance are equivalent. For simplicity, we will assume that n 3= . 81 If we denote our three numbers by x x,1 2 and x3, then x )2 Variance is defined by Var( x ) = , so if we expand the brackets and rearrange, we get 1 (= 3 x 1 + x 2 + 3 −  ) −  1 3 1 3 1 3  2 x 1 + 2 x 1 + Var + + xx x 1 2 x 2 − 2 xx xx 3 + x 2    + Note: the term in brackets is equal to x. 2 x Σ 3 − x 2, which is the alternative formula 2 x Σ n 2 − x in the case where n 3= . Try showing that the two formul
ae for variance are equivalent for the simple case where 4= or larger. Can you generalise this n argument to an arbitrary value of n? 2= , and then challenge yourself by taking on n Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Checklist of learning and understanding ● Commonly used measures of variation are the range, interquartile range and standard deviation. ● A box-and-whisker diagram shows the smallest and largest values, the lower and upper quartiles and the median of a set of data. ● For ungrouped data, the median Q2 is at the ● For grouped data with total frequency = Σn  th value. n 1 +  2 f , the quartiles are at the following values. ● Lower quartile Q1 is at or Σ . f ● Middle quartile Q2 is at or n 4 n 2 n3 or ● Upper quartile Q3 is at ● IQR = –3 Q Q 1 ● For ungrouped data: 82 E E E Standard deviation = Variance = ) , where x − = x Σ n . ● For grouped data: Standard deviation = Variance = ( x x Σ − f Σ ), where − x = xf Σ f Σ . ● For datasets x and y with nx and ny values, respectively: 2 Mean = and Variance =    . ● The formulae for ungrouped and grouped coded data can be summarised by: Var( x ) Var( – ) x b = and Var( x ) = 1 2 a × Var ( – ax b or ) Var ( – ax b ) = 2 a × Var ( x ) ● For ungrouped coded data −   ) ( x b Σ − n   2 and ( ax b − n −   Σ ( ) ax b − n   2     ● For grouped coded data −   ) ( x b f Σ − f Σ 2   and ( ax b − f Σ −   Σ ( ) ax b f − f Σ   2     Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation END-OF-CHAPTER REVIEW EXERCISE 3 1 Three boys and seven girls are asked how much money they have in their pockets. The boys have $2.50 each and the mean amount that the 10 children have is $3.90. a Show that the girls have a total of $31.50. b Given that the seven girls have equal amounts of money, find the standard deviation of the amounts that the 10 children have. [1] [3] P 2 Jean Luc was asked to record the times of 20 athletes in a long distance race. He started his stopwatch when the race began and then went to sit in the shade, where he fell asleep. On waking, he found that x athletes had already completed the race but he was able to record the times taken by all the others. State the possible value(s) of x if Jean Luc was able to use his data to calculate: a the variance of the times taken by the 20 athletes b the interquartile range of the times taken by the 20 athletes. [1] [2] P 3 The quiz marks of nine students are written down in ascending order and it is found that the range and interquartile range are equal. Find the greatest possible number of distinct marks that were obtained by the nine students. [2] 4 Two days before a skiing competition, the depths of snow, x metres, at 32 points on the course were measured and it was discovered that the numerical values of xΣ and x2Σ were equal. a Given that the mean depth of snow was 0.885 m, find the standard deviation of x. b Snow fell the day before the competition, increasing the depth over the whole course by 1.5cm. Explain what effect this had on the mean and on the standard deviation of x. [2] [2] 83 5 The following box plots summarise the percentage scores of a class of students in the three Mathematics tests they took this term. 30 40 50 Percentage scores (%) 60 70 80 90 100 first test second test third test a Describe the progress made by the class in Mathematics tests this term. b Which of the tests has produced the least skewed set of scores? c What type of skew do the scores in each of the other two tests have? 6 The following table shows the mean and standard deviation of the lengths of 75 adult puff adders (Bitis arietans), which are found in Africa and on the Arabian peninsula. Frequency Mean (cm) SD (cm) African Arabian 60 15 102.7 78.8 6.8 4.2 a Find the mean length of the 75 puff adders. b The lengths of individual African puff adders are denoted by x f and the lengths of individual Arabian puff adders by xb. By first finding x f 75 puff adders. 2Σ and xb 2Σ , calculate the standard deviation of the lengths of all [2] [1] [2] [3] [5] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 M 7 The scores obtained by 11 people throwing three darts each at a dartboard are 54, 46, 43, 52, 180, 50, 41, 56, 52, 49 and 54. a Find the range, the interquartile range and the standard deviation of these scores. b Which measure in part a best summarises the variation of the scores? Explain why you have chosen this particular measure. [4] [2] 8 The heights, x cm, of a group of 28 people were measured. The mean height was found to be 172.6cm and the standard deviation was found to be 4.58cm. A person whose height was 161.8cm left the group. i Find the mean height of the remaining group of 27 people. ii Find x2Σ for the original group of 28 people. Hence find the standard deviation of the heights of the remaining group of 27 people. [2] [4] 9 120 people were asked to read an article in a newspaper. The times taken, to the nearest second, by the people to read the article are summarised in the following table. Cambridge International AS & A Level Mathematics 9709 Paper 63 Q4 June 2014 Time (seconds) 1–25 26–35 36– 45 46–55 56–90 Number of people 4 24 38 34 20 84 Calculate estimates of the mean and standard deviation of the reading times. [5] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q2 June 2015 10 The weights, in kilograms, of the 15 basketball players in each of two squads, A and B, are shown below. Squad A Squad B 97 75 98 104 84 100 109 115 99 122 82 116 96 84 107 91 79 94 101 96 77 111 108 83 84 86 115 82 113 95 i Represent the data by drawing a back-to-back stem-and-leaf diagram with squad A on the left-hand side [4] of the diagram and squad B on the right-hand side. ii Find the interquartile range of the weights of the players in squad A. [2] iii A new player joins squad B. The mean weight of the 16 players in squad B is now 93.9kg. Find the weight [3] of the new player. Cambridge International AS & A Level Mathematics 9709 Paper 62 Q5 November 2015 [Adapted] 11 The heights, x cm, of a group of 82 children are summarised as follows. x( Σ − 130) = − 287 . , standard deviation of x 6.9= i Find the mean height. ii Find x( Σ − 130)2 . [2] [2] Cambridge International AS & A Level Mathematics 9709 Paper 63 Q2 June 2010 12 A sample of 36 data values, x, gave x( Σ − 45) = − 148 and x( Σ − 45) 2 = . 3089 i Find the mean and standard deviation of the 36 values. ii One extra data value of 29 was added to the sample. Find the standard deviation of all 37 values. [3] [4] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q3 June 2011 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 3: Measures of variation 13 The ages, x years, of 150 cars are summarised by x 645 Σ = 2Σ and x = 8287.5 . Find ( Σ − x x the mean of x. )2 , where x denotes [4] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q1 June 2012 M 14 A set of data values is 152, 164, 177, 191, 207, 250 and 258. Compare the proportional change in the standard deviation with the proportional change in the interquartile range when the value 250 in the data set is increased by 40%. [5] PS 15 A shop has in its stock 80 rectangular celebrity posters. All of these posters have a width to height ratio of 1: 2, and their mean perimeter is 231.8cm. Given that the sum of the squares of the widths is 200120 cm2, find the standard deviation of the widths of the [4] posters. P PS 16 At a village fair, visitors were asked to guess how many sweets are in a glass jar. The best six guesses were 180, 211, 230, 199, 214 and 166. a Show that the mean of these guesses is 200, and use SD = )2 ( x x Σ − n to calculate the standard deviation. [4] b The jar actually contained 202 sweets. Without further calculation, write down the mean and the standard deviation of the errors made by these six visitors. Explain why no further calculations are required to do this. [4] 17 The number of women in senior management positions at a number of companies was investigated. The number of women at each of the 25 service companies and at each of the 16 industrial companies are denoted by Sw and 2 5) − Σ Iw , respectively. The findings are summarised by the totals: 2 , w 3 ) = Σ I and w 3 ) ( IΣ , w 5 ( SΣ ) 15 = 4 = − . w( S 28 12 − − − = ( 85 [3] [5] [3] [3] [7] a Show that there are, on average, more than twice as many women in senior management positions at the service companies than at the industrial companies. 2 b Show that w S Σ ≠ Σ w( S ) 2 2 and that w I Σ ≠ Σ w( I ) 2 . c Find the standard deviation of the number of women in senior management positions at all of these service and industrial companies together. 18 The a
ges, a years, of the five members of the boy-band AlphaArise are such that a( Σ − 21) 2 = 11.46 and a( Σ − 21) = − . 6 The ages, b years, of the seven members of the boy-band BetaBeat are such that b( b( Σ − = . 0 18) Σ − 2 18) = 10.12 and a Show that the difference between the mean ages of the boys in the two bands is 1.8 years. b Find the variance of the ages of the 12 members of these two bands. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Cambridge International AS & A Level Mathematics: Probability & Statistics 1 CROSS-TOPIC REVIEW EXERCISE 1 1 Two players, A and B, both played seven matches to reach the final of a tennis tournament. The number of games that each of them won in these matches are given in the following back-to-back stem-and-leaf diagram. Player Player B 8 9 1 1 3 4 6 Key: 5 2 6 represents 25 games for A and 26 games for B a How many fewer games did player B win than player A? b Find the median number of games won by each player. c In a single stem-and-leaf diagram, show the number of games won by these two players in all of the 14 matches they played to reach the final. 2 A total of 112 candidates took a multiple-choice test that had 40 questions. The numbers of correct answers given by the candidates are shown in the following table. No. correct answers No. candidates 0–9 18 10–15 16–25 26–30 31–39 24 27 23 19 40 1 a State which class contains the lower quartile and which class contains the upper quartile. Hence, find the least possible value of the interquartile range. b Copy and complete the following table, which shows the numbers of incorrect answers given by 86 the candidates in the test. No. incorrect answers No. candidates 0 1 1–9 c Calculate an estimate of the mean number of incorrectly answered questions. 3 At a factory, 50-metre lengths of cotton thread are wound onto bobbins. Due to fraying, it is common for a length, l cm, of cotton to be removed after it has been wound onto a bobbin. The following table summarises the lengths of cotton thread removed from 200 bobbins. Length removed ( cm) l 0 < l 2.5< 2.5 < l 5.0< No. bobbins 137 49 5 < l 10< 14 a Calculate an estimate of the mean length of cotton removed. b Use your answer to part a to calculate, in metres, an estimate of the standard deviation of the length of cotton remaining on the 200 bobbins. 4 People applying to a Computing college are given an aptitude test. Those who are accepted take a progress test 3 months after the course has begun. The following table gives the aptitude test scores, x, and the progress test scores, y, for a random sample of eight students, A to H. x y A 61 53 B 80 77 C 74 61 D 60 70 E 83 81 F 92 54 G 71 63 H 67 85 a Find the interquartile range of these aptitude test scores. b Use the summary totals Σ =x 588, Σ , Σ =y to calculate the variance of the aptitude and progress test scores when they are considered together. 544 and Σ 44 080 38030 2 =x 2 =y [1] [2] [3] [3] [3] [3] [3] [4] [1] [3] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cross-topic review exercise 1 c The mean progress score for all the students at the college is 70 and the variance is 112. Any student who scores less than 1.5 standard deviations below the mean is sent a letter advising that improvement is needed. Which of the students A to H should the letter be sent to? [1] 5 The growth of 200 tomato plants, half of which were treated with a growth hormone, was monitored over a 5-day period and is summarised in the following graphs 100 80 60 40 20 0 Treated Untreated 1 2 3 4 5 6 7 8 9 10 Growth (cm) 87 Use the graphs to describe two advantages of treating these tomato plants with the growth hormone. [2] 6 A survey of a random sample of 23 people recorded the number of unwanted emails they received in a particular week. The results are given below. 9 18 13 18 21 17 22 27 8 11 26 26 32 17 31 20 36 15 13 25 35 29 14 a Represent the data in a stem-and-leaf diagram. b Draw, on graph paper, a box-and-whisker diagram to represent the data. 7 The volumes of water, x 106× 2.82, 2.50, 2.75, 3.14, 3.66 and 3.07. litres, needed to fill six Olympic-sized pools are a Find the value of Σ −x( 2) and of Σ −x( 2)2 . b Use your answers to part a to find the mean and the standard deviation of the volumes of water, giving both answers correct to the nearest litre. 8 The speeds of 72 coaches at a certain point on their journeys between two cities were recorded. The results are given in the following table. Speed (km/h) Cumulative frequency < 50 0 < 54 9 < 70 41 < 75 54 < 85 72 a State the number of coaches whose speeds were between 54 and 70 km/h. b A student has illustrated the data in a cumulative frequency polygon. Find the two speeds between which the polygon has the greatest gradient. c Calculate an estimate of the lower boundary of the speeds of the fastest 25 coaches. [3] [4] [2] [5] [1] [1] [3] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Cambridge International AS & A Level Mathematics: Probability & Statistics 1 9 The following table shows the masses, m grams, of 100 unsealed bags of plain potato crisps. Mass (m grams) No. bags ( )f 34.6 < m < 35.4 35.4 < m < 36.2 36.2 < m < 37.2 20 30 50 a Show that the heights of the columns in a histogram illustrating these data must be in the ratio 2 : 3 : 4. b Calculate estimates of the mean and of the standard deviation of the masses. c Before each bag is sealed, 0.05 grams of salt is added. Find the variance of the masses of the sealed bags of salted potato crisps. 10 The masses, in carats, of a sample of 200 pearls are summarised in the following cumulative frequency graph. One carat is equivalent to 200 milligrams. 88 ) 200 160 120 80 40 0 20 40 60 80 100 120 Mass (carats) a Use the graph to estimate, in carats: i the median mass of the pearls ii the interquartile range of the masses. b To qualify as a ‘paragon’, a pearl must be flawless and weigh at least 20 grams. Use the graph to estimate the largest possible number of paragons in the sample. 11 The amounts spent, S dollars, by six customers at a hairdressing salon yesterday were as follows. 12.50, 15.75, 41.30, 34.20, 10.80, 40.85. Each of the customers paid with a $50 note and each received the correct change, which is denoted by C$ . a Find, in dollars, the value of S C+ and of S C− . b Explain why the standard deviation of S and the standard deviation of C are identical. Copyright Material - Review Only - Not for Redistribution [2] [4] [1] [1] [2] [2] [3] [3] Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cross-topic review exercise 1 12 The numbers of goals, x, scored by a team in each of its previous 25 games are summarised by the totals x( Σ − 1) 2 = 30 and Σ − x( 1) 12. = a Find the mean number of goals that the team scored per game. b Find the value of Σx2. c Find the value of a and of b in the following table, which shows the frequencies of the numbers of goals scored by the team. No. goals No. games ( )> 0 13 The lengths of some insects of the same type from two countries, X and Y , were measured. [2] [3] [2] The stem-and-leaf diagram shows the results. Country X Country Y (10) (18) (16) (16) (11 80 81 82 83 84 85 86 13) (15) (17) (15) (12) (11) Key: 5 81 3 means an insect from country X has length 0.815 cm and an insect from country Y has length 0.813 cm. i Find the median and interquartile range of the lengths of the insects from country X . [2] 89 ii The interquartile range of the lengths of the insects from country Y is 0.028cm. Find the values of q and r. [2] iii Represent the data by means of a pair of box-and-whisker plots in a single diagram on graph paper. [4] iv Compare the lengths of the insects from the two countries. [2] Cambridge International AS & A Level Mathematics 9709 Paper 63 Q6 June 2010 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 90 Chapter 4 Probability In this chapter you will learn how to: elementary events ■ evaluate probabilities by means of enumeration of equiprobable (i.e. equally likely) ■ use addition and multiplication of probabilities appropriately ■ use the terms mutually exclusive and independent events ■ determine whether two events are independent ■ calculate and use conditional probabilities. 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versity Press - Review Copy Chapter 4: Probability PREREQUISITE KNOWLEDGE Where it comes from What you should be able to do Check your skills IGCSE / O Level Mathematics Calculate the probability of a single event as either a fraction, decimal or percentage. Understand and use the probability scale from 0 to 1. Understand relative frequency as an estimate of probability. Calculate the probability of simple combined events, using possibility diagrams and tree diagrams where appropriate. Use language, notation and Venn diagrams to describe sets and represent relationships between sets. 1 How many 6s are expected when an ordinary fair die is rolled 180 times? 2 Find the probability of obtaining a total of 4 when the scores on two ordinary fair dice are added together. 3 It is given that {2, 4, 5} =A and {1, 2, 5} ′ =A ′ =B a Venn diagram, or otherwise, find n( n( , {3, 4} . Using ) and A B ∪ ′ ′ ∩A B ). 91 If we do this, how likely is that? Probability measures the likelihood of an event occurring on a scale from 0 (i.e. impossible) to 1 (i.e. certain). We write this as P(name of event), and its value can be expressed as a fraction, decimal or percentage. The greater the probability, the more likely the event is to occur. Although we do not often calculate probabilities in our daily lives, we frequently assess and compare them, and this affects our behaviour. Do we have a better chance of performing well in an exam after a good night’s sleep or after revising late into the night? Should you visit a doctor or is your sore throat likely to heal by itself soon? Insurance is based on risk, which in turn is based on the probability of certain events occurring. Government spending is largely determined by the probable benefits it will bring to society. 4.1 Experiments, events and outcomes The result of an experiment is called an outcome or elementary event, and a combination of these is known simply as an event. Rolling an ordinary fair die is an experiment that has six possible outcomes: 1, 2, 3, 4, 5 or 6. Obtaining an odd number with the die is an event that has three favourable outcomes: 1, 3 or 5. Random selection and equiprobable events The purpose of selecting objects at random is to ensure that each has the same chance of being selected. This method of selection is called fair or unbiased, and the selection of any particular object is said to be equally likely or equiprobable. KEY POINT 4.1 When one object is randomly selected from n objects, P(selecting any particular object) 1 = . n Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 The probability that an event occurs is equal to the proportion of equally likely outcomes that are favourable to the event. KEY POINT 4.2 P(event) = Number of favourable equally likely outcomes Total number of equally likely outcomes Consider randomly selecting 1 student from a group of 19, where 11 are boys and eight are girls. There are 19 possible outcomes: 11 are favourable to the event selecting a boy and eight are favourable to the event selecting a girl, as shown in the following table. Event/outcome Probability Description TIP Selecting any particular boy Selecting any particular girl Selecting any particular student 92 Selecting a boy Selecting a girl 1 19 1 19 1 19 11 19 8 19 These three outcomes are equally likely. 11 of the 19 equally likely outcomes are favourable to this event. 8 of the 19 equally likely outcomes are favourable to this event. Exhaustive events A set of events that contains all the possible outcomes of an experiment is said to be exhaustive. In the special case of event A and its complement, not A, the sum of their probabilities is 1 because one of them is certain to occur. Recall that the notation used for the complement of set A is A′. The word particular specifies one object. It does not matter whether that object is a boy, a girl or a student; each has a 1 19 chance of being selected. KEY POINT 4.3 A ) 1 = P( A + ) P(not or A ) P( + A ) 1 ′ = Examples of complementary exhaustive events are shown in the following table. P( Experiment Exhaustive events A A′ Toss a fair coin heads tails Roll a fair die less than 2 Play a game of chess win 2 or more not win Probabilities (win) P(not win) 1 = + Trials and expectation Each repeat of an experiment is called a trial. The proportion of trials in which an event occurs is its relative frequency, and we can use this as an estimate of the probability that the event occurs. If we know the probability of an event occurring, we can estimate the number of times it is likely to occur in a series of trials. This is a statement of our expectation. Copyright Material - Review Only - Not for Redistribution KEY POINT 4.4 In n trials, event A is expected to occur n times. ) AP( × Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 4: Probability WORKED EXAMPLE 4.1 The probability of rain on any particular day in a mountain village is 0.2. On how many days is rain not expected in a year of 365 days? Answer 365 n = and P(does not rain) 1 – 0.2 = = 0.8 365 × 0.8 = 292 days We multiply the probability of the event by the number of days in a year. M EXPLORE 4.1 We can see how closely expectation matches with what happens in practice by conducting simple experiments using a fair coin (or an ordinary fair die). Toss the coin 10 times and note as a decimal the proportion of heads obtained. Repeat this and note the proportion of heads obtained in 20 trials. Continue doing this so that you have a series of decimals for the proportion of heads obtained in 10, 20, 30, 40, 50, … trials. Represent these proportions on a graph by plotting them against the total number of trials conducted. How do your results compare with the expected proportion of heads? For trials with a die, draw a graph to represent the proportions of odd numbers obtained. EXERCISE 4A 1 A teacher randomly selects one student from a group of 12 boys and 24 girls. Find the probability that the teacher selects: a a particular boy b a girl. 2 United’s manager estimates that the team has a 65% chance of winning any particular game and an 85% chance of not drawing any particular game. a What are the manager’s estimates most likely to be based on? b If the team plays 40 games this season, find the manager’s expectation of the number of games the team will lose. c If the team loses one game more than the manager expects this season, explain why this does not necessarily mean that they performed below expectation. 3 Katya randomly picks one of the 10 cards shown If she repeats this 40 times, how many times is Katya expected to pick a card that is not blue and does not have a letter B on it? Copyright Material - Review Only - Not for Redistribution 93 Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 4 A numbered wheel is divided into eight sectors of equal size, as shown. The wheel is spun until it stops with the arrow pointing at one of the numbers. Axel decides to spin the wheel 400 times. a Find the number of times the arrow is not expected to point at a 4. b How many more times must Axel spin the wheel so that the expected number of times that the arrow points at a 4 is at least 160 bag contains black and white counters, and the probability of selecting a black counter is 1 6 . a What is the smallest possible number of white counters in the bag? b Without replacement, three counters are taken from the bag and they are all black. What is the smallest possible number of white counters in the bag? 6 When a coin is randomly selected from a savings box, each coin has a 98% chance of not being selected. How many coins are in the savings box? PS 7 A set of data values is 8, 13, 17, 18, 24, 32, 34 and 38. Find the probability that a randomly selected value is more than one standard deviation from the mean. PS 8 One student is randomly selected from a school that has 837 boys. The probability that a girl is selected is 4 7 particular boy is selected. . Find the probability that a 94 REWIND We studied the mean in Chapter 2, Section 2.2 and standard deviation in Chapter 3, Section 3.3. 4.2 Mutually exclusive events and the addition law To find the probability that event A or event B occurs, we can simply add the probabilities of the two events together, but only if A and B are mutually exclusive. Mutually exclusive events have no common favourable outcomes, which means that it is not possible for both events to occur, so A P( and ) B 0. = For example, when we roll an ordinary die, the events ‘even number {2, 4, 6} of 5 {1, 5} = It is not possible to roll a number that is even and a factor of 5. We say that the intersection of these two sets is empty. Therefore: ’ are mutually exclusive because they have no common favourable outcomes. ’ and ‘factor = P(evenor factor of 5) P(even) P(factor of 5) = + Events are not mutually exclusive if they have at least one common favourable outcome, which means that it is possible for both events to occur, so P( and ) A B ≠ . 0 For example, when we roll an ordinary die, the events ‘odd number {1, 3, 5} ‘factor of 5 {1, 5} outcomes. It is possible to roll a number that is
odd and a factor of 5. We say that the intersection of these two sets is not empty. Therefore: ’ are not mutually exclusive because they do have common favourable ’ and = = P(odd or factor of 5) P(odd) P(factor of 5) ≠ + Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 4: Probability KEY POINT 4.5 The addition law for mutually exclusive events is A B P( or ) P( = A ) P( + B ) . This can be extended for any number of mutually exclusive events: P(A or B or C or …) B P( A C ) P( + ) P( + + … = ) Venn diagrams Venn diagrams are useful tools for solving problems in probability. We can use them to show favourable outcomes or the number of favourable outcomes or the probabilities of particular events. The number of outcomes favourable to event A is denoted by An( ). The set of outcomes that are not favourable to event A is the complement of A, denoted by ′A . The following Venn diagrams illustrate various sets and their complements. TIP The universal set ℰ represents the complete set of outcomes and is called the possibility space TIP A Aʹ not A A ∪ B A or B (A ∪ B)ʹ neither A nor B A ∩ B A and B (A ∩ B)ʹ not both A and B KEY POINT 4.6 ‘A or B’ means event A occurs or event B occurs or both occur. ∪A B means ‘A or B’ and ∩A B means ‘A and B’. 95 Using set notation, the addition law for two mutually exclusive events is A B P( ∪ = ) P( A ) P( + B . ) A and B are mutually exclusive when A B empty set). P( ∩ = ; that is, when A B∩ = ∅ (∅ means the ) 0 For non-mutually exclusive events, P( )∪A B can be found by enumerating (counting) the favourable equally likely outcomes, taking care not to count any of them twice. We show how this can be done in part b of the following example. WORKED EXAMPLE 4.2 One digit is randomly selected from 1, 2, 3, 4, 5, 6, 7, 8 and 9. Three possible events are: A: a multiple of 3 is selected. B: a factor of 8 is selected. C: a prime number is selected. a Show that the only pair of mutually exclusive events from A, B and C is A and B, and find P( )∪A B . b Find: i P( )∪A C ii P( )∪B C . Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Answer = ℰ {1, 2, 3, 4, 5, 6, 7, 8, 9} 3 = . 9 A {3, 6, 9} ) , so AP( = B {1, 2, 4, 8} = , so BP( ) C {2, 3, 5, 7} = , so CP∩ = ∅, so A and B are mutually exclusive. A C∩ ≠ ∅, so A and C are not mutually exclusive. B C∩ ≠ ∅, so B and C are not mutually exclusive. P( 96 A B ∪ = ) P( or A B ) B ) ) P( 3 9 7 9 By listing and counting the favourable outcomes, we can find the probability for each event. Outcomes favourable to pairs of events are shown in the three Venn diagrams opposite. Two events are mutually exclusive when they have no common favourable outcomes; that is, when their intersection is an empty set. We can use the addition law because A and B are mutually exclusive events. b Both parts of this question can be answered using the lists of elements or the previous Venn diagrams. i n( A C ∪ ) n( = A ) n( + C ) n( − A C ∩ ) P( A C ∪ ) P( = A ) P( + C ) P or 2 3 ii n( B ∪ C ) n( = B ) n( + C ) n( − B ∩ C ) P( B ∪ C ) P( = B ) P( + C ) P In the first part of b above, we subtracted ( common elements in ∩A C n( ) have been counted in An( ∩A C ) because the ) and in Cn( ). We follow the same steps when working directly with probabilities. This also applies to events that are mutually exclusive, where the number of common elements is equal to zero. So in general, for any two events A and B: n( A B ∪ = ) n( A ) n( + B ) n( − A B ∩ ) and Set A contains 3 of the 9 elements. Set C contains 4 of the 9 elements. Set A and set C have 1 of the 9 elements in common. Set B contains 4 of the 9 elements. Set C contains 4 of the 9 elements. Set B and set C have 1 of the 9 elements in common. P( A B ∪ = ) P( A ) P( + B ) P( − A B ∩ ). Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy WORKED EXAMPLE 4.3 In a survey, 50% of the participants own a desktop ( 15% own both. )D , 60% own a laptop ( )L and What percentage of the participants owns neither a desktop nor a laptop? Answer 1 0.5 D L p 0.15 q 0.6 x The Venn diagram shows the given information, where ,p q and x represent, respectively, the percentage that own a desktop only, a laptop only and neither of these. p = 0.5 – 0.15 = 0.35 q = 0.6 – 0.15 = 0.45 x 1 – (0.35 = + 0.15 + 0.45) = 0.05 or 5% 5%∴ of the participants own neither a desktop nor a laptop. Chapter 4: Probability TIP The symbol ∴ means ‘therefore’. WORKED EXAMPLE 4.4 97 Forty children were each asked which fruits they like from apples ( bananas ( )B and cherries ( )A , )C . The following Venn diagram shows the number of children that like each type of fruit. Find the probability that a randomly selected child likes apples or bananas. Answer 40 17 A 6 5 3 7 1 2 C 14 2 B 24 8 P( A B ∪ = ) P( A B ) – P( A B ∩ ) =An( ) 17, =Bn( ) 8 and n( A B ) ∩ = 4. ) P( + 8 40 − + = = 17 40 21 40 4 40 n( A B ∪ ≠ ) 17 + 8 because A B )∩ ≠ ∅. ( There are 17 like apples or bananas. 8 – 4 = + 21 children who Alternatively, we can add up the numbers in the A and B circles 21. 2 KEY POINT 4.7 For any two events, A and B, P( or A B ) P( = A B ∪ = ) P( A ) P( + B ) – P( A B ∩ ). Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXERCISE 4B 1 Find the probability that the number rolled with an ordinary fair die is: a a prime number or a 4 b a square number or a multiple of 3 c more than 3 or a factor of 8. 2 A group of 40 students took a test in Economics. The following Venn )B took the test and that seven students diagram shows that 19 boys ( failed the test ( )F . 40 B F a Describe the 21 students who are members of the set B′. 16 3 4 b Find the probability that a randomly selected student is a boy or someone who failed the test. 17 3 The following table gives information about all the animals on a farm. Male Female Goats Sheep 5 3 25 22 a Find the probability that a randomly selected animal is: i male or a goat ii a sheep or female. b Find a different way of describing each of the two types of animal in part a. 98 4 Two ordinary fair dice are rolled and three events are: X : the sum of the two numbers rolled is 6. Y : the difference between the two numbers rolled is zero. Z: both of the numbers rolled are even. a List the outcomes that are favourable to: i X and Y ii X and Z iii Y and Z. b What do your answers to part a tell you about the events ,X Y and Z? 5 The letters A, B, B, B, C, D, D and E are written onto eight cards and placed in a bag. Find the probability that the letter on a randomly selected card is: a a vowel or in the word DOMAIN b a consonant or in the word DOUBLE. 6 In a group of 25 boys, nine are members of the chess club ( are members of the debating club ( )D and 10 are members of neither of these clubs. This information is shown in the Venn diagram. )C , eight 25 9 C D a b c 8 a Find the values of ,a b and c. b Find the probability that a randomly selected boy is: 10 i a member of the chess club or the debating club ii a member of exactly one of these clubs. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 4: Probability 7 Forty girls were asked to name the capital of Cuba and of Hungary; 19 knew the capital of Cuba, 20 knew the capital of Hungary and seven knew both. a Draw a Venn diagram showing the number of girls who knew each of these capitals. b Find the probability that a randomly selected girl knew: i the capital of Cuba but not of Hungary ii just one of these capitals. 8 In a survey on pet ownership, 36% of the participants own a cat, 20% own a hamster but not a cat, and 8% own a hamster and a cat. What percentage of the participants owns neither a hamster nor a cat? 9 A garage repaired 132 vehicles last month. The number of vehicles that required electrical ( diagram opposite. )E , mechanical ( )M and bodywork ( )B repairs are given in the Find the probability that a randomly selected vehicle required: a mechanical or bodywork repairs b no bodywork repairs c exactly two types of repair. E 12 11 13 5 26 M 47 18 B P 33 S 17 32 7 11 M 99 10 The 100 students at a technical college must study at least one subject from Pure Mathematics ( studying these subjects are given in the diagram opposite. )S and Mechanics ( )P , Statistics ( )M . The numbers a Who does the number 17 in the diagram refer to? b Find the probability that a randomly selected student studies: i Pure Mathematics or Mechanics ii exactly two of these subjects. c List the three subjects in ascending order of popularity. 11 Events X and Y a
re such that P( X ) = 0.5, P( Y ) = 0.6 and P( X Y∩ = ) 0.2 . a State, giving a reason, whether events X and Y are mutually exclusive. b Using a Venn diagram, or otherwise, find P( X Y∪ . ) c Find the probability that X or Y, but not both, occurs. 12 ,A B and C are events where P( ) A = 0.3, P( B ) = 0.4, P( C ) = 0.3, P( A B ) ∩ = 0.12, P( A C ) ∩ = 0 and P( B C ) ∩ = 0.1. a State which pair of events from ,A B and C is mutually exclusive. b Using a Venn diagram, or otherwise, find P[( ∪ ∪ ′ , ) ] which is the probability that neither A nor B nor C occurs. A B C 13 The diagram opposite shows a 30 cm square board with two rectangular cards attached. The 15 cm by 20 cm card covers one-quarter of the 8cm by 12 cm card. A dart is randomly thrown at the board, so that it sticks within its perimeter. Use areas to calculate the probability that the dart pierces: 30 a both cards b at least one of the cards c exactly one of the cards. Copyright Material - Review Only - Not for Redistribution 8 12 15 20 30 Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 ) 14 Given that A P( = 0.4, P( B ) = 0.7 and that A B ) ∪ = P( 0.8 , find: a ∪ ′BAP( ) b P( ′ ∩A B ) PS 15 Each of 27 tourists was asked which of the countries Angola ( )A , Burundi ( )B and Cameroon ( )C they had visited. Of the group, 15 had visited Angola; 8 had visited Burundi; 12 had visited Cameroon; 2 had visited all three countries; and 21 had visited only one. Of those who had visited Angola, 4 had visited only one other country. Of those who had not visited Angola, 5 had visited Burundi only. All of the tourists had visited at least one of these countries. a Draw a fully labelled Venn diagram to illustrate this information. b Find the number of tourists in set B′ and describe them. c Describe the tourists in set ( ) A B C ∪ ∩ ′ and state how many there are. d Find the probability that a randomly selected tourist from this group had visited at least two of these three countries. 4.3 Independent events and the multiplication law Two events are said to be independent if either can occur without being affected by the occurrence of the other. Examples of this are making selections with replacement and performing separate actions, such as rolling two dice. KEY POINT 4.8 100 The multiplication law for independent events is A P( and ) P( B = A B ∩ ) P( = A ) P( × B . ) This can be extended for any number of independent events: P(A and B and C and …) = P( A B C ∩ ∩ ∩ = ...) P( A ) P( × B ) P( × C ) × ... Consider the following bag, which contains two blue balls ( and five white balls ( )W . )B We will select one ball at random, replace it and then select another ball. For the first selection: BP( ) = and WP( 2 7 ) 5 = . 7 For the second selection: BP( 5 = . 7 The tree diagram below shows how we can use the multiplication law to find probabilities. = and WP( 2 7 ) ) 1st 1st B W 2 7 5 7 2nd Events and probabilities ....... P(BB) B W ....... P(BW) ....... P(WB) B W ....... P(WW 49 10 49 10 49 25 49 2 7 5 7 2 7 5 7 Copyright Material - Review Only - Not for Redistribution TIP The first and second selections are made from the same seven balls, so probabilities are identical and independent. TIP We can denote the event ‘2 blue balls are selected’ by BB; B and B; &B B or ,B B. TIP , , BB BW WB Events and WW are exhaustive, so their probabilities sum to 1. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Multiplication of independent events is performed from left to right along the branches. Addition of mutually exclusive events is performed vertically. As examples: P(different colours) P( = or BW WB ) P( = BW ) P( + WB ) = P(same colours) P( = or BB WW ) P( = BB ) P( + WW ) = 2 ×  7 5 7   + 5 ×  7 2 7   2 ×  7 2 7   + 5 ×  7 5 7   = = 10 49 4 49 + + 10 49 25 49 = = 20 49 29 49 As an alternative to using a tree diagram, we can use a possibility diagram (or outcome space), as shown below. The diagram shows the 7 four mutually exclusive combined events 7 × = BB BW WB and WW . , equally likely outcomes and the 49 , BW BW WW WW WW WW WW BW BW WW WW WW WW WW BW BW WW WW WW WW WW BW BW WW WW WW WW WW BW BW WW WW WW WW WW BB BB WB WB WB WB WB BB BB WB WB WB WB WB B B WWWW W 1st selection To find probabilities for combined events, we count how many of the 49 outcomes are favourable. For example: P(at least 1 blue) P( = BB ) P( + BW ) P( + WB ) = 4 49 + 10 49 + 10 49 = 24 29 . Alternatively, P(at least 1 blue) 1 P( = − WW ) 1 = − 25 49 = 24 29 . Chapter 4: Probability TIP If we just used a 2 by 2 diagram, with B and W as the outcomes of each selection, we could not just count cells to find probabilities, because the events in those four cells would not be equally likely. TIP Probabilities are equal to the relative frequencies of the favourable outcomes. 101 WORKED EXAMPLE 4.5 Find the probability that the sum of the scores on three rolls of an ordinary fair die is less than 5. Answer P(sum<5) P(3) P(4) = + The lowest possible sum is 3. P(sum 3) = = P(sum 4) = = 1 216 3 216 P(sum 5) < = 1 216 + 3 216 = 1 54 Each roll has six equiprobable outcomes, so there are 6 216 possible outcomes, each with a probability 6 = 6 × × 1 216 . of There is one way to obtain a sum of 3: (1, 1, 1). There are three ways to obtain a sum of 4: (1, 1, 2), (1, 2, 1), (2, 1, 1). Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 4.6 Abha passes through three independent sets of traffic lights when she drives to work. The probability that she has to stop at any particular set of lights is 0.2. Find the probability that Abha: a first has to stop at the second set of lights b has to stop at exactly one set of lights c has to stop at any set of lights. Answer a P(XS) = 0.8 × 0.2 = 0.16 b P(has to stop at exactly 1 set of lights) P(SXX) P(XSX) P(XXS) + + = (0.2 × 0.8 × 0.8) 3 = × = 0.384 c P(has to stop) 1 – P (does not have to stop) = = 1 – P(XXX) = − 1 0.8 3 102 0.488 = Alternatively, P(has to stop) = EXERCISE 4C P(S) P(XS) P(XXS). + + We use S and X to represent ‘stopping’ and ‘not stopping’. For each set of lights, P(S) and P(X) . 0.8= 0.2= The three favourable outcomes, SXX, XSX and XXS, are equally likely. The events ‘has to stop’ and ‘does not have to stop’ are complementary. 1 Using a tree diagram, find the probability that exactly one head is obtained when two fair coins are tossed. 2 Two ordinary fair dice are rolled. Using a possibility diagram, find the probability of obtaining: a two 6s b two even numbers c two numbers whose product is 6. 3 It is known that 8% of all new FunX cars develop a mechanical fault within a year and that 15% independently develop an electrical fault within a year. Find the probability that within a year a new FunX car develops: a both types of fault b neither type of fault. 4 A certain horse has a 70% chance of winning any particular race. Find the probability that it wins exactly one of its next two races. 5 The probabilities that a team wins, draws or loses any particular game are 0.6, 0.1 and 0.3, respectively. a Find the probability that the team wins at least one of its next two games. b If 2 points are awarded for a win, 1 point for a draw and 0 points for a loss, find the probability that the team scores a total of more than 1 point in its next two games. 6 On any particular day, there is a 30% chance of snow in Slushly. Find the probability that it snows there on: a none of the next 3 days b exactly one of the next 3 days. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 4: Probability M 7 Fatima will enter three sporting events at the weekend. Her chances of winning each of them are shown in the following table. Event Shot put Javelin Discus Chance of winning 85% 40% 64% a Assuming that the three events are independent, find the probability that Fatima wins: i the shot put and discus ii the shot put and discus only iii exactly two of these events. b What does ‘the three events are independent’ mean here? Give a reason why this may not be true in real life. 8 A fair six-sided spinner, P, has edges marked 0, 1, 2, 2, 3 and 4. A fair four-sided spinner, Q, has edges marked 0, –1, –1 and –2. Each spinner is spun once and the numbers on which they come to rest are added together to give the score, S. Find: a SP( 2)= b SP( 2 = 1) M 9 Letters and packages can take up to 2 days to be delivered by Speedipost couriers. The following table shows the percentage of items delivered at certain times after sending. Same day After 1 day After 2 days Letter Package 40% 15% 50% 55% 10% 30% a Is there any truth in the statement ‘If you post 10 letters on Monday then only nine of them will be delivered before Wednesday’? Give a reason for your answer. b Find the probability that when three letters are posted on Monday, none of them are delivered on Tuesday. c Find the probability that when a letter and
a package are posted together, the letter arrives at least 1 day before the package. 103 10 The following histogram represents the results of a national survey on bus departure delay times. Two buses are selected at random. Calculate an estimate of the probability that: a both departures were delayed by less than 4 minutes b at least one of the buses departed more than 7 minutes late % ( 20 15 10 10 Delay time (min) 11 Praveen wants to speak on the telephone to his friend. When his friend’s phone rings, he answers it with constant probability 0.6. If Praveen’s friend doesn’t answer his phone, Praveen will call later, but he will only try four times altogether. Find the probability that Praveen speaks with his friend: a after making fewer than three calls b on the telephone on this occasion. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 12 Each morning, Ruma randomly selects and buys one of the four newspapers available at her local shop. Find the probability that she buys: a b the same newspaper on two consecutive mornings three different newspapers on three consecutive mornings. 13 A coin is biased such that the probability that three tosses all result in heads is obtaining no heads with three tosses of the coin. 125 512 . Find the probability of 14 In a group of five men and four women, there are three pairs of male and female business partners and three teachers, where no teacher is in a business partnership. One man and one woman are selected at random. Find the probability that they are: a both teachers b in a business partnership with each other c each in a business partnership but not with each other. PS 15 A biased die in the shape of a pyramid has five faces marked 1, 2, 3, 4 and 5. The possible scores are 1, 2, 3, 4 and 5 and x P( ) = k x − 25 , where k is a constant. a Find, in terms of k, the probability of scoring: i 5 ii less than 3. b The die is rolled three times and the scores are added together. Evaluate k and find the probability that the 104 sum of the three scores is less than 5. PS 16 A game board is shown in the diagram. Players take turns to roll an ordinary fair die, then move their counters forward from ‘start’ a number of squares equal to the number rolled with the die. If a player’s counter ends its move on a coloured square, then it is moved back to the start 18 17 15 14 13 10 11 a Find the probability that a player’s counter is on ‘start’ after rolling the die: i once ii twice. b Find the probability that after rolling the die three times, a player’s counter is on: i 18 ii 17 DID YOU KNOW? It has long been common practice to write or philosophical argument is complete. demonstrandum, meaning ‘which is what had to be shown’. . . .Q E D. at the point where a mathematical proof .Q E D. is an initialism of the Latin phrase quad erat Latin was used as the language of international communication, scholarship and science until well into the 18th century. Q.E.D. does not stand for Quite Easily Done! A popular modern alternative is to write 5W , an abbreviation of Which Was What Was Wanted. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 4: Probability Application of the multiplication law The multiplication law given in Key point 4.8 can be used to show whether or not events are independent. If we can show that A B independent, and vice versa. If, for example, XP( independent only if X Y ) P( × and YP( ∩ = 0.3= ) ) , then A and B are 0.4= ) P( . 0.12 B ) 0.4 0.3 P( P( A ) ∩ = × = , then X and Y are WORKED EXAMPLE 4.7 Events ,J K and L are independent. Given that P( J ) = 0.5, P( K ) = 0.6 and P( J L ) ∩ = 0.24 , find: a P( )∩J K b P( )L c P( )∩K L . Answer a P( J K∩ ) = 0.5 × 0.6 = 0.3 b 0.5 P( × L ) = 0.24 P( L ) = = 0.24 0.5 0.48 c P( K L ) ∩ = = 0.48 0.6 × 0.288 We know that P( J K ∩ ) P( = J ) P( × K . ) We know that P( J ) P( × L ) P( = J L ∩ . ) We know that K L P( ∩ = ) P( K ) P( × L . ) 105 Examples of independence and non-independence that you may have come across are: ● Enjoyment of sport is independent of gender if equal proportions of males and females enjoy sport. ● If unequal proportions of employed and unemployed people own cars, then car ownership is not independent of employment status – it is dependent on it. WORKED EXAMPLE 4.8 In a group of 60 students, 27 are male ( )H . The )M and 20 study History ( Venn diagram shows the numbers of students in these and other categories. One student is selected at random from the group. Show that the events ‘a male is selected’ and ‘a student who studies History is selected’ are independent. 60 27 M H 18 9 11 20 22 Answer Does M P( ) P( × H ) P( = M H ) ∩ ? If the multiplication law holds for the events M and H, then they are independent. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy TIP When we show that two events are independent (or not), it is important to state a conclusion in words after doing the mathematics. A short sentence like the last line of Worked example 4.8 is sufficient. Writing .Q E D., however, is optional. . Cambridge International AS & A Level Mathematics: Probability & Statistics 1 P( M ) = 27 60 , P( H ) = 20 60 and P( M H ∩ ) = 9 60 We state the three probabilities concerned. P( M ) P( × H ) = = = 27 60 9 60 P( M H ∩ ) × 20 60 P( Then we evaluate M show whether or not it is equal to P( )∩M H . ) P( × H to ) The multiplication law holds for events M and H, therefore they are independent. . .Q E D. EXPLORE 4.2 In Worked example 4.8, we showed that the events M and H are independent. For the 60 students in that particular group, we could show by similar methods whether or not the three pairs of events M and independent. ′M and H and ′M and ′H , ′H are Do this then discuss what you believe would be the most appropriate conclusion to write. 106 EXERCISE 4D 1 Y and Z are independent events. YP( ) 0.7= and ZP( ) 0.9= . Find P( )∩Y Z . 2 Two independent events are M and N. Given that MP( )N . , find P( 0.21 M N ∩ P( = ) ) = 0.75 and 3 Independent events S and T are such that P( 0.4=S ) and P( ) ′ =T 0.2 . Find: a P( )∩S T b P( ′ ∩S T . ) 4 ,A B and C are independent events, and it is given that A B∩ = P( ) 0.35, 5, P( B C∩ = ) 0.56 and P( A C ) ∩ = 0.4 . a Express P( )A in terms of: i P( )B ii P( )C . b Use your answers to part a to find: i P( )B ii P( )′A iii P( B C . ) ′ ∩ ′ 5 In a class of 28 children, 19 attend drama classes, 13 attend singing lessons, and six attend both drama classes and singing lessons. One child is chosen at random from the class. Event D is ‘a child who attends drama classes is chosen’. Event S is ‘a child who attends singing lessons is chosen’. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 4: Probability a Illustrate the data in an appropriate table or diagram. b Are events D and S independent? Give a reason for your answer. 6 Each child in a group of 80 was asked )R or )M . The results whether they regularly read ( regularly watch a movie ( are given in the Venn diagram opposite. One child is selected at random from the group. Event R is ‘a child who regularly reads is selected’ and event M is ‘a child who regularly watches a movie is selected’. 80 R M 12 20 30 18 107 Determine, with justification, whether events R and M are independent. P 7 Two fair 4-sided dice, both with faces marked 1, 2, 3 and 4, are rolled. Event A is ‘the sum of the numbers obtained is a prime number’. Event B is ‘the product of the numbers obtained is an even number’. a Find, in simplest form, the value of P( )A , of P( )B and of P( )∩A B . b Determine, with justification, whether events A and B are independent. c Give a reason why events A and B are not mutually exclusive. 8 Two ordinary fair dice are rolled. Event X is ‘the product of the two numbers obtained is odd’. Event Y is ‘the sum of the two numbers obtained is a multiple of 3’. a Determine, giving reasons for your answer, whether X and Y are independent. b Are events X and Y mutually exclusive? Justify your answer. 9 A fair 8-sided die has faces marked 1, 2, 3, 4, 5, 6, 7 and 8. The score when the die is rolled is the number on the face that the die lands on. The die is rolled twice. Event V is ‘one of the scores is exactly 4 less than the other score’. Event W is ‘the product of the scores is less than 13’. Determine whether events V and W are independent, justifying your answer. PS 10 Two hundred children are categorised by gender and by whether or not they own a bicycle. Of the 108 males, 60 own a bicycle, and altogether 90 children do not own a bicycle. a Tabulate these data. b Determine, giving reasons for your answer, whether ownership of a bicycle is independent of gender for these 200 children. c What percentage of the females and what percentage of the males own bicycles? Explain how your answers to part c confirm the result obtai
ned in part b. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PS 11 At an election, it was found that people voted for Party X independently of their income group. The following table shows that 12 400 people from three income groups voted altogether, and that 7440 of them voted for Party X. Low Medium High Totals Party X Totals a 3100 b 6820 c 2480 7440 12 400 Find the value of a, of b and of c. PS 12 The speed limit at a motorway junction is 120 km/h. Information about the speeds and directions in which 207 vehicles were being driven are shown in the following table. North East South Under limit Over limit 36 15 27 15 36 18 West 39 21 Providing evidence to support your answer, determine which vehicles’ speeds were independent of their direction of travel. FAST FORWARD We will study discrete random variables that arise from independent events in Chapter 6. 4.4 Conditional probability The word conditional is used to describe a probability that is dependent on some additional information given about an outcome or event. 108 For example, if your friend randomly selects a letter from the word ACE, then P(selects E) = . 1 3 However, if we are told that she selects a vowel, we now have a conditional probability that is not the same as P(selects E). This conditional probability is P(selects E, given that she selects a vowel) = . 1 2 Conditional probabilities are usually written using the symbol \| to mean given that. We read P( )A B as ‘the probability that A occurs, given that B occurs’. \| WORKED EXAMPLE 4.9 A child is selected at random from a group of 11 boys and nine girls, and one of the girls is called Rose. Find the probability that Rose is selected, given that a girl is selected. Answer P(Rose is selected \| a girl is selected) = 1 9 The additional information, ‘given that a girl is selected’, reduces the number of possible selections from 20 to 9, and Rose is one of those nine girls. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 4: Probability WORKED EXAMPLE 4.10 The following table shows the numbers of students in a class who study Biology ( )B and who study Chemistry ( )C . B ′B Totals Represent the data in a suitable Venn diagram, and find the probability that a randomly selected student: 9 7 16 8 1 9 C ′C Totals Answer 25 B 7 9 8 17 8 25 C 1 a P( \| C B ) = 9 16 b P WORKED EXAMPLE 4.11 a studies Chemistry, given that they study Biology b does not study Biology, given that they do not study Chemistry. The number of students in each category is shown in the Venn diagram. 16 study Biology and nine of these study Chemistry. Eight students do not study Chemistry and one of these does not study Biology. 109 Two children are selected at random from a group of five boys and seven girls. Find the probability that the second child selected is a boy, given that the first child selected is: a a boy b a girl. Answer a P(second is a boy \| first is a boy) = b P(second is a boy \| first is a girl) = 4 11 5 11 If a boy is selected first, then the second child is selected from four boys and seven girls. If a girl is selected first, then the second child is selected from five boys and six girls. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXERCISE 4E 1 One letter is randomly selected from the six letters in the word BANANA. Find the probability that: a an N is selected, given that an A is not selected b an A is selected, given that an N is not selected. 2 One hundred children were each asked whether they have brothers )S . Their responses are given in )B and whether they have sisters ( ( the Venn diagram opposite. 100 B S Find the probability that a randomly selected child has: 16 48 24 a sisters, given that they have brothers b brothers, given that they do not have sisters c sisters or brothers, given that they do not have both. 12 3 Two photographs are randomly selected from a pack of 12 colour and eight black and white photographs. Find the probability that the second photograph selected is colour, given that the first is: a colour b black and white. 8 4 3 1 5 D 11 2 H 4 The Venn diagram opposite shows the responses of 40 girls who were )D asked if they have an interest in a career in nursing ( or human rights ( )N , dentistry ( )H . 40 N 110 a Find the probability that a randomly selected girl has an interest in: i human rights, given that she has an interest in nursing ii nursing, given that she has no interest in dentistry. b Describe any group of girls for whom dentistry is the least popular 6 career of interest. 5 The quiz marks of 40 students are represented in the following bar chart 10 5 Marks Two students are selected at random from the group. Find the probability that the second student: a scored more than 5, given that the first student did not score more than 5 b scored more than 7, given that the first student scored more than 7. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 6 The histogram shown represents the times taken, in minutes, for 115 men to complete a task. Two men are selected at random from the group. Find the probability that the: a first man took less than 1 minute, given that he took less than 3 minutes b second man took less than 6 minutes, given that the first man took less than 1 minute 20 15 10 5 0 7 At an insurance company, 60% of the staff are male ( )M and )FT . The following Venn diagram shows 70% work full-time ( this and one other piece of information. Chapter 4: Probability 1 2 3 4 5 6 7 8 9 10 Time taken (min) 1 M FT a What information is given by the value 0.10 in the Venn 0.60 a b c 0.70 diagram? b Find the value of a, of b and of c. c An employee is randomly selected. Find: i P( M FT ) \| ii P( \| FT M )′ [ iii P ( ) \| ( M FT M FT ∪ ∩ 0.10 ) ] 8 Two fair triangular spinners, both with sides marked 1, 2 and 3, are spun. Given that the sum of the two numbers spun is even, find the probability that the two numbers are the same. 9 Two ordinary fair dice are rolled and the two numbers rolled are added together to give the score. Given that a player’s score is greater than 6, find the probability that it is not greater than 8. 111 10 The circular archery target shown, on which 1, 2, 3 or 5 points can be scored, is divided into four parts of unequal area by concentric circles. The radii of the circles are 3cm, 9cm, 15cm and 30 cm. You may assume that a randomly fired arrow pierces just one of the four areas and is equally likely to pierce any part of the target. a Show that the probability of scoring 5 points is 0.01. b Find the probability of scoring 3 points, 2 points and 1 point with an arrow. c Given that an arrow does not score 5 points, find the probability that it scores 1 point. d Given that a total score of 6 points is obtained with two randomly fired arrows, find the probability that neither arrow scores 1 point. 1 2 3 5 Independence and conditional probability At the beginning of Section 4.3, ‘independent’ was described in quite familiar terms. In general, we can use the multiplication law given in Key point 4.8 as the definition of ‘independent’. However, a more formal definition can now be given. Events X and Y are said to be independent if each is unaffected by the occurrence of the other. If this is the case then the probability that X occurs is the same in two complementary situations: (i) when Y occurs, and (ii) when Y does not occur. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 From these, we can now say that X and Y are independent if and only if X Y P( \| ) P( = \| X Y ) ′ . Consider rolling an ordinary fair die and the events X and Y, as defined below. X: the outcome is a square number (1 or 4). Y: the outcome is an odd number (1, 3 or 5). First note that 1 is odd and square, so X and Y are not mutually exclusive; but are they independent? When Y occurs, the die shows 1, 3 or 5, so P( \| X Y ) = 1 3 When Y does not occur, the die shows 2, 4 or 6, so P( X ) = 1 3 , whether Y occurs or not. \| X Y ) P( = P( Events X and Y are not mutually exclusive, but they are independent. ) ′ means that P(X) is unaffected by the occurrence of Y. \| X Y EXPLORE 4.3 1 Consider rolling an ordinary fair die. In each case below, determine whether the given events are mutually exclusive, and whether they are independent. a b ‘X: a number less than 3’, and ‘Y: an even number’. ‘A: a number that is 4 or more’, and ‘B: a number that is not more than 4’.
112 2 Now consider rolling a fair 12-sided die, numbered from 1 to 12. In each case below, determine whether the given events are mutually exclusive, and whether they are independent. a b c ‘X: an even number’, and ‘Y: a factor of 28’. ‘A: a prime number’, and ‘B: a multiple of 3’. ‘F: a factor of 12’, and ‘M: a multiple of 5’. 3 An integer from 1 and 20 inclusive is selected at random. Three events are defined as follows: A: the number is a multiple of 3. B: the number is a factor of 72. C: the number has exactly two digits, and at least one of those digits is a 1. Determine which of the three possible pairs of these events is independent. 4.5 Dependent events and conditional probability Two events are mutually dependent when neither can occur without being affected by the occurrence of the other. An example of this is when we make selections without replacement; that is, when probabilities for the second selection depend on the outcome of the first selection. The multiplication law for independent events (see Key point 4.8) is a special case of the multiplication law of probability. The multiplication law of probability is used to find the probability that ‘this and that’ occurs when the events involved might not be independent. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy KEY POINT 4.9 Chapter 4: Probability P( and ) P( A B = P( and ) P( B A ∩ = ) P( A ) P( × ) \| B A B ) P( × \| A B ) P( A ) P( × \| B A ) P( ≡ B ) P( × A B . \| ) WORKED EXAMPLE 4.12 Two children are randomly selected from 11 boys ( consists of: )B and 14 girls ( )G . Find the probability that the selection a two boys b a boy and a girl, in any order. Answer a P(2 boys) P( and ) ( × ) P( B × 1 11 10 24 25 11 60 b P (a boy and a girl) P= (B and G) P+ (G and B) ) P( + ) P( × ) P( = ) \| G B 1 2 14 25   +   × 14 24 × 11 24   B 1 11   25 77 150 = = The first child is selected from 25 but the second is selected from the remaining 24. This tells us that the selections are dependent and that we must use conditional probabilities (suffices mean 1st and 2nd). There are two different orders in which a boy and a girl can be selected, but note that P(B and G) = P(G and B). 113 WORKED EXAMPLE 4.13 Every Saturday, a man invites his sister to the theatre or to the cinema. 70% of his invitations are to the theatre and 90% of these are accepted. His sister rejects 40% of his invitations to the cinema. Find the probability that the brother’s invitation is accepted on any particular Saturday. Answer The probability that the sister accepts depends on where she is invited to go. This tells us that our calculations must involve conditional probabilities. 0.7 0.3 T C 0.9 0.1 0.6 0.4 accepts .... P(T and accepts) = 0.7 × 0.9 = 0.63 rejects ..... P(T and rejects) = 0.7 × 0.1 = 0.07 accepts .... P(C and accepts) = 0.3 × 0.6 = 0.18 The given information is shown in the tree diagram, where T and C stand for ‘theatre’ and ‘cinema’. rejects ..... P(C and rejects) = 0.3 × 0.4 = 0.12 P(accepts) P= (T and accepts) P+ (C and accepts) = = 0.63 0.18 + 0.81 The sister can accept an invitation to the theatre or to the cinema. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 We can use the multiplication law of probability to find conditional probabilities. We know that A B P( P( )A are known. ∩ = ) P( A ) P( × , so P( ) \| B A B A can be found when P( ) \| )∩A B and KEY POINT 4.10 P( ) \| B A = ) P( A B ∩ ) P( A and P( \| A B ) = WORKED EXAMPLE 4.14 , where A B P( ∩ ≡ ) P( ∩ . ) B A TIP The symbol ≡ means ‘is identical to’. P( ) B A ∩ P( B ) Given that A B ) ∩ = P( 0.36 and BP( ) 0.9= , find P(A \| B). Answer P( P( ) B A ∩ P( B A B ∩ P( B 0.36 0.9 0.4 ) 114 WORKED EXAMPLE 4.15 An ordinary fair die is rolled. Find the probability that the number obtained is prime, given that it is odd. Answer P(prime \| odd) = P(odd and prime) P(odd WORKED EXAMPLE 4.16 The odd numbers are 1, 3 and 5, so 3 6 P(odd) = are 3 and 5, so P(odd and prime) . The odd prime numbers 2 = . 6 TIP Alternatively, two of the three odd numbers on a die are prime, so P(prime \| odd) 2 = . 3 A boy walks to school (W) 60% of the time and cycles (C) 40% of the time. He is late to school (L), on 5% of the occasions that he walks, and he is late on 2% of the occasions that he cycles. Given that he is late to school, find the probability that he cycles; that is, find P( \| C L ). Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Answer 0.6 W 0.4 C 0.05 0.95 0.02 L ...... P(W and L) = 0.030 Lʹ ..... P(W and Lʹ) = 0.570 L ..... P(C and L) = 0.008 0.98 Lʹ ..... P(C and Lʹ) = 0.392 P( L ) P( = W L and ) P( and ) L C + 0.008 = = 0.030 + 0.038 Chapter 4: Probability The tree diagram shows the given information. The boy can arrive late by walking or by cycling. P( and ) L C P( L 0.008 0.038 4 19 or 0.211 EXERCISE 4F 115 1 Two ties are taken at random from a bag of three plain and five striped ties. By use of a tree diagram, or otherwise, find the probability that both ties are: a plain b striped. 2 There are four toffee sweets and seven nutty sweets in a girl’s pocket. Find the probability that two sweets, selected at random, one after the other, are not the same type. 3 On a library shelf there are seven novels, three dictionaries and two atlases. Two books are randomly selected without replacement from these. Find the probability that the selected books are: a both novels b both dictionaries or both atlases. 4 A woman travels to work by bicycle 70% of the time and by scooter 30% of the time. If she uses her bicycle she is late 3% of the time but if she uses her scooter she is late only 2% of the time. a Find the probability that the woman is late for work on any particular day. b Given that the woman expects not to be late on approximately 223 days in a year, find the number of days in a year on which she works. 5 Two children are randomly selected from a group of five boys and seven girls. Determine which is more likely to be selected: a b two boys or two girls? the two youngest girls or the two oldest boys? Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 6 A boy has five different pairs of shoes mixed up under his bed. Find the probability that when he selects two shoes at random they can be worn as a matching pair. 7 A bag contains five 4cm nails, six 7 cm nails and nine 10 cm nails. Find the probability that two randomly selected nails: a have a total length of 14cm b are both 7 cm long, given that they have a total length of 14cm. 8 Yvonne and Novac play two games of tennis every Saturday. Yvonne has a 65% chance of winning the first game and, if she wins it, her chances of winning the second game increase to 70%. However, if she loses the first game, then her chances of winning the second game decrease to 55%. Find the probability that Yvonne: a loses the second game b wins the first game, given that she loses the second game. 9 a Given X Y ) ∩ = P( 0.13 and XP( ) = 0.65 , find P( Y X . \| ) b Given M N P( ∩ ) = c Given P( V W ∩ ) = 0.27 and N MP( \| ) = 0.81 , find P( )M . 0.35 and P( ) ′ =W 0.60 , find P( V W . \| ) 116 10 When a customer at a furniture store makes a purchase, there is a 15% chance that they purchase a bed. Given that 4.2% of all customers at the store purchase a bed, find the probability that a customer does not make a purchase at the store. 11 A number between 10 and 100 inclusive is selected at random. Find the probability that the number is a multiple of 5, given that none of its digits is a 5. 12 Three of Mr Jumbillo’s seven children, who include one set of twins, are selected at random. a Calculate the probability that exactly one of the twins is selected. b Given that exactly three of the children are girls, find the probability that the selection of three children contains more girls than boys. 13 Anya calls Zara once each evening before she goes to bed. She calls Zara’s mobile phone with probability 0.8 or her landline. The probability that Zara answers her mobile phone is 0.74, and the probability that she answers her landline is y. This information is displayed in the tree diagram shown. 0.8 Calls mobile Calls landline 0.74 Answers Does not answer y Answers Does not answer a Given that Zara answers 68% of Anya’s calls, find the value of y. b Given that Anya’s call is not answered, find the probability that it is made to Zara’s landline. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 4: Probability FAST FORWARD We will study further techni
ques for calculating probabilities using permutations and combinations in Chapter 5, Section 5.4. 117 14 Every Friday, Arif offers to take his sons to the beach or to the park. The sons refuse an offer to the beach with probability 0.65 and accept an offer to the park with probability 0.85. The probability that Arif offers to take them to the beach is x. This information is shown in the tree diagram. a Find the value of x, given that 33% of Arif’s offers are refused. Beach x Park Accept 0.65 Refuse 0.85 Accept Refuse b Given that Arif’s offer is accepted, find the probability that he offers to take his sons to the park. PS 15 Two children are selected from a group in which there are 10 more boys than girls. Given that there are 756 equiprobable ordered selections that can occur, find the probability that two boys or two girls are selected. PS PS 16 There is a 43% chance that Riya meets her friend Jasmine when she travels to work. Given that Riya walks to work and does not meet Jasmine 30% of the time, and that she travels to work by a different method and meets Jasmine 25% of the time, find the probability that Riya walks to work on any particular day. 17 Aaliyah buys a randomly selected magazine that contains a crossword puzzle on five randomly chosen days of each week. On 84% of the occasions that she buys a magazine, she attempts its crossword, which she manages to complete 60% of the time. Find the probability that, on any particular day, Aaliyah does not complete the crossword in a magazine. Checklist of learning and understanding ● Probabilities are assigned on a scale from 0 (impossible) to 1 (certain). ● When one object is randomly selected from n objects, P(selecting any particular object) 1 = . n ● P(event) = Number of favourable equally likely outcomes Total number of equally likely outcomes ● P( A ) P(not + A ) = or A P( 1 ) P( + A ) ′ = 1 ● In n trials, event A is expected to occur n × AP( times. ) ● ∪A B means ‘A or B’ and ∩A B means ‘A and B’. ● Mutually exclusive events have no common favourable outcomes. For mutually exclusive events A and B, P( or A B ) P( = A B ∪ ) P( = A ) P( + B ) ● Non-mutually exclusive events have at least one common favourable outcome. For any two events A and B, P( or A B ) P( = A B ∪ ) P( = A ) P( + B ) P( − A B ∩ ) ● Independent events can occur without being affected by the occurrence of each other. Events A and B are independent if and only if A P(A \| B) = P(A \| B ′). P( and ) P( B = A B ∩ ) P( = A ) P( × B ) , such that ● For any two events A and B, A P( and ) P( B = A B ∩ ) P( = A ) P( × \| B A P( and B A ) \| ) = P( A B ∩ P( A ) ) . Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 END-OF-CHAPTER REVIEW EXERCISE 4 1 Four quality-control officers were asked to test 1214 randomly selected electronic components from a company’s production line, and to report the proportions that they found to be defective. The proportions reported were: 2 333 , 3 411 , 0 187 and 4 283 . These figures confirm what the manager thought; that about %k of the components produced are defective. Of the 7150 components that will be produced next month, approximately how many does the manager expect to be defective? [2] 2 Three referees are needed at an international tournament and there are 12 to choose from: three from Bosnia, four from Chad and five from Denmark. If the referees are selected at random, find the probability that at least two of them are from the same country. [3] 3 The diagram opposite gives details about a company’s 115 Part-time Full-time employees. For example, it employs four unqualified, part-time females. Male 10 Unqualified 42 Two employees are selected at random. Find the probability that: a one is a qualified male and the other is an unqualified female [2] b both are unqualified, given that neither is employed part-time. [3] 7 4 3 2 Female 16 33 4 The numbers of books read in the past 3 months by the members of a reading club are shown in the following table. 118 No. books No. members f( ) 2< 3 2– 4 5 5–7 22 8–9 7 10 2 10> 1 Find the probability that three randomly selected members have all read fewer than eight books, given that they have all read more than four books. [3] 5 When they are switched on, certain small devices independently produce outputs of 1, 2 or 3 volts with respective probabilities of 0.3, 0.6 and 0.1. Find the probability that three of these devices produce an output with a sum of 5 or 6 volts. [4] 6 One hundred people are attending a conference. The following Venn diagram shows how many are male ( )M , have brown eyes ( )BE and are right-handed ( )RH . M BE 100 47 39 RH 55 87 a Given that there are 43 males with brown eyes, 42 right-handed males and 46 right-handed people with brown eyes, copy and complete the Venn diagram. b Two attendees are selected at random. Find the probability that: i they are both females who are not right-handed ii exactly one of them is right-handed, given that neither of them have brown eyes. [3] [2] [3] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 4: Probability 7 A student travels to college by either of two routes, A or B. The probability that they use route A is 0.3, and the probability that they are passed by a bus on their way to college on any particular day is 0.034. They are twice as likely to be passed by a bus when they use route B as when they use route A. a Use the tree diagram opposite to form and solve a pair of simultaneous equations in x and y. b Find the probability that the student uses route B, given that they are not passed by a bus on their way to college. [3] [3] x y 0.3 0.7 A B 8 Two ordinary fair dice are rolled. If the first shows a number less than 3, then the score is the mean of the numbers obtained; otherwise the score is equal to half the absolute (non-negative) difference between the numbers obtained. Find the probability that the score is: a positive b greater than 1, given that it is less than 2 c less than 2, given that it is greater than 1. Bus No bus Bus No bus [3] [3] [3] 9 Three friends, Rick, Brenda and Ali, go to a football match but forget to say which entrance to the ground they will meet at. There are four entrances, A B C, • The probability that Rick chooses entrance A is 1 3 , are all equal. and D. Each friend chooses an entrance independently. . The probabilities that he chooses entrances B C, or D • Brenda is equally likely to choose any of the four entrances. • The probability that Ali chooses entrance C is 2 7 The probabilities that he chooses the other two entrances are equal. and the probability that he chooses entrance D is 3 5 i Find the probability that at least 2 friends will choose entrance B. ii Find the probability that the three friends will all choose the same entrance. 119 . [4] [4] Cambridge International AS & A Level Mathematics 9709 Paper 61 Q5 November 2010 10 Maria chooses toast for her breakfast with probability 0.85. If she does not choose toast then she has a bread roll. If she chooses toast then the probability that she will have jam on it is 0.8. If she has a bread roll then the probability that she will have jam on it is 0.4. i Draw a fully labelled tree diagram to show this information. ii Given that Maria did not have jam for breakfast, find the probability that she had toast. [2] [4] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q3 November 2009 11 Ronnie obtained data about the gross domestic product (GDP) and the birth rate for 170 countries. He classified each GDP and each birth rate as either ‘low’, ‘medium’ or ‘high’. The table shows the number of countries in each category. Birth rate Low Medium High Low GDP Medium High 3 20 35 5 42 8 45 12 0 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 One of these countries is chosen at random. i Find the probability that the country chosen has a medium GDP. ii Find the probability that the country chosen has a low birth rate, given that it does not have a medium GDP. iii State with a reason whether or not the events ‘the country chosen has a high GDP’ and ‘the country chosen has a high birth rate’ are exclusive. One country is chosen at random from those countries which have a medium GDP and then a different country is chosen at random from those which have a medium birth rate. [1] [2] [2] iv Find the probability that both countries chosen have a medium GDP and a medium birth rate. [3] Cambridge International AS & A Level Mathematics 9709 Paper 63 Q3 November 2012 12 Three boxes, A, B and C, each contain orange balls and blue balls, as shown. box A box B box C 120 probability that it is from box C. a A girl selects a ball at random from a randomly selected box. Given that she selects a blue ball, find the b A boy randomly selects one ball from each box. Given that he selects exactly one blue ball, find the probability that it is from box A. 13 A and B are independent events. If AP( ) = 0.45 and P( 14 Two ordinary fair dice are rolled. Event A is ‘the sum of the numbers rolled is
2, 3 or 4’. B = ) 0.64 P[( , find A B )∪ ′]. Event B is ‘the absolute difference between the numbers rolled is 2, 3 or 4’. a When the dice are rolled, they show a 1 and a 3. Explain why this result shows that events A and B are not mutually exclusive. b Explain how you know that A B P( ) ∩ = 1 18 . c Determine whether events A and B are independent. 15 A and B are independent events, where A B P( ∩ ′ = ) 0.14 , A B ′ ∩ = P( ) 0.39 and P( A B ∩ < ) 0.25 . Use a Venn diagram, or otherwise, to find A B P[( )∪ ′]. 16 In a survey, adults were asked to answer yes or no to the question ‘Do you regularly watch the evening TV news?’ Some of the results from the survey are detailed in the Venn diagram opposite. One adult is selected at random and it is found that the events ‘a female is selected’ and ‘a person who regularly watches the evening TV news is selected’ are independent. Find the number of adults questioned in the survey. [4] Male No 90 x 63 105 Copyright Material - Review Only - Not for Redistribution [3] [4] [2] [1] [2] [3] [4] Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 17 Bookings made at a hotel include a room plus any meal combination )L and supper ( of breakfast ( )S . The Venn diagram opposite shows the number of each type of booking made by 71 guests on Friday. a A guest who has not booked all three meals is selected at random. )B , lunch ( Find the probability that this guest: i has booked breakfast or supper [2] ii has not booked supper, given that they have booked lunch. [2] b Find the probability that two randomly selected guests have both booked lunch, given that they have both booked at least two meals. [3] Chapter 4: Probability 71 B L 24 6 3 7 5 16 1 9 S PS PS 18 Three strangers meet on a train. Assuming that a person is equally likely to be born in any of the 12 months of the year, find the probability that at least two of these three people were born in the same month of the year. 19 A box contains three black and four white chess pieces. Find the probability that a random selection of five chess pieces, taken one at a time without replacement, contains exactly two black pieces which are selected one immediately after the other. PS 20 The following table shows the numbers of IGCSE (I) and A Level (A) examinations passed by a group of university students. IGCSEs (I) 5 13 10 0 0 0 1 A Levels (A) a For a student selected at random, find: i P(I + A = 11 \| A < 4) ii P(I A > 5 \| I + A > 10) − b Six students who all have at least three A Level passes are selected at random. Find the greatest possible range of the total number of IGCSE passes that they could have. 121 [4] [4] [2] [3] [2] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 122 Chapter 5 Permutations and combinations In this chapter you will learn how to: ■ solve simple problems involving selections ■ solve problems about arrangements of objects in a line, including those involving repetition and ■ evaluate probabilities by calculation using permutations or combinations. restriction Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations PREREQUISITE KNOWLEDGE Where it comes from What you should be able to do Check your skills Chapter 4, Section 4.4. Recognise and calculate conditional probabilities. An experiment has 10 equally likely outcomes: three are favourable to event A, five are favourable to event B and four are favourable to neither A nor B. P( Find A B ) and B A P( ). \| \| Simple situations with millions of possibilities This topic is concerned with selections and arrangements of objects. Permutations and combinations appear in many complex modern applications: transport logistics; relationships between proteins in genetic engineering; sorting algorithms in computer science; and protecting computer passwords and e-commerce transactions in cryptography. A selection of objects is called a combination if the order of selection does not matter; however, if the order of selection does matter, then the selection is called a permutation. To make the difference between a permutation and a combination clear, we can describe them as follows. ● A combination is a way of selecting objects. There are three combinations of two letters from A, B and C. These are A and B, A and C, ● A permutation is a way of selecting objects and arranging them in a particular order. There are six permutations of two letters from A, B and C. These are and B and C. AB, BA, AC, CA, BC and CB. EXPLORE 5.1 123 Each letter A to Z is encrypted (or transformed) to a fixed distinct letter using its position in the alphabet (A 1, B 2, C 3, = the password SATURN is encrypted as ECHKBP. = … . By doing this, = ) This information gives us six clues to work out the method of encryption (e.g. S 3→ , and so on). 1 E→ means 19 5→ , A C→ means An Enigma Encryption Machine, circa 1940. Investigate the method of encryption and then find the password that is encrypted as UJSNOL. There are over 27 million possible passwords, so the probability that a random guess is correct is approximately zero. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 5.1 The factorial function In this chapter, we will frequently need to write and evaluate expressions such as 4 × × × . 3 2 1 A shorthand method of doing this is to use the factorial function: 4 factorial’ and is written 4! On most calculators, the factorial function appears as n! or x! × × × is called ‘four 3 2 1 As examples: 7! 6! 4 5040 = × = 30 5! − 4! = 4!(5 1) − = 4! 4 × = 96 3! 0 KEY POINT 5.1 ! n = ( n n – 1)( – 2)… 3 2 1, for any integer × × × n n 0> . 0! 1= The following figure shows values of n! in sequence, where the next term is obtained by division. From this it is clear that 0! must be equal to 1. 124 n! 7! 6! 5! 4! 3! 2 ÷ value 5040 720 120 24 6 2 1! 1 1 ÷ 0! 1 EXERCISE 5A 1 Without using a calculator, find the value of: a 5! 3! b 4! 2! 3!− c 7 × 4! 21 3! + × d 10! 8! + 9! 7! e 20! 18! − 13! 11! 2 Use your calculator to find the smallest value of n for which: a n! 1000 000 > b 5! 6! × < n ! c n( !)! 1020 > 3 Use your calculator to find the largest value of n for which: a n! 500 000 80< b 1.5 10 × 12 – ! 0 n > 4 Express, in as many different ways as possible, the numbers 144, 252 and 1 1 a b, or c is equal to 0 or to 1. 5 Express the area of a 53cm by 52 cm rectangle using factorials. c n − a 2 in the form n ( ! 2)! ! × ! c < 500 b ! , where none of 6 Two cubical boxes measure 25cm by 24cm by 23cm, and 8cm by 7 cm by 6cm. Express the difference between their volumes using factorials. 7 Eight children each have seven boxes of six eggs and each egg is worth $0.09. Write the total value of all these eggs in dollars, using factorials. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations DID YOU KNOW? There is a famous legend about the Grand Vizier in Persia who invented chess. The King was so delighted with the new game that he invited the Vizier to name his own reward. The Vizier replied that, being a modest man, he wanted only one grain of wheat on the first square of a chessboard, two grains on the second, four on the third, and so on, with twice as many grains on each square as on the previous square. The innumerate King agreed, not realising that the total number of grains on all 64 squares would be 2 – 1 wheat production for the next 150 years. 64 , or 1.84 1019 × , which is equivalent to the world’s present Although the number 2 64 – 1 is extremely large, it is only about one-third of 21! factorial. As a challenge, try showing that 2 0 1 2 + + 2 2 + … + 63 2 64 = 2 – 1 without using a formula. 5.2 Permutations We can make a permutation by taking a number of objects and arranging them in a line. For example, the two possible permutations of the digits 5 and 9 are the numbers 59 and 95. Although there are several methods that we can use to find the number of possible permutations of objects, all methods involve use of the factorial function. 125 Permutations of n distinct objects The number of permutations of n distinct objects is denoted by Pn n, and there are n! permutations that can be made. For example, there are P 2! 2 digits 5 and 9, as we have just seen. 2 = 2 = permutations of the two KEY POINT 5.2 The number of permutations of n distinct objects is P n integer n > 0)( n − 2)… 3 2 1, for × × × any Consider all the three-digit numbers that can be made by arranging the digits 5, 6 and 7. In this simple case, we can make a list to show there are six possible three-digit numbers. These are 567, 576, 657, 675, 756 and 765. The following
tree diagram gives another method of showing the six possible arrangements of the three digits. left middle right number ............ 567 ............ 576 ............ 657 ............ 675 ............ 756 ............ 765 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Unfortunately, writing out lists and constructing tree diagrams to find numbers of possible arrangements of objects are suitable methods only for small numbers of objects. Imagine listing all the possible arrangements of seven different letters; there would be over 5000 on the list and a tree diagram would have over 5000 branches at its right-hand side! Clearly, a more practical method for finding numbers of arrangements is needed. This is the primary use of the factorial function. We can show that six three-digit numbers can be made from 5, 6 and 7 by considering how many choices we have for the digit that we place in each position in the arrangement. If we first place a digit at the left side, we have three choices. Next, we place a digit in the middle (two choices). Finally, we place the remaining digit at the right side, as shown in the following diagram. 3 choices × 2 choices × 1 choices The numbers above the lines in the diagram are not the digits that are being arranged – they are the numbers of choices that we have for placing the three digits. The three digits can be arranged in 3 × × = 2 1 3! = 3 P 3 = ways. 6 The seven letters mentioned previously can be arranged in 7! 5040 7 = × × × × × × = ways. 4 5 1 3 6 2 WORKED EXAMPLE 5.1 126 In how many different ways can five boys be arranged in a row? TIP Answer ! = 5 P 5 = 120 ways. We multiply together the number of choices for each of the five positions, working from left to right. We could just as easily work from right to left, giving 1 5! 2 × × × × = ways. 4 5 3 WORKED EXAMPLE 5.2 In how many ways can nine elephants and four mice be arranged in a line? Answer 13! = 13 P 13 = 6227 020800 ways. The nine elephants and four mice are distinct, so we are arranging 13 different animals. TIP Large number answers can be given more accurately than to 3 significant figures. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations EXERCISE 5B 1 In how many ways can the six letters A, B, C, D, E and F be arranged in a row? 2 From a standard deck of 52 playing cards, find how many ways there are of arranging in a row: a all 52 cards b the four kings c the 13 diamonds. 3 In how many different ways can the following stand in a line? a two women b six men c eight adults. 4 In how many different ways can the following sit in a row on a bench? a four girls b three boys c four girls and three boys. 5 Seven cars and x vans can be parked in a line in 39 916 800 ways. Find the number of ways in which five cars and x 2+ vans can be parked in a line. 6 A woman has 10 children. She arranges 11 chairs in a row and sits on the chair in the middle. If her youngest child sits on the adjacent chair to her left, in how many ways can the remaining children be seated? PS 7 A group of n boys can be arranged in a line in a certain number of ways. By adding two more boys to the group, the number of possible arrangements increases by a factor of 420. Find the value of n. Permutations of n objects with repetitions When n objects include repetitions (i.e. when they are not all distinct), there will be fewer than Pn n permutations, so an adjustment to the use of the factorial function is needed. Consider making five-letter arrangements with A, A, B, C and D. To simplify the problem, we can distinguish the repeated A by writing the letters as A, A, B, C and D. AABCD is the same arrangement as AABCD. DACAB is the same arrangement as DACAB, and so on. Each time we swap A and A, we obtain the same arrangement. 127 If the five letters were distinct, there would be P 5! 120 arrangements, but the number 5 5 = =  is reduced to half  1 2!   of the total arrangements because the two repeated letters can be placed in 2! ways in any particular arrangement without changing that arrangement. Letters to be arranged 5= ; number of the same letter 2= . 5 There are 2 P 5 P 2 = 5! 2! = 60 five-letter arrangements that can be made. KEY POINT 5.3 The number of permutations of n objects, of which p are of one type, q are of another type, r are of another type, and so on, is = ! × ... p ! × q ! ! n × r ! × ... where . Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 5.3 The capital of Burkina Faso is OUAGADOUGOU. Find the number of distinct arrangements of all the letters in this word. Answer 11! 3! 3! 2! 2! × × × = 277200 11 letters are to be arranged, with repeats of three Os, three Us, two As and two Gs. In the formula of Key point 5.3, excluding 1! for the D in the denominator does not change our answer. EXERCISE 5C 1 Find the number of distinct arrangements of all the letters in these words: a TABLE b TABLET c COMMITTEE d MISSISSIPPI e HULLABALLOO. 2 Find how many six-digit numbers can be made from these sets of digits: a 1, 1, 1, 1, 1 and 3 b 2, 2, 2, 7, 7 and 7 c 5, 6, 6, 6, 7 and 7 d 8, 8, 9, 9, 9 and 9. 128 3 A girl has 20 plastic squares. There are five identical red squares, seven identical blue squares and eight identical green squares. By placing them in a row, joined edge-to-edge, find how many different arrangements she can make using: a one square of each colour b the five red squares only c all of the blue and green squares d all of the 20 squares. 4 Two students are asked to find how many ways there are to plant two trees and three bushes in a row. The first student gives 5! 120 , and the second gives = 5! 2! 3! × = 10 . Decide who you agree with and explain the error made by the other student. 5 Ten coins are placed in a row on a table, each showing a head or a tail. a How many different arrangements of heads and/or tails are possible? b Of the arrangements in part a, find how many have: i five heads and five tails showing ii more heads than tails showing. 6 There are 420 possible arrangements of all the letters in a particular seven-letter word. Give a description of the letters in this word. 7 Find the number of distinct five-letter arrangements that can be made from: a c two As and three Bs b two identical vowels and three Bs two identical vowels and any three identical consonants. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations EXPLORE 5.2 Consider the number of distinct arrangements of the 16 letters in the word COUNTERCLOCKWISE – there are close to 8.72 1011 numbers in our daily lives, so we are likely to see this as just a very large number whose true size we cannot really comprehend until it is put into some human context. For example, if everyone on Earth over the age of 14 (i.e. about 5.46 109 contributed one new arrangement of the word every day starting on 1st January, we would complete the list of arrangements around 9th June. . We rarely meet such people) × × The calculation for this is 11 8.72 10 × 5.46 10 × ≈ 160 days. 9 Devise a way of expressing the number of distinct arrangements of the letters in the word PNEUMONOULTRAMICROSCOPICSILICOVOLCANOCONIOSIS (which is the full name for the disease known as silicosis, and is the longest word in any major English language dictionary) in a way that is meaningful to human understanding. There are, for example, 3.15 107 × Earth, and the masses of the Earth and Sun are 5.97 1024 respectively. seconds in a year, about 7.48 109× people on 30 and 1.99 10 kg × × , WEB LINK For other options, perform a web search for large numbers. Permutations of n distinct objects with restrictions The number of possible arrangements of objects is reduced when restrictions are put in place. As a general rule, the number of choices for the restricted positions should be investigated first, and then the unrestricted positions can be attended to. 129 WORKED EXAMPLE 5.4 Find the number of ways of arranging six men in a line so that: a the oldest man is at the far-left side b the two youngest men are at the far-right side c the shortest man is at neither end of the line. Answer Without restrictions, the six men can be arranged in P 6! than 720 arrangements. 6 = 6 = 720 ways. So, with restrictions, there will be fewer ! 120 = arrangements 4P4 P = 2 4 P 4 × 2 4! 2! × = 2P2 arrangements 48 The oldest man must be at the far-left side (one choice), and the other five men can be arranged in the remaining five positions in P5 5 ways. The two spaces at the right are reserved for the two youngest men, who can be placed there in P2 2 ways. The other four men can be arranged in the remaining four positions in P4 4 ways, as shown. Copyright Material
- Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 4P4 ! 4 × = 480 arrangements. There are only five men who can be placed at the far-left side, so there are only four men who can be placed at the far right. The remaining four positions can be filled by any of the other four men (one of whom is the shortest man) in P4 4 ways, as shown. Alternatively, the shortest man can be placed in one of four positions, and the other five positions can be filled in 5P5 ways, so 4 × 5! = 480. WORKED EXAMPLE 5.5 How many odd four-digit numbers greater than 3000 can be made from the digits 1, 2, 3 and 4, each used once? Answer Restrictions affect the digits in the thousands column and in the units column. The digit at the far left (i.e. thousands column) can be only 3 or 4, and the digit at the far right (i.e. units column) can be only 1 or 3. The 3 can be placed in either of the restricted positions, so we can investigate separately the four-digit numbers that start with 3, and the four-digit numbers that start with 4. 130 1 × 2 × 1 × 1 2P2 × = numbers 2P2 × = numbers 2 4 1 × 2 P 2 Start with 3: We must place 1 at the far right (one choice), and the remaining two positions can be filled by the other two digits in P2 2 ways, as shown. Start with 4: We can place 1 or 3 at the far right (two choices), and the remaining two positions can be filled by the other two digits in P2 2 ways, as shown. 4 6 + = odd numbers greater 2 than 3000 can be made. TIP Alternatively, we could solve this problem by investigating separately the numbers that end with 1, and the numbers that end with 3. TIP These six numbers are 3241, 3421, 4123, 4213, 4213, 4231 and 4321. WORKED EXAMPLE 5.6 Find how many ways two mangoes M( a line if the five fruits are distinguishable and the mangoes: ) and three watermelons W( ) can be placed in a must not be separated b must be separated Answer a 2P2 M1 M2 W1 W2 W3 1 object 4P4 2 P 2 × 4 P 4 = 48 ways The two mangoes can be placed next to each other in P2 2 ways. This pair is now considered as a single object to be arranged with the three watermelons, giving a total of four objects to arrange, as shown. TIP Objects that must not be separated are treated as a single object when arranged with others. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations b 120 − 48 = 72 ways EXERCISE 5D With no restrictions, the five items can be 5 arranged in P 5! 120 = know that the mangoes are not separated in 48 of these. ways, and we 5 = 1 Find how many five-digit numbers can be made using the digits 2, 3, 4, 5 and 6 once each if: a b there are no restrictions the five-digit number must be: i odd ii even iii odd and less than 40 000. 2 Find how many ways four men and two women can stand in a line if: a b c d the two women must be at the front there must be a woman at the front and a man at the back the two women must be separated the four men must not be separated e no two men may stand next to each other. 3 Find the ratio of odd-to-even six-digit numbers that can be made using the digits 1, 2, 3, 4, 5 and 7. 131 4 Find how many ways 10 books can be arranged in a row on a shelf if: a b the two oldest books must be in the middle two positions the three newest books must not be separated. 5 Five cows and one set of twin calves can be housed separately in a row of seven stalls in P =5040 ways. Find 7 7 in how many of these arrangements: a b the two calves are not in adjacent stalls the two calves and their mother, who is one of the 5 cows, are in adjacent stalls c each calf is in a stall adjacent to its mother. 6 Find how many of the six-digit numbers that can be made from 1, 2, 2, 3, 3 and 3: a begin with a 2 b are not divisible by 2. 7 Find the number of distinct arrangements that can be made from all the letters in the word THEATRE when the arrangement: a begins with two Ts and ends with two Es b has H as its middle letter c ends with the three vowels E, A and E. PS 8 The following diagram shows a row of post boxes with the owners’ names beneath. Five parcels, one for the owner of each box, have arrived at the post office. If one parcel is randomly placed in each box, find the number of ways in which: Mr A Ms B Mr C Ms D Mrs E Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 a the five parcels can all be placed in the correct boxes b exactly one parcel can be placed in the wrong box c the correct parcels can be placed in Mr A’s and one other person’s box only d exactly two parcels can be placed in the correct boxes. P 9 There are x boys and y girls to be arranged in a line. Find the relationship between x and y if it is not possible to separate all the boys. Permutations of r objects from n objects So far, we have dealt only with permutations in which all of the objects are selected and arranged. We can now take this a step further and look at permutations in which only some of the objects are selected and arranged. When we select and arrange r objects in a particular order from n distinct objects, we call this a permutation of r from n. Suppose, for example, we wish to select and arrange three letters from the five letters A, B, C, D and E. We have five choices for the first letter, four for the second, and three choices for the third. This gives us a total of 5 permutations, which is effectively 5! but 2! are missing. 60 × × = 3 4 There are 5! (5 3)! − = 5! 2! = 60 permutations altogether. 132 WORKED EXAMPLE 5.7 How many three-digit numbers can be made from the seven digits 3, 4, 5, 6, 7, 8 and 9, if each is used at most once? Answer KEY POINT 5.4 n ( = There are P r ! n n r − permutations of r objects from n distinct objects. )! 7 P 3 = 7! (7 3)! − 7! 4! 7 6 5 = × × 210 = = three-digit numbers We select and arrange just three of the seven distinct digits (and ignore four of them). TIP The choices we have for the first, second and third digits are 7 × × = 210. 5 6 WORKED EXAMPLE 5.8 In how many ways can five playing cards from a standard deck of 52 cards be arranged in a row? Answer 52 P 5 = = 52! 47! 311875 200 ways We select and arrange five of the 52 playing cards (and ignore 47 of them). Copyright Material - Review Only - Not for Redistribution TIP The choices we have are 52 51 50 × × × 311 875200. 49 48 × = Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations WORKED EXAMPLE 5.9 In how many ways can 4 out of 18 girls sit on a four-seat sofa when the oldest girl must be given one of the seats? Answer 4 × 17 P 3 = 16320 ways WORKED EXAMPLE 5.10 Four ways for the oldest girl to occupy a seat, and P17P17 arrange three of the remaining 17 girls to sit with her. 3 ways to select and In how many ways can four boys and three girls stand in a row when no two girls are allowed to stand next to each other? Answer 4P4 ways to arrange 4 boys in a row B1 B2 B3 B4 Arrange the girls in 3 of these 5 spaces 5P3 4 5 P 4 × P 1440 = 3 ways 4 ways to arrange the four P4 boys in a row. 3 ways to select three of the P5 five spaces between or to the side of the boys and arrange the three girls in them, as shown. TIP Objects that must be separated are individually placed between or beyond the objects that can be separated. 133 EXERCISE 5E 1 Find how many permutations there are of: a five from seven distinct objects b four from nine distinct objects. 2 From 12 books, how many ways are there to select and arrange exactly half of them in a row on a shelf? 3 In how many ways can gold, silver and bronze medals be awarded for first, second and third places in a race between 20 athletes? You may assume that no two athletes tie in these positions. 4 a Find the number of ways in which Alvaro can paint his back door and his front door in a different colour if he has 14 colours of paint to choose from. b In how many ways could Alvaro do this if he also considered painting them the same colour? 5 Find how many of the arrangements of four letters from A, B, C, D, E and F: a begin with the letter A b contain the letter A. 6 From a group of 10 boys and seven girls, two are to be chosen to act as the hero and the villain in the school play. Find in how many ways this can be done if these two roles are to be played by: a any of the children b two girls or two boys c a boy and a girl. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 7 From a set of 10 rings, a jeweller wishes to display seven of them in their shop window. The formation of the display is s
hown in the diagram opposite. Find the number of possible displays if, from the set of 10: a b the ring with the largest diamond must go at the top of the display the most expensive ring must go at the top with the two least expensive rings adjacent to it. 8 Using each digit not more than once, how many even four-digit numbers can be made from the digits 1, 2, 3, 4, 5, 6 and 7? 9 Find how many three-digit numbers can be made from the digits 0, 1, 2, 3 and 4, used at most once each, if the three-digit number: a must be a multiple of 10 b cannot begin with zero. 10 Give an example of a practical situation where the calculation Pn r = 120 might arise. P 11 a Under what condition is Given that , find an expression for k in terms of n and r. 12 Five playing cards are randomly selected from a standard deck of 52 cards. These five cards are shuffled, and then the top three cards are placed in a row on a table. How many different arrangements of three of the 52 cards are possible? PS 13 Seven chairs, A to G, are arranged as shown. C D E 134 In how many ways can the chairs be occupied by 7 of a group of 12 people if three particular people are asked to sit on chairs B, D and F, in any order? PS 14 A minibus has 11 passenger seats. There are six seats in a row on the sunny side and five seats in a row on the shady side, as shown in the following diagram. B A F G Find how many ways eight passengers can be arranged in these seats if: a there are no restrictions b one particular passenger refuses to sit on the sunny side c two particular passengers refuse to sit in seats that are either next to each other or one directly in front of the other. shady sunny DID YOU KNOW? The rule to determine the number of permutations of n objects was known in Indian culture at least as early as 1150 and is explained in the Līlāvatī by Indian mathematician Bhaskara II. In his books Campanalogia and Tintinnalogia, Englishman Fabian Stedman in 1677 described factorials when explaining the number of permutations of the ringing of church bells. A complete peal of changes of n bells is made when they are rung in !n sequences without repetition. The speed at which church bells ring cannot be changed very much by the ringers and this may be why there are at most five bells in most churches. For example, 10 bells can be rung in 3628800 different sequences and it would take the ringers over 3 months to ring a complete peal of changes of 10 bells! Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations 5.3 Combinations A combination is simply a selection, where the order of selection is not important. Choosing strawberries and ice cream from a menu is the same combination as choosing ice cream and strawberries. When we select r objects in no particular order from n objects, we call this is a combination. KEY POINT 5.5 A combination of r objects which are then arranged in order is equivalent to a permutation. We write nCr to mean the number of combinations of r objects from n. Since there are rPr = r! ways of arranging the r objects, we have !( r n − ( )! r )! Suppose we wish to select three children from a group of five. We can view this task as ‘choosing three and ignoring two’ or as ‘choosing to ignore two and remaining with three’. Regardless of how we view it, choosing three from five and choosing two from five can be done in an equal number of ways, and so C 5 C . 5 = 3 2 The following three points should be noted. n n n C = r øC r n C n P r n r– C r = ! n !( r n − = r )! No. we select from! No. selected! No. not selected! × WORKED EXAMPLE 5.11 There are ! n !( r n − n C = r r )! combinations of r objects from n distinct objects. FAST FORWARD    n r   =  ! n !( r n − r )! We will use this more modern notation in Chapter 7. However, most calculators use the Cn r notation, so we will use this in the current chapter. 135 In how many ways can three fish be selected from a bowl containing seven fish and two potatoes? Answer 7 C 3 = 7! × 4! 3! = 35 ways The two potatoes are irrelevant. We select three fish from seven fish. WORKED EXAMPLE 5.12 In how many ways can five books and three magazines be selected from eight books and six magazines? Answer 8 6 C 56 5 = ways C 3 = 20 ways 8 C 5 × 6 C 3 = = 56 20 × 1120 ways We select five from eight books and we select three from six magazines. TIP The books and the magazines are selected independently, so we multiply the numbers of combinations. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 5.13 A team of five is to be chosen from six women and five men. Find the number of possible teams in which there will be more women than men. Answer From 6 women From 5 men No. teams 3 4 5 or or = 200 6 C 4 × 5 C 75 = 200 + 75 6 + = 281 teams with more women than men. The table shows the possible make-up of the team when it has more women than men in it; and also the number of ways in which those teams can be chosen. WORKED EXAMPLE 5.14 How many distinct three-digit numbers can be made from five cards, each with one of the digits 5, 5, 7, 8 and 9 written on it? Answer The 5 is a repeated digit, so we must investigate three situations separately. 136 No 5s selected: P = 6 numbers. 3 3 One 5 selected: C three-digit numbers. 2 × 3 Two 5s selected: C three-digit numbers. 1 × 3 three-digit The digits 7, 8 and 9 are selected and arranged. 3! 18 = 3! 2! 9 = Two digits from 7, 8 and 9 are selected and arranged with a 5. One digit from 7, 8 and 9 is selected and arranged with two 5s. 6 18 9 + = + be made. 33 three-digit numbers can EXERCISE 5F FAST FORWARD You will learn about probability distributions for the number of objects that can be selected in Chapter 6, such as the number of women selected for this team. TIP The selections in these three situations are mutually exclusive, so we add together the numbers of three-digit numbers. 1 Find the number of ways in which five apples can be selected from: a eight apples b nine apples and 12 oranges. 2 From seven men and eight women, find how many ways there are to select: a four men and five women b three men and six women c at least 13 people. 3 a How many different hands of five cards can be dealt from a standard deck of 52 playing cards? b How many of the hands in part a consist of three of the 26 red cards and two of the 26 black cards? 4 a From the 26 letters of the English alphabet, find how many ways there are to choose: i six different letters ii 20 different letters. b Use your results from part a to find the condition under which C x = y x C z , where x is a positive integer. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations 5 In a classroom there are four lights, each operated by a switch that has an on and an off position. How many possible lighting arrangements are there in the classroom? 6 From six boys and seven girls, find how many ways there are to select a group of three children that consists of more girls than boys. 7 A bag contains six red fuses, five blue fuses and four yellow fuses. Find how many ways there are to select: a three fuses of different colours b three fuses of the same colour c 10 fuses in exactly two colours d nine fuses in exactly two colours. 8 The diagram opposite shows the activities offered to children at a school camp. If children must choose three activities to fill their day, how many sets of three activities are there to choose from? Today’s Activities Morning: acting, painting or singing Afternoon: swimming, tennis, golf or cricket Evening: night-hike, star-gazing or drumming Afternoon swimming can be done at the pool or at the lake 9 Two taxis are hired to take a group of eight friends to the airport. One taxi can carry five passengers and the other can carry three passengers. What information is given in this situation by the fact that C 5 8 = 8 C 3 = ? 56 10 Ten cars are to be parked in a car park that has 20 parking spaces set out in two rows of 10. Find how many different patterns of unoccupied parking spaces are possible if: a b c d the cars can be parked in any of the 20 spaces the cars are parked in the same row the same number of cars are parked in each row two more cars are parked in one row than in the other. 137 11 A boy has eight pairs of trousers, seven shirts and six jackets. In how many ways can he dress in trousers, shirt and jacket if he refuses to wear a particular pair of red trousers with a particular red shirt? 12 A girl has 11 objects to arrange on a shelf but there is room for only seven of them. In how many ways can she arrange seven of the objects in a row along the shelf, if her clock must be included? 13 A Mathematics teacher has 10 different posters to pin up in their classroom but there is enough space for only five of them. They have three posters on algebra, two on calculus and five on trigonometry. In how many ways can they choose the five posters to pin up if: a b c there are no restrictions they decide not to pin up either of the calculus posters they decide to pin up at least one poster on each
of the three topics algebra, calculus and trigonometry? 14 As discussed at the beginning of this chapter in Explore 5.1 about encrypting letters, it states that there are over 27 million possibilities for the password encrypted as UJSNOL. How many possibilities are there? 15 How many distinct three-digit numbers can be made from 1, 2, 2, 3, 4 and 5, using each at most once? Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PS 16 From three sets of twins and four unrelated girls, find how many selections of five people can be made if exactly: a two sets of twins must be included b one set of twins must be included. EXPLORE 5.3 different ways. The Two women and three men can sit on a five-seater bicycle in 5! 120 photo shows an arrangement in which the two women are separated and the three men are also separated. = Consider, separately, the arrangements in which the women, and in which the men, are all separated from each other. 138 a Women separated from each other. Women next to each other 2!= Arrange three men with the women as a single object 4!= There are 2! the women are not separated. 72 (2! So there are 5! arrangements in which the women are separated from each other. 4!× arrangements in which 4!) × − = b Men separated from each other. Men next to each other 3!= Arrange two women with the men as a single object 3!= There are 3! 3!× arrangements in which the men are not separated. So there are 5! – (3! 3!) arrangements in which the men are separated from each other. 84 × = The calculations in a and b follow the same steps; however, the logic in one of them is flawed. Which of the two answers is correct? Can you explain why the other answer is not correct? 5.4 Problem solving with permutations and combinations Permutations and combinations can be used to find probabilities for certain events. If an event consists of a number of favourable permutations that are equiprobable, or a number of favourable combinations that are equiprobable, then P(event) = No. favourable permutations No. possible permutations P(event) = No. favourable combinations No. possible combinations Using either of the previous given methods can greatly reduce the amount of working required to solve problems in probability. Nevertheless, we must decide carefully which of them, if any, it is appropriate to use. Copyright Material - Review Only - Not for Redistribution REWIND From the introduction to this chapter, we know that the order of selection matters in a permutation but does not matter in a combination. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations WORKED EXAMPLE 5.15 There are 15 identical tins on a shelf. None of the tins are labelled but it is known that eight contain soup S( ) and three contain peas P( ), four contain beans B( ). If seven tins are randomly selected without replacement, find the probability that exactly five of them contain soup. Answer Favourable selections are when five tins of soup and two tins that are not soup are selected. (It is not important whether these two tins contain beans or peas.) We denote the 15 tins by S8 and S7 ′, where S ′ represents not soup. 8 C 5 × 7 C 2 favourable combinations Selecting S5 from S8 and S ′2 from S ′7 . C15 7 possible combinations Selecting seven from 15 tins. P(select 5tinsof soup 15 C 7 21 56 × 6435 392 2145 or 0.183. 2 TIP In the numerator we have 8 7 15 and + = + = . 7 5 2 WORKED EXAMPLE 5.16 139 A girl has a bag containing 13 red cherries R( five cherries from the bag at random. Find the probability that she takes more red cherries than black cherries. ) and seven black cherries B( ). She takes Answer From 13 red From 7 black Number of ways 5 4 3 or or 0 1 2 13 C 5 × 7 C 1287 = 0 13 C 4 × 7 C 5005 = 1 13 C 3 × 7 C 2 = 6006 Total 12298 = 20 C 15504 5 = ways P(more red than black) = 12298 15504 or 0.793. The table shows the possible make-up of the selected cherries when there are more red than black; and also the number of ways in which those cherries can be chosen. Selecting five from 20 cherries. REWIND From Chapter 4, Section 4.2, recall that ) P( or or A B C ) P( ) P( P( ) B A = + + for mutually exclusive events. C Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXPLORE 5.4 We can, of course, find the solution to Worked example 5.16 using conditional probabilities. ● There is one way to select R5 and B0 . ● There are five ways to select R4 and B1 . ● There are 10 ways to select R3 and B2 . Complete the calculations using conditional probabilities. Note how much working is involved and how long the calculations take. Compare the two approaches to solving this problem and decide for yourself which you prefer. REWIND We studied conditional probabilities in Chapter 4, Section 4.4. WORKED EXAMPLE 5.17 A minibus has seats for the driver (D) and seven passengers, as shown. When seven passengers are seated in random order, find the probability that two particular passengers, A and B, are sitting on: D a the same side of the minibus b opposite sides of the minibus. 140 Answer a P3 2 ways P4 2 ways P7 2 ways A and B both sitting on the driver’s side. A and B both not sitting on the driver’s side. A and B sitting in any two of the seven seats. P(same side) P(both ondriver’sside) P(both not on driver’s side + 12 42 7 3 P 2 P 2 6 42 3 7 b P(oppositesides) 1 P(same side The events ‘sitting on the same side’ and ‘sitting on opposite sides’ are complementary. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations EXERCISE 5G 1 Two children are selected at random from a group of six boys and four girls. Use combinations to find the probability of selecting: a two boys b two girls c one boy and one girl. 2 Three chocolates are selected at random from a box containing 10 milk chocolates and 15 dark chocolates. Find the probability of selecting exactly: a two dark chocolates b two milk chocolates c two dark chocolates or two milk chocolates. 3 Four bananas are randomly selected from a crate of 17 yellow and 23 green bananas. Find the probability that: a no green bananas are selected b less than half of those selected are green. 4 A curator has 36 paintings and 44 sculptures from which they will randomly select eight items to display in their gallery. Find the probability that the display consists of at least three more paintings than sculptures. 5 Five people are randomly selected from a group of 67 women and 33 men. Find the probability that the selection consists of an odd number of women. 6 In a toolbox there are 25 screwdrivers, 16 drill bits, 38 spanners and 11 chisels. Find the probability that a random selection of four tools contains no chisels. 7 Five clowns each have a red wig and a blue wig, which they are all equally likely to wear at any particular time. Find the probability that, at any particular time: a exactly two clowns are wearing red wigs b more clowns are wearing blue wigs than red wigs. 141 8 A gardener has nine rose bushes to plant: three have red flowers and six have yellow flowers. If they plant them in a row in random order, find the probability that: a a yellow rose bush is in the middle of the row b the three red rose bushes are not separated c no two red rose bushes are next to each other. 9 A farmer has 50 animals. They have 24 sheep, of which three are male, and they have 26 cattle, of which 20 are female. A veterinary surgeon wishes to test six randomly selected animals. Find the probability that the selection consists of: a equal numbers of cattle and sheep b more females than males. 10 a How many distinct arrangements of the letters in the word STATISTICS are there? b Find the probability that a randomly selected arrangement begins with: i three Ts ii three identical letters. 11 Three skirts, four blouses and two jackets are hung in random order on a clothes rail. Find the probability that: a b the three skirts occupy the middle section of the arrangement the two jackets are not separated. 12 In a group of 180 people, there are 88 males, nine of whom are left-handed, and there are 85 females who are not left-handed. If six people are selected randomly from the group, find the probability that exactly four of them are left-handed or female. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 13 A small library holds 1240 books: 312 of the 478 novels N( ) have hard 1240 covers H( ), and there are 440 books that do not
have hard covers. Some of this information is shown in the Venn diagram opposite. a Find the value of a, of b and of c. b A random selection of 25 of these books is to be donated to a charity group. The charity group hopes that at least 22 of the books will be novels or hard covers. Calculate the probability that the charity group gets what they hope for. 478 N H a 312 c b PS 14 A netball team of seven players is to be selected at random from five men and 10 women. Given that at least five women are selected for the team, find the probability that exactly two men are selected. PS 15 Two items are selected at random from a box that contains some tags and some labels. Selecting two tags is five times as likely as selecting two labels. Selecting one tag and one label is six times as likely as selecting two labels. Find the number of tags and the number of labels in the box. P 16 A photograph is to be taken of a pasta dish and n pizzas. The items are arranged in a line in random order. Event X is ‘the pasta dish is between two pizzas’. 142 a Investigate the value of XP( b Hence, express the value of for any value of n ⩾ 2? ) for values of n from 2 to 5. P( P( X ′ X ) ) in terms of n. Can you justify your answer WEB LINK You will find a range of interesting and challenging probability problems (with hints and solutions) in Module 16 on the NRICH website. Checklist of learning and understanding ● n ! = ( n n − 1)( n 0! 1= − … × × × , for any integer n 0> . 3 2 1 2) ● A key word that points to a permutation is arranged. A permutation is a way of selecting and arranging objects in a particular order. ● Key words that point to a combination are chosen and selected. A combination is a way of selecting objects in no particular order. ● From n distinct objects, there are: n P n != permutations of all n objects )! ( n permutations of r objects !( r n − permutations in which there are p q r , , , … of each type. ... combinations of r objects. r )! Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations END-OF-CHAPTER REVIEW EXERCISE 5 1 The word MARMALADE contains four vowels and five consonants. Find the number of possible arrangements of its nine letters if: a there are no restrictions on the order b the arrangement must begin with the four vowels. 2 Five men, four children and two women are asked to stand in a queue at the post office. Find how many ways they can do this if: a the women must be separated b all of the children must be separated from each other. 3 Find the probability that a randomly selected arrangement of all the letters in the word PALLETTE begins and ends with the same letter. 4 Eight-digit mobile phone numbers issued by the Lemon Network all begin with 79. a How many different phone numbers can the network issue? b Find the probability that a randomly selected number issued by this network: i ends with the digits 97 ii reads the same left to right as right to left. 5 There are 12 books on a shelf. Five books are 15cm tall; four are 20 cm tall and three are 25cm tall. Find the number of ways that the books can be arranged on the shelf so that none of them is shorter than the book directly to its right. 6 The 11 letters of the word REMEMBRANCE are arranged in a line. i Find the number of different arrangements if there are no restrictions. ii Find the number of different arrangements which start and finish with the letter M. iii Find the number of different arrangements which do not have all 4 vowels (E, E, A, E) next to each other. 4 letters from the letters of the word REMEMBRANCE are chosen. iv Find the number of different selections which contain no Ms and no Rs and at least 2 Es. [1] [2] [2] [3] [3] [1] [2] [2] [2] [1] [2] [3] [3] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q6 November 2013 7 Find how many ways 15 children can be divided into three groups of five if: a there are no restrictions b two of the children are brothers who must be in the same group. 8 An entertainer has been asked to give a performance consisting of four items. They know three songs, five jokes, two juggling tricks and can play one tune on the mandolin. Find how many different ways there are for them to choose the four items if: a there are no restrictions on their performance b they decide not to sing any songs c they are not allowed to tell more than two jokes. 9 From a group of nine people, five are to be chosen at random to serve on a committee. In how many ways can this be done if two particular people refuse to serve on the committee together? [2] [3] [1] [2] [3] [3] Copyright Material - Review Only - Not for Redistribution 143 Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 10 Twenty teams have entered a tournament. In order to reduce the number of teams to eight, they are put into groups of five and the teams in each group play each other twice. The top two teams in each group progress to the next round. From this point on, teams are paired up, playing each other once with the losing team being eliminated. How many games are played during the whole tournament? [3 11 A bank provides each account holder with a nine-digit card number that is arranged in three blocks, as shown in the example opposite. Find, in index form, the number of card numbers available if: a there are no restrictions on the digits used b none of the three blocks can begin with 0 c the two digits in the second block must not be the same d the three-, two- and four-digit numbers on the card are even, odd and even, respectively. 12 A basket holds nine flowers: two are pink, three are yellow and four are red. Four of these flowers are chosen at random. Find the probability that at least two of them are red. 13 Find the number of ways in which 11 different pieces of fruit can be shared between three boys so that each boy receives an odd number of pieces of fruit. PS 14 A bakery wishes to display seven of its 14 types of cake in a row in its shop window. There are six types of 144 sponge cake, five types of cheesecake and three types of fruitcake. Find the number of possible displays that can be made if the bakery places: a a sponge cake at each end of the row and includes no fruitcakes in the display b a fruitcake at one end of the row with sponge cakes and cheesecakes placed alternately in the remainder of the row. PS 15 Five cards, each marked with a different single-digit number from 3 to 7, are randomly placed in a row. Find the probability that the first card in the row is odd and that the three cards in the middle of the row have a sum of 15. PS 16 Two ordinary fair dice are rolled and the two faces on which they come to rest are hidden by holding the dice together, as shown, and lifted off the table. The sum of the numbers on the 10 visible faces of the dice is denoted by T. a Find the number of possible values of T , and find the most likely value of T . b Calculate the probability that øT 38. PS 17 Three ordinary fair dice are rolled. Find the number of ways in which the number rolled with the first die can exceed the sum of the numbers rolled with the second and third dice. Hence, find the probability that this event does not occur in two successive rolls of the three dice. Copyright Material - Review Only - Not for Redistribution [1] [2] [2] [3] [4] [5] [2] [4] [4] [4] [3] [6] Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations PS 18 How many even four-digit numbers can be made from the digits 0, 2, 3, 4, 5 and 7, each used at most once, when the first digit cannot be zero? 19 a i Find how many numbers there are between 100 and 999 in which all three digits are different. ii Find how many of the numbers in part i are odd numbers greater than 700. b A bunch of flowers consists of a mixture of roses, tulips and daffodils. Tom orders a bunch of 7 flowers from a shop to give to a friend. There must be at least 2 of each type of flower. The shop has 6 roses, 5 tulips and 4 daffodils, all different from each other. Find the number of different bunches of flowers that are possible. [4] [3] [4] [4] Cambridge International AS & A Level Mathematics 9709 Paper 61 Q6 June 2016 20 Three identical cans of cola, 2 identical cans of green tea and 2 identical cans of orange juice are arranged in a row. Calculate the number of arrangements if i ii the first and last cans in the row are the same type of drink, the 3 cans of cola are all next to each other and the 2 cans of green tea are not next to each other. [3] [5] Cambridge International AS & A Level Mathematics 9709 Paper 63 Q4 June 2010 145 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 CROSS-TOPIC REVIEW EXERCISE 2 1 Each of the eight players in a chess team plays 12 games against opponents from other teams. The total number of wins, draws and losses for th
e whole team are denoted by X Y, and Z, respectively. a State the value of X Y Z + . + b Find the least possible value of Z X− , given that Y 25= . c Given that none of the players drew any of their games and that X Z 50 , find the exact mean number − = of games won by the players. 2 Six books are randomly given to two girls so that each receives at least one book. a In how many ways can this be done? [1] [1] [2] [3] 3 4 b Are both girls more likely to receive an odd number or an even number of books? Give a reason for your answer. [2] The 60 members of a ballroom dance society wish to participate in a competition but the coach that has been hired has seats for only 57 people. In how many ways can 57 members be selected if the society’s president and vice president must be included? [2] Four discs in two colours and in four sizes are placed in any order on either of two sticks. The following illustration shows one possible arrangement of the four discs. 146 a Find the number of ways in which the four discs can be arranged so that: i they are all on the same stick ii there are two discs on each stick. b In how many ways can the discs be placed if there are no restrictions? [2] [2] [2] 5 A fair triangular spinner with sides numbered 1, 2 and 3 is spun three times and the numbers that it comes to rest on are written down from left to right to form a three-digit number. a How many possible three-digit numbers are there? b Find the probability that the three-digit number is: i even ii odd and greater than 200. 6 A book of poetry contains seven poems, three of which are illustrated. In how many different orders can all the poems be read if no two illustrated poems are read one after the other? 7 Find the number of ways that seven goats and four sheep can sleep in a row if: a all the goats must sleep next to each other b no two sheep may sleep next to each other. 8 A teacher is looking for 6 pupils to appear in the school play and has decided to select them at random from a group of 11 girls and 13 boys. a Find the number of ways in which the teacher can select the 6 pupils. [1] [1] [2] [3] [2] [3] [1] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cross-topic review exercise 2 b Two roles in the play must be played by girls; three roles must be played by boys, but the fool can be played by a girl or by a boy. If the first pupil selected is to play the role of the fool, find the probability that the fool is played by i a particular girl ii a boy. c If instead, the pupil who is to play the fool is the last of the six pupils selected, investigate what effect this change in the order of selection has on the probability that the fool is played by: i a particular girl ii a boy. [1] [1] [3] [3] 9 A radio presenter has enough time at the end of their show to play five songs. She has 13 songs by four groups to choose from: five songs by The Anvils, four by The Braziers, three by The Chisels and one by The Dustbins. Find the number of ways she can choose five songs to play if she decides: a that there should be no restrictions b to play all three songs by The Chisels c to play at least one song by each of the four groups. [1] [2] [4] 10 Students enrolling at an A Level college must select three different subjects to study from the six that are available. One subject must be chosen from each of the option groups A, B and C, as shown in the following table. Group A Physics Chemistry History Group B Biology Physics Group C Mathematics Biology Mathematics Computing a One student has chosen to study History and Mathematics. How many subjects do they have to choose from to complete their selection? b How many combinations of three subjects are available to a student who enrols at this college? 11 Four ordinary fair dice are arranged in a row. Find the number of ways in which this can be done if the four numbers showing on top of the dice: a are all odd b have a sum that is less than 7. 12 At company V, 12.5% of the employees have a university degree. At company W, 85% of the employees do not have a university degree. There are 112 employees at company V and 120 employees at company W. a One employee is randomly selected. Find the probability that they: i work for company V ii have a university degree. b Five employees from company W are selected at random. Find the probability that none of them has a university degree. 13 One hundred qualified drivers are selected at random. Out of these 100 drivers, of the 40 drivers who wear spectacles, 30 passed their driving test at the first attempt. Altogether, 25 of the drivers did not pass at their first attempt. a Show the data given about the drivers in a clearly labelled table or diagram. b Did these drivers pass the test at their first attempt independently of whether or not they wear spectacles? Explain your answer. Copyright Material - Review Only - Not for Redistribution 147 [1] [2] [1] [3] [1] [2] [2] [3] [3] Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 14 A conference hall has 24 overhead lights. Pairs of lights are operated by switches next to the main entrance, and each switch has three numbered settings: 0 (off), 1 (dim), 2 (bright). Find the number of possible lighting arrangements in the hall if: a there are no restrictions b two particular pairs of lights must be on setting 2 c three lights that are not operated by the same switch and five pairs of lights that are operated by the same switch are not working. 15 Twelve chairs in two colours are arranged, as shown. A B C D 1 2 3 Find in how many ways nine people can sit on these chairs if: a the two blue chairs in column C must remain unoccupied b all of the green chairs must be occupied c more blue chairs than green chairs must be occupied 148 d at least one of the chairs in row 2 must remain unoccupied. 16 In a certain country, vehicle registration plates consist of seven characters: a letter, followed by a three-digit number, followed by three letters. [1] [1] [2] [2] [3] [2] [5] For example: B 474 PQR The first letter cannot be a vowel; the three-digit number cannot begin with 0; and the first of the last three letters cannot be a vowel or any of the letters X, Y or Z. a Find the number of registration plates available. [2] b Find the probability that a randomly selected registration plate is unassigned, given that there are 48.6 million vehicle owners in the country, and that each owns, on average, 1.183 registered vehicles. [3] 17 Seats for the guests at an awards ceremony are arranged in two rows of eight and ten, divided by an aisle, as shown. e l s i a Seats are randomly allocated to 18 guests. a Find the probability that two particular guests are allocated seats: i on the same side of the aisle ii in the same row iii on the same side of the aisle and in the same row. [3] [3] [4] b Give a reason why the answer to part a iii is not equal to the product of the answers to part a i and part a ii. [1] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6 Probability distributions In this chapter you will learn how to: 149 ■ identify and use a discrete random variable ■ construct a probability distribution table that relates to a given situation involving a discrete random variable, X , and calculate its expectation, XE( ), and its variance, Var( )X . Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PREREQUISITE KNOWLEDGE Where it comes from What you should be able to do Check your skills IGCSE / O Level Mathematics Use the fact that A P( ) 1 P( = − A ′ . ) Chapters 4 and 5, sections 4.3, 4.4, 4.5 and 5.4 Distinguish between independent and dependent events, and calculate probabilities accordingly. 1 A game can be won W( Given that W ′ = )D . find P( P( ), lost ( )L or drawn ( , 0.65 ) ′ = )D . ) 0.46 and LP( 2 Two cubes are selected at random from a bag of three red cubes and three blue cubes. Show that the selected cubes are more likely to both be red when the selections are made with replacement than when the selections are made without replacement. Tools of the trade Suppose a trading company is planning a new marketing campaign. The campaign will probably go ahead only if the most likely outcome is that sales will increase. However, the company also needs to be aware of worst-case and best-case outcomes, as sales may decrease or decrease dramatically, stay the same or increase dramatically. The company will be able to make informed decisions based on its estimates of the probabilities of these possible outcomes. The likelihood of these outcomes will be based on an analysis of a probability distribution for the changes in sales. The probability distribution described above acts as a prediction for future sales and the risks involved. Suppose the company is considering entering a new line of business b
ut needs to generate at least $50 000 in revenue before it starts to make a profit. If their probability distribution tells them that there is a 40 % chance that revenues will be less than $50 000, then the company knows roughly what level of risk it is facing by entering that new line of business. 150 6.1 Discrete random variables A variable is said to be discrete and random if it can take only certain values that occur by chance. For example, when we buy a carton of six eggs, some may be broken; the number of broken eggs in a carton is a discrete random variable that can take values 0, 1, 2, 3, 4, 5 or 6. Discrete random variables may arise from independent trials. For example, if we roll four . dice then the number of 6s obtained, S, is a discrete random variable with S {0, 1, 2, 3, 4} ∈ Situations where selections are made without replacement, can also generate discrete random variables. For example, if we randomly select three children from a group of four boys and two girls, the number of boys selected, B, and the number of girls selected, G, are discrete random variables with B {1, 2, 3} . and G {0, 1, 2} ∈ ∈ 6.2 Probability distributions The probability distribution of a discrete random variable is a display of all its possible values and their corresponding probabilities. The usual method of display is by tabulation in a probability distribution table. The probability distribution also can be represented in a vertical line graph or in a bar chart. Copyright Material - Review Only - Not for Redistribution TIP A variable is denoted by an upper-case letter and its possible values by the same lower-case letter. If X can take values of 1, 2 and 3, we write X ∈ {1, 2, 3}, where the symbol ∈ means ‘is an element of’. REWIND We learnt how to find probabilities for selections with and without replacement in Chapters 4 and 5, Sections 4.3, 4.4, 4.5 and 5.4. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions Consider tossing two fair coins, where we can obtain 0, 1 or 2 heads. . The number of heads obtained in each trial, X, is a discrete random variable and X {0, 1, 2} ∈ XP( = 0) P(tails and tails) = = 0.5 0.5 × = 0.25 XP( = 1) P(heads and tails) P(tails and heads) + = = (0.5 0.5) × + (0.5 0.5) × = 0.5 XP( = 2) P(heads and heads) = = 0.5 0.5 × = 0.25 The probability distribution for X is displayed in the following table. x P( )==X x 0 0.25 1 0.5 2 0.25 The probabilities for the possible values of X are equal to the relative frequencies of the values. We would expect 25% of the tosses to produce zero heads; 50% to produce one head and 25% to produce two heads. WORKED EXAMPLE 6.1 A fair square spinner with sides labelled 1, 2, 3 and 4 is spun twice. The two scores obtained are added together to give the total, X . Draw up the probability distribution table for X . Answer 1st spin The grid shows the 16 equally likely outcomes for the discrete random variable X , where . X {2, 3, 4, 5, 6, 7, 8} ∈ 2 1 16 3 2 16 4 3 16 5 4 16 6 3 16 7 2 16 8 1 16 Sum 1= The probability distribution for X is shown in the table. P( )== X x 151 TIP P( )=X x is equal to the relative frequency of each particular value of X . TIP Note that Σ P( X x = = . ) 1 WORKED EXAMPLE 6.2 The following table shows the probability distribution for the random variable V. v P( )==V v 2 0.05 3 2c 4 5 6 0.1+c 2 +c 0.05 0.16 Find the value of the constant c and find P( V > . 4) Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Answer 0.05 + c 2 c + + 0.1 2 + c ( c − c 0.2= or = − c 3.2 0.05 0.16 1 = + + 2 c + 0.2)( 3 – 0.64 c c + 3.2) The valid solution is c 0.2= . ∴ V P( > 4) P( = = V 5) P( + (2 0.2) 0.05 0.16 6) V = + + = = × 0.61 = = 0 0 Σ = to form and solve We use p 1 an equation in c. = − Note that if c P( = and VP( 3.2 , then 3) 10.24, P( 4) V = . 6.35 5) = = = − V = − 3.1 KEY POINT 6.1 A probability distribution shows all the possible values of a variable and the sum of the probabilities is Σ = p 1 TIP Do check whether the solutions are valid. Remember that a probability cannot be less than 0 or greater than 1. WORKED EXAMPLE 6.3 There are spaces for three more passengers on a bus, but eight youths, one man and one woman wish to board. The bus driver decides to select three of these people at random and allow them to board. Draw up the probability distribution table for Y , the number of youths selected. 152 Answer )= Selections are made without replacement, so we can use combinations to find Y y P( . Possible values of Y are 1, 2 and 3 At least one youth will be selected because there are only two non-youths, who we denote by Y ′. C10 3 possible selections. Selecting three from 10 people. P( Y = 1) = 2 C 8 C × 1 10 C 3 2 = 1 15 Selecting one from Y8 , and two from Y2 ′. P( Y = 2) = 8 C × 2 10 C 2 C 1 3 = 7 15 P( Y = 3) = 8 C × 3 10 C 2 C 0 3 = 7 15 y P( )==Y y 1 1 15 2 7 15 3 7 15 Selecting two from Y8 , and one from Y2 ′. Selecting three from Y8 , and none from Y2 ′. The table shows the probability distribution for Y . TIP Always check that 1pΣ = . Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions EXERCISE 6A 1 The discrete random variable V is such that V {1, 2, 3} . Given that V P( ∈ = 1) P( = V = 2) = × 2 P( V = 3) , draw up the probability distribution table for V . 2 The probability distribution for the random variable X is given in the following table. x P( )==X x 2 p 3 p2 4 1 2 p 5 p3 Find the value of p and work out P(2 < X < . 5) 3 The probability distribution for the random variable W is given in the following table. w P( )==W w 3 k2 6 k2 9 k 2 12 4 5 3k− 15 13 50 a Form an equation using k, then solve it. b Explain why only one of your solutions is valid. c Find øW < P(6 10) . 4 The probability that a boy succeeds with each basketball shot is random variable S represents the number of successful shots. 7 9 . He takes two shots and the discrete 153 Show that SP( = 0) = 4 81 and draw up the probability distribution table for S. 5 At a garden centre, there is a display of roses: 25 are red, 20 are white, 15 are pink and 5 are orange. Three roses are chosen at random. a Show that the probability of selecting three red roses is approximately 0.0527. b Draw up the probability distribution table for the number of red roses selected. c Find the probability that at least one red rose is selected. 6 Three vehicles from a company’s six trucks, five vans, three cars and one motorbike are randomly selected and tested for roadworthiness. a Show that the probability of selecting three vans is 2 . 91 b Draw up the probability distribution table for the number of vans selected. c Find the probability that, at most, one van is selected. 7 Five grapes are randomly selected without replacement from a bag containing one red grape and six green grapes. Name and list the possible values of two discrete random variables in this situation. State the relationship between the values of your two variables. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 8 A pack of five DVDs contains three movies and two documentaries. Three DVDs are selected and the following table shows the probability distribution for M, the number of movies selected. m 1 2 3 P( ==M m ) 0.3 0.6 0.1 Draw up the probability distribution table for D, the number of documentaries selected. 9 In a particular country, 90% of the population is right-handed and 40% of the population has red hair. Two people are randomly selected from the population. Draw up the probability distribution for X , the number of right-handed, red-haired people selected, and state what assumption must be made in order to do this. 10 A fair 4-sided die, numbered 1, 2, 3 and 5, is rolled twice. The random variable X is the sum of the two numbers on which the die comes to rest. a Show that XP( = 8) 1 = . 8 b Draw up the probability distribution table for X , and find XP( 6)> . 11 There are eight letters in a post box, and five of them are addressed to Mr Nut. Mr Nut removes four letters at random from the box. a Find the probability that none of the selected letters are addressed to Mr Nut. b Draw up the probability distribution table for N , the number of selected letters that are addressed 154 to Mr Nut. c Describe one significant feature of a vertical line graph or bar chart that could be used to represent the probability distribution for N . 12 A discrete random variable Y is such that Y {8, 9, 10} ∈ )= . Given that Y y P( = ky , find the value of the constant k. . 13 Q is a discrete random variable and Q {3, 4, 5, 6} ∈ a Given that Q q P( = ) = cq 2 , find the value of the constant c. b Hence, find QP( 4)> . 14 Four books are randomly selected from a box containing 10 novels, 10 reference books and 5 dictionaries. The random variable N represent
s the number of novels selected. a Find the value of NP( 2)= , correct to 3 significant figures. b Without further calculation, state which of N 0= or N 4= is more likely. Explain the reasons for your answer. 15 In a game, a fair 4-sided spinner with edges labelled 0, 1, 2 and 3 is spun. If a player spins 1, 2 or 3, then that is their score. If a player scores 0, then they spin a fair triangular spinner with edges labelled 0, 1 and 2, and the number they spin is their score. Let the variable X represent a player’s score. a Show that XP( = 0) = 1 12 . b Draw up the probability distribution table for X , and find the probability that X is a prime number. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions 16 A biased coin is tossed three times. The probability distribution for H, the number of heads obtained, is shown in the following table. h 0 1 2 P( )==H h 0.512 0.384 0.096 3 a a Find the probability of obtaining a head each time the coin is tossed. b Give another discrete random variable that is related to these trials, and calculate the probability that its value is greater than the value of H. 17 Two ordinary fair dice are rolled. A score of 3 points is awarded if exactly one die shows an odd number and there is also a difference of 1 between the two numbers obtained. A player who rolls two even numbers is awarded a score of 2 points, otherwise a player scores 1 point. a Draw up the probability distribution table for S, the number of points awarded. b Find the probability that a player scores 3 points, given that the sum of the numbers on their two dice is greater than 9. 18 The discrete random variable R is such that R {1, 3, 5, 7} . 1) ( k r + 2 r + a Given that R r , find the value of the constant k. P( ∈ ) = = 4)øR . b Hence, find P( EXPLORE 6.1 Consider the probability distribution for X , the number of heads obtained when two fair coins are tossed, which was given in the table presented in the introduction of Section 6.2. Sketch or simply describe the shape of a bar chart (or vertical line graph) that can be used to represent this distribution. In this activity, you will investigate how the shape of the distribution of X is altered when two unfair coins are tossed; that is, when the probability of obtaining heads is p . 0.5≠ Consider the case in which p probability distribution of X , the number of heads obtained. 0.4= for both coins. Draw a bar chart to represent the Next consider the case in which p represent the probability distribution of X . 0.6= for both coins, and draw a bar chart to What do you notice about the bar charts for p 0.4= and p ? 0.6= Investigate other pairs of probability distributions for which the values of p add up to 1, such as p results. . Make general comments to summarise your and p 0.3= 0.7= M Investigate how the value of XP( represent this graphically. On the same diagram, show how the values of XP( and XP( changes as p increases from 0 to 1, and then 0)= change as p increases from 0 to 1. 1)= 2)= Copyright Material - Review Only - Not for Redistribution 155 WEB LINK This can be done manually or using the Coin Flip Simulation on the GeoGebra website. FAST FORWARD We will learn how to extend this Explore activity to more than two coins in Chapter 7. We will see how to represent the probability distribution for a continuous random variable in Chapter 8, Section 8.1. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 DID YOU KNOW? n The simple conjecture of Fermat’s Last Theorem, which is that x n 2> , defeated the greatest mathematicians for 350 years. has no positive integer solutions for any integer = + y z n n The theorem is simple in that it says ‘a square can be divided into two squares, but a cube cannot be divided into two cubes, nor a fourth power into two fourth powers, and so on’. Pierre de Fermat himself claimed to have a proof but only wrote in his notebook that ‘this margin is too narrow to contain it’! Fermat’s correspondence with the French mathematician, physicist, inventor and philosopher Blaise Pascal helped to develop a very important concept in basic probability that was revolutionary at the time; namely, the idea of equally likely outcomes and expected values. Since his death in 1665, substantial prizes have been offered for a proof, which was finally delivered by Briton Andrew Wiles in 1995. Wiles’ proof used highly advanced 20th century mathematics (i.e. functions of complex numbers in hyperbolic space and the doughnut-shaped solutions of elliptic curves!) that was not available to Fermat. 156 Pierre de Fermat, 1607–1665. Andrew Wiles 6.3 Expectation and variance of a discrete random variable Values of a discrete random variable with high probabilities are expected to occur more frequently than values with low probabilities. When a number of trials are carried out, a frequency distribution of values is produced, and this distribution has a mean or expected value. Expectation The mean of a discrete random variable X is referred to as its expectation, and is written XE( ). Suppose we have a biased spinner with which we can score 0, 1, 2 or 3. The probabilities for these scores, X , are as given in the following table and are also represented in the graph. x P( )==X x 0 0.1 1 0.3 2 0.4 3 0.2 However many times it is spun, we expect to score 0 with 10% of the spins; 1 with 30%; 2 with 40% and 3 with 20%. The expected frequencies of the scores in 1600 trials are shown in the following table. P(X = x) 0.4 0.2 0 0 1 2 x Expected frequency ( )f 0 1 2 0.1 1600 160 × = 0.3 1600 × = 480 0.4 1600 × = 640 0.2 1600 × = 320 Copyright Material - Review Only - Not for Redistribution x 3 3 TIP We can think of XE( ) as being the long-term average value of X over a large number of trials. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions From this table of expected frequencies, we can calculate the mean (expected) score in 1600 trials. KEY POINT 6.2 The expectation of a discrete random ) variable is E( = Σ X xp TIP The denominator is Σ =p 1, so we can omit it from our calculation of XE( ). TIP An alternative way to write the formula for expectation is E( X x = = Σ P( X x × ) [ ) ] . 157 REWIND We can remember variance from Chapter 3, Section 3.3 as ‘mean of the squares minus square of the mean’. Mean E( = X ) = (0 160) × + xf Σ f Σ = (1 480) × (2 640) × + 1600 + (3 320) × = 1.7 We obtain the same value for XE( instead of frequencies. ) if relative frequencies (i.e. probabilities) are used Mean E( = X ) = xp Σ p Σ = (0 0.1) × + (1 0.3) × (2 0.4) × + (3 0.2) × = 1.7 + 1 EXPLORE 6.2 Adam and Priya each have a bag of five cards, numbered 1, 2, 3,4 and 5. They simultaneously select a card at random from their bag and place it face-up on a table. The numerical difference between the numbers on their cards, X , is recorded, where X {0, 1, 2, 3, 4} probability distribution table for X . . They repeat this 200 times and use their results to draw up a ∈ Adam suggests a new experiment in which the procedure will be the same, except that each of them can choose the card that they place on the table. He says the probability distribution for X will be very different because the cards are not selected at random. Priya disagrees, saying that it will be very similar, or may even be exactly the same. Do you agree with Adam or Priya? Explain your reasoning. Variance The variance and standard deviation of a discrete random variable give a measure of the spread of values around the mean, XE( using probabilities in place of frequencies. ). These measures, like XE( ), can be calculated If we replace f by p and replace x by XE( ) in the second of the two formulae for variance , we obtain Σ 2 x p p Σ − X{E( KEY POINT 6.3 )} 2, which simplifies to Σ 2 x p − X{E( )} 2 because Σ =p 1. The variance of a discrete random variable is VarE( X 2 . )} WORKED EXAMPLE 6.4 The following table shows the probability distribution for X . Find its expectation, variance and standard deviation. x P( )==X x 0 1 12 5 3 12 15 5 12 20 3 12 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Answer XE( ) = = = = × × 0  1 12 1 12 12.5 × 1 12   + × 5  3 12   +   15 × 5 12   +   20 × 3 12   [(0 1) × + (5 3) × + (15 5) × + (20 3)] × 150 1 12   +   2 5 × 3 12   +   2 15 × 5 12   +   2 20 × 3 12   2 {12.5} − [(0 1) × + (25 3) × + (225 5) × + (400 3)] 156.25 − × Var 12 2400 12 43.75 − 156.25 XSD( ) = 43.75 = 6.61 , correct to 3 significant figures. TIP The working is simpler when all fractions have the same denominator. TIP Remember to subtract the square of XE( when calculating variance. ) Substitute values of X and X x )= P( into the formula ) E( X = Σ xp . Substitute into the formula for Var(X). Take the square root of the variance. EXERCISE 6B 158 1 The probability distributio
n for the random variable X is given in the following table. x 0 1 2 3 P( )==X x 0.10 0.12 0.36 0.42 Calculate XE( ) and XVar( ). 2 The probability distribution for the random variable Y is given in the following table. y 0 P( )==Y y 0.03 1 2 p 2 0.32 3 p 4 0.05 a Find the value of p. b Calculate YE( ) and the standard deviation of Y . 3 The random variable T is such that T {1, 3, 6, 10} . Given that the four possible values of T are equiprobable, ∈ find TE( ) and TVar( ). 4 The following table shows the probability distribution for the random variable V . v 1 3 9 m P( )==V v 0.4 0.28 0.14 0.18 Given that VE( ) = 5.38 , find the value of m and calculate VVar( ). 5 R is a random variable such that R {10, 20, 70, 100} ∈ . Given that R r P( )= is proportional to r, show that RE( ) 77= and find RVar( ). Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions 6 The probability distribution for the random variable W is given in the following table. w 2 7 a P( )==W w 0.3 0.3 0.1 24 0.3 Given that W a ) = , find a and evaluate WVar( E( ). M 7 The possible outcomes from a business venture are graded from 5 to 1, as shown in the following table. Grade 5 4 3 2 1 Outcome High profit Fair profit No loss Small loss Heavy loss Probability 0.24 0.33 0.24 0.11 0.08 a Calculate the expected grade and use it to describe the expected outcome of the venture. Find the standard deviation and explain what it gives a measure of in this case. b Investigate the expected outcome and the standard deviation when the grading is reversed (i.e. high profit is graded 1, and so on). Compare these outcomes with those from part a. 8 Two ordinary fair dice are rolled. The discrete random variable X is the lowest common multiple of the two numbers rolled. a Draw up the probability distribution table for X . b Find XE( ) and X P c Calculate XVar( [ ). > E( X ) ] . 159 9 In a game, a player attempts to hit a target by throwing three darts. With each throw, a player has a 30% chance of hitting the target. a Draw up the probability distribution table for H, the number of times the target is hit in a game. b How many times is the target expected to be hit in 1000 games? 10 Two students are randomly selected from a class of 12 girls and 18 boys. a Find the expected number of girls and the expected number of boys. b Write the ratio of the expected number of girls to the expected number of boys in simplified form. What do you notice about this ratio? c Calculate the variance of the number of girls selected. 11 A sewing basket contains eight reels of cotton: four are green, three are red and one is yellow. Three reels of cotton are randomly selected from the basket. a Show that the expected number of yellow cotton reels is 0.375. b Find the expected number of red cotton reels. c Hence, state the expected number of green cotton reels. 12 A company offers a $1000 cash loan to anyone earning a monthly salary of at least $2000. To secure the loan, the borrower signs a contract with a promise to repay the $1000 plus a fixed fee before 3 months have elapsed. Failure to do this gives the company a legal right to take $1540 from the borrower’s next salary before returning any amount that has been repaid. From past experience, the company predicts that 70% of borrowers succeed in repaying the loan plus the fixed fee before 3 months have elapsed. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 a Calculate the fixed fee that ensures the company an expected 40% profit from each $1000 loan. b Assuming that the company charges the fee found in part a, how would it be possible, without changing the loan conditions, for the company’s expected profit from each $1000 loan to be greater than 40%? 13 When a scout group of 8 juniors and 12 seniors meets on a Monday evening, one scout is randomly selected to hoist a flag. Let the variable X represent the number of juniors selected over n consecutive Monday evenings. a By drawing up the probability distribution table for X , or otherwise, show that XE( ) 1.2= when n 3= . b Find the number of Monday evenings over which 14 juniors are expected to be selected to hoist the flag. 14 An ordinary fair die is rolled. If the die shows an odd number then S, the score awarded, is equal to that number. If the die shows an even number, then the die is rolled again. If on the second roll it shows an odd number, then that is the score awarded. If the die shows an even number on the second roll, the score awarded is equal to half of that even number. a List the possible values of S and draw up a probability distribution table. [ P b Find S E S > ( ) ] . c Calculate the exact value of SVar( ). PS 15 A fair 4-sided spinner with sides labelled A B B B , , , is spun four times. a Show that there are six equally likely ways to obtain exactly two Bs with the four spins. 160 b By drawing up the probability distribution table for X , the number of times the spinner comes to rest on B, find the value of Var( X ) E( X ) . c What, in the context of this question, does the value found in part b represent? EXPLORE 6.3 In this activity we will investigate a series of trials in which each can result in one of two possible outcomes. A ball is dropped into the top of the device shown in the diagram. When a ball hits a nail (which is shown as a red dot), there are two equally likely outcomes: it can fall to the left or it can fall to the right. Using L and R (to indicate left and right), list all the ways that a ball can fall into each of the cups A B, and C. Use your lists to tabulate the probabilities of a ball falling into each of the cups. Give all probabilities with denominator 4. A B C FAST FORWARD We will study the expectation of two special discrete random variables, and the variance of one of them, in Chapter 7, Sections 7.1 and 7.2. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions The diagram shows a similar device with four cups labelled A to D. List all the ways that a ball can fall into each of the cups. Use your lists to tabulate the probabilities of a ball falling into each of the cups. Give all probabilities with denominator 8. A B C D Can you explain how and why the values probabilities in your tables? 2   11 2   and 3   11 2   are connected with the The next device in the sequence has 10 nails on four rows. Tabulate the probabilities of a ball falling into each of its five cups, A to E. FAST FORWARD We will study independent trials that have only two possible outcomes, such as left or right and success or failure, in Chapter 7, Sections 7.1 and 7.2. Checklist of learning and understanding ● A discrete random variable can take only certain values and those values occur in a certain random manner. ● A probability distribution for a discrete random variable is a display of all its possible values 161 and their corresponding probabilities. ● For the discrete random variable X : Σ = p 1 E( X ) = Σ xp VarE( X 2 )} Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 END-OF-CHAPTER REVIEW EXERCISE 6 1 Find the mean and the variance of the discrete random variable X , whose probability distribution is given in the [3] following table. x 1 2 3 4 P( )==X x 1 – k 2 – 3k 3 – 4k 4 – 6k 2 The following table shows the probability distribution for the random variable Y . y P( )==Y y 1 0.2 10 0.4 q 0.2 101 0.2 a Given that YVar( ) 1385.2 = , show that q 2 – 61 q + 624 = and solve this equation. 0 b Find the greatest possible value of YE( ). [4] [2] 3 An investment company has produced the following table, which shows the probabilities of various percentage profits on money invested over a period of 3 years. Profit ( )% 1 5 10 15 20 30 40 45 50 Probability 0.05 0.10 0.50 0.20 0.05 0.04 0.03 0.02 0.01 a Calculate the expected profit on an investment of $50 000. 162 b A woman considers investing $50 000 with the company, but decides that her money is likely to earn more when invested over the same period in a savings account that pays r% compound interest per annum. Calculate, correct to 2 decimal places, the least possible value of r. [3] [3] 4 A chef wishes to decorate each of four cupcakes with one randomly selected sweet. They choose the sweets at random from eight toffees, three chocolates and one jelly. Find the variance of the number of cupcakes that will [6] be decorated with a chocolate sweet. 5 The faces of a biased die are numbered 1, 2, 3, 4, 5 and 6. The random variable X is the score when the die is thrown. The probability distribution table for X is given. x P( )==.2 6 0.2 The die is thrown 3 times. Find the probability that the score is at least 4 on at least 1 of the 3 throws. [5] Cambridge international AS & A
Level Mathematics 9709 Paper 61 Q2 June 2016 [Adapted] 6 A picnic basket contains five jars: one of marmalade, two of peanut butter and two of jam. A boy removes one jar at random from the basket and then his sister takes two jars, both selected at random. a Find the probability that the sister selects her jars from a basket that contains: i exactly one jar of jam ii exactly two jars of jam. [1] [1] b Draw up the probability distribution table for J , the number of jars of jam selected by the sister, and show that JE( ) . 0.8= [4] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions 7 Two ordinary fair dice are rolled. The product and the sum of the two numbers obtained are calculated. The score awarded, S, is equal to the absolute (i.e. non-negative) difference between the product and the sum. For example, if 5 and 3 are rolled, then S (5 3) × = − (5 3) + = . 7 a State the value of S when 1 and 4 are rolled. b Draw up a table showing the probability distribution for the 14 possible values of S, and use it to calculate SE( ). 8 A fair triangular spinner has sides labelled 0, 1 and 2, and another fair triangular spinner has sides labelled –1, 0 and 1. The score, X , is equal to the sum of the squares of the two numbers on which the spinners come to rest. a List the five possible values of X . b Draw up the probability distribution table for X . c Given that X 4< , find the probability that a score of 1 is obtained with at least one of the spinners. d Find the exact value of a, such that the standard deviation of X is 1 a × E( X ) . 9 A discrete random variable X , where X {2, 3, 4, 5} , is such that X x P − 30 a Calculate the two possible values of b. b Hence, find P(2 < X < . 5) 10 Set A consists of the ten digits 0, 0, 0, 0, 0, 0, 2, 2, 2, 4. Set B consists of the seven digits 0, 0, 0, 0, 2, 2, 2. One digit is chosen at random from each set. The random variable X is defined as the sum of these two digits. i Show that XP( = 2) 3 = . 7 ii Tabulate the probability distribution of X . iii Find XE( ) and XVar( ). iv Given that X 2= , find the probability that the digit chosen from set A was 2. [1] [5] [1] [3] [2] [3] [3] [2] [2] [2] [3] [2] 163 Cambridge International AS & A Level Mathematics 9709 Paper 63 Q5 June 2010 PS PS 11 The discrete random variable Y is such that Y {4, 5, 8, 14, 17} ∈ )= and Y y P( is directly proportional to Find YP( 4)> . 12 X is a discrete random variable and X {0, 1, 2, 3} , find ∈ . 0)> 0 or 2) øX 0.62 XP( P( 2 \| X = = . Given that XP( > 1) = 0.24 , P(0 < X < 3) = 0.5 and 1 . 1+ y [4] [5] 13 Four students are to be selected at random from a group that consists of seven boys and x girls. The variables B and G are, respectively, the number of boys selected and the number of girls selected. a Given that B P( = 1) P( = B = 2) , find the value of x. b Given that G 3≠ , find the probability that G 4= . [3] [3] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 14 A box contains 2 green apples and 2 red apples. Apples are taken from the box, one at a time, without replacement. When both red apples have been taken, the process stops. The random variable X is the number of apples which have been taken when the process stops. 1 3 ii Draw up the probability distribution table for X . i Show that XP( [3] [3] 3) = = . Another box contains 2 yellow peppers and 5 orange peppers. Three peppers are taken from the box without replacement. iii Given that at least 2 of the peppers taken from the box are orange, find the probability that all 3 peppers are orange. [5] Cambridge International AS & A Level Mathematics 9709 Paper 63 Q7 November 2014 120 15 In a particular discrete probability distribution the random variable X takes the value r r 45 , where r takes all integer values from 1 to 9 inclusive. i Show that XP( 1 15 ii Construct the probability distribution table for X . 40) = = . iii Which is the modal value of X ? iv Find the probability that X lies between 18 and 100. with probability [2] [3] [1] [2] 164 Cambridge International AS & A Level Mathematics 9709 Paper 62 Q5 November 2009 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7 The binomial and geometric distributions 165 In this chapter you will learn how to: ■ use formulae for probabilities for the binomial and geometric distributions, and recognise ■ use formulae for the expectation and variance of the binomial distribution and for the expectation practical situations in which these distributions are suitable models of the geometric distribution. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PREREQUISITE KNOWLEDGE Where it comes from What you should be able to do Check your skills Chapter 4 Calculate expectation in a fixed number of repeated independent trials, given the probability that a particular event occurs. 1 Two ordinary fair dice are rolled 378 times. How many times can we expect the sum of the two numbers rolled to be greater than 8? IGCSE / O Level Mathematics Pure Mathematics 1 Expand products of algebraic expressions. Use the expansion of ( + ) where n is a positive integer. a + b n ab a b + 2 Given that ( 2 b a + , find the four fractions in the 3  and confirm  1 +  4 that their sum is equal to 1. expansion of 3 4 3 a 2 Two special discrete distributions Seen in very simple terms, all experiments have just two possible outcomes: success or failure. A business investment can make a profit or a loss; the defendant in a court case is found innocent or guilty; and a batter in a cricket match is either out or not! In most real-life situations, however, there are many possibilities between success and failure, but taking this yes/no view of the outcomes does allow us to describe certain situations using a mathematical model. 166 Two such situations concern discrete random variables that arise as a result of repeated independent trials, where the probability of success in each trial is constant. of independent trials. • A binomial distribution can be used to model the number of successes in a fixed number • A geometric distribution can be used to model the number of trials up to and including the first success in an infinite number of independent trials. 7.1 The binomial distribution Consider an experiment in which we roll four ordinary fair dice. In each independent trial, we can obtain zero, one, two, three or four 6s . Let the variable R be the number of 6s rolled, then ∈{0, 1, 2, 3, 4} . R To find the probability distribution for R, we must calculate P( possible values. )=R r for all of its Using 6 to represent a success and X to represent a failure in each trial, we have: P(success) P(6) = = 1 6 and P(failure) P(X) = = 5 6 . Calculations to find P( R r are shown in the following table. ) = Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions r Ways to obtain r successes No. ways 0 (XXXX) 1 (6XXX), (X6XX), (XX6X), (XXX6) 2 3 (66XX), (6X6X), (6XX6), (X66X), (X6X6), (XX66) (666X), (66X6), (6X66), (X666) 4 0C = 1 4 1C = 4 4 2C = 6 4 3C = 4 4 4 =C 1 P( )== R r 4 C 0   1 6   0 4   5 6   3 1 1 6     5 6   4 C 1 4 C 2     1 6   4 C 3   1 6   4 C 4   1 6   2 3 4 2 1 0   5 6     5 6     5 6   4 Cr . These probabilities are the terms in the 4 (6666) In the table, we see that P( R r = ) = 1 + binomial expansion of  6 5 6   4 . notation, the five probabilities shown in the previous table are given by  Using the    R r = P( = ) n r   4   discrete random variable that meets the following criteria is said to have a binomial distribution and it is defined by its two parameters, n and p. • There are n repeated independent trials. • n is finite. • There are just two possible outcomes for each trial (i.e. success or failure). • The probability of success in each trial, p, is constant. The random variable is the number of trials that result in a success. A discrete random variable, X , that has a binomial distribution is denoted by X ~ B( , n p . ) KEY POINT 7.1 If ~ B( , n p then the probability of r successes is X ) p r = n r     r p (1 − p ) n r − . TIP  Values of  n r   are the coefficients of the terms in a binomial expansion, and give the number of ways of obtaining r successes in n trials. r p (1 – ) – p n r is the probability for each way of obtaining r successes and ( – ) n r failures. For example, if the variable
X ~ B(3, probabilities. p , then X {0, 1, 2, 3} ) ∈ , and we have the following Copyright Material - Review Only - Not for Redistribution TIP Each way of obtaining a particular number of 6s has the same probability. TIP n is For work involving binomial expansions, the notation Cr rarely used nowadays. Your calculator may use this notation but it has mostly been 167 replaced by   n r   . REWIND We met a series of independent events with just two possible outcomes in the Explore 6.3 activity in Chapter 6, Section 6.3. TIP Coefficients for power 3 are 1, 3, 3 and 1. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 P( X = 0) = P( X = 2) = 3 0   ×   3 2   ×   ( X = 1( X = 3) = 3   × 1   3   × 3   = The coefficients in all binomial expansions are symmetric strings of integers. When arranged in rows, they form what has come to be known as Pascal’s triangle (named after the French thinker Blaise Pascal). Part of this arrangement is shown in the following diagram, which includes the coefficient for power 0 for completeness. 1 2 6 1 3 1 4 11 3 1 4 10 10 15 20 15 power 0 1 2 3 4 5 6 WORKED EXAMPLE 7.1 168 A regular pentagonal spinner is shown. Find the probability that 10 spins produce exactly three As. Answer XP( = 3) =   10 3   × 10! × 7! 3! = = 3 0.4 7 0.6 × 3 0.4 × × 7 0.6 0.215 to 3 significant figures. Let the random variable X be the number of As obtained. We have 10 independent trials with a constant 0.4. P(A) probability of a success, So , and we require ~ B(10, 0.4) three successes and seven failures. X = WORKED EXAMPLE 7.2 ~ B(8, 0.7) , find P( X > 6) , correct to 3 significant figures. Given that X Answer P( X > 6) P( = X = 7) P( + X = 8) X ~ B(8, 0.7) tells us that = = = 8 7 0.7    ×  0.197650 8 1 0.3 +    ×  0.057648 … 8 7 × … + 0.7 8 0 0.3 × q p = = 8, , 0.3 0.7, n = and that . X {0, 1, 2, 3, 4, 5, 6, 7, 8} ∈ 0.255 Copyright Material - Review Only - Not for Redistribution REWIND We saw in Chapter 5, Section 5.3 that n r   =   n C r = ! n !( r n r − . )! A B D C A TIP Remember that X represents the number of successes, so it can take integer values from 0 to n. TIP Premature rounding of probabilities in the working may lead to an incorrect final answer. Here, 0.198 + 0.0576 = 0.256. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions WORKED EXAMPLE 7.3 In a particular country, 85% of the population has rhesus-positive +(R ) blood. Find the probability that fewer than 39 people in a random sample of 40 have rhesus-positive blood. Answer P( X < 39) 1– [P( = X = 39) P( + X = 40)] 40   39 1 –   × 1– 0.010604    [ 0.988 = = = 0.85 39 1 0.15 × + 40 40   ×   0.85 40 0.15 0 ×    … + 0.001502 ] … Let the random variable X be the number in the sample with R+ blood, then X . ~ B(40, 0.85) REWIND Recall from Chapter 4, Section 4.1 that P( A ′ 1 P( − ) A ). = EXPLORE 7.1 Binomial distributions can be investigated using the Binomial Distribution resource on the GeoGebra website. We could, for example, check our answer to Worked example 7.3 as follows. 169 Click on the distribution tab and select binomial from the pop-up menu at the bottomleft. Select the parameters p = probability distribution will be generated. , and a bar chart representing the and 0.85 n = 40 < 39) , enter into the boxes To find XP( 38 ) and, by tapping the chart, P( 0 the value for this probability is displayed. (At the right-hand side you will see a list of the probabilities for the 41 possible values of X in this distribution.) ø øX WORKED EXAMPLE 7.4 Given that ~ B( , 0.4) X n and that XP( = 0) < 0.1 , find the least possible value of n. TIP Answer P ( = X 0 ) = n 0   ×   0 0.4 × n 0.6 n 0.6 = So we need 0.6 n < 0.1 n log 0.6 log 0.1 < log 0.6 log 0.1 < n n > log 0.1 log 0.6 n > 4.50... n = 5 is the least possible value of . n We first express P( in terms of n. X = 0) This leads to an inequality, which we can solve using base 10 logarithms. a> , . We must Recall from IGCSE / O Level that if x then x –< – a reverse the inequality sign when we multiply or divide by a negative number, such as 0.6 log . 10 TIP n takes integer values only. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 3… = 0.6 0.216; 0.6 4 = 5 0.1296; 0.6 = 0.07776 . The least possible value is =n 5. Alternatively, we can solve 0.6 n < 0.1 by trial and improvement. We know that n is an integer, so we 3 1 evaluate 0.6 , 0.6 , 0.6 , 2 … up to the first one whose value is less than 0.1. EXERCISE 7A 1 The variable X has a binomial distribution with n = and 4 p = 0.2 . Find: a P( X = 4) b P( X = 0) c P( X = 3) d P( X = 3 or 4). 2 Given that ~ B(7, 0.6) Y , find: a P( Y = 7) b P( Y = 5) c P( Y ≠ 4) d P(3 < < . 6) Y 3 Given that W ~ B(9, 0.32) , find: a P( 5)W = b P( 5)W ≠ c WP( 2)< d WP(0 < 9) < . 170 4 Given that ~ B 8, V   2 7   , find: a P( V = 4) b ùVP( 7) c VP( 2)ø d P(3 ø , 6) V e P(V is an odd number). 5 Find the probability that each of the following events occur. a Exactly five heads are obtained when a fair coin is tossed nine times. b Exactly two 6s are obtained with 11 rolls of a fair die. 6 A man has five packets and each contains three brown sugar cubes and one white sugar cube. He randomly selects one cube from each packet. Find the probability that he selects exactly one brown sugar cube. 7 A driving test is passed by 70% of people at their first attempt. Find the probability that exactly five out of eight randomly selected people pass at their first attempt. 8 Research shows that the owners of 63% of all saloon cars are male. Find the probability that exactly 20 out of 30 randomly selected saloon cars are owned by: a males b females. 9 In a particular country, 58% of the adult population is married. Find the probability that exactly 12 out of 20 randomly selected adults are married. 10 A footballer has a 95% chance of scoring each penalty kick that she takes. Find the probability that she: a b scores from all of her next 10 penalty kicks fails to score from exactly one of her next seven penalty kicks. 11 On average, 13% of all tomato seeds of a particular variety fail to germinate within 10 days of planting. Find the probability that 34 or 35 out of 40 randomly selected seeds succeed in germinating within 10 days of planting. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions 12 There is a 15% chance of rain on any particular day during the next 14 days. Find the probability that, during the next 14 days, it rains on: a exactly 2 days b at most 2 days. 13 A factory makes electronic circuit boards and, on average, 0.3% of them have a minor fault. Find the probability that a random sample of 200 circuit boards contains: a exactly one with a minor fault b fewer than two with a minor fault. 14 There is a 50% chance that a six-year-old child drops an ice cream that they are eating. Ice creams are given to 5 six-year-old children. a Find the probability that exactly one ice cream is dropped. b 45 six-year-old children are divided into nine groups of five and each child is given an ice cream. Calculate the probability that exactly one of the children in at most one of the groups drops their ice cream. 15 A coin is biased such that heads is three times as likely as tails on each toss. The coin is tossed 12 times. The variables H and T are, respectively, the number of heads and the number of tails obtained. Find the value of P( P( H T = = 7) 7) . 16 Given that ~ B( , 0.3) Q n and that P( Q = 0) 0.1 > , find the greatest possible value of n. 17 The variable ~ B( , 0.96) T n and it is given that P( T n= ) 0.5 > . Find the greatest possible value of n. 18 Given that ~ B( , 0.8) R n and that P( R n> − < 1) 0.006 , find the least possible value of n. M 19 The number of damaged eggs, D, in cartons of six eggs have been recorded by an inspector at a packing depot. The following table shows the frequency distribution of some of the numbers of damaged eggs in 150 000 boxes. 171 No. damaged eggs ( )D 0 1 No. cartons ( )f 141 393 8396 The distribution of D is to be modelled by ~ B(6, D p . ) a Estimate a suitable value for p, correct to 4 decimal places. b Calculate estimates for the value of a and of b. c Calculate an estimate for the least number of additional cartons that would need to be inspected for there to be at least 8400 cartons containing one damaged egg. M 20 The number of months during the 4-month monsoon season (June to September) in which the total rainfall was greater than 5 metres, R, has been recorded at a location in Meghalaya for the past 32 years, and is shown in the following table. No. months No. years ( )f 0 2 1 8 2 12 3 8 4 2 The distribution of R is to be modelled by ~ B(4, R p . ) a Find the value of p, and state clearly what this val
ue represents. b Give a reason why, in real life, it is unlikely that a binomial distribution could be used to model these data accurately. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 21 In a particular country, 90% of both females and males drink tea. Of those who drink tea, 40% of the females and 60% of the males drink it with sugar. Find the probability that in a random selection of two females and two males: a all four people drink tea b an equal number of females and males drink tea with sugar. PS 22 It is estimated that 0.5% of all left-handed people and 0.4% of all right-handed people suffer from some form of colour-blindness. A random sample of 200 left-handed and 300 right-handed people is taken. Find the probability that there is exactly one person in the sample that suffers from colour-blindness. DID YOU KNOW? Although Pascal’s triangle is named after the 17th century French thinker Blaise Pascal, it was known about in China and in Persia as early as the 11th century. The earliest surviving display is of Jia Xian’s triangle in a work compiled in 1261 by Yang Hui, as shown in the photo. 172 EXPLORE 7.2 A frog sits on the bottom-left square of a 5 by 5 grid. In each of the other 24 squares there is a lily pad and four of these have pink flowers growing from them, as shown in the image. 1 The frog can jump onto an adjacent lily pad but it can only jump northwards (N) or eastwards (E). The four numbers on the grid represent the number of different routes the frog can take to get to those particular lily pads. For example, there are three routes to the lily pad with the number 3, and these routes are EEN, ENE and NEE . 3 5 1 N Sketch a 5 by 5 grid and write onto it the number of routes to all 24 lily pads. Describe any patterns that you find in the numbers on your grid. The numbers on the lily pads with pink flowers form a sequence. Can you continue this sequence and find an expression for its nth term? Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions Expectation and variance of the binomial distribution Expectation and standard deviation give a measure of central tendency and a measure of variation for the binomial distribution. We can calculate these, along with the variance, from the parameters n and p. Consider the variable following table. X ~ B(2, 0.6) , whose probability distribution is shown in the REWIND We saw in Chapter 6, Section 6.3 that expectation is a variable’s long-term average value.16 1 0.48 2 0.36 Applying the formulae for E( )X and Var X( ) gives the following results. E( X ) = Σ xp = (0 × 0.16) + (1 0.48) × + (2 × 0.36) 1.2 = 2 x p ) – {E( Var( X = Σ Our experiment consists of should not be surprised to find that E( = × n = trials with a probability of success 2 = × 0.6 1.2 = np.16) 0.48) (0 (2 X 0.36) – 1.2 p = 2 )} 2 2 (1 2 2 = 0.6 0.48 in each, so we REWIND We saw in Chapter 4, Section 4.1 that event A is expected to occur times. ) P( A× n What may be surprising (and a very convenient result), is that the variance of X also can be found from the values of the parameters n and p. Var( X ) = np (1 – ) p 2 = × 0.6 × 0.4 = 0.48 KEY POINT 7.2 TIP The mean and variance of X ~ B( , n p ) are given by npµ= and 2 σ = np (1 – ) p = npq . WORKED EXAMPLE 7.5 Given that X ~ B(12, 0.3) , find the mean, the variance and the standard deviation of X . Answer Note that and 2 σ = µ= Var( )X E( . ) X 173 E( X ) np= 12 = × 3.6 0.3 = ) X Var( SD( X ) = = = = = = np 12 (1 – ) p 0.3 × × 2.52 0.7 TIP np (1 − p ) 2.52 1.59 to 3 significant figures Copyright Material - Review Only - Not for Redistribution We can also write our answers as 2 3.6, σ µ = . 1.59 σ = 2.52 = and Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 7.6 The random variable X ~ B( , n p . Given that E( ) a the value of n and of p b P( X = . 11) Answer a q = p = 7.5 12 1 – = 0.625 q = 0.375 n = 12 0.375 = 32 X = ) 12 and Var( ) X = 7.5 , find: We use q = npq np = ) Var( X ) E( X to find p. E( ) =X np, so =n ) E( X p 0.375 11 × 0.625 21 X ~ B(32, 0.375) 32 11    ×  0.138 b P( X = 11) = = EXERCISE 7B 1 Calculate the expectation, variance and standard deviation of each of the following discrete random 174 variables. Give non-exact answers correct to 3 significant figures. a V ~ B(5, 0.2) b W ~ B(24, 0.55) c X ~ B(365, 0.18) d Y ~ B(20, 0.5 ) 2 Given that X ~ B(8,0.25), calculate: a XE( ) and XVar( ) b P[ =X E( X )] c P[ <X E( X )]. 3 Given that Y ~ B(11, 0.23), calculate: a ≠YP( 3) b P[ <Y E( Y )]. 4 Given that X ~ B( , n p ), E( X ) = 20 and =XVar( ) 12, find: a the value of n and of p b =XP( 21). 5 Given that G ~ B( , n p ), E( G ) = 24 1 2 and GVar( ) 10 5 = 24 , find: a the parameters of the distribution of G b =GP( 20). 6 W has a binomial distribution, where =WE( ) 2.7 and =WVar( ) 0.27. Find the values of n and p and use them to draw up the probability distribution table for W . 7 Give a reason why a binomial distribution would not be a suitable model for the distribution of X in each of the following situations. a X is the height of the tallest person selected when three people are randomly chosen from a group of 10. b X is the number of girls selected when two children are chosen at random from a group containing one girl and three boys. c X is the number of motorbikes selected when four vehicles are randomly picked from a car park containing 134 cars, 17 buses and nine bicycles. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions , and its standard deviation is one-third of its mean. Calculate the non-zero value of   8 The variable Q n and find P(5 <  ~ B , n  <Q 1 3 8). 9 The random variable H ~ B(192, p p and find the value of k, given that ), and HE( 2) H P( = ) is 24 times the standard deviation of H . Calculate the value of k = × 2–379. 10 It is estimated that 1.3% of the matches produced at a factory are damaged in some way. A household box contains 462 matches. a Calculate the expected number of damaged matches in a household box. b Find the variance of the number of damaged matches and the variance of the number of undamaged matches in a household box. c Show that approximately 10.4% of the household boxes are expected to contain exactly eight damaged matches. d Calculate the probability that at least one from a sample of two household boxes contains exactly eight damaged matches. 11 On average, 8% of the candidates sitting an examination are awarded a merit. Groups of 50 candidates are selected at random. a How many candidates in each group are not expected to be awarded a merit? b Calculate the variance of the number of merits in the groups of 50. c Find the probability that: i ii three, four or five candidates in a group of 50 are awarded merits 175 three, four or five candidates in both of two groups of 50 are awarded merits. 7.2 The geometric distribution Consider a situation in which we are attempting to roll a 6 with an ordinary fair die. How likely are we to get our first 6 on the first roll; on the second roll; on the third roll, and so on? We can answer these questions using the constant probabilities of success and failure: p and 1 – p. P(first 6 on first roll) = → success. a p P(first 6 on second roll) = (1 – ) p p → failure followed by a success. a P(first 6 on third roll) = (1 – ) 2 →p p two failures followed by a success. The distribution of X , the number of trials up to and including the first success in a series of repeated independent trials, is a discrete random variable whose distribution is called a geometric distribution. The following table shows the probability that the first success occurs on the rth trial. r P( )== X r 1 p 2 3 4 (1 – ) p p (1 – )2 p p (1 – )3 p p .... .... n (1 – ) p 1 p n− .... .... Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 P( =X r The values of with first term p and common ratio to infinity of the GP. ) in the previous table are the terms of a geometric progression (GP) p1 – . The sum of the probabilities is equal to the sum ∑ [P( X r = )] = S ∞ = first term 1 common ratio − = p 1 (1 − − p ) = 1 . The sum of the probabilities in a geometric probability distribution is equal to 1. A discrete random variable, X , is said to have a geometric distribution, and is defined by its parameter p, if it meets the following criteria. • The repeated trials are independent. • The repeated tria
ls can be infinite in number. • There are just two possible outcomes for each trial (i.e. success or failure). • The probability of success in each trial, p, is constant. KEY POINT 7.3 A random variable X that has a geometric distribution is denoted by probability that the first success occurs on the rth trial is X ~ Geo( p , and the ) =p r (1 – ) –1 p p r for = r 1, 2, 3, … 176 The binomial and geometric distributions arise in very similar situations. The significant difference is that the number of trials in a binomial distribution is fixed from the start and the number of successes are counted, whereas, in a geometric distribution, trials are repeated as many times as necessary until the first success occurs. For X ~ B( , n p  ), there are  n r   ways to obtain r successes. ~ Geo( For X is when there are r – 1 failures followed by a success. p ), there is only one way to obtain the first success on the rth trial, and that REWIND We saw in Chapter 6, Section 6.2 that 1 Σ =p for a probability distribution. You will also have seen geometric progressions and geometric series in Pure Mathematics 1, Chapter 6. TIP r q –1 = , p X r = An alternative form of this formula, P( ) × where p = 1 − q, reminds us that the – 1r failures occur before the first success. REWIND Recall from Section 7.1 that     = n C !( )! WORKED EXAMPLE 7.7 Repeated independent trials are carried out in which the probability of success in each trial is 0.66. Correct to 3 significant figures, find the probability that the first success occurs: a on the third trial b on or before the second trial c after the third trial. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions X = 2) Let X represent the number of trials up to and including the first success, then X ~ Geo(0.66), 0.66 and where =p =p1 – 0.34. Answer a P( X = 3) = (.66 0.34 × 0.0763 b øP( X 2) P( = X 1) P( + = (1 – ) p p p + 0.884 = = c P( X > 3) 1 – P( = X ø 3) = = = 1 – [P( X = 1) P( + X = X = 3)] 2) P( + 2 (1 – ) ] p p (.0393 Probabilities that involve inequalities can be found by summation for small values of r, as in parts b and c of Worked example 7.7. However, for larger values of r, the following results will be useful. P( X rø = ) P(success on one of the first r trials) = 1 – P(failure on the first r trials) P( X r> ) P(first success after the rth trial) = P(failure on the first r trials) = These two results are written in terms of q in Key point 7.4. WORKED EXAMPLE 7.8 KEY POINT 7.4 ~ Geo( p ) p, then When and X 1 –=q =<P( X r • ) 1 – • P( )> X r = qr qr 177 In a particular country, 18% of adults wear contact lenses. Adults are randomly selected and interviewed one at a time. Find the probability that the first adult who wears contact lenses is: a one of the first 15 interviewed b not one of the first nine interviewed. Answer a øP( X 15) 1 – = q 15 = − 1 0.82 15 = 0.949 b P( X > 9) = 9 q 0.82 9 0.168 = = Let X represent the number of adults interviewed up to and including the first one who wears contact lenses, then X ~ Geo(0.18) and = 1 – 0.18 0.82. = q Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 7.9 A coin is biased such that the probability of obtaining heads with each toss is equal to . The coin is tossed until the first head is obtained. Find the probability that the 5 11 coin is tossed: a at least six times b fewer than eight times. Answer a ùP( X 6) P( = X > 5) 5 q = 5 6  =    11 0.0483 = b P( X < 8) P( = X ø 7 11 = −  1  0.986 = 178 EXERCISE 7C Let X represent the number of times the coin is tossed up to and including the first heads, then 5   11 and   X ~ Geo 6 . 11 =q TIP ‘At least six times’ has the same meaning as ‘more than five times’. TIP ‘Fewer than eight times’ has the same meaning as ‘seven or fewer times’. 1 Given the discrete random variable X ~ Geo(0.2), find: a =XP( 7) b P( X ≠ 5) c >XP( 4) . 2 Given that T ~ Geo(0.32), find: a =TP( 3) b øTP( 6) c >TP( 7) . 3 The probability that Mike is shown a yellow card in any football match that he plays is probability that Mike is next shown a yellow card: 1 2 . Find the a in the third match that he plays b before the fourth match that he plays. 4 On average, Diya concedes one penalty in every six hockey matches that she plays. Find the probability that Diya next concedes a penalty: a in the eighth match that she plays b after the fourth match that she plays. 5 The sides of a fair 5-sided spinner are marked 1, 1, 2, 3 and 4. It is spun until the first score of 1 is obtained. Find the probability that it is spun: a exactly twice b at most five times c at least eight times. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions 6 It is known that 80% of the customers at a DIY store own a discount card. Customers queuing at a checkout are asked if they own a discount card. a Find the probability that the first customer who owns a discount card is: i the third customer asked ii not one of the first four customers asked. b Given that 10% of the customers with discount cards forget to bring them to the store, find the probability that the first customer who owns a discount card and remembered to bring it to the store is the second customer asked. 7 In a manufacturing process, the probability that an item is faulty is 0.07. Items from those produced are selected at random and tested. a Find the probability that the first faulty item is: i the 12th item tested ii not one of the first 10 items tested iii one of the first eight items tested. b What assumptions have you made about the occurrence of faults in the items so that you can calculate the probabilities in part a? 8 Two independent random variables are X ~ Geo(0.3) and Y ~ Geo(0.7). Find: a =XP( 2) b =YP( 2) c P( X = 1 and Y = 1) . 9 On average, 14% of the vehicles being driven along a stretch of road are heavy goods vehicles (HGVs). A girl stands on a footbridge above the road and counts the number of vehicles, up to and including the first HGV that passes. Find the probability that she counts: a at most three vehicles b at least five vehicles. 179 10 The probability that a woman can connect to her home Wi-Fi at each attempt is 0.44. Find the probability that she fails to connect until her fifth attempt. 11 Decide whether or not it would be appropriate to model the distribution of X by a geometric distribution in the following situations. In those cases for which it is not appropriate, give a reason. a A bag contains two red sweets and many more green sweets. A child selects a sweet at random and eats it, selects another and eats it, and so on. X is the number of sweets selected and eaten, up to and including the first red sweet. b A monkey sits in front of a laptop with a blank word processing document on its screen. X is the number of keys pressed by the monkey, up to and including the first key pressed that completes a row of three letters that form a meaningful three-letter word. c X is the number of times that a grain of rice is dropped from a height of 2 metres onto a chessboard, up to and including the first time that it comes to rest on a white square. d X is the number of races in which an athlete competes during a year, up to and including the first race that he wins. 12 The random variable T has a geometric distribution and it is given that P( P( T T = = 2) 5) = 15.625. Find =TP( 3). PS PS 13 X ~ Geo( p ) and XP( = 2) = 0.2464. Given that <p 0.5, find >XP( 3). 14 Given that X ~ Geo( p ) and that øXP( 4) = 2385 2401 , find P(1 ø X 4)< . Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PS 15 Two ordinary fair dice are rolled simultaneously. Find the probability of obtaining: a b the first double on the fourth roll the first pair of numbers with a sum of more than 10 before the 10th roll. PS 16 X ~ Geo(0.24) and Y ~ Geo(0.25) are two independent random variables. Find the probability that Mode of the geometric distribution All geometric distributions have two features in common. These are clear to see when bar charts or vertical line graphs are used to represent values of ) for different values of the parameter p. You can do this manually or using a graphing tool such as GeoGebra. =X r P( X Y 4. + = =XP( The first common feature is that 1) has the greatest probability in all geometric distributions. This means that the most likely value of X is 1, so the first success is most likely to occur on the first trial. Secondly, the value of =X r ( This is because the common ratio between the probabilities q (1 – ) p p p ) decreases as r increases. = is less than 1: (1 – ) p p (1 – ) p p (( 3 2 4 . > … > > > > The following table shows some probabilities for the distr
ibutions X ~ Geo(0.2) and X ~ Geo(0.7). In both distributions, we can see that probabilities decrease as the value of X increases. P( X == 1) P( X == 2) P( X == 3) P( X == 4) P( X == 5) Geo(0.2) 180 Geo(0.7) 0.2 0.7 0.16 0.21 0.128 0.063 0.1024 0.0189 0.08192 0.00567 Expectation of the geometric distribution Recall that the expectation or mean of a discrete random variable is its long-term average, which is given by E( X ) . µ= = Σ xpx Applying this to the geometric distribution Geo(p), it turns out we find that the mean is equal to 1 p , the reciprocal of p. EXPLORE 7.3 KEY POINT 7.5 The mode of all geometric distributions is 1. KEY POINT 7.6 p then ) If X ~ Geo( 1 p µ = . Using algebra, we can prove that the mean of the geometric distribution is equal to For X ~ Geo(p), we have X {1, 2, 3, 4,...} and ∈ =p x { , , p pq pq 2 , 3 pq ,...} . Step 1 of the proof is to form an equation that expresses µ in terms of p and q. To do this we use µ = Σ xpx . 1 p . REWIND We studied the expectation of a discrete random variable in Chapter 6, Section 6.3. There are three more steps required to complete the proof, which you might like to try without any further assistance. However, some guidance is given below if needed. Step 2: Multiply the equation obtained in step 1 throughout by q to obtain a second equation. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions Step 3: Subtract one equation from the other. Step 4: If you have successfully managed steps 1, 2 and 3, you should need no help completing the proof! WORKED EXAMPLE 7.10 One in four boxes of Zingo breakfast cereal contains a free toy. Let the random variable X be the number of boxes that a child opens, up to and including the one in which they find their first toy. a Find the mode and the expectation of X . b Interpret the two values found in part a in the context of this question. Answer a The mode of X is 1 . E child is most likely to find their first toy in the first box they open but, on average, a child will find their first toy in the fourth box that they open. WORKED EXAMPLE 7.11 The variable is X ~ Geo   1 4   . TIP An answer written ‘in context’ must refer to a specific situation; in this case, the situation described in the question. 181 The variable X follows a geometric distribution. Given that X = E( ) 3 1 2 , find >XP( 6). Answer E( X ) = 1 p = 7 2 , so p = 2 7 q = 1 – P( X > 2 7 6.133 = We find the parameter p and then we find q. We use P( X r > ) = qr from Key point 7.4. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 7.12 Given that X ~ Geo( p ) and that øXP( 3) = 819 1331 , find: a >XP( 3) b P(1 < ø X 3) . Answer a P( X > 3) 1 P( = − ø 3) X 819 1331 1 = − = 512 1331 b 1 – q 3 = P( X ø 3 11 819 1331 819 1331 − and p = 3 11 r 1 – We use q find q and p. P( < X r ) to = TIP 182 P(1 ø< X 3) P( = X = 2) P( + X = 3) = = pq + 456 1331 2 pq or 0.343 = = = Alternatively, we can use < < P(1 3) X < − 3 11 P( X 819 1331 456 1331 3) P( − X = 1) EXERCISE 7D 1 Given that X ~ Geo(0.36), find the exact value of XE( ). 2 The random variable Y follows a geometric distribution. Given that YP( = 1) = 0.2, find YE( ). 3 Given that S ~ Geo( p ) and that SE( ) 4 1 2= , find =SP( 2). 4 Let T be the number of times that a fair coin is tossed, up to and including the toss on which the first tail is obtained. Find the mode and the mean of T . 5 Let X be the number of times an ordinary fair die is rolled, up to and including the roll on which the first 6 is obtained. Find XE( ) and evaluate P[ >X E( X )]. 6 A biased 4-sided die is numbered 1, 3, 5 and 7. The probability of obtaining each score is proportional to that score. a Find the expected number of times that the die will be rolled, up to and including the roll on which the first non-prime number is obtained. b Find the probability that the first prime number is obtained on the third roll of the die. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions 7 Sylvie and Thierry are members of a choir. The probabilities that they can sing a perfect high C note on each attempt are and , respectively. 4 7 5 8 a Who is expected to fail fewer times before singing a high C note for the first time? b Find the probability that both Sylvie and Thierry succeed in singing a high C note on their second attempts. PS 8 A standard deck of 52 playing cards has an equal number of hearts, spades, clubs and diamonds. A deck is shuffled and a card is randomly selected. Let X be the number of cards selected, up to and including the first diamond. a Given that X follows a geometric distribution, describe the way in which the cards are selected, and give the reason for your answer. b Find the probability that: i X is equal to XE( ) ii neither of the first two cards selected is a heart and the first diamond is the third card selected. PS 9 A study reports that a particular gene in 0.2% of all people is defective. X is the number of randomly selected people, up to and including the first person that has this defective gene. Given that X b >øP( XE( ) and find the smallest possible value of b. ) 0.865, find PS 10 Anouar and Zane play a game in which they take turns at tossing a fair coin. The first person to toss heads is the winner. Anouar tosses the coin first, and the probability that he wins the game is 1 0.5 5 0.5 0.5 0. Describe the sequence of results represented by the value 0.55 in this series. b Find, in a similar form, the probability that Zane wins the game. c Find the probability that Anouar wins the game. EXPLORE 7.4 In a game for two people that cannot be drawn, you are the stronger player with a 60% chance of winning each game. The probability distributions for the number of games won by you and those won by your opponent when a single game is played, X and Y , are shown. x P( X x== ) 0 0.40 1 0.60 y 0 P( = ) Y y 0.60 1 0.40 Investigate the probability distributions for X and Y in a best-of-three contest, where the first player to win two games wins the contest. Who gains the advantage as the number of games played in a contest increases? What evidence do you have to support your answer? PS How likely are you to win a best-of-five contest? Copyright Material - Review Only - Not for Redistribution 183 Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Checklist of learning and understanding ● A binomial distribution can be used to model the number of successes in a series of n repeated independent trials where the probability of success on each trial, p, is constant. If X ~ B( , n p ) then p r = E( X ) µ= = np n r     r p (1 – ) – . n r p Var( X ) 2 σ= = np (1 – ) p = npq , where geometric distribution can be used to model the number of trials up to and including the first success in a series of repeated independent trials where the probability of success on each trial, p, is constant. ● ● If X ~ Geo( XE( ) µ= = ) then p 1 p =p r (1 – ) –1 for r = p p r 1, 2, 3, … ● P X r =ø( ) 1 – qr and P( X r > ) = qr , where =q p1 – . ● The mode of all geometric distributions is 1. 184 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions END-OF-CHAPTER REVIEW EXERCISE 7 1 Given that , ~ X B n   1 n   find an expression for =XP( 1) in terms of n. 2 A family has booked a long holiday in Skragness, where the probability of rain on any particular day is 0.3. Find the probability that: a b the first day of rain is on the third day of their holiday it does not rain for the first 2 weeks of their holiday. [2] [1] [2] 3 One plastic robot is given away free inside each packet of a certain brand of biscuits. There are four colours of plastic robot (red, yellow, blue and green) and each colour is equally likely to occur. Nick buys some packets of these biscuits. Find the probability that i he gets a green robot on opening his first packet, ii he gets his first green robot on opening his fifth packet. Nick’s friend Amos is also collecting robots. iii Find the probability that the first four packets Amos opens all contain different coloured robots. [1] [2] [3] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q3 November 2015 4 Weiqi has two fair triangular spinners. The sides of one spinner are labelled 1, 2, 3, and the sides of the other are labelled 2, 3, 4. Weiqi spins them simultaneously and notes the two numbers on which they come to rest. a Find the probability that these two numbers
differ by 1. b Weiqi spins both spinners simultaneously on 15 occasions. Find the probability that the numbers on which they come to rest do not differ by 1 on exactly eight or nine of the 15 occasions. 5 A computer generates random numbers using any of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The numbers appear on the screen in blocks of five digits, such as 50119 26317 40068 ....... Find the probability that: a there are no 7s in the first block b the first zero appears in the first block c the first 9 appears in the second block. 6 Four ordinary fair dice are rolled. a In how many ways can the four numbers obtained have a sum of 22? b Find the probability that the four numbers obtained have a sum of 22. c The four dice are rolled on eight occasions. Find the probability that the four numbers obtained have a sum of 22 on at least two of these occasions. 185 [2] [3] [1] [1] [2] [2] [2] [3] 7 When a certain driver parks their car in the evenings, they are equally likely to remember or to forget to switch off the headlights. Giving your answers in their simplest index form, find the probability that on the next 16 occasions that they park their car in the evening, they forget to switch off the headlights: a 14 more times than they remember to switch them off b at least 12 more times than they remember to switch them off. [2] [3] M 8 Gina has been observing students at a university. Her data indicate that 60% of the males and 70% of the females are wearing earphones at any given time. She decides to interview randomly selected students and to interview males and females alternately. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 a Use Gina’s observation data to find the probability that the first person not wearing earphones is the third male interviewed, given that she first interviews: i a male ii a female iii a male who is wearing earphones. b State any assumptions made about the wearing of earphones in your calculations for part a. 9 In Restaurant Bijoux 13% of customers rated the food as ‘poor’, 22% of customers rated the food as ‘satisfactory’ and 65% rated it as ‘good’. A random sample of 12 customers who went for a meal at Restaurant Bijoux was taken. i Find the probability that more than 2 and fewer than 12 of them rated the food as ‘good’. On a separate occasion, a random sample of n customers who went for a meal at the restaurant was taken. ii Find the smallest value of n for which the probability that at least 1 person will rate the food as ‘poor’ is greater than 0.95. [2] [2] [2] [1] [3] [3] 10 A biased coin is four times as likely to land heads up compared with tails up. The coin is tossed k times so that Cambridge International AS & A Level Mathematics 9709 Paper 62 Q3 June 2012 the probability that it lands tails up on at least one occasion is greater than 99%. Find the least possible value of k. 186 PS 11 Given that X 1) k = × . find the smallest value of n for which k 25> ~ B( , 0.4) and that X P( = n P( X n = – 1) , express the constant k in terms of n, and PS 12 A book publisher has noted that, on average, one page in eight contains at least one spelling error, one page in five contains at least one punctuation error, and that these errors occur independently and at random. The publisher checks 480 randomly selected pages from various books for errors. a How many pages are expected to contain at least one of both types of error? b Find the probability that: i ii the first spelling error occurs after the 10th page the first punctuation error occurs before the 10th page iii the 10th page is the first to contain both types of error. 13 Robert uses his calculator to generate 5 random integers between 1 and 9 inclusive. i Find the probability that at least 2 of the 5 integers are less than or equal to 4. Robert now generates n random integers between 1 and 9 inclusive. The random variable X is the number of these n integers which are less than or equal to a certain integer k between 1 and 9 inclusive. It is given that the mean of X is 96 and the variance of X is 32. ii Find the values of n and k. [4] [5] [2] [2] [2] [2] [3] [4] PS 14 Anna, Bel and Chai take turns, in that order, at rolling an ordinary fair die. The first person to roll a 6 wins the game. Cambridge International AS & A Level Mathematics 9709 Paper 62 Q4 June 2013 Find the ratio P(Anna wins) : P(Bel wins) : P(Chai wins), giving your answer in its simplest form. [7] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8 The normal distribution In this chapter you will learn how to: 187 distribution tables ■ sketch normal curves to illustrate distributions or probabilities ■ use a normal distribution to model a continuous random variable and use normal ■ solve problems concerning a normally distributed variable ■ recognise conditions under which the normal distribution can be used as an approximation to the binomial distribution, and use this approximation, with a continuity correction, in solving problems. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PREREQUISITE KNOWLEDGE Where it comes from What you should be able to do Check your skills Chapter 7 Find and calculate with the expectation and variance of a binomial distribution. 1 Given XVar( X ~ B 45, 0.52 , find XE( ). ( ) ) and 2 Given that X follows a binomial distribution with XE( Var( the distribution of X . 7.28 X = ) ) 11.2 , find the parameters of and = Why are errors quite normal? If you study any of the sciences, you will be required at some time to measure a quantity as part of an experiment. That quantity could be a measurement of time, mass, distance, volume and so on. Whatever it is, any measurement you make of a continuous quantity such as these will be subject to error. The very nature of continuous quantities means that they cannot be measured precisely and, no matter how hard we try, inaccuracy is also likely because our tools lack perfect calibration and we, as human beings, add in a certain amount of unreliability. 188 However, small errors are more likely than large errors and our measurements are usually just as likely to be underestimates as overestimates. When repeated measurements are taken, errors are likely to cancel each other out, so the average error is close to zero and the average of the measurements is virtually error-free. This chapter serves as an introduction to the idea of a continuous random variable and the method used to display its probability distribution. We will later focus our attention on one particular type of continuous random variable, namely a normal random variable. The normal distribution was discovered in the late 18th century by the German mathematician Carl Friedrich Gauss through research into the measurement errors made in astronomical observations. Some key properties of the normal distribution are that values close to the average are most likely; the further values are from the average, the less likely they are to occur, and the distribution is symmetrical about the average. 8.1 Continuous random variables A continuous random variable is a quantity that is liable to change and whose infinite number of possible values are the numerical outcomes of a random phenomenon. Examples include the amount of sugar in an orange, the time required to run a marathon, measurements of height and temperature and so on. A continuous random variable is not defined for specific values. Instead, it is defined over an interval of values. Consider the mass of an apple, denoted by X grams. Within the range of possible masses, X can take any value, such as 111.2233…, or 137.8642…, or 145.2897…, or …. The probability that X takes a particular value is necessarily equal to 0, since the number of values that it can take is infinite. However, there will be a countable number of values in any chosen interval, such as < <X be found. 140, so a probability for each and every interval can 130 The probability distribution of a discrete random variable shows its specific values and their probabilities, as we saw in Chapters 6 and 7. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy The probability distribution of a continuous random variable shows its range of values and the probabilities for intervals within that range. Chapter 8: The normal distribution ● When X is a discrete random variable, we can represent X r ● When X is a continuous random variable, we can represent P( )= . ø ø a X b . ) P( Before looking at probability distributions for continuous random variables in detail, we will consider how we can represent the probability distribution of a set of collected or observed continuous data. Representation of a probability distribution A set of continuous data can be illustrated
in a histogram, where column areas are proportional to frequencies. To illustrate the probability distribution of a set of data, we draw a graph that is based on the shape of a histogram, as we now describe. (relative frequency density If we change the frequency density values on the vertical axis to relative frequency density relative frequency values will represent relative frequencies, which are estimates of probabilities. The vertical axis of the diagram can now be labeled ‘probability density’. class width) then column areas = ÷ For equal-width class intervals, the process described above has no effect on the ‘shape’ of fΣ ’ to 1, which the diagram. The result is that the total area of the columns changes from ‘ is the sum of the probabilities of all the possible values. So we can draw a curved graph over the columns of an equal-width interval histogram (preferably one displaying large amounts of data with many classes) to model the probability distribution of a set of continuous data. In the case of a random variable, such a curved graph represents a function, = f( and is called a probability density function, abbreviated to PDF or pdf. The area under the graph of the PDF is also equal to 1. x , ) y A curved graph is sketched over each of the histograms in the diagram below area under curve = 1 y = f(x area under curve = 1 y = f(x) x If you were asked to describe these two curves, you may be tempted to say that the curve on the right is ‘a bit odd’ and that the curve on the left is ‘a bit more normal’… and you would be quite right in doing so, as you will see shortly. Copyright Material - Review Only - Not for Redistribution REWIND We saw how to display continuous data in a histogram in Chapter 1, Section 1.3. 189 TIP The word function should only be used when referring to a random variable. For data, we should rather use curve and/or graph. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy TIP The mode is located at the graph’s peak. The median is at the value where the area under the graph is divided into two equal parts; this value can be found by calculation from the histogram or estimated from a cumulative frequency graph. Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Three commonly occurring types of curved graph are shown in the diagrams below. negatively skewed symmetric positively skewed longer tail to the left even tails longer tail to the right EXPLORE 8.1 Three frequency distributions are shown in the tables below. Use a histogram to sketch a graph representing the probability distribution for each of w, x and y. w f x f y f w3 ø < 6 6 9<ø w 9 12<ø w 12 15<ø w 15 18<ø w 18 21<ø w 21 24<ø w 13 13 13 13 13 13 13 3 6<ø x 6 9<ø x 9 12<ø x 12 15<ø x 15 18<ø x 18 21<ø x 21 24<ø x 3 9 18 24 18 9 3 3 6<ø y 6 9<ø y 9 12<ø y 12 15<ø y 15 18<ø y 18 21<ø y 21 24<ø y 8 19 10 4 10 19 8 Discuss and describe the shapes of the three graphs. What feature do they have in common? Compare the measures of central tendency (averages) for w, x and y. 190 The normal curve The frequency distribution of x in the Explore 8.1 activity produces a special type of curved graph. It is a symmetric, bell-shaped curve, known as a normal curve. If a probability distribution is represented by a normal curve, then: ● Mean median mode = = ● The peak of the curve is at the mean ( )µ , and this is where we find the curve’s line of symmetry ● Probability density decreases as we move away from the mean on both sides, so the further the values are from the mean, the less likely they are to occur ● An increase in the standard deviation σ( ) means that values become more spread out from the mean. This results in the curve’s width increasing and its height decreasing, so that the area under the graph is kept at a constant value of 1. Graphs that represent probability distributions of related sets of data, such as the heights of the boys and the heights of the girls at your school, can be represented on the same diagram, so that comparisons can be made. The following diagram shows two pairs of normal curves with their means and standard deviations compared. Note that the areas under the graphs in each pair are equal. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution As we can see, A and B have the same mean, but the shapes of the normal curves are different because they do not have the same standard deviation. Curve B is obtained from curve A by stretching it both vertically (from the horizontal axis) and horizontally (from the line of symmetry). X and Y have identically-shaped normal curves because they have the same standard deviation, but their positions or locations are different because they have different means. Each curve can be obtained from the other by a horizontal translation. EXPLORE 8.2 191 You can investigate the effect of altering the mean and/or standard deviation on the location and shape of a normal curve by visiting the Density Curve of Normal Distribution resource on the GeoGebra website. Note that the area under the curve is always equal to 1, whatever the values of µ and σ. EXERCISE 8A 1 The probability distributions for A and B are represented in the diagram. Indicate whether each of the following statements is true or false. a µ µ>Α Β b σ σ<Α B c A and B have the same range of values. d σ 2 2 σ=Α Β e At least half of the values in B are greater than µΑ. f At most half of the values in A are less than µΒ Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 2 The diagram shows normal curves for the probability distributions of P and Q, that each contain n values. a Write down a statement comparing: i σP and σQ ii iii the median value for P and the median value for Q the interquartile range for P and the interquartile range for Q The datasets P and Q are merged to form a new dataset denoted by W . i Describe the range of W . ii Is the probability distribution for W a normal curve? Explain your answer. iii Copy the diagram above and sketch onto it a curved graph representing the probability distribution for W . Mark the relative positions of µP, µQ and µW along the horizontal axis. 192 3 The distributions of the heights of 1000 women and of 1000 men both produce normal curves, as shown. The mean height of the women is 160 cm and the mean height of the men is 180 cm. The heights of these women and men are now combined to form a new set of data. Assuming that the combined heights also produce a normal curve, copy the graph opposite and sketch onto it the curve for the combined heights of the 2000 women and men. 4 Probability distributions for the quantity of apple juice in 500 apple juice tins and for the quantity of peach juice in 500 peach juice tins are both represented by normal curves. The mean quantity of apple juice is 340 ml with variance 4 ml2, and the mean quantity of peach juice is 340 ml with standard deviation 4 ml. a Copy the diagram and sketch onto it the normal curve for the quantity of peach juice in the peach juice tins. b Describe the curves’ differences and similarities women men 160 180 Height (cm) apple juice 340 Volume (ml) 5 The masses of 444 newborn babies in the USA and 888 newborn babies in the UK both produce normal curves. For the USA babies, µ = 3.4 kg and σ = 200 g; for the UK babies, µ = 3.3 kg and σ = 36100 g 2. 2 a On a single diagram, sketch and label these two normal curves. b Describe the curves’ differences and similarities. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution 6 The values in two datasets, whose probability distributions are both normal curves, are summarised by the following totals: 2 =x Σ 2 =y Σ 35 000 , Σ =x 12 000 and n = 5000 . 72 000 , Σ =y 26 000 and =n 10 000. a Show that the centre of the curve for y is located to the right of the curve for x. b On the same diagram, sketch a normal curve for each dataset. 8.2 The normal distribution In Section 8.1, we saw how a curved graph can be used to represent the probability distribution of a set of continuous data. A curved graph that represents the probability distribution of a continuous random variable, as stated previously, is called a probability density function or PDF. If we collect data on, say, the masses of a randomly selected sample of 1000 pineapples, we can produce a curved graph to illustrate the probabilities for the full and limited range of these masses. If there are no pineapples with masses under 0.2 kg or over 6 kg, then our 0 and graph will indicate that P(mass P(mass 0.2) 6) 0. > = = < However, the continuous random variable ‘the possible mass of a pineapple’ is a theoretical model for the probability distribution. In the model, masses of less than 0.2 kg and masses of more than 6 kg would be shown to be extremely unlikely, but not impossible. 0 and The continuous random var
iable would, therefore, indicate that P(mass P(mass 0.2) 6) 0. < > > > [Incidentally, the greatest ever recorded mass of a pineapple is 8.28 kg!] The probability distribution of a continuous random variable is a mathematical function that provides a method of determining probabilities for the occurrence of different outcomes or observations. If the random variable X is normally distributed with mean µ and variance σ2, then its equation is TIP 193 f( x ) = 1 2 π σ exp { − ( 2 x − ) µ 2 2 σ , for all real values of x. } The parameters that define a normally distributed random variable are its mean µ and its variance σ2. To describe the normally distributed random variable X , we write X ~ N , µ σ ) ( 2 . KEY POINT 8.1 X ~ N , ( 2µ σ ) describes a normally distributed random variable. We read this as ‘X has a normal distribution with mean µ and variance σ2’ The probability that X takes a value between a and b is equal to the area under the curve between the x-axis and the boundary lines =x a and =x b. b The area under the graph of = f( y x can be found by integration: ) P( ø ø a X b ) f( x ) d x ∫= a Copyright Material - Review Only - Not for Redistribution exp { } means the number e 2.71828 = raised to the power in the bracket, and pe for any power p. …, 0> TIP The area under any part of the curve is the same, whether or not the boundary values are included. ø ø , 7) P(3 X ø < P(3 7) X , 7)ø< P(3 X and P(3 are 7) X < < indistinguishable. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Unfortunately, it is not possible to perform this integration accurately but, as we will see later, mathematicians have found ways to handle this challenge. Normal distributions have many interesting properties, some of which are detailed in the following table. Properties Half of the values are less than the mean. Half of the values are greater than the mean. Approximately 68.26% of the values lie within 1 standard deviation of the mean. Approximately 95.44% of the values lie within 2 standard deviations of the mean. Approximately 99.72% of the values lie within 3 standard deviations of the mean. Probabilities P( P( P.5 = 0.5 P( – µ σ ø ø X µ σ+ ) = 0.6826 P+ = 0.9544 P+ = 0.9972 In the following diagrams, the values 0, ±1 and ±2 represent numbers of standard deviations from the mean. 0.6826 0.8413 0.9544 0.9772 194 –1 –σμ 0 1 +σμμ 10 +σμμ –2 – 2σμ 0 2 + 2σμμ –2 – 2σμ 0 μ We can use the curve’s symmetry, along with the table and diagrams above, to find other probabilities, such as: We know that P(–1 ø ø X 1) = 0.6826 , so TIP The probability that the values in a normal distribution lie within a certain number of standard deviations of the mean is fixed. 0.6826 + 0.5 ) ø ø = ) 0.5 2) X + 0.9544 P( X ø 1) (= 1 2 × We know that P(–2 P( X ø 2) (= 1 2 × EXPLORE 8.3 = 0.8413 = P( X ù − 1) . 0.9544 , so = 0.9772 = P( X ù − 2) . Calculated estimates of the mean and variance of the continuous random variables A, B and C are given in the following table. Mean Variance A 40 64 B 72 144 C 123 121 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution Random observations from each distribution were made with the following results: For A: 8060 out of 13 120 observations lie in the interval from 32 to 48. For B: 8475 out of 12 420 observations lie in the interval from 60 to 84. For C: 8013 out of 10 974 observations lie in the interval from 112 to 134. Investigate this information (using the previous table showing properties and probabilities for normal distributions) and comment on the statement ‘The distributions of A, B and C are all normal’. The standard normal variable Z There are clearly an infinite number of values for the parameters of a normally distributed random variable. Nevertheless, most problems can be solved by transforming the random variable into a standard normal variable, which is denoted by Z, and which has mean 0 and variance 1. 0µ = and σ = 1 By substituting can find the equation of the PDF for Z ~ N(0, 1). This is denoted by φ( )z and its equation is into the equation for the normal distribution PDF, we z is shown below. . The graph of exp ( ) z y = φ( ) − = φ 2 1 2 π { 2z 2 } 0.4 0.2 y = ϕ(z) –3 –2 –1 0 1 2 3 z The mean of Z is 0. The axis of symmetry is a vertical line through the mean, as with every normal distribution. Z has a variance of 1 and, therefore, a standard deviation of 1. , ±2 and ±3 represent values that are 1, 2 and 3 standard deviations above or below = ±1 z the mean. Any < 0 z represents a value that is less the mean. Any > 0 z represents a value that is greater the mean. For > 3 z and for < −3 z , φ ( ) z ≈ 0 . The area under the graph of = φ( ) y z is equal to 1. A vertical line drawn at any value of Z divides the area under the curve into two parts: one representing Z z )ø and the other representing Z z . ) > P( P( FAST FORWARD Later in this section, we will see how any normal variable can be transformed to the standard normal variable by coding. KEY POINT 8.2 The standard normal (0, 1) variable is Z N~ 195 TIP φ and Φ are the lower and upper-case Greek letter phi. The value of Z z values by integration. Tables showing the value of )ø is denoted by P( z( )Φ and, as mentioned earlier, we do not find such z( )Φ for different values of z have been Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 compiled and appear in the Standard normal distribution function table at the end of the book. In addition, some modern calculators are able to give the value of inverse function z( )Φ and the directly. z( ) 1Φ− 0ù ) appear in the tables, the graph’s Although only zero and positive values of Z (i.e. z symmetry allows us to use the tables for positive and for negative values of z, as you will see after Worked example 8.2. = 2.999 Values of the standard normal variable appear as 4-figure numbers from = 0.000 in the tables. The first and second figures of z appear in the left-hand column; z the third and fourth figures appear in the top row. The numbers in the ‘ADD’ column for the fourth figures indicate what we should add to the value of z( )Φ in the body of the table. to z zΦ( ) zΦ( ) can be found for any given value of z, and z can be found for any given value of by using the tables in reverse (as shown in Worked example 8.4). In the critical values table, values for zΦ( ) are denoted by p. A section of the tables, from which we will find the value of Φ(0.274), is shown below. First and second figures Third figure Fourth figure .0 0.1 0.2 0.3 0.5000 0.5398 0.5793 0.6179 0.5040 0.5438 0.5832 0.6217 0.5080 0.5478 0.5871 0.6255 0.5120 0.5517 0.5910 0.6293 0.5160 0.5557 0.5949 0.6331 0.5199 0.5596 0.5987 0.6368 0.5239 0.5636 0.6026 0.6406 0.5279 0.5675 0.6064 0.6443 0.5319 0.5714 0.6103 0.6480 0.5359 0.5753 0.6141 0.6517 1 4 4 4 4 4 432 8765 9 ADD 8 8 8 8 7 12 12 12 12 11 16 16 16 15 15 20 20 20 19 19 24 24 24 23 22 28 28 28 27 26 32 32 32 31 30 36 36 36 35 34 TIP Critical values refer to probabilities of 75%, 90%, 95%, … and their complements 25%, 10%, 5%, … and so on. 196 We locate the first and second figures of z (namely 0.2) in the left-hand column. TIP We then locate the third figure of z (namely 7) along the top row… this tells us that Φ 0.6064. (0.27) = Next we locate the fourth figure of z (namely 4) at the top-right. In line with 0.6064, we see ‘ADD 15’, which means that we must add 15 to the last two figures of 0.6064 to obtain the value of Φ(0.274). Φ (0.274) = 0.6064 + 0.0015 = 0.6079 WORKED EXAMPLE 8.1 Given that Z ~ N(0, 1), find P( Z < 1.23) and ZP( ù 1.23) . 0.6079 (0.274) Φ = can be expressed using inverse notation as –1Φ 0.274 (0.6079) = . Answer 0.8907 1 – 0.8907 = 0.1093 (1.23) Φ left of z = = is the area to the 0.8907 . 1.23 Φ (1.23) 1 – = the right of z the graphs. 0.1093 1.23 = is the area to , as shown in 0 1.23 z 0 1.23 z ∴ < ZP( 1.23) = 0.8907 and ZP( ù 1.23) = 0.1093 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution WORKED EXAMPLE 8.2 Given that Z ~ N(0, 1), find P(0.4 ø Z < 1.7) correct to 3 decimal places. Answer 0.4 1.7 z Φ (1.7) = 0.9554 and Φ (0.4) = 0.6554 P(0.4 ø < Z 1.7) = P( 1.7) – P( Z < 0.4) Z < (1.7) – = Φ Φ (0.4) The required probability is equal to the difference between the area to the left of z left of z and the area to the , as illustrated. 1.7= 0.4 = Φ (1.70) (0.40) in the main body of the We find the values of Φ table, which means that we do not need to use the ADD section. and = = 0.9554 – 0.6554 0.300 As noted previously, the normal distribution function tables do not show values for < 0 However, we can use the symmetry properties of the normal curve, and the fact that the area under the curve is equal to 1, to find values of zΦ( ) when z is negative. z . 197 Situations in which > 0 z , and in wh
ich < 0 z , are illustrated in the two diagrams below. For a positive value, z b= : For a negative value, z a–= : The shaded area in this graph represents the value of bΦ( ). The shaded area in this graph represents the value of aΦ( ). Φ( b) = P(Z b)> Φ(a) = P(Z –a)> 0 b z –a 0 z Φ ( ) b = P( Z b )ø and 1 – Φ ( ) b = P( Z b )ù . Φ ( ) a = P( Z – )ù a and 1 – Φ ( ) a = P( Z – )ø a . From the tables, the one piece of information, Φ probabilities: (0.11) = 0.5438, actually tells us four P( Z ø P( Z ù 0.11) = 0.5438 and ZP( ù –0.11) = . 0.5438 0.11) = 1 – 0.5438 = 0.4562 and ZP( ø –0.11) = 1 – 0.5438 = 0.4562 . Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Information given about probabilities in a normal distribution should always be transferred to a sketched graph. Useful information, such as whether a particular value of z is positive or negative, will then be easy to see. This could, of course, also be determined by considering inequalities. If, for example, Z z P( ù > ) 0.5 P( , then Z z ø < ) 0.5 and, therefore, < 0 z . WORKED EXAMPLE 8.3 Given that Z ~ N(0, 1), find P(–1 ø < Z 2.115) correct to 3 significant figures. Answer The required probability is given by the difference between the area to the left of . –1= z and the area to the left of z 2.115 = The first of these areas is greater than 0.5 and the second is less than 0.5. –1 0 2.115 z = Φ (2.115) and ZP( < –1) = 1 – Φ . (1) ZP( P(–1 198 2.115) < ø < Z 2.115) = Φ (2.115) – [1 – (1)] Φ (1) – 1 + Φ 0.8413 – 1 = Φ (2.115) = = 0.9828 + 0.824 WORKED EXAMPLE 8.4 Given that Z ~ N(0, 1), find the value of a such that P( Z a < ) = 0.9072 . Answer 0.9066 = Φ (1.32) 0.004 a = = 1.32 + 1.324 0.9072) (0.9066 + 0.0006) (0.9066) + 0.004 0.004 = = 1.32 + 1.324 To find a value of z, we use the tables in reverse and search for the value closest to 0.9072, which is 0.9066. z( )Φ For our value of 0.9072, we need to add 0.0006 to 0.9066, so ‘ADD 6’ is required – this will be done if 1.32 is given a 4th figure of 4. We can check the value obtained for a by reading the tables in the usual way. (1.324) Φ ‘ADD 6’ (1.32) = Φ + = = 0.9066 + 0.0006 0.9072 Alternatively, we can show our working using inverse notation. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy WORKED EXAMPLE 8.5 Given that Z ~ N(0, 1), find the value of b such that Z b P( ≥ Chapter 8: The normal distribution ) = . 0.7713 Answer 0.7713 0.7713 –a 0 z 0 a z P( Z a ø = ) 0.7713 , so 0.7713) (0.7704 + 0.0009) (0.7704) 0.003 + 0.003 = = = 0.740 + 0.743 –0.743 ∴ = b a– 0.5 ù > ) Z b tells us that b is P( negative, so on our diagram we can replace b by a– , resulting in the two situations shown. The value closest to 0.7713 in the tables is 0.7704. This requires the addition of 0.0009 to bring it up to 0.7713, and 9 is in the column headed ‘ADD 3’. EXERCISE 8B 199 1 Given that Z ~ N(0, 1), find the following probabilities correct to 3 significant figures. a e i P( Z < 0.567) P( Z > 0.817) P( Z < 1.96) b ZP( ø 2.468) f j ZP( ù 2.009) P( Z > 2.576) c g P( Z P( Z > < –1.53) –1.75) d h ZP( ZP( ù ø –0.077) –0.013) 2 The random variable Z is normally distributed with mean 0 and variance 1. Find the following probabilities, correct to 3 significant figures. a d g j P(1.5 < Z < 2.5) P(–2.807 < Z < –1.282) P(–1.2 ( P 2 < Z < ø <Z 1.2) ) 5 b e h P(0.046 < Z < 1.272) c P(1.645 < Z < 2.326) P(–1.777 < Z < –0.746) P(–1.667 < Z < 2.667) f i P( –1.008 0.337) < 8 5 ) 3 Given that Z ~ N(0, 1), find the value of k, given that: a e i P( Z k < ) = 0.9087 b P( Z k < ) = 0.5442 P( Z k < ) = 0.25 P(– k Z k < < ) = 0.9128 f j P( Z k < ) = 0.3552 P(– k Z k < < ) = 0.6994 c g P( Z k > ) = 0.2743 P( Z k > ) = 0.9296 d h P( Z k > ) = 0.0298 P( Z k > ) = 0.648 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 4 Find the value of c in each of the following where Z has a normal distribution with . 0µ = and ( c Z < < 1.638) = 0.2673 P(1 < Z c < ) = 0.1408 P( c Z < < 2) = 0.6687 P(–1.221 < Z c < ) = 0.888 P(–2.63 < Z c < ) = 0.6861 b d f h j P( c Z < < 2.878) = 0.4968 P(0.109 < Z c < ) = 0.35 P( c Z < < 1.85) = 0.9516 P(–0.674 < Z c < ) = 0.725 P(–2.7 < Z c < ) = 0.0252 Standardising a normal distribution The probability distribution of a normally distributed random variable is represented by a normal curve. This curve is centred on the mean µ; the area under the curve is equal to 1, and its height is determined by the standard deviation σ. We already have a method for finding probabilities involving the standard normal variable Z ~ N(0, 1) using the normal distribution function tables. Fortunately, this same set of tables can be used to find probabilities involving any normal random variable, no matter what the values of µ and σ2. Although we have only learnt about coding data, it turns out that coding works in exactly the same way for normally distributed random variables: they behave in the way that we expect and remain normal after coding. If we code X by subtracting µ, then the PDF is translated horizontally by µ– units and is has mean 0 and standard deviation σ. now centred on 0. The new random variable µ–X µ–X by multiplying by If we now code deviation (and variance) will be equal to 1, while the mean remains 0. The coded random variable X µ (i.e. dividing by σ) then the standard − is normally distributed with mean 0 and variance 1. σ 1 σ Coding the random variable X in this way is called standardising, because it transforms the distribution )2 to Z ~ N(0, 1). X ~ N( , µ σ KEY POINT 8.3 200 When X ~ N( , µ σ )2 then Z = A standardised value z = x µ − σ has a standard normal distribution. X µ − σ tells us how many standard deviations x is from the mean. Probabilities involving values of X are equal to probabilities involving the corresponding values of Z, which can be found from the normal distribution function tables for Z ~ N(0, 1). For example, if X ~ N(20, 9), then P( X < 23) = P   Z < 23 20 − 9   . Copyright Material - Review Only - Not for Redistribution REWIND In Chapter 2, Section 2.2 and in Chapter 3, Section 3.3, we saw how the coding of data by addition and/or multiplication affects the mean and the standard deviation. FAST FORWARD We will learn more about coding random variables in the Probability & Statistics 2 Coursebook, Chapter 3. REWIND In the table showing properties and probabilities of normal distributions prior to Explore 8.3, we saw that probabilities are determined by the number of standard deviations from the mean. The properties given in that table apply to all normal random variables. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution WORKED EXAMPLE 8.6 Given that X ~ N(11, 25) , find P( X < 18) correct to 3 significant figures. Answer z = P( X < 18) = 18 P( 11 − 25 Z < (1.4) = Φ 1.4) = 1.4 We standardise x 18= and find that it is 1.4σ above the mean of 11. = 0.919 WORKED EXAMPLE 8.7 Given that X ~ N(20, 7) , find P( X < 16.6) correct to 3 significant figures. Answer 16.6 z = P( X ø 16.6) = = = − 7 ø 20 = –1.285 – 1.285) P( Z (1.285) 1 – Φ 0.0994 We standardise x below the mean of 20. = 16.6 and find that it is 1.285σ 201 WORKED EXAMPLE 8.8 Given that X ~ N(5, 5) , find P(2 9)<Xø correct to 3 significant figures. Answer For x = z2, = For x = 91.342 = 1.789 The required area is shown in two parts in the diagram. 2 –1.342 5 0 9 1.789 x z Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Area to the right of z Φ (1.789) – Φ (0) = Φ 0= is (1.789) − = 0.4633 Area to the left of z = Φ (–1.342) (0) – Φ Total area X< ∴ P(2 = = 0.4633 < 9) = 0= is [ 0.5 – 1 – 0.4102 0.4102 + 0.874 0.5 Φ (1.342) ] = 0.8735 Here, we find the two areas separately then add them to obtain our final answer, which is where we round to the accuracy specified in the question. TIP Where possible, always use a 4-figure value for z. Try solving this problem by the method shown in Worked example 8.3 (using the areas to the left of both z –1.342 z approach you prefer. and ) and decide which 1.789 = = Some useful results from previous worked examples are detailed in the following graphs. 0 For < a < b For – a < 0 < b For – a < 0 < a 202 P( 0 a b z –a 0 b z –(– (– WORKED EXAMPLE 8.9 Given that ~ N( , Y 2µ σ ), P( Y < 10) = 0.75 and YP( ù 12) = 0.1 , find the values of µ and σ. Answer P( Y ù 12) < 0.5, so Y < P( 12) > 0.5, which means that µ>12 . P( Y < 10) > 0.5, which means also that . µ>10 0.75 0.90 μ 0 10 za y z μ 0 12 zb y z 0.674 gives 10 – µ = 0.674 σ
…… [1] 1.282 gives 12 – µ = 1.282 σ …… [2] µ = µ = 10 12 − σ − σ These simple sketch graphs allow us to locate the values 10 and 12 relative to µ. za zb = Φ = Φ –1 –1 (0.75) (0.90) = = 0.674 1.282 and Note that both 0.75 and 0.90 are critical values. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution We subtract equation 1[ ] from 2[ ] to solve this pair of simultaneous equations. [2] [1] . 7.78 12 – µ = 1.282 σ 10 – µ = 0.674 σ 2 ∴σ = = 3.29 0.608 σ and µ = EXERCISE 8C 1 Standardise the appropriate value(s) of the normal variable X represented in each diagram, and find the required probabilities correct to 3 significant figures. a Find XP( 11)ø , given that X . ~ N(8, 25) b Find P( X < 69.1) , given that X . ~ N(72, 11) 8 0 11 ...... x z 203 69.1 ....... 72 0 c Find P(3 < <X 7) , given that X . ~ N(5, 5) 3 .... 5 0 7 .... x z x z From this point in the exercise, you are strongly advised to sketch a diagram to help answer each question. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 2 Calculate the required probabilities correct to 3 significant figures. a Find XP( 9.7)ø and P( X > 9.7) , given that X ~ N(6.2, 6.25). b Find XP( 5)ø and P( X > 5) , given that X ~ N(3, 49). c Find P( X d Find P( X > < 33.4) and XP( 13.5) and XP( ø ù 33.4) , given that X ~ N(37, 4). 13.5) , given that X ~ N(20, 15). 91(1 P(2 and XP( 91)ø , given that X ~ N(80, 375). 21) , given that X ~ N(11, 25). 5) , given that X ~ N(3, 7). 8.8) , given that X ~ N(7, 1.44). [Read carefully.] 28) , given that X ~ N(25, 6). e Find P( f Find g Find h Find i Find P(6.2 P(26 Find P(8 10) , given that X ~ N(12, 2.56). 3 a Find a, given that X ~ N(30, 16) and that P( X a< ) = 0.8944. b Find b, given that X ~ N(12, 4) and that P( X b< ) = 0.9599. c Find c, given that X ~ N(23, 9) and that P( X c > ) = 0.9332. d Find d , given that X ~ N(17, 25) and that P( X d > ) = 0.0951. e Find e, given that X ~ N(100, 64) and that P( X e > ) = 0.95. 204 4 a Find f , given that X ~ N(10, 7) and that P( f Xø < 13.3) = 0.1922. b Find g, given that X ~ N(45, 50) and that P( g Xø < 55) = 0.5486. c Find h, given that X ~ N(7, 2) and that P(8 d Find j, given that X ~ N(20, 11) and that P( ø X h < j Xø < ) = 0.216. 22) = 0.5. 5 X is normally distributed with mean 4 and variance 6. Find the probability that X takes a negative value. 6 Given that X 2) ( µ µ ~ N , 4 9 where 0µ > , find XP( . 2 )µ< 7 If ~ N(10, T )2σ and P( T > 14.7) = 0.04 , find the value of σ. 8 It is given that ~ N( , 13)µ V and P( V < 15) = 0.75 . Find the value of µ. 9 The variable W ~ N( , )2µ σ . Given that 4µ σ= and P( W < 83) = 0.95 , find the value of µ and of σ. 10 X has a normal distribution in which σ µ= – 30 and XP( ù 12) = 0.9 . Find the value of µ and of σ. 11 The variable )2µ σ Q and of σ, and calculate the ~ N( , . Given that ø < Q P(4 P( Q . 5) < 1.288) = 0.281 and P( Q < 6.472) = 0.591 , find the value of µ 12 For the variable ~ N( , V )2µ σ , it is given that P( V = 0.7509 and P( V > 9.2) = 0.1385 . Find the value of µ and of σ, and calculate VP( 8.4) < 10)ø . 13 Find the value of µ and of σ and calculate 0.6858 = and WP( 4.75) 2.25) WP( ø ù = 6.48) for the variable W ~ N( , )2µ σ , given that P( W > . 0.0489 14 X has a normal distribution, such that P( X > 147.0) = 0.0136 and XP( ø 59.0) = 0.0038 . Use this information to calculate the probability that 80.0 ø < X . 130.0 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution 8.3 Modelling with the normal distribution The German mathematician Carl Friedrich Gauss showed that measurement errors made in astronomical observations were well modelled by a normal distribution, and the Belgian statistician and sociologist Adolphe Quételet later applied this to human characteristics when he saw that distributions of such things as height, weight, girth and strength were approximately normal. We are now in a position to apply our knowledge to real-life situations, and to solve more advanced problems involving the normal distribution. WORKED EXAMPLE 8.10 The mass of a newborn baby in a certain region is normally distributed with mean 3.35 kg and variance 0.0858 kg2. Estimate how many of the 1356 babies born last year had masses of less than 3.5 kg. TIP We cannot know the exact number of newborn babies from the model because it only gives estimates. However, we do know that the number of babies must be an integer. 205 Answer Φ Z( ) = Φ = Φ − 3.35 0.0858 3.5   (0.512) = 0.6957   We standardise the mass of 3.5 kg. 0.6957 is a relative frequency equal to 69.57% < P(mass 69.57% of 1356 ∴ There were about 943 newborn babies. 0.6957 = 943.3692 3.5 kg) = WORKED EXAMPLE 8.11 A factory produces half-litre tins of oil. The volume of oil in a tin is normally distributed with mean 506.18 ml and standard deviation 2.96 ml. a What percentage of the tins contain less than half a litre of oil? b Find the probability that exactly 1 out of 3 randomly selected tins contains less than half a litre of oil. Answer a z = 500 506.18 − 2.96 = –2.088 Let X represent the amount of oil in a tin, then X ~ N(506.18, 2.96 )2 . The graph shows the probability distribution for the amount of oil in a tin. 500 –2.088 506.18 0 x z Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 P( X < 500) = = = = P( Z –2.088) < (2.088) 1 – Φ 1 – 0.9816 0.0184 ∴ 1.84% of the tins contain less than half a litre of oil. b YP( = 1) = = 3 1    ×  0.0532 1 0.0184 × 0.9816 2 EXERCISE 8D Let the discrete random variable Y be the number of tins containing less than half a litre of oil, then ~ B(3, 0.0184) Y . TIP A probability obtained from a normal distribution can be used as the parameter p in a binomial distribution. 1 The length of a bolt produced by a machine is normally distributed with mean 18.5 cm and variance 0.7 cm2. Find the probability that a randomly selected bolt is less than 18.85 cm long. 2 The waiting times, in minutes, for patients at a clinic are normally distributed with mean 13 and variance 16. a Calculate the probability that a randomly selected patient has to wait for more than 16.5 minutes. b Last month 468 patients attended the clinic. Calculate an estimate of the number who waited for less than 206 9 minutes. 3 Tomatoes from a certain producer have masses which are normally distributed with mean 90 grams and standard deviation 17.7 grams. The tomatoes are sorted into three categories by mass, as follows: Small: under 80 g; Medium: 80 g to 104 g; Large: over 104 g. a Find, correct to 2 decimal places, the percentage of tomatoes in each of the three categories. b Find the value of k such that P( k X ø < 104) = 0.75 , where X is the mass of a tomato in grams. 4 The heights, in metres, of the trees in a forest are normally distributed with mean µ and standard deviation 3.6. Given that 75% of the trees are less than 10 m high, find the value of µ. 5 The mass of a certain species of fish caught at sea is normally distributed with mean 5.73 kg and variance 2.56 kg2. Find the probability that a randomly selected fish caught at sea has a mass that is: a less than 6.0 kg b more than 3.9 kg c between 7.0 and 8.0 kg 6 The distance that children at a large school can hop in 15 minutes is normally distributed with mean 199 m and variance 3700 m2. a Calculate an estimate of b, given that only 25% of the children hopped further than b metres. b Find an estimate of the interquartile range of the distances hopped. 7 The daily percentage change in the value of a company’s shares is expected to be normally distributed with mean 0 and standard deviation 0.51. On how many of the next 365 working days should the company expect the value of its shares to fall by more than 1%? 8 The masses, w grams, of a large sample of apples are normally distributed with mean 200 and variance 169. , calculate an estimate of the number of 213 <w< Given that the masses of 3413 apples are in the range 187 apples in the sample. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution 9 The ages of the children in a gymnastics club are normally distributed with mean 15.2 years and standard deviation σ. Find the value of σ given that 30.5% of the children are less than 13.5 years of age. 10 The speeds, in kmh–1, of vehicles passing a particular point on a rural road are normally distributed with mean µ and standard deviation 20. Find the value of µ and find what percentage of the v
ehicles are being driven at under 80 kmh–1, given that 33% of the vehicles are being driven at over 100 kmh–1. 11 Coffee beans are packed into bags by the workers on a farm, and each bag claims to contain 200 g. The actual mass of coffee beans in a bag is normally distributed with mean 210 g and standard deviation σ. The farm owner informs the workers that they must repack any bag containing less than 200 g of coffee beans. Find the value of σ, given that 0.5% of the bags must be repacked. 12 Colleen exercises at home every day. The length of time she does this is normally distributed with mean 12.8 minutes and standard deviation σ. She exercises for more than 15 minutes on 42 days in a year of 365 days. a Calculate the value of σ. b On how many days in a year would you expect Colleen to exercise for less than 10 minutes? 13 The times taken by 15-year-olds to solve a certain puzzle are normally distributed with mean µ and standard deviation 7.42 minutes. a Find the value of µ, given that three-quarters of all 15-year-olds take over 20 minutes to solve the puzzle. b Calculate an estimate of the value of n, given that 250 children in a random sample of n 15-year-olds fail to solve the puzzle in less than 30 minutes. 14 The lengths, cmX , of the leaves of a particular species of tree are normally distributed with mean µ and variance σ2. 207 a Find P Find the probability that a randomly selected leaf from this species has a length that is more than 2 standard deviations from the mean. c Find the value of µ and of σ, given that XP( < 7.5) = 0.75 and XP( < 8.5) = . 0.90 15 The time taken in seconds for Ginger’s computer to open a specific large document is normally distributed with mean 9 and variance 5.91. a Find the probability that it takes exactly 5 seconds or more to open the document. b Ginger opens the document on her computer on n occasions. The probability that it fails to open in less than exactly 5 seconds on at least one occasion is greater than 0.5. Find the least possible value of n. 16 The masses of all the different pies sold at a market are normally distributed with mean 400 g and standard deviation 61g. Find the probability that: a the mass of a randomly selected pie is less than 425 g b 4 randomly selected pies all have masses of less than 425 g c exactly 7 out of 10 randomly selected pies have masses of less than 425 g. 17 The height of a female university student is normally distributed with mean 1.74 m and standard deviation 12.3 cm. Find the probability that: a a randomly selected female student is between 1.71 and 1.80 metres tall b 3 randomly selected female students are all between 1.71 and 1.80 m tall c exactly 15 out of 50 randomly selected female students are between 1.71 and 1.80 metres tall. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 8.4 The normal approximation to the binomial distribution In Chapter 7, Section 7.1, we saw that the binomial distribution can be used to solve problems such as ‘Find the probability of obtaining exactly 60 heads with 100 tosses of a fair coin’, and that this is equal to   100 60   × 60 0.5 × 0.5 40 . Therefore, to find the probability of obtaining 60 or more heads, we must find the probability for 60 heads, for 61 heads, for 62 heads and so on, and add them all together. Imagine how long it took to calculate binomial probabilities before calculators and computers! However, in certain situations, we can approximate a probability such as this by a method that involves far fewer calculations using the normal distribution. EXPLORE 8.4 Binomial probability distributions for 2 , 4, and 12 tosses of a fair coin are shown in the following diagrams. Notice that, as the number of coin tosses increases, the shape of the probability distribution becomes increasingly normal. 0.5 p = 0.5, n = 2 0.4 p = 0.5, n = 4 0.25 p = 0.5, n = 12 208 10 12 Does the binomial probability distribution maintain its normal shape for large values of n when p varies? Find out using the Binomial Distribution resource on the GeoGebra website. Select any ùn note of when the distribution loses its normal shape. Repeat this for other values of n. 20 then use the pause/play button or the slider to vary the value of p. Take Can you generalise as to when the binomial distribution begins to lose its normal shape? DID YOU KNOW? FAST FORWARD Abraham de Moivre, the 18th century statistician and consultant, was often asked to make long calculations concerning games of chance. He noted that when the number of events increased, the shape of the binomial distribution approached a very smooth curve, and saw that he would be able to solve these long calculation problems if he could find a mathematical expression for this curve: this is exactly what he did. The curve he discovered is now called the normal curve. Before the late 1870s, when the term normal was coined independently by Peirce, Galton and Lexis, this distribution was known – and still is by some – as the Gaussian distribution after the German mathematician Carl Friedrich Gauss. The word normal is not meant to suggest that all other distributions are abnormal! Abraham de Moivre 1667–1754 Copyright Material - Review Only - Not for Redistribution n = de Moivre’s theorem, sin ) (cos i θ θ + cos sin n n i θ θ + links trigonometry with complex numbers – a topic that we cover in the Pure Mathematics 2 & 3 Coursebook, Chapter 11. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution The following diagrams show the shapes of four binomial distributions for n 25= . p = 0.15, q = 0.85 p = 0.35, q = 0.65 p = 0.75, q = 0.25 p = 0.95, q = 0.05 0 25 0 25 0 25 0 25 np = 3.75, nq = 21.25 np = 8.75, nq = 16.25 np = 18.75, nq = 6.25 np = 23.75, nq = 1.25 As you can see, the binomial distribution loses its normal shape when p is small and also when q is small. KEY POINT 8.4 ) ~ B( , n p can be X approximated by )2µ σ , where N( , 2σ = npq , µ = np and provided that n is large enough to ensure that np and nq > 5 > 5 . 209 KEY POINT 8.5 Continuity corrections must be made when a discrete distribution is approximated by a continuous distribution. A more detailed investigation shows that the binomial distribution has an approximately normal shape if np and nq are both greater than 5. These are the values that we use to decide whether a binomial distribution can be well-approximated by a normal distribution. The larger the values of np and nq, the more accurate the approximation will be. As we can see from the above diagrams, the approximation is adequate (but not very good) when np and nq = 18.75 = 6.25 . The distribution . because nq < 5 X ~ B(40, 0.9) cannot be well-approximated by a normal distribution The distribution = 50 because np X and ~ B(250, 0.2) = 200 nq can be well-approximated by a normal distribution , both of which are substantially greater than 5. = 13 13.5 must be treated as being represented by the class of continuous values . For this reason, When we approximate a discrete distribution by a continuous distribution, a discrete value such as X ø X 12.5 < in our probability calculations. Making this replacement is known as ‘making a = 13.5 X continuity correction’. Deciding whether to use depends on whether or not is included in the probability that we wish to find. must be replaced by either = 12.5 = 13.5 = 12.5 or by = 13 or X X X X X = 13 For example, if we wish to find . = 12.5 using X P( X < 13) , where X = 13 is not included, we calculate If we wish to find øXP( 13), where X = 13 is included, we calculate using X . = 13.5 Further details of continuity corrections are given in Worked example 8.12. WORKED EXAMPLE 8.12 Given that ~ B(100, 0.4) X to find: a P( X < 43) b P( X > 43) , use a suitable approximation and continuity correction Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Answer 40 =np and µ= X approximated by N(40, 24). ~ B(100, 0.4) 2σ = can be npq = 24. The conditions for approximating a binomial distribution by a normal distribution are met because np and nq than 5. 40= , which are both greater 60 42 43 44 x 42.5 43.5 In the continuous distribution N(40, 24), 43 is represented by the class of continuous values 42.5 X ≤ diagram. , as shown in the 43.5 < Possible continuity corrections for a discrete value of 43 are given below: For X < P( <XP( For 43), we would use the lower boundary value 42.5…… [part a] 43), we would use the upper boundary value 43.5 For XP( For > >XP( , we would use the upper boundary value 43.5 …… [part b] 43) 43), we would use the lower boundary value 42.5 210 = Φ a P( X < 43) ≈ P( 0.510) Z < (0.510) = 43) 0.6950 ≈ 0.695 X∴ P( < b P( X > 43) ≈ = = 43) X∴ P( > P(Z 0.714) > (0.714) 1 – Φ 0.2377 ≈ 0.238 For x = 42.5, z = 40 42.5 − 24 = 0.5103 For x = 43.5, z = 40 43.5 − 24 = 0.7144 FAST FORWARD We will also make continuity corrections when using the normal distribution as an approximation to the Poisson distribution in the Probability & Statistics 2 Coursebook, Chapter 2. TIP X a< means ‘X is fewer/less than a’. X a> means ‘X is more/greater t
han a’. X a< means ‘X is at most a’ and ‘X is not more than a’ and ‘X is a or less’. X a> means ‘X is at least a’ and ‘X is not less than a’ and ‘X is a or more’. WORKED EXAMPLE 8.13 Boxes are packed with 8000 randomly selected items. It is known that 0.2% of the items are yellow. Find, using a suitable approximation, the probability that: a a box contains fewer than 20 yellow items b exactly 2 out of 3 randomly selected boxes contain fewer than 20 yellow items. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution Answer a X ~ B(8000, 0.002) Let X be the number of yellow items in a box. X ~ B(8000, 0.002) can be approximated by N(16, 15.968). P( X < 20) P ≈ Z   P( Z < 19.5 16 − 15.968   ) … 0.87587 = = Φ < (0.876) np = 16 and nq = 7984 are both greater than 5, so we can approximate the binomial distribution by a normal distribution using npµ= = 16 and = 15.968 . 2 σ = npq P( To find X < calculate with the value x = 19.5. 20), we must = 0.8094 ∴ The probability that a box contains fewer than 20 yellow items is approximately 0.809. b P( Y = 2) = = 3 2    ×  0.375 0.8094 2 1 0.1906 × Let Y be the number of boxes containing fewer than 20 yellow items, then Y ~ B(3, 0.8094). TIP Do not forget to make the continuity correction! TIP Although our answer to part a is only an approximation, we should not use a rounded probability, such as 0.8, in further calculations. 211 WORKED EXAMPLE 8.14 A fair coin is tossed 888 times. Find, by use of a suitable approximation, the probability that the coin lands heads-up at most 450 times. Answer X ~ B(888, 0.5) Let X represent the number of times the coin lands heads-up. X ~ B(888, 0.5) can be approximated by N(444, 222). np = 444, nq = 444 are both greater than 5, and npq = 222. P( X < 450) P ≈ Z   P( Z = < 0.436) < 450.5 444 − 222   To find X øP( 450) we calculate with x = 450.5. = Φ (0.436) = 0.6686 ∴ P(at most 450 heads) ≈ 0.669 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXPLORE 8.5 By visiting the Binomial and Normal resource on the GeoGebra website you will get a clear picture of how the normal approximation to the binomial distribution works. Select values of n and p so that np and (1 – ) n p are both greater than 5. The binomial probability distribution is displayed with an overlaid normal curve ) p are displayed in red at the top-right). If (the value of µ = np and of you then check the probability box, adjustable values of =x a and =x b appear on the diagram, with the area between them shaded. Remember that a discrete variable is being approximated by a continuous variable, so appropriate continuity corrections are needed to find the best probability estimates. σ = np (1 − EXERCISE 8E 1 Decide whether or not each of the following binomial distributions can be well-approximated by a normal distribution. For those that can, state the values of the parameters µ and 2σ . For those that cannot, state the reason. a B(20, 0.6) b B(30, 0.95) c B(40, 0.13) d B(50, 0.06) 212 2 Find the smallest possible value of n for which the following binomial distributions can be well-approximated by a normal distribution. a B( , 0.024) n b B( , 0.15) n c B( , 0.52) n d B( , 0.7) n 3 Describe the binomial distribution that can be approximated by the normal distribution N(14, 10.5). 4 By first evaluating np and npq, use a suitable approximation and continuity correction to find . the discrete random variable ~ B(100, 0.7) X P( X < 75) for 5 The discrete random variable ~ B(50, 0.6) Y . Use a suitable approximation and continuity correction to find P( Y > 26) . 6 A biased coin is tossed 160 times. The number of heads obtained, H, follows a binomial distribution where E( H ) = 100 . Find: a b the value of p and the variance of H the approximate probability of obtaining more than 110 heads. 7 One card is selected at random from each of 40 packs. Each pack contains 52 cards and includes 13 clubs. Let C be the number of clubs selected from the 40 packs. a Show that the variance of C is 7.5. b Obtain an approximation for the value of <P( C 8) , and justify the use of this approximation. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution 8 In a large survey, 55% of the people questioned are in full-time employment. In a random sample of 80 of these people, find: a b c the expected number in full-time employment the standard deviation of the number in full-time employment the approximate probability that fewer than half of the sample are in full-time employment. 9 A company manufactures rubber and plastic washers in the ratio 4:1. The washers are randomly packed into boxes of 25. a Find the probability that a randomly selected box contains: i exactly 21 rubber washers ii exactly 10 plastic washers. b A retail pack contains 2000 washers. Find the expectation and variance of the number of rubber washers in a retail pack. c Using a suitable approximation, find the probability that a retail pack contains at most 1620 rubber washers. 10 In a certain town, 63% of homes have an internet connection. a In a random sample of 20 homes in this town, find the probability that: i ii exactly 15 have an internet connection exactly nine do not have an internet connection. b Use a suitable approximation to find the probability that more than 65% of a random sample of 600 homes in this town have an internet connection. 213 11 17% of the people interviewed in a survey said they watch more than two hours of TV per day. A random sample of 300 of those who were interviewed is taken. Find an approximate value for the probability that at least one-fifth of those in the sample watch more than two hours of TV per day. 12 An opinion poll was taken before an election. The table shows the percentage of voters who said they would vote for parties A, B and C. Party Votes (%) A 36 B 41 C 23 Find an approximation for the probability that, in a random sample of 120 of these voters: a exactly 50 said they would vote for party B b more than 70 but fewer than 90 said they would vote for party B or party C. 13 Boxes containing 24 floor tiles are loaded into vans for distribution. In a load of 80 boxes there are, on average, three damaged floor tiles. Find, approximately, the probability that: a b there are more than 65 damaged tiles in a load of 1600 boxes in five loads, each containing 1600 boxes, exactly three loads contain more than 65 damaged tiles. 14 It is known that 2% of the cheapest memory sticks on the market are defective. a In a random sample of 400 of these memory sticks, find approximately the probability that at least five but at most 11 are defective. b Ten samples of 400 memory sticks are tested. Find an approximate value for the probability that there are fewer than 12 defective memory sticks in more than seven of the samples. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PS 15 Randomly selected members of the public were asked whether they approved of plans to build a new sports centre and 57% said they approved. Find approximately the probability that more than 75 out of 120 people said they approved, given that at least 60 said they approved. PS 16 A fair coin is tossed 400 times. Given that it shows a head on more than 205 occasions, find an approximate value for the probability that it shows a head on fewer than 215 occasions. PS 17 An ordinary fair die is rolled 450 times. Given that a 6 is rolled on fewer than 80 occasions, find approximately the probability that a 6 is rolled on at least 70 occasions. Checklist of learning and understanding ● A continuous random variable can take any value, possibly within a range, and those values occur by chance in a certain random manner. ● The probability distribution of a continuous random variable is represented by a function called a probability density function or PDF. ● A normally distributed random variable X is described by its mean and variance as X ~ N( )2 µ σ . , ● The standard normal random variable is ~ N(0, 1) Z . ● When X ~ N( )2 µ σ then , Z = X µ has a standard normal distribution, and the standardised value − σ z = x µ − tells us how many standard deviations x is from the mean. σ ~ B( , , where =µ np and σ =2 )2σµ , ) X enough to ensure that n p can be approximated by N( . > 5 np and > 5 nq ● 214 npq, provided that n is large ● > 5 and np of np and nq result in better approximations. > 5 nq are the necessary conditions for making this approximation, and larger values ● Continuity corrections must be made when a discrete distribution is approximated by a continuous distribution. 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bridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution END-OF-CHAPTER REVIEW EXERCISE 8 1 A continuous random variable, X , has a normal distribution with mean 8 and standard deviation σ. . Given that 0.9772 , find 9.5) P( P( 5) X X > = < 2 The variable Y is normally distributed. Given that σ 10 µ= 3 and P( Y < 10) = 0.75 , find P( Y > 6). 3 In Scotland, in November, on average 80% of days are cloudy. Assume that the weather on any one day is independent of the weather on other days. i Use a normal approximation to find the probability of there being fewer than 25 cloudy days in Scotland in November (30 days). ii Give a reason why the use of a normal approximation is justified. [3] [4] [4] [1] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q2 June 2011 4 At a store, it is known that 1 out of every 9 customers uses a gift voucher in part-payment for purchases. A randomly selected sample of 72 customers is taken. Use a suitable approximation and continuity correction to find the probability that at most 6 of these customers use a gift voucher in part-payment for their [5] purchases. 5 A survey shows that 54% of parents believe mathematics to be the most important subject that their children study. Use a suitable approximation to find the probability that at least 30 out of a sample of 50 parents believe mathematics to be the most important subject studied. [5] 6 Two normally distributed continuous random variables are X and Y . It is given that and . On the same diagram, sketch graphs showing the probability density functions of X that ~ N(2.0, 0.5 )2 and of Y . Indicate the line of symmetry of each clearly labelled graph. ~ N(1.5, 0.2 )2 Y X 7 The random variable X is such that ~ N(82, 126) X . i A value of X is chosen at random and rounded to the nearest whole number. Find the probability that this whole number is 84. ii Five independent observations of X are taken. Find the probability that at most one of them is greater than 87. iii Find the value of k such that P(87 < X k < ) = 0.3 . [3] 215 [3] [4] [5] Cambridge International AS & A Level Mathematics 9709 Paper 63 Q5 November 2012 8 a A petrol station finds that its daily sales, in litres, are normally distributed with mean 4520 and standard deviation 560. i Find on how many days of the year (365 days) the daily sales can be expected to exceed 3900 litres. [4] The daily sales at another petrol station are X litres, where X is normally distributed with mean m and standard deviation 560. It is given that . 0.122 8000) P( X > = ii Find the value of m. iii Find the probability that daily sales at this petrol station exceed 8000 litres on fewer than 2 of 6 randomly chosen days. b The random variable Y is normally distributed with mean µ and standard deviation σ. Given that σ = 2 3 µ , find the probability that a random value of Y is less than µ2 . [3] [3] [3] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q7 November 2015 9 V and W are continuous random variables. ~ N(9, 16) 2 P( × given that W P( 8) 8) . V V < < = and W ~ N(6, )2σ . Find the value of σ, [4] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 10 The masses, in kilograms, of ‘giant Botswana cabbages’ have a normal distribution with mean µ and standard deviation 0.75. It is given that 35.2% of the cabbages have a mass of less than 3 kg. Find the value of µ and the percentage of cabbages with masses of less than 3.5 kg. 11 The ages of the vehicles owned by a large fleet-hire company are normally distributed with mean 43 months 6 years old is 0.28. and standard deviation σ. The probability that a randomly chosen vehicle is more than 4 1 Find what percentage of the company’s vehicles are less than two years old. 12 The weights, X grams, of bars of soap are normally distributed with mean 125 grams and standard deviation 4.2 grams. i Find the probability that a randomly chosen bar of soap weighs more than 128 grams. ii Find the value of k such that P( k X < < 128) = 0.7465 . iii Five bars of soap are chosen at random. Find the probability that more than two of the bars each weigh more than 128 grams. [5] [5] [3] [4] [4] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q7 November 2009 PS 13 Crates of tea should contain 200 kg, but it is known that 1 out of 45 crates, on average, is underweight. A sample of 630 crates is selected at random. a Find the probability that more than 12 but fewer than 17 crates are underweight. b Given that more than 12 but fewer than 17 crates are underweight, find the probability that more than 14 crates are underweight. 216 PS 14 Once a week, Haziq rows his boat from the island where he lives to the mainland. The journey time, X minutes, is normally distributed with mean µ and variance σ2. a Given that P(20 ø < X 30) = 0.32 and that P( X < 20) = 0.63 , find the values of µ and σ2. b The time taken for Haziq to row back home, Y minutes, is normally distributed and P( Y < 20) = 0.6532 . Given that the variances of X and Y are equal, calculate: [5] [5] [4] [3] the mean time taken by Haziq to row back home i ii the expected number of days over a period of five years (each of 52 weeks) on which Haziq takes more [3] than 25 minutes to row back home. PS 15 The time taken, T seconds, to open a graphics programme on a computer is normally distributed with mean 20 and standard deviation σ. Given that P( T > 13 \| T ø 27) = 0.8 , find the value of: a σ b k for which P( T > k ) = 0.75 . PS 16 A law firm has found that their assistants make, on average, one error on every 36 pages that they type. A random sample of 90 typed documents, with a mean of 62 pages per document, is selected. Given that there are more than 140 typing errors in these documents, find an estimate of the probability that there are fewer than 175 typing errors. [5] [3] [6] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cross-topic review exercise 3 CROS-TPI EVW X CROSS-TOPIC REVISION EXERCISE 3 1 a The following table shows the probability distribution for the random variable X . x P X x =( ) 0 1 k 1 3 10 2 3 20 3 1 20 i Show that k 2= . ii Calculate XE( ) and XVar( ). iii Find the probability that two independent observations of X have a sum of less than 6. b The following table shows the probability distribution for the random variable .1 1 0.2 2 0.3 3 0.4 = . If one independent observation of each random variable is made, find the probability that X Y 3 + The random variable X has a geometric distribution such that P( P( X X = = 2) 5) = 3 3 8 . Find XP( 3)< . The variable X has a normal distribution with mean µ and standard deviation XP( , find the value of µ and of and that ùXP( 27.45) 32.83) 0.409 0.834 = = < .σ .σ Given that The length of time, in seconds, that it takes to transfer a photograph from a camera to a computer can be modelled by a normal distribution with mean 4.7 and variance 0.7225. Find the probability that a photograph can be transferred in less than 3 seconds. The mass of a berry from a particular type of bush is normally distributed with mean 7.08 grams and standard deviation σ. It is known that 5% of the berries have a mass of exactly 12 grams or more. a Find the value of .σ b Find the proportion of berries that have a mass of between 6 and 8 grams. The time taken, in minutes, to fit a new windscreen to a car is normally distributed with mean µ and standard deviation 16.32. Given that three-quarters of all windscreens are fitted in less than 45 minutes, find: a the value of µ b the proportion of windscreens that are fitted in 35 to 40 minutes. 2 3 4 5 6 [2] [3] [2] [3] [3] [4] [3] [3] [3] [3] [3] 217217 7 The mid-day wind speed, in knots, at a coastal resort is normally distributed with mean 12.8 and standard deviation .σ a Given that 15% of the recorded wind speeds are less than 10 knots, find the value of σ. b Calculate the probability that exactly two out of 10 randomly selected recordings are less than 10 knots. c Using a suitable approximation, calculate an estimate of the probability that at least 13 out of 100 randomly selected recordings are over 15.5 knots. [3] [3] [4] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 8 A technical manual contains 10 pages of text, 7 pages of diagrams and 3 pages of colour illustrations. Four different pages are selected at random from the manual. Let X be the number of pages of colour illustrations selected. a Draw up the probability distribution table for X . b Find: i XE( ) ii the probability that fewer than two pages with colour illustrations are selected, given that at least one page with colour illustrations is selected. 9 A fair six-sided die is numbered 1, 1, 2, 3, 5 and 8. The die is rolled twice and the two numbers obtained are added together to give the score, X . a Find XE( ). b Given that the first number rolled is odd, find
the probability that X is an even number. 10 The following table shows the probability distribution table for the random variable Q. q P )= 18 3 x x − + 3 4 218218 a Find the value of x. b Evaluate QVar( ). 11 Research shows that 17% of children are absent from school on at least five days during winter because of ill health. A random sample of 55 children is taken. a Find the probability that exactly 10 of the children in the sample are absent from school on at least five days during winter because of ill health. [4] [2] [2] [4] [2] [3] [2] [2] b Use a suitable approximation to find the probability that at most seven children in the sample are absent from school on at least 5 days during winter because of ill health. c Justify the approximation made in part b. [4] [1] 12 The ratio of adult males to adult females living in a certain town is 17 : 18, and gender, do not have a driving license. of these adults, independent of 2 9 a Show that the probability that a randomly selected adult in this town is male and has a driving license is equal to 17 45 . [1] b Find the probability that, in a randomly selected sample of 25 adults from this town, from 8 to 10 inclusive are females who have a driving license. 13 A fair eight-sided die is numbered 2, 2, 3, 3, 3, 4, 5 and 6. The die is rolled up to and including the roll on which the first 2 is obtained. Let X represent the number of times the die is rolled. a Find XE( ). b Show that P( X > 4) = 27 64 . [4] [1] [1] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cross-topic review exercise 3 c The die is rolled up to and including the roll on which the first 2 is obtained on 20 occasions. Find, by use of a suitable approximation, the probability that 4X > on at least half of these 20 occasions. d Fully justify the approximation used in part c. 14 A student wishes to approximate the distribution of ~ B(240, X p by a continuous random variable Y that ) has a normal distribution. a Find the values of p for which: i approximating X by Y can be justified ii YVar( ) . 45< b Find the range of values of p for which both the approximation is justified and YVar( ) . 45< [4] [1] [3] [3] [2] 219219 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PRACTICE EXAM-STYLE PAPER Time allowed is 1 hour and 15 minutes (50 marks) 1 A mixed hockey team consists of five men and six women. The heights of individual men are denoted by hm metres and the heights of individual women are denoted by hw metres. It is given that =hm Σ 2Σ =hw 16.25 and 9.08 . Σ =hw 9.84 , a Calculate the mean height of the 11 team members. 2Σ . b Given that the variance of the heights of the 11 team members is 0.0416 m2, evaluate hm 2 A and B are events such that ′ ∩ = ) A B ′ ) ] = 0.364, as shown in the Venn diagram opposite. , A B ∩ ′ = 0.196 P( P( ) P[( A ∪ B 0.286 and 1 A a Find the value of x and state what it represents. 0.196 x 0.286 b Explain how you know that events A and B are not mutually exclusive. c Show that events A and B are independent. B 0.364 3 Meng buys a packet of nine different bracelets. She takes two for herself and then shares the remainder at random between her two best friends. a How many ways are there for Meng to select two bracelets? b If the two friends receive at least one bracelet each, find the probability that one friend receives exactly one bracelet more than the other. 220 4 Every Friday evening Sunil either cooks a meal for Mina or buys her a take-away meal. The probability that he buys a take-away meal is 0.24. If Sunil cooks the meal, the probability that Mina enjoys it is 0.75, and if he buys her a take-away meal, the probability that she does not enjoy it is x. This information is shown in the following tree diagram. The probability that Sunil buys a take-away meal and Mina enjoys it is 0.156. 0.75 Enjoys the meal Cooks a meal a Find the value of x. b Given that Mina does not enjoy her Friday meal, find the probability that Sunil cooked it. 0.24 Buys a meal Does not enjoy the meal Enjoys the meal x Does not enjoy the meal 5 The following histogram summarises the total distance covered on each of 123 taxi journeys provided for customers of Jollicabs during the weekend. 30 20 10 10 8 Distance (km) 11 12 13 14 15 a Find the upper boundary of the range of distances covered on these journeys. b Estimate the number of journeys that covered a total distance from 8 to 13 kilometres. c Calculate an estimate of the mean distance covered on these 123 journeys. Copyright Material - Review Only - Not for Redistribution [2] [3] [2] [1] [2] [1] [4] [2] [3] [1] [2] [3] Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Practice exam-style paper 6 In each of a series of independent trials, a success occurs with a constant probability of 0.9. a The probability that none of the first n trials results in a failure is less than 0.3. Find the least possible value of n. b State the most likely trial in which the first success will occur. c Use a suitable approximation to calculate an estimate of the probability that fewer than 70 successes occur in 80 trials. 7 The following stem-and-leaf diagram shows the number of shots taken by 10 players to complete a round of golf Key: 6 1 represents 61 shots a Given that the median number of shots is 74.5 and that the mean number of shots is 75.4, find the value of x and of y. The numbers of golf shots are summarised in a box-and-whisker diagram, as shown. 16.8 cm b cm b Given that the whisker is 16.8 cm long, find the value of b, if the width of the box is b cm. c Explain why the mode would be the least appropriate measure of central tendency to use as the average value for this set of data. 8 To conduct an experiment, a student must fit three capacitors into a circuit. He has eight to choose from but, unknown to him, two are damaged. He fits three randomly selected capacitors into the circuit. The random variable X is the number of damaged capacitors in the circuit. a Draw up the probability distribution table for X . b Calculate XVar( ). c The student discovers that exactly one of the capacitors in the circuit is damaged but he does not know which one. He removes one capacitor from the circuit and replaces it with one from the box, both selected at random. Find the probability that the circuit now has at least one damaged capacitor in it. [2] [1] [4] [3] [3] [1] [3] [3] [4] 221 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 THE STANDARD NORMAL DISTRIBUTION FUNCTION If Z is normally distributed with mean 0 and variance 1, the table gives the value of ( )zΦ for each value of z, where Φ (z) Φ ( ) z = P( Z < z . ) Use Φ − ( z ) = − Φ for negative values of z.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 222 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 1.5 1.6 1.7 1.8 1.9 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9
979 0.9980 0.9981 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986 Critical values for the normal distribution The table gives the value of z such that P( <Z ) = , where Z ~ N(0, 1). ADD 12 12 12 11 11 10 10 16 16 15 15 14 14 13 12 11 10 20 20 19 19 18 17 16 15 14 13 12 10 24 24 23 22 22 20 19 18 16 15 14 12 11 10 28 28 27 26 25 24 23 21 19 18 16 14 13 11 10 32 32 31 30 29 27 26 24 22 20 19 16 15 13 11 10 36 36 35 34 32 31 29 27 25 23 21 18 17 14 13 11 .75 0.90 0.95 0.975 0.99 0.995 0.9975 0.999 0.9995 0.674 1.282 1.645 1.960 2.326 2.576 2.807 3.090 3.291 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Answers 1 Representation of data Prerequisite knowledge 1 a 62 m b 256.25 m2 2 3 Frequencies are equal because areas are equal. a 27 b 6 Exercise 1A Key: 1 0 represents 10 visits 2 a 15 16 17 Key: 15 0 represents 150 coins b $1615 3 a 18 d i 30–39 a 88 4 5 b 8 ii 10–19 b $10.80 c 20% c 0 and 3 Answers 3 a 33 b Boundaries at 1.2, 1.3, 1.6, 1.8, 1.9 m. Densities ∝ 170, 110, 210, 80. c 29 4 a Any u from 35 to 50. Densities 12.8, 23.2, 16, ∝ b Boundaries at 0, 5, 15, 30, cmu 64 30 u − ii 246 i 456 c . . 5 a 2.85 – 2.55 = 0.3 b Boundaries at 2.55, 2.85, 3.05, 3.25, 3.75 min. Densities ∝ 50, 125, 100, 20. c 2 min 45s or 165s. d i 3.5 6 a 324 b i 30 c Proof ii 3.01 ii 92 d 440; Population and sample proportions are the same. a Batsman P Batsman ; scored more runs Key: 6 3 1 represents 36 runs for P and 31 runs for Q 7 8 9 a 480 a 17 23 a 12 : 8 : 3 b 130 b 399 c 110 c 12.6cm 223 b =n 1 150 ii 36 0.720 mm c d i 210 < 0.215 <k ii P; scores are less spread out. 6 a Wrens (10) Dunnocks (10 Key: 8 1 9 represents 18 eggs for a wren and 19 eggs for a dunnock b 218 c 93% 7 The girl who scored 92%; 5 boys. Exercise 1B 1 a 175 and 325 years b All 150 years c Boundaries at 25, 175, 325, 475, 625 years. Densities ∝ 15, 18, 12, 6 (such as 0.1, 0.12, 0.08, 0.04). d 15 a 70 2 b Boundaries at 4, 12, 24, 28 grams. Densities ∝ 28, 33, 17.5. c 310 We can be certain only that 0.1 0.4 . 0.8 0.4<<a and that b< < 159, b = 636 b 23.5kg 10 11 = a a 4 hd 5 n 12 33cm 13 p = q29, = 94 Exercise 1C 1 a Points plotted at (1.5, 0), (3, 3), (4.5, 8), (6.5, 32), (8.5, 54), (11, 62), (13, 66). b i 23 ii 7.8s 2 a 19.5cm b Width (cm) No. books ) cf( <9.5 <14.5 <19.5 < 29.5 < 39.5 < 44.5 0 3 16 41 65 70 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Points plotted at (9.5, 0), (14.5, 3), (19.5, 16), 11 a Points plotted at (1.0, 0), (1.5, 60), (29.5, 41), (39.5, 65), (44.5, 70). i 34 or 35 ii ≈ 33.25 to 44.5cm c 3 a Points plotted at (0.10, 0), (0.35, 16), (0.60, 84), (0.85, 134), (1.20, 156) for A. Points plotted at (0.10, 0), (0.35, 8), (0.60, 52), (0.85, 120), (1.20, 156) for B. b i ≈ 107 for engine A; ≈ 87 for engine B. ii ≈ 108 c ≈ 42 a 17; cfs 20 and 37 are precise. 4 b c i 12 =k 4.7 to 4.8 ii 28 d It has the highest frequency density. 5 a i 64 ii 76 6 7 8 224 b ≈ 7.4g c (12, 304) a = 32, b = 45, c = 15, d = 33 a 65 b 24 a Ratio of under 155cm to over 155cm is 3:1 for boys and 1:3 for girls. b 81 or 82 c There are equal numbers of boys and girls below and above this height. d Polygon or curve through (140, 0), (155, 25), (175, 50). 9 a Points plotted at (18, 0), (20, 27), (22, 78), (25, 89), (29, 94), (36, 98), (45, 100). b 27 years and 4 or 5 months c i 1000 ii All age groups are equally likely to find employment. Either with valid reasoning; e.g. underestimate because older graduates with work experience are more attractive to employers. 10 Points plotted at (4.4, 0), (6.6, 5), (8.8, 12), (12.1, 64), (15.4, 76), (18.7, 80) for new cars. Points plotted at (4, 0), (6, 5), (8, 12), (11, 64), (14, 76), (17, 80) for > 100 000 km. Polygons 17 cars; curves ≈ 16 cars. (2.0, 182), (2.5, 222), (3.0, 242) for diameters. Points plotted at (2.0, 0), (2.5, 8), (3.0, 40), (3.5, 110), (4.0, 216), (4.5, 242) for lengths. b Least =n c 0; greatest =n 28. Diameter and length for individual pegs are not shown. Best estimate is ‘between 171 and 198 inclusive’. The length and diameter of each peg should be recorded together, then the company can decide whether each is acceptable or not. Exercise 1D 1 a Any suitable for qualitative data. b Pie chart, as circle easily recognised, or a sectional percentage bar chart. 3 4 2 Histogram; area of middle three columns > half total column area. 3 a Numbers can be shown in compact form on three rows; bar chart requires 17 bars, all with frequencies 0 or 1. b Sum 100 = be offered for sale. shows that 11 boxes of 100 tiles could 4 a 7 months b a 5 Percentage cf graph; passes below the point (12, 100). Histogram: Frequency density may be mistaken for frequency. Pie chart: does not show numbers of trees. b Pictogram: short, medium, tall; two, three and four symbols, each for six trees, plus a key. Shows 12, 18, 24 and a total of 54 trees. 6 a 1. Score (%) Frequency 30–39 40–49 50–59 60–69 70–79 80–89 90–99 3 5 6 15 5 4 2 b 2. Grade Frequency C 8 B 26 A 6 Any three valid, non-zero frequencies that sum to 40. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy c Raw: stem-and-leaf diagram is appropriate. Tables 1 and 2 do not show raw marks, so these diagrams are not appropriate. Table 1: Any suitable for grouped discrete data; e.g. histogram. Table 2: Any suitable for qualitative data. E.g. He worked for less than 34 hours in 49 weeks, and for more than 34 hours in 3 weeks. It may appear that Tom worked for more than 34 hours in a significant number of weeks. Histogram: boundaries at 9, 34 and 44; densities ∝ 98 and 15. 7 a b c Pie chart: sector angles ≈ Bar chart: frequencies 49 and 3. 339.2 and ° 20.8 . ° Sectional percentage bar chart: ≈ 94.2 and 5.8%. a Some classes overlap (are not continuous). 8 b Refer to focal lengths as, say, A to E in a key. Pie chart: sector angles 77.1 , 128.6 ,77.1 , ° ° ° ° 51.4 , 25.7 . ° Bar chart or vertical line graph: heights 18, 30, 18, 12, 6. Pictogram: symbol for 1, 3 or 6 lenses. 9 Country C SL Ma G Mo % of population 14.4 8.9 3.8 17.7 27.4 Answers End-of-chapter review exercise 1 1 i 50 ii Boundaries at 20, 30, 40, 45, 50, 60, 70 g. Frequency densities ∝ 2, 3, 10, 12, 5, 1. 2 3 16.5, 3 and 18cm a 6 b Quantitative and continuous 4 a 6 b Five additional rows for classes 0–4, 20–24, 25–29, 30–34, 35–39. 5 6 a = b9, = 2 a 48 b 0.7 cm 7 a 120, 180 and 90 b 6.75cm c There is a class between them (not continuous). 8 a 30 days for region A, 31 days for region B. 225 b Bindu: unlikely to be true but we cannot tell, as the amount of sunshine on any particular day is not shown. Janet: true (max. for region A is 106h; min. for region B is 138h). 0 in hundred thousands 10 5 People living in poverty as % of country population 15 20 25 30 Chile Sri Lanka Malaysia Georgia Mongolia Mongolia, for example, has the lowest number, but the highest percentage, of people living in poverty. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 9 i Points plotted at: (20.5, 10), (40.5, 42), (50.5, 104), (60.5, 154), (70.5, 182), (90.5, 200) or (20, 10), (40, 42), (50, 104), (60, 154), (70, 182), (90, 200) or (21, 10), (41, 42), (51, 104), (61, 154), (71, 182), (91, 200). ii 174 to 180 iii 58,59 or 60 2 Measures of central tendency Prerequisite knowledge 1 Mean 5, median 4.3, mode 3.9. = = = 2 1.94 Exercise 2A 1 a No mode b 16, 19 and 21 2 4 ‘The’ is the mode. 3 7 for x; −2 for y. 14–20 for x; 3–6 for y. 5 Most popular size(s) can be pre-cut to serve 226 customers quickly, which may result in less wastage of materials. 6 216 7 69 8 73 c 4 13 40 c 88 Exercise 2B a 50 1 b 7.1 7 a = ±p a 23.25 9 or −10 b =q b 1062 d 12 a 19 12 =a a 4.1 73.8% e 113.67 b 3.68825 b 24.925 8 $1846 2 3 4 5 6 7 9 12 9 cm2 9 13 a 54.6 b 59.0 c The scales may have underestimated or overestimated masses. Not all tomatoes may have been sold (i.e. some damaged and not arrived at market). 14 a 1.5 b c i 1.96 ii 3.48 For example, bar chart with four groups of four bars, or separate tables for boys and girls. 15 12 n = None of the 120 refrigerators have been removed from the warehouse. 16 One more day required. He works at the same rate or remaining rooms take a similar amount of time (are of a similar size). 17 a b i 5.89 cm 152.0 ° Exercise 2C a 74 1 ii 5.76 cm b 94 c 64 3 204 4 40.35 mm 18 –0.8 a To show whether the cards fit ( ( x > b 2% . 0) x < 0) , or not c –0.0535 24 + = 23.9465 mm Fidel; Fidel’s deviations 0< . Ramon’s deviations 0> , 63.5s; accurate to 1 decimal place. a 3 ; 60n ° 10 3.48 b n90 11 $1.19 2 5 6 7 8 9 30 years; the given means may only be accurate to the nearest month. Actual age could be any from 28 yr 8.5 m to Exercise 2D 1 5700; the total mass of the objects, in grams. to 31 yr 3.5 m. 10 a Mean ($10) is not a good average; 36 of the 37 employees ear
n less than this. b $7.25 11 a $143282 41% means from 1495 to 1531 passengers. 29% means from 1052 to 1088 passengers. b =k 252 2 3 4 5 a b 5∑ x or 5∑ x ∑ 0.001 x or 0.001∑ x or 0.01 w∑ 0.01w ∑ 3.6 a Calculate estimate in mph, then multiply by or 1.6. b 19.7625 1.6 × = 31.62 km/h 8 5 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Answers 6 7 8 x = 12.4; b = –3 a = b4, = 5 a (–1.8, 2.8) b (26, –6) c TE(5.2, –1.2) → (19, –2) ET(5.2, –1.2) → (–9, 14) Location is dependent on order of transformations. 9 p = q40; = 12 000; $75000 8 ‘Average’ could refer to the mean, the median or the mode. Median > 150; Mean < 150. 150 is close to lower boundary of modal class. Claim can be neither supported nor refuted. 9 There is no mode. Mean ($1000 000) is distorted by the expensive home. Median ($239000) is the most useful. Appears unfair; the smaller the amount invested, the higher the percentage profit. 281 g/cm2 x 10 10 a b p i r 54 40, 12, q = = = Reflection in a horizontal line through cf value of 30. Exercise 2E a 15 1 b Median; it is greater than the mean (12.4). c For example, being unable to pay a bill because of low earnings. 2 a 11.5 b Negatively skewed; Median 6; mode = = 3 a b 10.9 median < . t = 8 Median is central to the values but occurs less frequently than all others. Mode is the most frequently occurring value but is also the highest value. 4 5 c Two incorrect a ≈ 4.4 min b 2.8 and 6.4 min Points plotted at (0, 0), (0.2, 16), (0.3, 28), (0.5, 120), (0.7, 144), (0.8, 148). Median 0.4kg = a 92 b 32 6 Mode (15) and median (16) unaffected; mean 7 decreases from 16 to 14.75. a Points plotted at (85, 0), (105, 12), (125, 40), (145, 94), (165, 157), (195, 198), (225, 214), (265, 220). Polygon and curve give median ≈ 150 days. b Likely to use whichever is the greatest. Estimate of mean (152.84) appears advantageous. (They could consider using the greatest possible mean, 164.41.) ii Median safe current = median unsafe current 11 a First-half median is in 1–2; second-half median is in 4 – 5. b i 3 ii First half data are positively skewed (least possible mean is 100.8 s). 12 a Points plotted at (0, 0), (26, 15), (36, 35), (50, 60), (64, 75), (80, 80). Median 39%= New points at (0, 0), (16, 9.23), (26, 15), (40, 42.1), (54, 64.2), (80, 80). 227 b 20 13 a i f 5 6 7 8 9 10 11 x ii Mode mean median 8. = = = b c No effect on mode or median. Mean increases to 9. Curve positively skewed. – 11; No effect on mode or median. Curve =b negatively skewed. 14 Any symmetrical curve with any number of modes (or uniform). 15 a Symmetrical; mean median mode = = b Chemistry: negatively skewed; Physics: positively skewed. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 End-of-chapter review exercise 2 a Mean medianand mode 1 . < b Mean medianand mode > . 2 3 c n = a b =Mean medianand mode. 22; 623g =Mode 13 Median 28 = c 55 4 a 15.15 b 13.3 5 a 6 b i 14 ii 25 25 a Proof 6 7 b $0.18; it is an estimate of the mean amount paid. 8 a Mode indicates the most common response. Median indicates a central response (one of the options or half-way between a pair). 228 b Allows for a mean response, which indicates which option the average is closest to. 9 a 1 b No; it is the smallest value and not at all central. c 11 d Positively skewed; mode median mean. < < 10 i Boundaries at 0.05, 0.55, 1.05, 2.05, 3.05, 4.55h. Frequency densities ∝ 22, 30, 18, 30, 14. ii 2.1h 11 16.4 12 81 13 14 15 a b Mode = 0, mean 1, median 0 = = Mean; others might suggest that none of the items are damaged. a 4006 − 2980 = $1026 b $3664 1.95 3 Measures of variation Prerequisite knowledge 1 16cm 2 a 4.5 b 27.3 Exercise 3A Box plots given by: smallest … … … … Item (units), as appropriate. Q2 3 Q Q 1 largest / b 35 and 20 d 96 and 59 1 a 25 and 17 c 65 and 25 e 8.5 and 5.6 2 a Range = 3.3; IQR 1.75 = b Negative 3 a 41 and 18 b 9 ... 28 ... 37 ... 46 ... 50 / Marks. c =Q 3 –2 2 Q Q 1 4 a Yes, if the range alone is considered. b Hockey: 11 ... 13 ... 17... 20 ... 24 / Fouls. Football: 10 ... 18.5 ... 20 ... 22.5 ... 23 / Fouls. with the same scale. Fewer fouls on average in hockey but the numbers varied more than in football. 5 a b Ranges and IQRs are the same (35 and 18) but their marks are quite different. One of median (33/72) or mean (33/72) and one of range or IQR. 6 a Points plotted at (35, 0), (40, 20), (45, 85), (50, 195), (55, 222), (70, 235), (75, 240). 35 ... 43.1 ... 46.6 ... 49.3 ... 75 / Speed (km/h), parallel to speed axis. b Positive skew 7 a i Males: 0 ... 0 ... 3 ... 14 ... 39 / Trips abroad. Females: 3 ... 5 ... 12 ... 20 ... 22 / Trips abroad. Same scale. ii Males: range = 39; IQR 14; median 3 = = Females: range 19; IQR 15; median 12 = = = On average, females made more trips abroad than males. Excluding the male who made 39 trips, variation for males and females is similar. b No, there are no data on the number of different countries visited. 8 9 a Ω 0.130 th percentile ≈ 68≈ c a 52 cm2 b c 15.2 to 16.0 cm2 b d ≈ ≈ 0.345 0.095 Ω Ω 4.0 ... 25.8 ... 33.2 ... 38.8 ... 56.0 / Area (cm2). d Area < 6.3cm2 or area 58.3cm2 . > Estimate ≈ 8 (any from 0 to 15) Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 10 a Points plotted at (–1.5, 0), (–1.0, 24), (–0.5, 70), 11 x = 12, y = 18 Answers (0, 131), (0.5, 165), (1.0, 199), (1.5, 219), (2.5, 236). Gudrun is 22 years old. Variance increases from 69.12 to 72.88 years2. None of the original 50 staff have been replaced. b 30 12 a Mean decreases by 11.6cm. b 89.9 and ° °1.3 c 18% a 10 11 12 a Points plotted at (0, 0), (4, 2), (11, 21), b c d (17, 44), (20, 47), (30, 50). i ≈ 0.06g ii ≈ 0.12 g 40 ≈n Variation is quite dramatic (from 0 up to a possible 3% of mass). Mushrooms are notoriously difficult to identify (samples may not all be of the same type). Toxicity varies by season. 13 Should compare averages and variation (and skewness) and assess effectiveness in reducing pollution level for health benefits. Exercise 3B a 1 Mean 37.5, SD 12.4 = = 2 3 4 b a Mean 0.45, SD 9.23 = = Var(B) Var(C) Var(P) = = = 96 b The three values are identical. No; mean marks are not identical (B 33, C 53 and P 63). = = = Mean 1 or1.69; variance 1.64 = = 24 35 a b Mean 2; SD 0.803 = = . Q Q 2, soIQR 0 = = = 3 1 That the middle 50% of the values are identical. 5 a Girls: mean 40, SD 13.0 min Boys: mean 40, SD 16.3min = = = = b i On average, the times spent were very similar. ii Times spent by boys are more varied than times spent by girls. 5.94cm k = 6;Var( x ) = 2.72 a a = b13, = 40 b 6.23cm 43; SD 12.5km; IQR 24km; IQR 2 SD = ≈ × = Mean 0.97t; SD 0.44t = = 6 7 8 9 10 k = a b Median decreases by 40.3cm. b SD increases by 116cm. IQR increases by 216cm. (Range increases by 344cm.) c Discs get closer to P, but distances become more varied. d Proof Exercise 3C a 65.375 1 c 120 b 9.17 d 161 800 e 28 n = x20; = 11 2.15 Mean 60.2 kg; SD 14.1kg = = a Proof a ∑ = y 2 35 b =n 687.5 b 27.2 psi 229 2 52 n − 616549 4915 n + n + 29 8 31 Proof S 2 or 10 Exercise 3D 1 Men 8kg; women 6kg 2 3 4 5 1.5 7.92 mm and 24009.8 8 =n 15 6 Mean is not valid (it is 165cm); standard deviation is valid. 7 8 Mean 4h 20 min; SD 7.3min = = If 10-minute departure delay avoids busy traffic conditions. 0.96cm2 Mean decreases to 0.73t; SD increases to 0.57t. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 9 10 11 12 a c Mean 8; SD 4 = = b Mean 7; SD 4 = = 1 ; variance of the first n positive odd integers. 2 −n 3 a 43 a Proof b Proof c 1.179 2 ∑ = 15 416 y c 20.1376 y 1104, b ∑ = a 162.14 cm. 2 b ( ∑ x = 5 720 640, 2 ∑ = 7 445 100); y Var( X ) = 42.1004 cm 2 Exercise 3E $0.64 1 2 8.5 3 2.64 4 a 133 and 2673 b 0.457 C° , using 133 and 2673. c 0.209 ( C)2 ° 5 6 7 230 $75600 °27 F a b Mean 12.5 C; SD 4.5 C ° = = ° a b c Fruit & veg; mean unchanged, so total unchanged. Tinned food; mean increased but standard deviation unchanged. Bakery; mean and standard deviation decreased by 10%. 8 26 m 9 18.0% increase End-of-chapter review exercise 3 1 a Proof b $0.917 or $0.92 2 a 0 3 4 5 b 0, 1, 2, 3 or 4 Five a 0.319 m b a Mean increased by 1.5cm (to 90 cm); SD unchanged. Marks are improving and becoming more varied. b Third test c First test positive; second test negative 6 a 97.92 cm b 11.5cm 7 a Range 139; IQR 8;SD 37.7 = = = b IQR; unaffected by extreme value (180). 8 i 173cm ii 834728.6 and 4.16cm Squad Key: 1 9 4 represents 91 kg for squad A and 94 kg for squad B 9 10 45.8 and 14.9s i Squad 10 11 12 ii 18kg iii 103.4kg 11 i 126.5cm ii 4908.52 cm2 12 i =Mean 40.9 or 40 8 9; =SD 8.30 ii 8.41 13 5514 14 SD increases by 68.4%. IQR increases by 9.30% (or 6.90%, depending on method). Proportional change in SD is much greater than in IQR. 15 14.0 cm 16 a =SD 21.5 b Mean –2; SD 21.5 = = Mean is affected by addition of –202 but SD
is unaffected. 17 18 2.75 b a 5.6 2 > × a 19.8 − 18 = 1.8 b ∑ = 1964.46, a 2 2 b ∑ = 2278.12; 1.99 years 2 803 140 ; 132 ≠ 2 ≠ 44 2 c 1.63 Cross-topic review exercise 1 1 a 25 b Player =A 25, player =B 21 Key: 1 8 represents 18 games 2 a 10–15 and 26–30 26 – 15 11 = b No. incorrect answers No. candidates 0 1 c 19.6 3 a 2.3cm b 0.0178 m 1–9 10–14 15–24 25–30 31– 40 19 23 27 24 18 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 4 a 17.5 b 126.3125 c Student A and student F 5 Higher average and less varied growth Key: 1 1 represents 11 unwanted emails b 8 ... 14 ... 20 ... 27 ... 36 / Unwanted emails. 7 a 5.94 and 6.685 b =Mean 2990 000 or =SD 366 151 or 2.99 106 × 3.66151 105 × 8 a 32 b 70 and 75km/h c 72.3km/h 9 a Proof b 36.09g and 0.67 g c 0.4489g2 10 a i 75 ii ≈ 69 b 12 a 50 and 1.80 S 50 – =C b SD( C ) SD( = aS b + ) 11 12 Answers 3 ℰ A 2 5 B 1 3 n( ) 4 A B∪ ′ = and n( A B′ ∩ ) 1 = 4 Exercise 4A 36 b 2 3 a The team’s previous results. b 8 They may win some of the games that they are expected to draw. c 12 a 300 a 5 50 3 8 1 1953 b At least 240 b 15 231 –1 and =b a = a 1.48 50 Exercise 4B b 79 c i 13 a = b7, = 9 Median 0.825cm; IQR 0.019cm = = ii q = r4, = 2 iii X : 0.802 ... 0.814 ... 0.825 ... 0.833 ... 0.848 / Length (cm) Y : 0.811 ... 0.824 ... 0.837 ... 0.852 ... 0.869 / Length (cm) Same scale iv Longer on average in Y ; less varied in X . 4 Probability Prerequisite knowledge 1 30 2 1 12 Girls who took the test. b 23 40 i a 3 5 10 11 b Not a female sheep. Not a male goat. ii 4 a i (3, 3) ii (2, 4) and (4, 2) iii (2, 2), (4, 4), (6, 6) b X Y, and Z are not mutually exclusive, b = 2, c = 6 3 5 ii 13 25 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 7 a C H 12 7 13 3 10 b i 44% a 10 11 8 9 ii 5 8 b 6 11 8 c 7 22 10 a Students who study Pure Mathematics and Statistics but not Mechanics. b i 89 100 ii 6 25 c Mechanics, Statistics, Pure Mathematics 11 a No; P( X Y ) ∩ ≠ b 0.9 12 a A and C 13 a 2 75 14 15 a 0.6 a A 232 0 or equivalent. c 0.7 b 0.22 b 31 75 b 0.4 B c 29 75 9 3 1 2 7 5 0 C b 19; they had not visited Burundi. c d They had visited Angola or Burundi but not Cameroon; 15. 2 9 1 Exercise 4C 1 2 a 1 36 a 0.012 2 3 4 5 6 0.42 a 0.84 a 0.343 b 0.85 b 0.441 b 1 4 c 1 9 b 0.782 7 8 9 a b a a i 0.544 ii 0.3264 iii 0.4872 The result in any event has no effect on probabilities in other events. E.g. winning one event may increase an athlete’s confidence in others. 3 5 8 24 b Untrue. Any number from 0 to 10 may be delivered; 9 is the average. b 0.125 c 0.49 10 a 9 25 or 0.36 b 111 400 or 0.2775 11 a 0.84 12 a 1 4 13 14 27 512 a 0.1 15 a i −k 5 25 16 49 625 b a =k 8; 1 6 i b i 0 Exercise 4D 0.63 1 2 3 4 5 0.28 a 0.32 0.35 ) P( B 0.7 a i b i a 28 b 0.9744 b 3 8 b 0.15 c 0.3 ii 3 −k2 25 ii ii 2 9 1 108 b 0.48 ii 0.4 P( C ) ii 0.5 iii 0.06 D S 13 6 7 2 S ′S Totals 6 7 13 13 2 15 19 9 28 D ′D Totals Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy b No; 6 28 ≠ 19 28 × 13 28 6 Yes; 20 80 = 7 a P( ) A = b No; ≠ 1 2 × 32 80 9 16 9 16 50 80 × 3 4 , P( B ) = 3 4 , P and B both occur when, for example, 1 and 2 are rolled; P( 0 8 a P( Y , P( and ) X Y = 1 12 Yes; 1 12 = 1 4 × 1 3 b No; X and Y both occur when, for example, 1 ≠∩X Y and 5 are rolled; 27 64 V W ∩ 1 16 , P( , P( 1 8 W P( V No; 1 16 ≠ 1 8 × 27 64 10 a B B′ Totals 60 50 110 48 42 90 108 92 200 M M ′ Totals b Ownership is not independent of gender; e.g. for M and B: 60 200 ≠ 108 200 × 110 200 . c Females 54.3%, males 55.6%. If ownership were independent of gender, these percentages would be equal. 11 a = 1860, b = 4092, c = 1488 12 Southbound vehicles; 54 207 69 207 18 207 × = 36 207 = 54 207 × 138 207 or Exercise 4E 10 13 2 3 3 4 11 19 3 4 4 7 12 19 12 23 Those who expressed an interest in exactly two (or more than one) career, or any other appropriate description. 5 16 ii b i Answers a a 20 39 1 5 b b 8 39 47 57 or 0.825 a 10% of the staff are part-time females. b a = 0.2, b = 0.4, c = 0.3 4 7 i c 3 5 a Proof ii 3 4 iii 4 9 9 11 21 b P(3) = 0.08, P(2) = 0.16, P(1) = 0.75 5 6 7 8 10 c d 25 33 32 107 or 0.758 or 0.299 Exercise 4F a 28 55 a 3 28 7 22 b 5 14 b 2 33 > 42 132 a 0.027 a Two girls; b 229 or 230 20 132 b Equally likely; both 1 9 a or 0.316 1 66 10 11 12 or 0.3525 6 19 141 400 1 5 7 8 or 0.72 or 0.2 or 0.875 or 0.123 4 7 or 0.571 a a c 18 25 9 73 a 13 a =y 0.44 14 15 16 17 =x 0.36 or 0.548 a 23 42 0.48 16 25 or 0.64 233 or 0.553 1 4 26 47 1 3 b b or 0.333 b b 13 35 7 20 or 0.371 or 0.35 b 0.812 or 272 335 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 End-of-chapter review exercise 4 1 53 8 11 or 0.727 104 2185 1 316 or 0.0476 or 0.00316 or 0.310 or 0.675 a b 77 248 27 40 a 2 3 4 5 6 M 1 BE 5 7 4 39 38 3 3 RH iii Yes; P(high GDP and high a iv or 0.431 birth rate) = 0 287 666 4 15 9 19 0.198 b a b c A and B both occur or it shows that P( A B ) ∩ ≠ 0. Only two of the 36 outcomes, (1, 3) and (3, 1), are favourable to A and to B. No; 1 18 ≠ 1 6 × 1 2 to show P( A B ∩ ≠ ) P( A ) P( × B ). 0.26 i a =x 54; 312 adults 29 34 3 4 ii 12 13 14 15 16 17 or 0.00566 or 0.166 0.034 and =y 0.04 234 ii i ii or 0.696 14 2475 82 495 0.3 0.7 y x + = 0.02, y x = = 16 23 8 9 1 4 1 3 8 105 2 35 or 0.0571 or 0.0762 x2 b 18 17 72 8 35 20 a 19 13 30 i ii 1 8 1 4 b 22 5 Permutations and combinations Prerequisite knowledge P( \| B A ) = 2 3 10 i 0.85, 0.15 / 0.8, 0.2 / 0.4, 0.6 on branches with labels T, B / J, X / J, X. ii 11 i ii 17 26 37 85 19 48 or 0.654 or 0.435 or 0.396 Exercise 5A a 20 1 d 162 a 10 a 11 E.g. 144 53! 51! cm2 2 3 4 5 b 6 e 224 b 9 b 15 c 294 c 4 c 22 = 9! 2! × 7! ; 252 = 7! 3! × 5! ; 1 1 2 = 15! 4! × 16! Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 6 7 − 25! 22! E.g. $ 8! 5! cm3 5!(5! − 9! + 4! 2! 2!) + Exercise 5B 720 1 2 a 8.07 1067 × b 24 c 6227 020800 3 4 5 6 7 a 2 a 24 39 916 800 362 880 =n 19 Exercise 5C 1 a 120 b 720 b 6 c 40320 c 5040 b 360 c 45360 d 34650 e 415800 2 a 6 c 60 a 6 c 6435 3 w 4 b 20 d 15 b 1 d 99768240 First student is correct. Second student has treated them as two identical trees and three identical bushes. 5 a 1024 b i 252 ii 386 6 One letter appears three times; another appears twice, and two other letters appear once each (e.g. pontoon, feeless, seekers, orderer). 7 a 10 b 50 c 1050 Exercise 5D a 120 1 b i 48 ii 72 2 a 48 d 144 2 : 1 a 80640 a 3600 a 20 a 6 3 4 5 6 7 b 192 e 0 b 241 920 b 720 b 40 b 180 iii 18 c 480 c 240 c 36 Answers 8 9 a 1 c 8 b 0 d 20 x y > + 1 or x ù y + 2 or equivalent Exercise 5E a 2520 1 2 3 4 5 6 7 8 9 665280 6840 a 182 a 60 a 272 a 60 480 360 a 12 b 3024 b 196 b 240 b 132 b 1680 b 48 c 140 10 120 ways for =r( ) 3 passengers to sit in empty seats on a train, or use of n( , or ! n !( r n − ) 6 120 1. P r )! 235 11 a r 1 n> 2 12 132600 13 18144 14 a 6652800 b 3024000 c 4959360 Exercise 5F a 56 1 a 1960 b 126 b 980 c 121 ii 230230 2 3 4 5 6 7 8 9 a 2598960 b 845000 i 230230 x y = + z a b 16 161 a 120 c 12 45 b 34 d 66 They can share the taxis in 56 ways, no matter which is occupied first. 10 a 184756 c 63504 11 330 12 1058400 13 a 252 b Two d 88200 b 56 c 175 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 14 27907200 15 72 16 a 18 Exercise 5G 1 2 3 4 5 6 7 8 9 236 or 0.457 or 0.293 a a 1 3 21 46 27 92 3 4 a 0.0260 b c 0.0773 0.501 0.588 a 5 16 a 2 3 a 0.331 10 a 50 400 b i 1 120 a 11 1 84 12 0.290 or 0.0119 c 8 15 c 5 12 b 132 b 2 15 b 0.197 b 1 2 b 1 12 b 0.937 ii 1 60 b 2 9 4 a 1000 000 5 6 b i 0.01 ii 0.0001 17 280 i 1 663 200 ii 30 240 iii 1 622 880 iv 10 7 a 756 756 b 72 072 8 a 330 b 70 c 265 9 91 10 87 11 12 13 14 15 16 a 10 or 1 10 × 9 9 6 10 or 7.29 10 × 8 8 6 10 or 1.25 10 × b c 3 9 × ×9 108 3 × d 5 9 14 44 286 a 453 600 b 86 400 1 10 a 11 values; 35 13 a a = 166, b = 274, c = 488 b 0.162 28 41 or 0.683 14 15 Six tags and three labels. 16 a b , 3 5 and 2 3 for =n 2, 3, 4 and 5. , 1 3 1 2 2 −n 1 End-of-chapter review exercise 5 1 a 30 240 b 240 2 a 32 659200 b 8467200 3 28 3 11 12 2401 2916 b 17 20; 18 156 or 0.823 19 a i 648 ii 104 b 2700 20 i 50 ii 18 Cross-topic review exercise 2 1 a 96 b −71 c 9 1 8 or 9.125 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge
University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Answers 16 31 > 15 31 c 39 916 800 d 59 512 320 16 a 229 975200 3 4 6 7 8 2 a 62 5 a 27 b Odd; 30856 a i 48 ii 24 b 120 1 3 4 9 b i ii 1440 a 604 800 b 8467200 i b a 134596 1 24 1 19 10 19 ii c i ii 13 24 ; more likely. ; less likely. 9 a 1287 10 a 3 b 45 c 270 b 15 11 a 81 b 15 12 a i ii 14 29 4 29 or 0.483 or 0.138 b 0.437 13 a S S′ P P′ 30 45 75 10 15 25 40 60 100 or appropriate Venn diagram. b 0.75 17 a i ii iii 8 17 73 153 32 153 or 0.471 or 0.477 or 0.209 b The events ‘being on the same side’ and ‘being in the same row’ are not independent. 6 Probability distributions Prerequisite knowledge 0.11 1 ) =DP( 2 With replacement: P (both red) = 3 6 × Without replacement: P (both red) = or 0.20 Exercise 6A 1 v =P( V v ) 1 0.4 2 0.4 2 3 =p a b ; 2 13 50 k 5 13 2 – 25 k 3 + = 0; k = 0.2, k = 0.3 k = 0.3 gives P( W = 12) = – 0.1 . c 0.14 4 s =P( S s ) 5 a Proof b r 0 4 81 1 28 81 237 or 0.25 = 1 5 3 0.2 2 49 81 0 1 2 3 × 40 100 ) P( S × 75 100 ). P = P( and ) P( S b Yes; e.g. 30 100 P = a 312 or 531 441 b 310 or 59 049 c 37 or 2187 14 15 a 3 628 800 b 7 257 600 to show =P( R r ) 0.226 0.446 0.275 0.0527 c 0.774 6 a Proof b v =P( V v ) c 69 91 0 24 91 1 45 91 2 20 91 3 2 91 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 7 Number of red grapes selected ( Number of green grapes selected ( {0, 1} ); R R ∈ ); G G ∈ {4, 5} 18 a =k 315 1012 b 21 46 R G 5 + = 8 d P( = D d ) 9 x 0 0.1 0 1 0.6 1 2 0.3 2 P( = X x ) 0.4096 0.4608 0.1296 Hair colour and handedness are independent. a Proof 10 2 1 16 3 2 16 4 3 16 5 2 16 6 3 16 7 2 16 8 2 16 10 1 16 1 1 14 2 6 14 3 6 14 4 1 14 b x P( = X x ) P( X > 6) = 5 16 11 a 0 b n P( = N n ) c Symmetrical 12 =k 1 27 13 a =c 1 86 b 61 86 14 a 0.374 238 b =N 0 is more likely than P( each time a book is selected. ) =N 4; ) P( ′ > N N 15 a Proof b x P( = X x ) XP( is prime) = 16 a P(heads) 0.2= 0 1 12 7 12 1 4 12 2 4 12 3 3 12 b The number of tails obtained, but many others are possible, such as 2H and 0.5H. P( T H> ) = 0.896 17 a s =P( S s ) b 1 3 1 17 36 2 9 36 3 10 36 Exercise 6B E( 1 X ) = =p E( Y a b 0.2 2.1; Var( X ) = 0.93 2 3 4 5 6 7 ) 1.84; SD( = Y ) = 0.946 E( T ) = 5, Var( T ) 11.5 = m = 16; Var( V ) = 31.3956 831 ) =RVar( a 11; Var( = W ) = 79.8 a E(grade) profit. = 3.54; SD(grade) 1.20; A smallish = =SD 1.20; variability of the profit. E(grade) 2.46, SD(grade) 1.20 = = b Both are unchanged. 8 a x P( = X x ) 1 1 36 2 3 36 3 3 36 4 5 36 5 3 36 6 10 12 15 20 30 9 36 2 36 4 36 2 36 2 36 2 36 b E( X ) 8 = X > E( X )] = 1 3 ; P[ 5 12 49 41 48 ) X = c Var( 9 a h or 49.9 0 1 2 3 P( = H h ) 0.343 0.441 0.189 0.027 b 900 times E( G ) = 0.8; E( B ) 1.2 = 10 a b 2 : 3; It is the same as the ratio for the number of girls to boys in the class. 336 725 c =GVar( ) 0.463 or 11 a Proof b c ) 1.125 =RE( =GE( ) 1.5 12 a $340 b If the successful repayment rate is below 70%. 13 14 a Proof a 1, 2, 3, 5. b =n 35 s =P( S s ) 1 4 12 2 1 12 3 4 12 5 3 12 b P( S > 3 2 ) 4 = 7 12 c =SVar( ) 2 17 48 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 15 a Proof b x P( = X x ) c ) Var( X ) E( X = 1 4 0 1 256 1 12 256 2 54 256 3 108 256 4 81 256 ; the probability of not obtaining B with each spin. End-of-chapter review exercise 6 1 2 X 9 14 ; E( =  k   ) 1 45 =XVar( 196 ) = 2 or 2.36 5 14 or 1.23 13 or =q 48 a =q b 34 3 a $6675 4 5 6 7 b 8 9 b 4.27 6 11 0.909 2 5 3 5 a i ii b j =P( J j ) a =S 1 0 0.3 1 2 0.6 0.1 s =P( S s ) 0 1 1 36 13 36 2 2 36 3 3 36 4 5 7 8 9 11 14 15 19 24 2 36 2 36 2 36 1 36 2 36 2 36 2 36 1 36 2 36 1 36 ) =SE( 5 31 36 a 0, 1, 2, 4, 5. or 5.86 b x Pb 1 or =b 13 30 6 Answers 10 i Proof ii Score P(Score) 0 24 70 2 30 70 4 13 70 6 3 70 11 12 13 14 15 and 2.78 or 0.4 iv iii 1 6 7 2 5 = k  5 3   ; P Y( > 4 ) = 2 3 25 28 a =x 11 22 127 i Proof b ii x P( = X x ) iii 1 3 i Proof ii x P 120 60 40 30 24 20 171 7 15 13 1 3 1 45 2 45 3 45 4 45 5 45 6 45 7 45 8 45 9 45 239 iii 13 1 4 9 iv 3 or 13.3 or 0.444 7 The binomial and geometric distributions Prerequisite knowledge 1 105 2 1 64 + 9 64 + 27 64 + 27 64 = 1 Exercise 7A a 0.0016 1 c 0.0256 2 a 0.0280 c 0.710 3 a 0.0904 c 0.163 b 0.4096 d 0.0272 b 0.261 d 0.552 b 0.910 d 0.969 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 4 a 0.121 b 0.000933 c 0.588 10 a 6.006 d 0.403 a 0.246 0.0146 0.254 a 0.140 0.177 a 0.599 0.349 a 0.291 a 0.330 a 0.15625 or 5 6 7 8 9 10 11 12 13 14 15 9 17 16 e 0.499 b 0.296 b 0.000684 b 0.257 b 0.648 b 0.878 16 6 18 23 5 32 b 0.578 19 a 0.0098 b a = 208, b = 3 c 68 20 a =p 0.5; the probability of more than 5 m of rainfall in any given month of the monsoon season. b The probability of more than 5 m of rainfall in any given month in the monsoon season is unlikely to be constant or Whether one month has more than 5 m of rainfall is unlikely to be independent of whether another has. 21 a 0.6561 b 0.227 22 0.244 240 Exercise 7B 1 a 1, 0.8 and 0.894 b 13.2, 5.94 and 2.44 2 3 4 5 6 c 65.7, 53.874 and 7.34 d 14.1, 4.14 and 2.04 a 2 and 1.5 a 0.752 b 0.311 b 0.519 c 0.367 n = 50, p = n = p3, 42, p = = 0.9 a a n = w 0.4 b 0.109 7 12 b 0.0462 0 1 2 3 P( = )W w 0.001 0.027 0.243 0.729 7 a E.g. X is not a discrete variable or there are more than two possible outcomes. b E.g. Selections are not independent. c E.g. X can only take the value 0 or X is not a variable. 8 9 =n 18; 0.364 0.75, p k = = 5157 b 5.93 and 5.93 c Proof d 0.197 11 a 46 b 3.68 c i 0.566 ii 0.320 Exercise 7C a 0.0524 1 a 0.148 a 0.125 a 0.0465 a 0.24 a a i 0.032 i 0.0315 2 3 4 5 6 7 8 9 10 11 b 0.91808 b 0.901 b 0.875 b 0.482 b 0.922 ii 0.0016 ii 0.484 c 0.4096 c 0.0672 c 0.0280 b 0.2016 iii 0.440 b Faults occur independently and at random. a 0.21 a 0.364 0.0433 b 0.21 b 0.547 c 0.21 a Not suitable; trials not identical (p not constant). b Not suitable; success dependent on previous two letters typed or X cannot be equal to 1 or 2 or p is not constant. c It is suitable. d Not suitable; trials not identical (p not constant). 12 0.096 13 0.176 14 0.977 or 335 343 15 16 a 0.0965 or 0.103 125 1296 b 0.543 Exercise 7D 14 81 Mode 1, mean 2 = = 6 and 0.335 a 16 b 0.00366 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Answers − 2 ; n = 10 7 a Thierry 10 or 0.0574 45 784 With replacement, so that selections are independent. 8 b a b i or 0.105 ii or 0.0625 27 256 1 16 ) 500; = 9 E( X b = 1001 21 =k n3  =    2 a 12 11 k 12 b i 0.263 ii 0.866 iii 0.0199 13 i 0.735 ii n = 144; k = 6 10 a Any representation of the following sequence. 14 36 : 30 : 25 1st toss 2nd toss 3rd toss T T T T H 2 + 4 0.5 + 6 0.5 8 0.5 + + ... Anouar Zane b c 0.5 2 3 End-of-chapter review exercise 7 − 1 n 1 n −    n a 0.147 1 2 b 0.00678 3 i 1 4 ii 0.0791 or iii 0.09375 or 81 1024 3 32 4 a 4 9 b 0.394 5 648 5 a 0.59049 b 0.40951 c 0.242 6 a 10 7 8 b 0.00772 or c 0.00162 a 2–12 b a 137 × 2–16 i 0.0706 ii 0.0494 iii 0.1176 b The students wear earphones independently and at random. 9 i 0.993 ii =n 22 8 The normal distribution Prerequisite knowledge 1 23.4 and 11.232 2 n = 32, p = 0.35 Exercise 8A a False 1 d False 2 a i σ σ> P Q b True e True c False f False 241 ii Median for P < median for Q. iii IQR for for Q. IQR P > b i Same as range of P. ii iii No; High values of W are more likely than low values or negatively skewed Pμ μQμW Women & men 170 Height (cm) Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics Apple juice Peach juice 340 Volume (ml) b Peach juice curve wider and shorter than apple juice curve; equal areas; both symmetrical; both centred on 340 ml UK USA 3.3 3.4 Mass (kg) b USA curve wider, shorter and centred to the right of UK curve; equal areas; both symmetrical. 242 6 a Proof b σ X = 1.11 > σ Y = 0.663 Exercise 8B a 0.715 1 c 0.937 e 0.207 g 0.0401 i 0.975 2 a 0.0606 c 0.0400 e 0.190 g 0.770 i 0.719 X Y 2.4 2.6 b 0.993 d 0.531 f 0.0224 h 0.495 j 0.005 b 0.380 d 0.0975 f 0.211 h 0.948 j 0.066 k 1.333 0.600 =k =k =k =k 1.71 0.473 =c – 0.674 – 1.473 =c =c =c =c 2.10 – 0.500 3.09 0.497 .371 – 0.380 0.111 =k =k 1.884 =k =k =k 1.035 0.003 =c =c 1.245 – 2.14 =c =c =c –1.90 1.96 Exercise 8C a 0.726 1 b 0.191 c 0.629 2 a 0.919 and 0.0808 b 0.613 and 0.387 c 0.964 and 0.0359 d 0.0467 and 0.953 f 0.954 h 0.319 j b d b d 0.0994
=b 15.5 = 23.6 d =g =j 42.7 17.5 e 0.285 and 0.715 g 0.423 0.231 i a c 35.0 18.5 =a =c e e a = 86.8 11.4 =f =h 0.0513 9.80 c 0.933 3 4 5 6 7 σ = 2.68 12.6 8 µ= 9 µ = 58.8, σ = 14.7 10 µ = 93.8, σ = 63.8 = 5.00, σ = 6.40; 0.0620 = = 7.08, σ = 1.95; 0.933 5.78, σ = 2.13; 0.372 11 µ 12 µ 13 µ 14 0.831 Exercise 8D 0.662 1 2 3 a 0.191 b 74 a Small = 28.60%; medium 49.95%; = = large 21.45% =k 58.0 or 58.1 b Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 4 5 6 7 8 µ= 7.57 a 0.567 240 9 days a =b →9.09 5000 b 0.874 b 82.0 m c 0.136 c 0.257 c 0.118 b =n 14 b 0.189 b 0.0228 4.8 9 σ = 3.33 10 µ= 91.2; 28.8% 11 σ = 3.88 12 a σ = 1.83 a µ= 25.0 a 0.683 1.64, c = σ a 0.950 a 0.659 a 0.284 13 14 15 16 17 b 23 b =n 1000 b 0.0456 µ = 6.39 Exercise 8E a Yes; µ 1 b No; nq c Yes; µ d No; np = 2 12, σ= = <1.5 5 2 5.2, = σ = <3 5 = 4.524 2 3 4 5 6 a 209 =n =n 11 B(56, 0.25) c b d 34 =n =n 17 0.837 0.844 a p = 0.625; Var( H ) = 37.5 b 0.0432 7 a Proof b 0.292; np = 10 5 and > nq = 30 5 > b 4.45 ii 0.0118 8 9 a 44 a b i 0.187 E( X ) 1600; Var( = X ) = 320 c 0.874 10 a i 0.105 ii 0.135 b 0.145 0.0958 a 0.0729 a 0.239 a 0.789 11 12 13 14 15 0.100 16 17 0.748 0.660 b 0.877 b 0.0787 b 0.920 Answers 243 End-of-chapter review exercise 8 1 0.841 2 3 4 5 6 0.824 i 0.590 ii np = 24 5 and > nq = > 6 5 0.287 0.239 .5 2.0 7 i 0.035 ii 0.471 iii a =k 103 i 315 or 316 8 ii 7350 iii 0.840 b 0.933 9 σ = 2.35 10 µ = 3.285; 61.3% 11 5.69% 12 i 0.238 ii =k 116 iii 0.0910 13 a 0.408 b 0.483 c 0.156 14 a µ = 17.5, 2 σ = 58.0 b i 17.0 min ii →38.2 38 days 15 a σ = 7.24 b =k 15.1 16 0.936 Cross-topic review exercise 3 1 i Proof a , Var( ) = 3 4 or 0.9975 ii E( X iii 399 400 X ) = 63 80 or 0.7875 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Practice exam-style paper 1 a 1.72 m b 16.75 2 a b =x 0.154; the value of A )∩ or A B ) ∩ ≠ 0 or equivalent. A B P( P( P( and ) B c Proof 3 a 36 b 5 9 a 0.35 or 0.693 b 95 137 a 11km 4 5 b 61 c 10.8km 6 a 12 b First trial c 0.176 7 a b c 4 x y7, = 6.6 = =b It is neither central nor representative or 8 of the 10 values are less than 89. 0 10 28 1 15 28 2 3 28 8 a x P( = X x ) or 0.402 b c 45 112 11 15 2 3 4 5 6 7 or 0.325 b 13 40 19 27 µ = 25.8, σ = 7.27 0.0228 a σ = 2.99 b 26.1% or 26.2% a µ= 34.0 b 11.9% a σ = 2.70 b 0.276 c 0.822 8 a x P( = X x ) 0 140 285 1 120 285 2 24 285 3 1 285 b i ii 3 5 24 29 or 0.828 244 9 10 11 b a a 6 2 3 2 3 =x 5 209 324 a 0.135 b or 0.645 b 0.253 c np> and 5 nq > 5 12 a Proof b 0.432 13 a =XE( ) 4 b Proof c 0.315 d np = 8.4375 5 and > nq = 11.5625 5 > 14 a i 1 48 p< < ii 0 < p < 1 4 47 48 or b 1 48 p< < 1 4 or 3 4 3 4 < p < 1 p< < 47 48 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy G ssary The following abbreviation and symbols are used in this book. No. ≈ ≠ ∝ ∴ ≡ Meaning Number of is approximately equal to is not equal to is proportional to therefore is identical to A Arrangements: see permutations Average: any of the measures of central tendency, including the mean, median and mode B Binomial distribution: a discrete probability distribution of the possible number of successful outcomes in a finite number of independent trials, where the probability of success in each trial is the same C Categorical data: see qualitative data Class: a set of values between a lower boundary and an upper boundary Class boundaries: the two values (lower and upper) between which all the values in a class of data lie Class interval: the range of values from the lower boundary to the upper boundary of a class Class mid-value (or midpoint): the value exactly half-way between the lower boundary and the upper boundary of a class Class width: the difference between the upper boundary and the lower boundary of a class Coded: adjusted throughout by the same amount and/or by the same factor Combinations: the different selections that can be made from a set of objects Complement: a number or quantity of something required to make a complete set Continuity correction: an adjustment made when a discrete distribution is approximated by a continuous distribution Continuous data: data that can take any value, possibly within a limited range Cumulative frequency: the total frequency of all values less than a particular value Cumulative frequency graph: a graphical representation of the number of readings below a given value made by plotting cumulative frequencies against upper class boundaries for all intervals D Dependent (events): events that cannot occur without being affected by the occurrence of each other Discrete data: data that can take only certain values E Elementary event: an outcome of an experiment Equiprobable: events or outcomes that are equally likely to occur Expectation: the expected number of times an event occurs Extreme value: an observation that lies an abnormal distance from other values in a set of data F Factorial: the product of all positive integers less than or equal to any chosen positive integer Fair: not favouring any particular outcome, object or person Favourable: leading to the occurrence of a required event Frequency: the number of times a particular value occurs Frequency density: frequency per standard interval G Geometric distribution: a discrete probability distribution of the possible number of trials required to obtain the first successful outcome in an infinite number of independent trials, where the probability of success in each trial is the same Grouped frequency table: a frequency table in which values are grouped into classes H Histogram: a diagram consisting of touching columns whose areas are proportional to frequencies I Independent (events): events that can occur without being affected by the occurrence of each other 245 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Interquartile range: the range of the middle half of the values in a set of data; the numerical difference between the upper quartile and the lower quartile K Key: a note that explains the meaning of each value in a diagram L Lower and upper boundary: the smallest and largest values that can exist in a class of continuous data M Mathematical model: a description of a system using mathematical concepts and language Mean: the sum of a set of values divided by the number of values Median: the number in the middle of an ordered set of values Modal class: the class of values with the highest frequency density Mode: the value that occurs most frequently Mutually exclusive (events): events that cannot occur at the same time because they have no common favourable outcomes N Normal curve: a symmetrical, bell-shaped curve Normal distribution: a function that represents the probability distribution of particular continuous random variables as a symmetrical bell-shaped graph O Ordered data: data arranged from smallest to largest (ascending) or largest to smallest (descending) Outliers: extreme values; observations that lie an abnormal distance from other values in a set of data P Parameters: the fixed values that define the distribution of a variable PDF: see probability density function Permutations: the different orders in which objects can be selected and placed Probabilities: measurements on a scale of 0 to 1 of the likelihood that an event occurs Probability density function (PDF): a graph illustrating the probabilities for values of a continuous random variable Probability distribution: a display of all the possible values of a variable and their corresponding probabilities Q Qualitative data: data that take non-numerical values Quantitative data: data that take numerical values Quartile: any of three measures that divide a set of data into four equal parts R Random: occurring by chance and without bias Range: the numerical difference between the largest and smallest values in a set of data Raw data: numerical facts and other pieces of information in their original form Relative frequency: the proportion of trials in which a particular event occurs S Selection: an item or number of items that are chosen Skewed: unsymmetrical Standard deviation: a measure of spread based on how far the data values are from the mean; the square root of the variance Standard normal variable: the normally distributed variable, Z, with mean 0 and variance 1 Stem-and-leaf diagram: a type of table for displaying ordered discrete data in rows with intervals of equal widths Summarise: to give an accurate general description T Trial: one of a number of repeated experiments U Unbiased: not favouring any particular outcome, object or person V Variance: the mean squared deviation from the mean; the square of the standard deviation Variation: dispersion; a measure of how widely spread out a set of data values is 246 Copyright Material - Review Only - Not for Redistributio
n Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Index Φ(z) 195–9 table of values 222 addition law, mutually exclusive events 94–5 cumulative frequency 13 cumulative frequency graphs 13–15 estimation of the median 43–4 interquartile range 58–9 averages see measures of central data representation box-and-whisker diagrams 60 comparing different methods 20 cumulative frequency graphs 13–15 histograms 6–9 stem-and-leaf diagrams 3–4 data types 2–3 de Moivre, Abraham 208 dependent events 112–15 deviation 65 Index grouped data mean 30–1 variance and standard deviation 66, 67–8 grouped frequency tables 6 estimation of the mean 32–3 height variation 64 histograms 6–9 modal class 28 use in image processing 13 independent events 100–2 application of the multiplication law 105–6 and conditional probability 111–12 see also standard deviation interquartile range 56 discrete data 2–3 stem-and-leaf diagrams 3–4 discrete random variables 150 binomial distribution 166–74 expectation 156–7, 158 geometric distribution 166, 175–82 probability distributions 150–2 variance 157–8 elementary events (outcomes) 91 equiprobable events 91 errors 188 events 91 dependent 112–15 exhaustive 92 independent 100–2, 111–12 mutually exclusive 94–5 box-and-whisker diagrams 60 comparison with standard deviation 69 grouped data 58–9 ungrouped data 56–7 mathematical models 166 using the normal distribution 205–6 mean 27, 30–1, 44 of the binomial distribution 173–4 of coded data 37–8, 40–1 of combined datasets 31–2, 73 of a discrete random variable 156–7, 158 of the geometric distribution 180–1 from grouped frequency tables 32–3 expectation 92–3 of a normal distribution 190–1, 193 of the binomial distribution 173–4 of a discrete random variable measures of central tendency 27 choosing an appropriate 247 tendency bar charts 13 binomial distribution 166 expectation 173–4 normal approximation 208–12 variance 173–4 binomial expansions 167–8 box-and-whisker diagrams (box plots) 60 categorical (qualitative) data 2 data representation 20 central tendency, measures of see measures of central tendency class boundaries 6, 7 class frequency 7, 8 class widths 6, 7 classes histograms 6 stem-and-leaf diagrams 3, 4 coded data 37–8, 40–1 standardising a normal distribution 200–3 variance and standard deviation 75–80 combinations 123, 135 nCr notation 135–6 problem solving 138–40 combined datasets mean 31–2, 73 variance and standard deviation 72–3 complement of an event 92 conditional probability 108–9 and dependent events 112–15 and independence 111–12 continuity corrections 209–10 continuous data 3 156–7, 158 of the geometric distribution 180–1 see also mean factorial function 124–5 fair (unbiased) selection 91 Fermat's Last Theorem 156 frequency density 7–9 average 44–6 effect of extreme values 45 historical background 44 mean 30–41 median 42–4 mode and modal class 28–9 for skewed data 45–6 see also mean; median; mode measures of variation 55 coded data 75–80 interquartile range and percentiles 56–9 range 55–6 cumulative frequency graphs 13–15 histograms 6–9 continuous random variables 188–9 normal distribution 190–206 probability density functions 189–90 Gauss, Carl Friedrich 188, 205, 208 geometric distribution 166, 175–8 expectation 180–1 mode 180 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 measures of variation (Cont.) possibility diagrams (outcome variance and standard deviation spaces) 101 65–9, 72–3 see also standard deviation; variance possibility space 95 probability 91 median 27, 42–3, 44, 56 estimation from a cumulative frequency graph 43–4 modal class 28–9 mode 27, 44 addition law 94–5 conditional 108–9, 111–15 dependent events 112–15 experiments, events and outcomes 91–3 of the geometric distribution 180 multiplication law for independent independent events 100–2, 111–12 multiplication law for independent events 100–2 application of 105–6 multiplication law of probability 112–15 mutually exclusive events 94–5 normal curve 190–1 normal distribution 188, 193 approximation to the binomial distribution 208–12 modelling with 205–6 properties of 194 standard normal variable (Z) 195–9 standardising 200–3 tables of values 222 248 parameters of a binomial distribution 167 of a geometric distribution 176 of a normal distribution 193 Pascal's triangle 168, 172 percentiles 58–9 permutations 123, 134 of n distinct objects 125–6 of n distinct objects with restrictions 129–31 of n objects with repetitions 127–8 nPn notation 125 nPr notation 132 problem solving 138–40 of r objects from n objects 132–3 events 100–2, 105–6 multiplication law of probability 112–15 mutually exclusive events 94–5 Venn diagrams 95–7 probability density functions (PDFs) 189–90 probability distributions 150–2 binomial distribution 166–74 geometric distribution 175–82 normal distribution 190–206 qualitative (categorical) data 2 data representation 20 quantitative data 2–3 quartiles 56 grouped data 58–9 ungrouped data 56–7 random selection 91–2 range 55–6 repetitions, permutations with 127–8 restrictions, permutations with 129–31 selection, random 91–2 set notation 95 sigma (Σ) notation 30 skewed data skewed distributions 190 standard deviation 65–8 of the binomial distribution 173–4 calculation from totals 72 coded data 75–80 of combined datasets 72–3 comparison with interquartile range 69 of a discrete random variable 158 of a normal distribution 190–1, 194 standard normal variable (Z) 195–9 standardising a normal distribution 200–3 stem-and-leaf diagrams 3–4 interquartile range 57 median 42 tree diagrams for independent events 100–1 for permutations 125–6 trials 92–3 variables notation 150 see also continuous random variables; discrete random variables variance 65–8 of the binomial distribution 173–4 calculation from totals 72 of coded data 75–80 of combined datasets 72–3 of a discrete random variable 157–8 equivalence of two formulae for 81 of a normal distribution 193 variation 55 see also measures of variation Venn diagrams 95–7 Wiles, Andrew 156 box-and-whisker diagrams 60 measures of central tendency 45–6 Z (standard normal variable) 195–9 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for Redistribution R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for Redistribution R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for Redistribution R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for Redistribution R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for Redistribution R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for Redistribution
ike surds so that bn is the same for all the surds. 1.4.2 Rationalising Denominators It is useful to work with fractions which have rational denominators instead of surd denominators. It is possible to rewrite any fraction which has a surd in the denominator as a fraction which has a rational denominator. We will now see how this can be achieved. Any expression of the form pa + pb (where a and b are rational) can be pb changed into a rational number by multiplying by pa can be rationalised by multiplying by pa + pb). This is because pb (similarly pa ¡ ¡ (pa + pb)(pa pb) = a b ¡ ¡ (1.60) which is rational (since a and b are rational). If we have a fraction which has a denominator which looks like pa+pb, then pb achieving a rational we can simply multiply both top and bottom by pa denominator. ¡ or similarly c pa + pb c ¡ pb pa = = = = c pa + pb ¡ ¡ pa pa cpa a pb pb £ cpb b ¡ ¡ pa + pb pa + pb £ cpa + cpb c ¡ pb pa a b ¡ (1.61) (1.62) 1.4.3 Estimating a Surd (NOTE: has anyone got a better way to do this?) It is sometimes useful to know the approximate value of a surd without having to use a calculator. This involves knowing some roots which have integer solutions. For a surd npa, ﬂnd an integer smaller than a with an integer as its nth root and then ﬂnd the next highest integer (which should also be larger than a) with an integer nth root. The surd which you are trying to estimate will be between those two integers. (NOTE: this paragraph sucks, rewrite it so that it can be understood.) For example, when given the surd 3p52 you should be able to tell that it lies somewhere between 3 and 4, because 3p27 = 3 and 3p64 = 4 and 52 is between 27 and 64. In fact 3p52 = 3:73 : : : which is indeed between 3 and 4. 18 The easiest arithmetic procedure10 to ﬂnd the square root of any number N is to choose a number x that is close to the square root, ﬂnd N x and then use x0 = x+ N for the next choice of x. x converges rapidly towards the actual value of the square root - the number of signiﬂcant digits doubles each time. (NOTE: arithmetic? procedure? converges? rapidly? this language is not basic enough!) 2 x We will now use this method to ﬂnd p55. We know that 72 = 49, therefore the square root of 55 must be close to 7. Let x = 7 then, 55 7 7 + 7;8571 : : : 2 55 7:4285 : : : = 7;8571 : : : = 7;4285 : : : = 7;4038 : : : ) x0 = (1.63) (1.64) (1.65) (1.66) ) x0 = 7;4285 : : : + 7;4038 : : : 2 = 7;4162 : : : (1.67) Using a calculator we ﬂnd that p55 = 7;416198 : : :, which is very close to our approximation. 1.5 Accuracy The syllabus requires: write irrational (and rational) solutions rounded to a specified degree of accuracy know when to approximate an irrational by a terminating rational, and when not to express large and small numbers in scientific or engineering notation † † † We already mentioned in section 1.2.2 that certain numbers may take an inﬂnite amount of paper and ink to write out. Not only is that impossible, but writing numbers out to a high accuracy (too many decimal places) is very inconvenient and rarely gives better answers. For this reason we often estimate the number to a certain number of decimal places or to a given number of signiﬂcant ﬂgures, which is even better. (NOTE: the notes on rounding need to be better. this is not very good.) Approximating a decimal number to a given number of decimal places is the quickest way to approximate a number. Just count along the number of places you have been asked to approximate the number to and then forget all the numbers after that point. You round up the ﬂnal digit if the number you cut oﬁ was greater or equal to 5 and round down (leave the digit alone) otherwise. For example, approximating 2;6525272 to 3 decimal places is 2;653 because the ﬂnal digit is rounded up. 10This procedure is known as Heron’s Method, which was used by the Babylonians over 4000 years ago. 19 (NOTE: more on the diﬁerence between DP and SF needed) In a number, each non-zero digit is a signiﬂcant ﬂgure. Zeroes are only counted if they are between two non-zero digits or are at the end of the decimal part. For example, the number 2000 has 1 signiﬂcant ﬂgure, but 2000;0 has 5 signiﬂcant ﬂgures. Estimating a number works by removing signiﬂcant ﬂgures from your number (starting from the right) until you have the desired number of signiﬂcant ﬂgures, rounding as you go. For example 6;827 has 4 signiﬂcant ﬂgures, but if you wish to write it to 3 signiﬂcant ﬂgures it would mean removing the 7 and rounding up, so it would be 6;83. It is important to know when to estimate a number and when not to. It is usually good practise to only estimate numbers when it is absolutely necessary, and to instead use symbols to represent certain irrational numbers (such as …); If it is necessary approximating them only at the very end of a calculation. to approximate a number in the middle of a calculation, then it is often good enough to approximate to a few decimal places. 1.5.1 Scientiﬂc Notation In science one often needs to work with very large or very small numbers. These can be written more easily in scientiﬂc notation, which has the general form 10m a £ (1.68) where a is a decimal number between 1 and 10. The m is an integer and if it is positive it represents how many zeros should appear to the right of a. If m is negative then it represents how many times the decimal place in a should 3 10¡ be moved to the left. For example 3;2 represents 0;0032. 103 represents 32000 and 3;2 £ £ If a number must be converted into scientiﬂc notation, we need to work out how many times the number must be multiplied or divided by 10 to make it into a number between 1 and 10 (i.e. we need to work out the value of the exponent m) and what this number is (the value of a). We do this by counting the number of decimal places the decimal point must move. It is usually enough to estimate a to only a few decimal places. 1.5.2 Worked Examples Worked Example 1 : Manipulating Rational Numbers Question: Simplify the following expressions a) 7 8 + 5 2 b) 11 27 £ 20 3 c) 73 69 ¥ 73 69 Answer: a) Step 1 : Rule of addition Write out the rule of addition for rational numbers (1.25) a b + c d = ad + bc bd 20 Step 2 : Fill in the values Fill in the values for a;b;c and d. Here you can read oﬁ that a = 7, b = 8, c = 5 and £ = 54 16 £ Step 3 : Minimise the denominator 54 16 is the correct answer, but it is not the simplest way to write it. We can see that both 54 and 16 can be divided by 2, so we divide both by 2 and get 27 8 , which cannot be simpliﬂed any further. b) Step 1 : Rule of multiplication Write out the rule of multiplication for rational numbers (1.26) a b £ c d = ac bd Step 2 : Fill in the values Fill in the values for a;b;c and d. Here you can read oﬁ that a = 11, b = 27, c = 20 and d = 3 11 27 £ 20 3 = 11 27 20 3 = 220 81 £ £ There is no number which will divide into both 220 and 81, so 220 81 is the simplest form of the answer. c) Step 1 : Use the division rule Calculate the reciprocal of 73 (1.28) 69 = 69 73 , and write out the division rule = ad bc Step 2 : Fill in the values We can read oﬁ that a = c = 73 and b = d = 69 so 73 69 69 73 £ £ = 5037 5037 = 1 This question could also have been answered in one single line by noticing that the two fractions are the same, and any number divided by itself is one. 1.5.3 Exercises TODO 21 Chapter 2 Patterns in Numbers (NOTE: SH notes:at the moment, this whole chapter needs a lot more inline examples. i think it is too complicated for 16 year olds without examples. even the use of indices is also some of the equations might be over their heads. inconsistent... the use of letters like i, n, m should not be interchanged so much as it is only leading to confusion.) (NOTE: also, i think we are aiming too high. it is possible that when the syllabus says \prove", it really means \show explicitly the ﬂrst few terms and assume the rest of the sequence is the same". so perhaps we should drop a few of the proofs.) 2.1 Sequences The syllabus requires: investigate number patterns, be able to conjecture a pattern and prove those conjectures recognise a linear pattern when there is a constant difference between consecutive terms recognise a quadratic pattern when there is a constant 2nd difference identify ‘‘not real’’ numbers and how they occur (NOTE: i think this would be best taught in quadratic equations as that is the only place they occur in this syllabus) (grade 12) arithmetic and geometric sequences † † † † † Can you spot any patterns in the following lists of numbers? 2;4;6;8; : : : 1;2;4;7; : : : 1;4;9;16; : : : 5;10;20;40; : : : 3;1;4;1;5;9;2; : : : 22 (2.1) (2.2) (2.3) (2.4) (2.5) The ﬂrst is a list of the even numbers, the numbers in the second list ﬂrst diﬁer by one, then by two, then by three. The third list contains the squares of all the integers. In the fourth list, every term is equal to the previous term times two and the last list contains the digits of the number …. These lists are all examples of sequences. In this section we will be studying sequences and how they can be described mathematically. (NOTE: a few real world examples here wouldn’t go amiss.) A sequence is a list of objects (in our case numbers) which have been ordered. We could take as an example a sequence of books. If you put all your books in alphabetical order by the author, that would be a sequence because it is a list of things in order. Someone could look at the sequence and work out how you ordered them if they knew the alphabet. You could rearrange the collection so that it was ordered alphabetically by title. That would be a diﬁerent sequence because the order is diﬁerent. Similarly the sequence of numbers 1;2;3 is diﬁerent to 3;2;1. You could even shu†e up all the books so that the order they were in didn’t follow a pattern, but they would still make a sequence. Notice that not all sequences have to continue forever - what characterises a sequence is that it is a list which is ordered. In the alphabetised
books example, someone who didn’t know the alphabet would not be able to work out how you had ordered the books. How would you be able to ﬂnd your seats at the theatre or at a stadium if the seats were not ordered ? Likewise if you are shown a sequence of numbers, you may not be able to work out what pattern relates them. That might be because there is no pattern, or it might just be that you can’t see it straight away. We will be thinking about sequences in this chapter and it is useful to be able to talk in general about them. We will want to talk about the numbers in the sequence, so rather than having to say something longwinded like \the second term in the sequence is related to the ﬂrst term by this rule....", we give each term in the sequence a name. The ﬂrst term of a sequence is named a1, the second term is named a2 and the nth term is named an. Now we can say \a2 is related to a1 by this rule...". The small n or number like 1 or 2 beside the letter is called a subscript or index but we will refer to it as the subscript It helps us keep everything tidy by using the same letter (in this example, a) for all the terms in a sequence. A sequence does not have to follow a pattern, but when it does we can often write down a formula for the nth term, an. In the example above, 2.3 where the sequence was of all square numbers, the formula for the nth term is an = n2. You can check this by looking at a1 = 12 = 1; a2 = 22 = 4; a3 = 32 = 9; : : : 2.1.1 Arithmetic Sequences Deﬂnition: A linear arithmetic sequence is a sequence in which each successive term diﬁers by the same amount. Each term is equal to the previous term plus a constant number. Tn = 1 + k1 where k1 is some constant. We will see in the next example what this Tn constant is and how to determine it. ¡ Say you and 3 friends decide to study for maths and you are seated at a square table. A few minutes later, 2 other friends join you and would like to sit at your table and help you study. Naturally you move another table and add it 23 Figure 2.1: Tables moved together to the existing one. Now six of you sit at the table. Another two of your friends join your table and you take a third table and add it to the existing tables. Now 8 of you can sit comfortably. Let assume this pattern continues and we tabulate what is happening. (NOTE: Insert pictures here.) No. Tables (n) No of people seated 1 2 3 4 ... n 4 = 4 4+2 = 6 4+2+2 = 8 4+2+2+2 = 10 ... 4+2+2+2+. . . +2 Formula = 4 + 2(0) = 4 + 2(1) = 4 + 2(2) = 4 + 2(3) ... = 4 + 2(n - 1) We can see for 3 tables we can seat 8 people, for 4 tables we can seat 10 people and so on. We started out with 4 people and added two the whole time. Thus for each table added, the number of persons increase with two. Thus, 4;6;8;::: is a sequence and each term (table added), diﬁers by the same amount (two). More formally, the number we start out with is called a1 and the diﬁerence between each successive term is d. Now our equation for the nth term will be: an = a1 + d(n 1) ¡ (2.6) The general linear sequence looks like a1; a1 + d; a1 + 2d; a1 + 3d; : : :, using the general formula ?? How many people can sit in this case around 12 tables ? By simply using the derived equation we are looking for where n = 12 and thus a12 an = a1 + d(n a12 = 4 + 2(12 = 4 + 2(11) ¡ ¡ 1) 1) OR = 4 + 22 = 26 24 How many tables would you need for 20 people ? an = a1 + d(n 20 = 4 + 2(n 1) ¡ 1) ¡ 20 16 ¡ ¥ 4 = 2(n 2 = n ¡ 1 1 a2 = d A simple test for an arithmetic sequence is to check that a2¡ This is quite an important equation and is a deﬂnitive test for an arithmetic sequence. If this condition does not hold, the sequence is not an arithmetic sequence. It is also important to note the diﬁerence between n and an. n can be compared to a place holder while an is the value at the place ’held’ by n. Like our study table above.Table 1 holds 4 people thus at place n=1 the value of a1 = 4. a1 = a3¡ n an 1 4 2 6 3 8 4 10 . . . . . . 2.1.2 Quadratic Sequences (NOTE: maybe put in a note about the quadratic equations section, and 1st/2nd diﬁerences in terms of diﬁerentiating wrt n.) A quadratic sequence is a sequence in which the diﬁerences between each consecutive term diﬁer by the In the example sequences same amount, called a constant second diﬁerence. in the introduction, equation (2.2) is a quadratic sequence because the diﬁerence between each term diﬁers by one each time. We can look at the diﬁerence between each term and see that the diﬁerences form a linear sequence: a a2 ¡ 3 ¡ ¡ a4 ¡ a1 = 2 a2 = 4 a3 = Here you can see clearly that the diﬁerence between each diﬁerence is 1. We call this the constant second diﬁerence and in this case is = 1. The general form of this example of a quadratic sequence (a sequence with constant second diﬁerence) is an = 1 2 (n2 1) ¡ ¡ 1 2 (n ¡ 1) + 1 (2.7) For a general quadratic sequence with constant second diﬁerence D the formula for an is an = D 2 (n2 1) + d(n ¡ ¡ 1 is Dn + d. 1) + a1 (2.8) The diﬁerence between an and an Check for yourself that an ¡ an then again for an an ¡ diﬁerence not equal to 1. ¡ an 1 (setting n = n ¡ ¡ 1 = Dn + d. (Use the formula for an and 1 in the formula) and then work out what 1 is.) Make up your own quadratic sequences with a constant second ¡ ¡ 25 Figure 2.2: Tree diagram of series 2.1.3 Geometric Sequences Deﬂnition: A geometric sequence is a sequence in which every number in the sequence is equal to the previous number in the sequence, multiplied by another constant number. This means that the ratio between consecutive numbers in the sequence is a constant. We will explain what we mean by ratio after looking at this example. What is inuenza (u)? Inuenza, commonly called ’the u’, is caused by the inuenza virus, which infects the respiratory tract (nose, throat, lungs). It can cause mild to severe illness, that most of us get during winter time The main way that inuenza viruses are spread is from person to person in respiratory droplets of coughs and sneezes. (This is called ’droplet spread.’) This can happen when droplets from a cough or sneeze of an infected person are propelled (generally up to 3 feet) through the air and deposited on the mouth or nose of people nearby. It is good practise to cover your mouth when you cough or sneeze to not infect others around you when you have the u. Lets assume you have the u virus and you forgot to cover your mouth when two friends came to visit while you were sick in bed. They leave and the next day, they also have the u. Lets assume that they in turn spread the virus to two of their friends by the same droplet spread the following day. Lets assume this pattern continues and each person infected, infects 2 other friends. We can represent these events in the following manner: (NOTE: Insert pictures here.) Again we can tabulate the events and formulate an equation for the general case: # day (n) 1 2 3 4 5 ... n # Carrier You spread virus 2 4 8 16 ... . . . # Recipients/Carrier 2 4 8 16 32 ... . . . Formula 21 8 = 2 x 4 = 2 x 2x2 = 2 x 22 16 = 2 x 8 = 2 x 2x2x2 = 2 x 23 32 = 2 x 16 = 2 x 2x2x2x2 = 2 x 24 ... = 2 x 2x2x2x...x2 = 2 x 2n 1 ¡ You sneeze and the virus is carried over to 2 people who start the chain (a1 = 2). The next day, each one then infects 2 of their friends. Now 4 people are infected. Each of them infects 2 people the third day and 8 people are infected etc. These events can be written as a geometric sequence: 2;4;8;16;32;::: Note the common factor between the events. Recall from the linear arithmetic sequence how the common diﬁerence between terms were established. In the 26 geometric sequence we can determine the common factor, r by Or more general a2 a1 = a3 a2 = r an+1 an = an+2 an+1 = r (2.9) (2.10) a2 a1 is called the ratio and is used to describe the ’factor diﬁerence’ between the elements of the series. i.e. The ratio between a1 and a2 is 2 From the question in the above example we know a1 = 2 and r = 2 and we 1. Thus in have seen from the table that the nth term is given by an = 2 general, 2n £ ¡ an = a1rn ¡ 1 (2.11) So if we want to know how many people has been infected after 10 days, we need to work out a10 1 ¡ 1 an = a1rn ¡ 210 a10 = 2 29 = 2 512 = 2 £ £ = 1024 £ Or, how many days would pass before 16384 people are infected with the u virus ? (NOTE: I’m not sure if SURDs and exponents have been done at this stage. check ﬂrst. This chapter should be taught AFTER exponents and SURDs because the techniques are used !!) 16384 1 1 ¡ an = a1rn ¡ 2n 1 16384 = 2 £ 2 = 2n ¡ ¥ 8192 = 2n 213 = 2n 13 = n n = 14 ¡ 1 ¡ ¡ 1 1 2.1.4 Recursive Equations for sequences When discussing linear and quadratic sequences we noticed that the diﬁerence between two consecutive terms in the sequence could be written in a general way. For linear sequences, where the constant diﬁerence between two consecutive 1 = d for any term in terms was d, we can write this information as an ¡ the sequence. We can rearrange this to an = an 1 + d. This is an expression ¡ for an in terms of an 1, which is called a recursive equation. So the recursive equation for a linear sequence of constant diﬁerence d is an ¡ ¡ an ¡ an 1 = d ¡ (2.12) 27 We can do the same thing for quadratic sequences. There we noticed that 1 = Dn + d. Then the recursive equation for a quadratic sequence with an ¡ constant second derivative D is an ¡ an ¡ an 1 = Dn + d ¡ (2.13) an 2 wouldn’t include d.) (NOTE: Here we haven’t said explicitly what d is or how to work it out. This bothers me. I think to be honest that you need more information. like maybe an ¡ It is not always possible to ﬂnd a recursive equation for a sequence, even when you know the general way to write down any term an. Can you ﬂnd a recursive equation for a geometric sequence? This is not supposed to be easy! It’s just to get you to have a go at working things out. ¡ Recursive equations are extremely powerful: you can work out every term in the series just by knowing the previous one, and as you can see for the example above,
working out an from an 1 can be a much simpler computation than working out an from scratch using a general formula. This means that using a recursive formula when programming a computer to work out a sequence would mean the computer would ﬂnish its calculations signiﬂcantly quicker. (NOTE: Real world example of this?) ¡ 2.1.5 Extra (NOTE: Jacques had sections on arithmetic/geometric means. i want to add that back in, but make it clear it is non-syllabus. No questions are usually asked in the ﬂnal exam as far as I can see, but it is part of the syllabus.) 2.2 Series (Grade 12) The syllabus requires: (grade 12) prove and calculate the following sums † 1 = n n Xi=1 n i2 = Xi=1 n(2n + 1)(n + 1) 6 n Xi=1 a + (i ¡ 1)d = n 2 (2a + (n 1)) ¡ n Xi=1 1 Xi=1 a:ri ¡ 1 = a:ri ¡ 1 = 1) a(rn NOTE: equation 3 is not correct in the official syllabus. is a d missing.) there 28 When we sum terms in a sequence, we get what is called a series. If we only sum a ﬂnite amount of terms, we get a ﬂnite series. We use the symbol Sn to mean the sum of the ﬂrst n terms of a sequence. For example, the sequence of numbers 3;1;4;1;5;9;2; : : : has a ﬂnite series S4 which is simply the ﬂrst 4 terms added together. If we sum inﬂnitely many terms of a sequence, we get an inﬂnite series. A sum may be written out using the summation symbol (Sigma). This symbol is the capital \S" (for Sum) in the Greek alphabet. It indicates that you must sum the expression to the right of it P n Xi=m ai = am + am+1 + : : : + an 1 + an ¡ (2.14) ai are the terms in a sequence and here we sum from i = m (as indicated below the summation symbol) up until i = n (as indicated above). We usually just sum from n = 1, which is the ﬂrst term in the sequence. In which case we can notation since they mean the same thing use either Sn or P n Sn = ai = a1 + a2 + : : : + an (2.15) Xi=1 For example, in the following sum 5 i Xi=1 (2.16) we have to add together all the terms in the sequence ai = i from i = 1 up until = 15 (2.17) which gives us 15. Xi=1 2.2.1 Finite Arithmetic Series When we sum a ﬂnite number of terms in an arithmetic sequence, we get a ﬂnite arithmetic series. The simplest arithmetic sequence is when a1 = 1 and d = 0 in the general form (??), in other words all the terms in the sequence are one. ai = d(i = 0(i = 1 ¡ ¡ 1) + a1 1) + 1 a = 1;1;1;1;1; : : : (2.18) If we wish to sum this sequence from i = 1 to any integer n, we would write n Xi= times (2.19) 29 Since all the terms are equal to one, it means that if we sum to an integer n we will be adding n number of ones together, which is equal to n. n Xi=1 1 = n (2.20) Another simple arithmetic sequence is when a1 = 1 and d = 1, which is the sequence of positive integers ai = d(i = (i 1) + a1 ¡ 1) + 1 ¡ = i (2.21) a = 1;2;3;4;5; : : : If we wish to sum this sequence from i = 1 to any integer n, we would write n Xi=2.22) This is an equation with a very important solution as it gives the answer to the sum of positive integers1. We notice that the largest number may be added to the smallest, then the second largest added to second smallest, giving the same number. If we keep doing this we ﬂnd that all the numbers may be paired up together like this until we reach the middle and there are no more numbers left to pair oﬁ a1 + a2 + : : : + an = (a1 + an) + (a2 + an 1) + : : : ¡ n 2 times (2.23) If there are an odd number of numbers, then we must not forget to add the unpaired number to the answer at the end. For example 1 + 5) + (2 + 4) + 3 (2.24) = 6 + 6 + 3 = (3 + 3) + (3 + 3) + 3 = 15 We can write this down in general as n Xi=1 i = n 2 (n + 1) (2.25) If we wish to sum any arithmetic sequence, there is no need to work it out term for term as we just have for these examples. We will now show what the general form of a ﬂnite arithmetic series is by starting with the general form of an arithmetic sequence and summing it from i = 1 to any integer n. 1A famous mathematician named Carl Friedrich Gauss discovered this proof when he was only 8 years old. His teacher had decided to give his class a problem which would distract them for the entire day by asking them to add all the numbers from 1 to 100. Young Carl realised how to do this almost instantaneously and shocked the teacher with the correct answer, 5050. 30 Writing out the sum of a sequence and then substituting in the general form for an arithmetic sequence gives us n n ai = Xi=1 Xi=1 d(i ¡ 1) + a1 (2.26) If there is a sum inside a sum, we can break it into two separate sums and calculate each part separately. n Xi=1 d(i ¡ n 1) + a1 = di + (a1 ¡ d) (2.27) Xi=1 left \|{z} right \| {z } If a sum is multiplied by a constant, we can take the constant outside of the d, which is a constant, so we may . The term on the right is a sum of a1 ¡ rewrite that term as P n Xi=1 a1 ¡ d = (a1 ¡ = (a1 ¡ n d) 1 Xi=1 d)n (2.28) Here we used equation (2.20) to arrive at the solution. The term on the left of equation (2.27) is also quite simple. Firstly we can take the constant d out of the sum n n and then we can use equation (2.25) to ﬂnd di = d i Xi=1 Xi=1 n d i = Xi=1 dn 2 (n + 1) (2.29) (2.30) Adding together the solutions to the left and right terms (equations (2.28) and (2.30)) we get the general form of a ﬂnite arithmetic series n Xi=1 d(i ¡ 1) + a1 = n 2 (2a1 + d(n 1)) ¡ (2.31) For example, if we wish to know the series S20 for the arithmetic sequence 1) + 3, we could either calculate each term individually and sum them ai = 7(i ¡ 20 Xi=1 7(i ¡ 1) + 3 = 3 + 10 + 17 + 24 + 31 + 38 + 45 + 52 +59 + 66 + 73 + 80 + 87 + 94 + 101 +108 + 115 + 122 + 129 + 136 = 1390 (2.32) or more sensibly, we could use equation (2.31) noting that d = 7, a1 = 3 and n = 20 so that 20 Xi=1 7(i ¡ 1) + 3 = 20 2 (2 £ 3 + 7 19) £ = 1390 (2.33) In this example, it is clear that using (2.31) is beneﬂcial. 31 2.2.2 Finite Squared Series When we sum a ﬂnite number of terms in a quadratic sequence, we get a ﬂnite quadratic series. The general form of a quadratic series is quite complicated, so we will only look at the simple case when D = 1 and d = a0 = 0 in the general form (??). This is the sequence of squares of the integers ai = i2 = 12;22;32;42;52;62; : : : = 1;4;9;16;25;36 : : : (2.34) If we wish to sum this sequence and create a series, then we write Sn = n Xi=1 i2 = 1 + 4 + 9 + : : : + n2 (2.35) which can be written in general as (NOTE: the syllabus requires that we prove this result! any ideas, without confusing the hell out of a 16 year old? i thought even the other ones were a bit too hard for this level, to be honest.) n Xi=1 i2 = n(2n + 1)(n + 1) 6 (2.36) 2.2.3 Finite Geometric Series When we sum a ﬂnite number of terms in a geometric sequence, we get a ﬂnite geometric series. We know from (??) that we can write out each term of a geometric sequence in a general form. By simply adding together the ﬂrst n terms in the general form we are actually writing out the series Sn = a1 + a1r + a1r2 + : : : + a1rn ¡ 1 We may multiply this by r on both sides, giving us rSn = a1r + a1r2 + a1r3 + : : : + a1rn (2.37) (2.38) You may notice that all the terms are the same in (2.37) and (2.38), except the ﬂrst and last. If we subtract (2.37) from (2.38) we are left with just rSn ¡ Sn(r ¡ Sn = a1 + a1rn 1) = a1(1 + rn) (2.39) dividing by (r sequence since Sn = ¡ n i=1 a:ri ¡ 1 1) on both sides, we have the general form of a geometric 1) (2.40) a(rn r ¡ 1 ¡ P n Xi=1 a:ri ¡ 1 = 32 2.2.4 Inﬂnite Series Thus far we have been working only with ﬂnite sums, meaning that whenever we determined the sum of a series, we only considered the sum of the ﬂrst n terms. It is the subject of this section to consider what happens when we add inﬂnitely many terms together. You might think that this is a silly question - surely one will get to inﬂnity when one sums inﬂnitely many numbers, no matter how small they are? The surprising answer is that in some cases one will reach inﬂnity (like when you try to add all the integers together), but in some cases one will get a ﬂnite answer. If you don’t believe this, try doing the following sum on your calculator or computer: 1 32 + ::: . You might think that if you keep adding more and more terms you will eventually get larger and larger numbers, but in fact you won’t even get past 1 - try it and see for yourself! 16 + There is a special sigma notation for inﬂnite series: we write 1i=1 i to indicate the inﬂnite sum 1 + 2 + 3 + 4 + ::::. When we sum the terms of a series, and the answer we get after each summation gets closer and closer to some number, we say that the series converges. If a series does not converge, we say that it diverges. P There is a rule for knowing instantly which geometric series converge and which diverge. When r, the common ratio, is strictly between -1 and 1, i.e. 1 < r < 1, the inﬂnite series will converge, otherwise it will diverge. There ¡ is also a formula for working out what the series converges to. The sum of an inﬂnite series, symbolised by S , is given by the formula 1 S 1 = 1 Xi=1 a1:ri ¡ 1 = a1 ¡ 1 r 1 < r < 1 ¡ (2.41) where a1 is the ﬂrst term of the series, and r is the common ratio. (NOTE: the syllabus requires us to PROVE this series! how can we do that without a notion of a limit? again the syllabus talks nonsense.) We can see how this comes about by looking at (2.40), as . We can ignore the rn term since a small number raised to the power of inﬂnity is inﬂnitely small. Try this yourself by typing in a number between -1 and 1 into your calculator and square it, continuing to square the answers thereafter; your calculator will eventually decide that the answer is zero. 1 < r < 1 and n = 1 ¡ 2.3 Worked Examples (NOTE: I think maybe the worked examples should follow the relevant section. Check if the general layout is set.) 1. Classify the following as arithmetic sequence or geometric sequence: 15;19;23; : : : For arithmetic sequence, We have to check for a common diﬁerence or a common ratio. a2 ¡ a2 ¡ a3 ¡ a1 = a3 ¡ a1 = 19 ¡ a2 = 23 ¡ a2 = d 15 = 4 19 = 4 33
Thus, a2 ¡ sequence and d = 4 a1 = a3 ¡ a2 = 4 and we can say that 15;19;23; : : : is an arithmetic 2. Classify the following as arithmetic sequence or geometric sequence: 5;10;20; : : : For arithmetic sequence, We have to check for a common diﬁerence or a common ratio. a2 ¡ a2 ¡ a3 ¡ Thus, a2 ¡ a1 = a3 ¡ a1 = 10 ¡ a2 = 20 ¡ = a3 ¡ a1 6 a2 = d 5 = 5 10 = 10 sequence. a2 and we can say that 5;10;20; : : : is not an arithmetic Test for geometric sequence: a2 a1 a2 a1 a3 a2 = a3 = r a2 = 10 5 = 2 = 20 10 = 2 Thus, a2 a1 sequence. = a3 a2 and r = 2 and we can say that 5;10;20; : : : is a geometric 3. Determine d and a9 for the following arithmetic sequence: 17;14;11; : : : It is given that 17;14;11; : : : is an arithmetic sequence, thus a2 ¡ 14 ¡ d = a2 = d 14 = 3 ¡ a1 = a3 ¡ 17 = 11 ¡ 3 ¡ To determine a9 we use an = a1 + d(n Thus: ¡ 1) with n = 9 1) ¡ 3)(9 an = a1 + d(n a9 = 17 + ( = 17 ¡ = 17 ¡ 7 = ¡ 4. Determine r and a7 for the following geometric sequence: 81; ¡ 3(8) 24 1) ¡ 27; 9; : : : ¡ a2 a1 = a3 a2 = ¡ = 27 81 9 27 ¡ = = = r 1 3 1 3 ¡ ¡ a2 a1 a3 a2 34 To determine a7 we use an = a1rn Thus: 1 with n = 7 ¡ an = a1rn ¡ 1 a7 = (81)( 1 )7 ¡ 1(6) 1 3 ¡ (34)(3¡ (34:3¡ 6) 6) (34 ¡ 2) (3. The third term of a geometric sequence is equal to 1 and the 5th term is 16. Find r and the seventh term Given: a3 = 1 and a5 = 16 We also know an = a1rn Thus: ¡ 1 a3 = a1r3 ¡ 1 = a1r2 1 a5 = a1r5 ¡ 16 = a1r4 1 Also: Dividing (2) by (1): a1r4 a1r2 = 16 1 16 = r2 r = 4 To ﬂnd a7 we use a7 = a1r7 1 but ﬂrst we need a1. From (1) we know: ¡ 35 1 = a1r2 1 = a1:(4)2 a1 = 1 16 1 a7 = a1r7 ¡ :46 = 1 16 4096 16 = 256 = 6. The fourth term of an arithmetic sequence is 1 1 2 and the the 8th term is 1 2 . Find the second term. Given: a4 = 3 Thus we can use an = a1 + d(n 2 , a8 = 1 2 and this is an arithmetic sequence. 1) ¡ a4 = 3 a8 = 1 2 = a1 + d(4 2 = a1 + d(8 ¡ ¡ 1) = a1 + 3d and 1) = a1 + 7d Subtract the one equation from the other to get rid of a1 and solve for da1 + 3d) 4d 1 4 (a1 + 7d) ¡ ¡ ¡ 1 4 )(7) 1 2 = a1 + ( ¡ 7 1 4 ) 2 = a1 ¡ a1 = 9 4 a2 = 9 a2 = 8 ( 1 4 )(2 4 ¡ 4 = 2 1) ¡ 2.4 Exercises 1. Classify the following as arithmetic sequence or geometric sequence: 1 3 ¡ ; : : : i) ii) iii) iv) v) ; ; ;0;1; : : : 1; 3 1 2 6 22;2;1. Find a7 for each of the series above. 3. Determine which term in the series 14;8;2; : : : is equal to 34 ? 4. Which ¡ 36 term in 2;6;18; : : : is equal to 486 ? 5. In a geometric series, a4 = 2 6. In an arithmetic series, a3 = 7. The third term of a Geometric series is equal to minus three eights and the seventh term is equal to 3 8. Insert 4 numbers between 4 and 972 to form a geometric series. 3 and a6 = 3 ¡ 2 and a8 = 23. Determine a1 and d. 128 . Find a5. 2 . Find a2. 37 Chapter 3 Functions 3.1 Functions and Graphs The syllabus requires: † † † † † (grade 12) formal definition of the function concept able to switch between words, tables, graphs and formula to represent the relation between variables generate graphs using point to point plotting to test conjectures on relations between x and y for the situations (NOTE: this list we do the trig has changed in the latest syllabus... functions in the trig section. it makes more sense this way.) be warned. y = ax + b y = ax + b a > 0; = ax+b + c (NOTE: page 25 of the syllabus lists some more situations, but they are corrupted. the syllabus and add them to this list) we need to get an uncorrupted version of identify the domain and range, axes intercepts, turning points (max/min), asymptotes, shape and symmetry, periodicity and amplitude, rates of change, increasing/decreasing ranges and continuity. and can sketch graphs using these characteristics (grade 12) can generate graphs of function inverses. in particular a > 0; a = 1 y = ax + b y = ax y = ax2 y = sin(x) 38 6 6 † (grade 12) decide which inverses are functions and if necessary the restriction to make it a function (NOTE: functions are neither conceptually simple nor very interesting - so this intro needs to be very sexy (and i know mine probably isn’t, so rewriting is good). try to reword so as to not use 1st and 3rd person.) Most people don’t know it but they’ve come across functions all their lives. In fact, our very existence is tied to certain, very special functions called the laws of nature. Even ignoring those, though, it would be di–cult to go through a day without coming into contact with all sorts of functions. We can say that the idea of a function is one of the most basic and powerful ideas in the mathematics. Functions everywhere? But who’s ever heard of such a thing? Where does the word even come from? Well, they are everywhere, and, once you begin to see them, functions will be the easiest concept in mathematics. Where are these functions? Well, the menu in a restaurant is a function. So are the prices in a supermarket. Today’s temperature. Your height, your age, your weight. These are all functions. A function is just a way of attaching or relating one thing to another. A menu attaches prices to the food in a restaurant, and a supermarket attaches prices to the things it sells. We need to notice one very important fact, which is that these functions can give only one price to each item. We would certainly get angry if a restaurant charged two diﬁerent prices for the same dish. However, it’s perfectly natural for a restaurant to charge the same price for diﬁerent dishes. Similarly, one person cannot have two diﬁerent heights, but two people can have the same height. 3.1.1 Variables, Constants and Relations A variable is a label which we allow to change and become any element of some set of numbers. For example, on a menu in a restaurant \price" is a variable on the set of real numbers, since for any menu item the manager can choose any price he or she feels like (with the aim of staying in business). Most often, a variable will be a letter which can take on any value in some set of numbers. In this textbook we will only use real variables, which may take on the value of any real number. Though a variable is free to vary, if we wish we can specify that the variable takes on a speciﬂc value, in which case we say that we assign a value to the variable. In fact, we do this all the time when working with variables. When we say \what if we set the price to R50", we are just assigning the value \R50" to the variable \price". You have probably already done this quite frequently in algebra, when you say \let x be 1". A constant is a variable which is ﬂxed. We may not know the value of this constant, but this is a number which does not change throughout any problem. The \speed of light" is a variable which is always 300 000km per second, i.e. it is a constant. Such constant variables occur most frequently in the laws of physics. Variables on their own are very abstract, so don’t worry if it is slightly confusing. They become much more understandable when we start to relate them to each other. \Price" on a menu may not be a constant, but it must be tied to the items on that menu. For each item, we have a speciﬂed price. We can 39 think of \item" as a variable in its own right, and then the menu does nothing but tell us the relationship between the two variables \item" and \price". In general, a relation is an equation which relates two variables. For example, y = 5x and y2 + x2 = 5 are relations. In both examples x and y are variables and 5 is a constant, but for a given value of x the value of y will be very diﬁerent in each relation. Our example of a restaurant menu shows that relations between variables take on varied representations. Besides writing them as formulae, we most often come across relations in words, tables and graphs. Instead of writing y = 5x, we could also say \y is always ﬂve times as big as x". We could also give the following table: (NOTE: Working on a Latex-less machine, so table will come later) (I put in a table but not sure if it’s ok - Jothi) x 2 6 8 13 15 y = 5x 10 30 40 65 75 Some of you may object that this table isn’t very satisfactory, as the same table could represent almost any relation between x and y. However, when using tables we normally cheat and just assume that the obvious relationship in the table is the relationship. Finally, we look at graphs (NOTE: surely thisneeds to wait until later? sorry - structuring major headache here) 3.1.2 Deﬂnition of a Function (grade 12) A function is a relation for which there is only one value of y corresponding to any value of x. We sometimes write y = f (x), which is notation meaning ’y is a function of x’. This deﬂnition makes complete sense when compared to our real world examples \| each person has only one height, so height is a function of people; on each day, in a speciﬂc town, there is only one average temperature. However, some very common mathematical constructions are not functions. For example, consider the relation x2 + y2 = 4. This relation describes a circle of radius 2 centred at the origin, as in ﬂgure 3.1. If we let x = 0, we see that y2 = 4 and thus either y = 2 or y = 2. Since there are two y values which are ¡ possible for the same x value, the relation x2 + y2 = 4 is not a function. There is a simple test to check if a relation is a function, by looking at its graph. This test is called the vertical line test. If it is possible to draw any vertical line (a line of constant x) which crosses the relation more than once, then the relation is not a function. If more than one intersection point exists, then the intersections correspond to multiple values of y for a single value of x. We can see this with our previous example of the circle by looking at its graph again in ﬂgure 3.1. We see that we can draw a vertical line, for example the dotted line in the drawing, which cuts the circle more than once. Therefore this is not a function. 40 Figure 3.1: Graph of y2 + x2 = 4 In a function y = f (x), y is called the dependent variable, because the value of y depends on what you choose as x. We say x is the indepe
ndent variable, since we can choose x to be any number. 3.1.3 Domain and Range of a Relation The domain of a relation is the set of all the x values for which there exists at least one y value according to that relation. The range is the set of all the y values, which can be obtained using at least one x value. If the relation is of height to people, then the domain is all living people, while the range would be about 0:1 to 3 metres \| no living person can have a height of 0m, and while strictly it’s not impossible to be taller than 3 metres, no one alive is. An important aspect of this range is that it does not contain all the numbers between 0.1 and 3, but only six billion of them (as many as there are people). As another example, suppose x and y are real valued variables, and we have the relation y = 2x. Then for any value of x, there is a value of y, so the domain of this relation is the whole set of real numbers. However, we know that no matter what value of x we choose, 2x can never be less than or equal to 0. Hence the range of this function is all the real numbers strictly greater than zero. These are two ways of writing the domain and range of a function, set notation and interval notation. (NOTE: the syllabus does not say which notation method to use. we should ﬂnd out, and only use the one if possible. there is no need to add further confusion. then move the unused notation to Extra.) Set Notation First we introduce the symbols > ; < ; . > means ’is greater than’ and • means ’is greater than or equal to’. So if we write x > 5, we say that x is ‚ greater than 5 and if we write x y, we mean that x can be greater than or means ’is less than or equal equal to y. Similarly, < means ’is less than’ and to’. Instead of saying that x is between 6 and 10, we often write 6 < x < 10. This directly means ’six is less than x which in turn is less than ten’. ‚ ‚ • ; 41 A set of certain x values has the following form: x : conditions, more conditions f g (3.1) We read this notation as \the set of all x values where all the conditions are satisﬂed". For example, the set of all positive real numbers can be written as which reads as \the set of all x values where x is a real number x : x f and is greater than zero". (NOTE: have we even explained what > ; < ; ‚ mean yet? remember... this book must not assume that anyone has seen this stuﬁ before.) R; x > 0 g • 2 ; We use the same notation (with the letter y instead of x) for the range of the function. Interval Notation (NOTE: rewrite this subsubsection. ﬂrst describe what the brackets mean, and then introduce the concept of a union. all these concepts are new... so we must describe everything in detail.) Here we write an interval in the form ’lower bracket, lower number, comma, upper number, upper bracket’. We can use two types of brackets, square ones [; ] or round ones (; ). A square bracket means including the number at the end of the interval whereas a round bracket means excluding the number at the end of the interval. It is important to note that this notation can only be used for all real numbers in an interval. It cannot be used to describe integers in an interval or rational numbers in an interval. So if x is a real number greater than 2 and less than or equal to 8, then x is any number in the interval (2;8] (3.2) It is obvious that 2 is the lower number and 8 the upper number. The round bracket means ’excluding 2’, since x is greater than 2, and the square bracket means ’including 8’ as x is less than or equal to 8. Now we come to the idea of a union, which is used to combine things. The . Here we use it to combine two or more intervals. For x < 10, then the set symbol for union is example, if x is a real number such that 1 < x of all the possible x values is 3 or 6 • • [ (1;3] [ [6;10) (3.3) sign means the union(or combination) of the two intervals. We use where the the set and interval notation and the symbols described because it is easier than having to write everything out in words. [ 3.1.4 Example Functions In this section we will look at several examples of functions. Here we will let go of our real-world examples, and look exclusively at real valued functions, because only in such cases do we see the full use and power of functional mathematics. While it is instructive to see a menu or people’s height as a function, it is not very interesting. On the other hand, all of advanced physics and statistics depend on real valued functions. Very little is more important than gaining an 42 intuitive grasp of real functions, and we will spend the remainder of this chapter doing just that. When considering real valued functions, our major tool is drawing graphs. In the ﬂrst place, if we have two real variables, x and y, then we can assign values to them simultaneously. That is, we can say \let x be 5 and y be 3". Just as we write \let x = 5" for \let x be 5", we have the shorthand notation \let (x; y) = (5; 3)" for \let x be 5 and y be 3". We usually think of the real numbers as an inﬂnitely long line, and picking a number as putting a dot on that line. If we want to pick two numbers at the same time, we can do something similar, but now we must use two dimensions. What we do is use two lines, one for x and one for y, and rotate the one for y, as in diagram (NOTE: insert diagram). We call this the Cartesian plane. (NOTE: This whole y and f (x) thing needs to be cleared up \| I would do it here, but then it’s also discussed above in the deﬂnition of a function. I think the problem comes with the varying uses of y, and I think a physicist would be better than a mathematician to clear this up. Personally, rigorously, I don’t really know what’s going on with this notation.) The great beauty of doing this is that it allows us to \draw" functions, in a very abstract way. Let’s say that we were investigating the function f (x) = 2x. We could then consider all the points (x; y) such that y = f (x), i.e. y = 2x. For example, (1; 2); (2:5; 5); and (3; 6) would all be such points, whereas (3; 5) 3. If we put a dot at each of those points, and then at would not since 5 every similar one for all possible values of x, we would obtain the graph shown in (NOTE: put in). = 2 £ The form of this graph is very pleasing \| it is a simple straight line through the middle of the plane. Now some of you may have guessed this graph long before we plotted it, but the point is that the technique of \plotting", which we have followed here, is the key element in understanding functions. To show you why, we will now consider whole classes of functions, and we will relate them by the simple fact that their graphs are nearly identical. Straight Line Functions These functions have the general form f (x) = ax + b (3.4) where a and b are constants. The value of a is called the gradient or slope and tells us how steep the line is (the larger the number, the steeper the line). If a is greater than zero it means the line increases from left to right (slopes upwards), if it is smaller than zero the line increases from right to left (slopes downwards). b is called the y-intercept and tells us where the line goes through the y-axis. For example the function f (x) = 2x + 3 has a gradient of 2 and a y-intercept of 3. This means that the line cuts through the y-axis at a value of 3 and slopes upwards. We can calculate the values of y for certain values of x and then plot them in a graph (see ﬂgure 3.2). x : y = 2x + 3 : -5 -7 -4 -5 -3 -3 -2 -1 11 5 13 43 6 12 Figure 3.2: Graph of f (x) = 2x + 3 However we only need two points to plot a straight line graph. The easiest points to use are the x-intercept (where the line cuts the x-axis) and the ya intercept. The x-intercept occurs when y = 0, so it is always equal to b . So if asked to plot a straight line, there is no need to calculate lots of y values, you just need to ﬂnd the x and y intercepts and draw a line through them. ¡ Parabolic Functions A parabola looks like a hill, either upside down (for a \positive" parabola) or right way up (for a \negative" one), which is the same on both sides, as in the diagrams (NOTE: put in): You may have noted that when we say the parabola is \the same on both sides", we are just stating that these functions are horizontally symmetric. This means that if you ip them from left to right along a speciﬂc line, which is called the line of symmetry, they look the same. This line of symmetry is sometimes called the axis of symmetry. Parabolic functions are functions of the form f (x) = ax2 + bx + c (3.5) where a, b and c are constants. The a involves the shape of the parabola and says how steep the curves are. If a is positive, then the hill is upside-down. If a is negative, then the hill is the right way up. c is the y-intercept, which is where the parabola cuts the y axis. b has to do with the shift in the parabola to the left or the right. Two important features of the parabola are its turning point and line of symmetry(described above). The turning point says how high the hill is. If the hill is the right way up, then the turning point is the maximum value of the parabola, and if it is upside-down, then the turning point is the minimum value. Now the above form of the parabola is the standard form. It can also be 44 written in the form f (x) = a(x p)2 + q ¡ (3.6) where the two new constants p and q give the turning point (p;q) of the parabola. This form of the parabola can be obtained from the standard form by completing the square (See algebra). The value p of the turning point is actually the line of symmetry. So if p = 3, then x = 3 is the line of symmetry(which is always vertical for the parabola). q is the maximum or minimum value of the function(that is the maximum or minimum value of y). At ﬂrst it might seem di–cult to sketch the graph of a parabola but once a simple procedure is followed, then it becomes easier. When sketching the graph, we need to use some information about it. The only information we have are its
shape, x and y-intercepts and its turning point. We start oﬁ by seeing whether the parabola is an hill that is the right way up or upside-down. Recall that we can ﬂnd this out from the sign of a. Next we calculate the x-intercepts by setting y = 0 and solving the equation 0 = ax2 + bx + c (3.7) which doesn’t always have a solution, meaning that not all parabolas cut the x-axis. These would be hills which never quite make it to the x-axis, or upsidedown hills which are never low enough to touch the x-axis. However, if there is one solution, and it is not zero, then because of the symmetry there must be two solutions which can be both positive, or negative or a plus and a minus one. (NOTE: max/min turning points.) The y-intercept is just c. Last we ﬂnd the turning point of the parabola. One way is to write the equation of the parabola in the form 3.6 and we have b 2a , the line of symmetry and found p and q. Another way is to calculate x = ¡ also the value of p. q is found by putting p in the equation of the parabola. So now we are able to plot some points and join them up to form a parabola. We can create a table of x and y values for the parabola f (x) = x2 9 and then plot them (see ﬂgure 3.3). Note that you could spot the symmetry of the graph by examining the table alone, where we see that x = 1 and x = 1 give the same value for y. Note also that for this parabola b = 0, so the line of the symmetry is the y-axis since the parabola looks the same on both sides of the y-axis. ¡ ¡ x : y = x2 ¡ 9 : -4 7 -3 0 -2 -5 -1 -8 0 -9 1 -8 2 -5 3 0 4 7 Hyperbolic Functions Hyperbolas look like 2 parabolas on their side which are mirror reections of each other around the diagonal (NOTE: sketch). Hyperbolic functions look like f (x) = a x + b (3.8) where a and b are constants. Just like for parabolas, a tells us how steep the curves are and b tells us how high the curves are. 45 Figure 3.3: Graph of the parabola f (x) = x2 9 ¡ Since we cannot divide by zero1, it is not possible to have x = 0, so there is no y-intercept. When you go far enough away from the y-axes, the curves start to look like straight lines, and we call them asymptotes. For example we can create a table of x and y values for the hyperbolic function f (x) = 4 x and plot them (see ﬂgure 3.4) x : f (x) = 4 x : -8 - 1 2 -4 -1 -2 -2 -1 -4 - 1 2 - Exponential Functions y = abx + c b > 0 (3.9) 3.2 Exponentials and Logarithms The syllabus requires: (grade 12) switch between log and exp form of an equation (grade 12) derive and use the laws of logs † † (NOTE: need an intro. this should have lots of stuﬁ about how people used exp/logs to multiply numbers by adding them. with a few examples... to show that you don’t need a calculator.) 1(NOTE: i’m sure there are interesting facts about dividing by zero (not that i know of \| luke)) 46 Figure 3.4: Graph of the hyperbola f (x) = 4 x 3.2.1 Exponential Functions (NOTE: need an intro. we already covered exponentials in \numbers", but maybe we should move it here instead.) 3.2.2 Logarithmic Functions (NOTE: these Laws need introduced properly with more detailed derivations and examples of their use, highlight each ones importance. rewrite the intro to not include so many new terms... and to read better for a 16 year old.) Logarithms, commonly referred to as Logs, are the algebraic inverse of exponents. When we say \inverse function" we mean that the answer becomes the question and the question becomes the answer. For example, in the expression ab = x the \question" is \what is a raised to the b power." The answer is \x." The inverse function would be logax = b or \by what power must we raise a to obtain x." The answer is \b." Many students ﬂnd logarithms di–cult. For now you can be successful if you learn the terminology and come to understand the relationships of the terms. (NOTE: this next graph needs more explanation) Law 1 Since a0 = 1, loga1 = 0 47 Figure 3.5: The Exponential Function f (x) = ex Law 2 Since a1 = a, logaa = 1 Law 3 This one is a bit trickier to see. The law is that logaax = x. If we re-write it as loga(ax) = x we can see that it is ax = (ax), which is, of course, true. We can also then say that logaax = x logaa = x(1) = x. The upshot being that any exponent of the (operand?) can simply be moved to simple multiplication by the log. ¢ Law 4 The laws of exponents am logarithms loga(m ¢ n) = logam+logan and loga( m an = am+n and a m an = am ¡ n ) = logam n translate to the laws of logan respectively. ¡ ¢ Base In the previous examples a is the base. We generally use the \common" base, 10, or the natural base, e. The number e is an irrational number between 2:71 and 2:72. It comes up surprisingly often in Mathematics, but for now su–ce it to say that it is one of the two common bases. While the notation log10(x) and loge(x) may be used, log10(x) is often styled log(x) in Science and loge(x) is normally written as ln(x) in both Science and Mathematics. 48 Figure 3.6: The Logarithmic Function f (x) = ln(x) It is often necessary or convenient to convert a log from one base to another. An Engineer might need an approximate solution to a log in a base for which he does not have a table or calculator function, or it may be algebraically convenient to have two logs in the same base. To aﬁect a change of base, apply the change of base formula: logax = logbx logba (3.10) where b is any base you ﬂnd convenient. Normally a and b are known, therefore logba is normally a known, if irrational, number. 3.3 Extra (NOTE: this is non-syllabus content on absolute value functions, but perhaps the absolute operator should be worked into the main text and this section deleted, as it is quite important.) 3.3.1 Absolute Value Functions (NOTE: i’m pretty sure this is not on the syllabus) The absolute value of x has the following deﬂnition = x j j ‰ x x ¡ 0 if x ‚ if x < 0 (3.11) 49 Figure 3.7: The Functions f (x) = ln(x) and f (x) = ex are symmetrical about the origin. In other words, the absolute value sign makes the term inside this sign positive. If it is already positive, then there is no change, and otherwise the sign of this term changes. Now an absolute value function has the following general form where a, b and c are constants. x f (x3.12) Let us again consider an absolute value function with the general form y = + c. We must consider two cases separately: b: x a j b j ‚ ¡ x Now, since x is positive and therefore ‚ b and thus , the term inside the absolute value sign b. Thus y = a(x ¡ b) + c = ax + (c ab) ¡ In other words, this is a straight line with slope a and y-intercept c x < b: In this case, the term in the absolute value sign is negative and thus (x ¡ ¡ b) = x + b. Therefore ¡ y = a( x + b) + c = ¡ ax + (c + ab) ¡ which is a straight line with slope a and y-intercept c + ab. ¡ 50 (3.13) ab. ¡ x b j ¡ = (3.14) j 0 Now at x = b the function value is y = a j j consists of half of two straight lines with slopes turning point (b;c). + c = c. Therefore function a and a, which meet at the ¡ The function has the axis of symmetry x = b. In other words, the part of the function on one side of the vertical line x = b is the same as the reection about this line of the part of the function on the other side. We can see this as follows: Consider the function values at the points x = b + z and x = b z, where z > 0 (these are two point the same distance from the line x = b). Now the function values at these two points are ¡ and (b + z) f (b + z = az + c j + c b j ¡ f (b ¡ z) = a (b ¡ j z = a j j ¡ = az + c + c b j z) ¡ + c (3.15) (3.16) (3.17) (3.18) (3.19) (3.20) These function values are the same. Therefore, whether we move to the left or the right of the line x = b, the function values remain the same. Therefore x = b is an axis of symmetry. x = b z z az + c (b;c) b + z z b ¡ Figure 3.8: Graph of f (b + z) and f (b of symmetry x = b x z) where f (x) = a j j ¡ + c; with the line Now let us consider two cases: a < 0 and a > 0. If a is positive, then the line on the left of the turning point (with slope a) will have a negative slope and the line on the right (with slope a) will have a positive slope. Thus the graph will be shaped like a V. ¡ 51 Otherwise, if a is negative, then the line on the left has the positive slope a and the line on the right has the negative slope a. Therefore the graph is ¡ an upsidedown V. (b;c) a < 0 (b;c) a > 0 Figure 3.9: Graph of f (b + z) and f (b for a > 0 the other for a < 0. x z) where f (x) = a j j ¡ + c. One case is Notice also that an absolute value function does not necessarily have xIt these do exist, then they will be the x-intercepts of the two intercepts. straight lines making up the absolute value function. 52 Chapter 4 Numerics 4.1 Optimisation The syllabus requires: Linear Programming (Grade 11) † 1. Solve linear programming problems by optimising a function in two variables, subject to one or more linear constraints, by numerical search along the boundary of the feasible region. 2. Solve a system of linear equations to find the co-ordinates of the vertices of the feasible region. In everyday life people are interested in knowing the most e–cient way of carrying out a task or achieving a goal. For example, a farmer might want to know how many crops to plant during a season in order to maximise yield (produce) or a stock broker might want to know how much to invest in stocks in order to maximise proﬂt. These are examples of optimisation problems, where by optimising we mean ﬂnding the maxima or minima of a function. This function we wish to optimise (i.e. maximise or minimise) is called the objective function (we will only be looking at objective functions which are functions of two variables). In the case of the farmer, the objective function is the yield and it is dependent on the amount of crops planted. If the farmer has two crops then we can express the yield as f (x;y) where the variable x represents the amount of the ﬂrst crop planted and y the amount of the second crop planted. For
the stock broker, assuming that there are two stocks to invest in, f (x;y) is the amount of proﬂt earned by investing x rand in the ﬂrst stock and y rand in the second. In practice it is often that constraints, or restrictions, are placed on x and y. The most common of these constraints is the non-negativity constraint. That is, we might require that x 0. For the farmer, it would make little ‚ sense if we were to speak of planting a negative amount of crops and so when 0 must be considered. Other optimising f (x;y) the constraints x constraints might be that the farmer cannot plant more of the second crop than the ﬂrst crop and that no more than 20 units of the ﬂrst crop can be planted; 20. Constraints these constraints translate into the inequalities x y and x 0 and y 0 and y ‚ ‚ ‚ ‚ • 53 mean that we can’t just take any x and y when looking for the x and y that optimise our objective function. If we think of the variables x and y as a point (x;y) in the xy plane then we call the set of all points in the xy plane that satisfy our constraints the feasible region. Any point in the feasible region is called a feasible point. y 20 15 10 5 5 10 15 20 x Figure 4.1: The feasible region corresponding to the constraints x x y and x 20. 0, y 0, ‚ ‚ ‚ • For example, the non-negativity constraints x 0 mean that every (x;y) we can consider must lie in the ﬂrst quadrant of the xy plane. The y means that every (x;y) must lie on or below the line y = x constraint x 20 means that x must lie on or to the left of the line x = 20. For these and x constraints the feasibility region is illustrated as the shaded region in Figure 4.1. 0 and y ‚ ‚ ‚ • Constraints that have the form ax + by c or ax + by = c are called linear p2; constraints. Examples of linear constraints are x + y a constraint being linear just means that it requires that any feasible point (x;y) lies on one side of or on a line. Interpreting constraints as graphs in the xy plane is very important since it allows us to construct the feasible region such as in Figure 4.1. We have the following rule for any linear constraint: 2x = 7 and y • • • ¡ 0, ax + by = c ax + by c • If b = 0, feasible points must lie on the line y = a b If b = 0, feasible points must lie on the line x = c=a ¡ x + c b : If b = 0, feasible points must lie on or below the line y = a b ¡ x + c b : If b = 0, feasible points must lie on or to the left of the line x = c=a Once we have determined the feasible region the solution of our problem will be the feasible point where the objective function is a maximum/ minimum. Sometimes there will be more than one feasible point where the objective function is a maximum/minimum \| in this case we have more than one solution. 54 6 6 4.1.1 Linear Programming The objective function is called linear if it looks like f (x;y) = ax + by where the coe–cients a and b are real numbers. For example, f (x;y) = 10x y is a linear objective function. If the objective function and all of the constraints are linear then we call the problem of optimising the objective function subject to these constraints a linear program. All optimisation problems we will look at will be linear programs. ¡ The major consequence of the constraints being linear is that the feasible region is always a polygon. This is evident since the constraints that deﬂne the feasible region all contribute a line segment to its boundary (see Figure 4.1). It is also always true that the feasible region is a convex polygon. The objective function being linear means that the feasible point(s) that gives the solution of a linear program always lies on one of the vertices of the feasible region. This is very important since, as we will soon see, it gives us a way of solving linear programs. (NOTE: Should I mention that a linear objective function deﬂnes a plane? This is crucial to the fact that optimal solutions are obtained at the vertices, though. Do Grade 11s know the equation of a plane? I would like to use the idea that the level sets of planes are lines and in so doing justify the \ruler" method.) We will now see why the solutions of a linear program always lie on the vertices of the feasible region. Firstly, note that if we think of f (x;y) as lying on the z axis, then the function f (x;y) = ax+by (where a and b are real numbers) is the deﬂnition of a plane. If we solve for y in the equation deﬂning the objective function then f (x;y) = ax + by x;y) b (4.1) If we consider Equation 4.1 corresponding to f (x;y) = What this means is that if we ﬂnd all the points where f (x;y) = c for any real number c (i.e. f (x;y) is constant with a value of c), then we have the equation of a line. This line we call a level line of the objective function (NOTE: Should I use this terminology?). Consider again the feasible region described in Figure 2y with this 4.1. Lets say that we have the objective function f (x;y) = x feasible region. 20 ¡ then we we get the level line y = 1 2 x + 10 which has been drawn in Figure 4.2. 2 + 5), f (x;y) = 0 (y = x Level lines corresponding to f (x;y) = 2 ), f (x;y) = 10 (y = x 10) have also been drawn in. It is very important to realise that these aren’t the only level lines; in fact, there are inﬂnitely many of them and they are all parallel to each other. Remember that if we look at any one level line f (x;y) has the same value for every point (x;y) that lies on that line. Also, f (x;y) will always have diﬁerent values on diﬁerent level lines. 5) and f (x;y) = 20 (y = x 10 (y = x 2 ¡ 2 ¡ ¡ ¡ If a ruler is placed on the level line corresponding to f (x;y) = 20 in Figure 4.2 and moved down the page parallel to this line then it is clear that the ruler will be moving over level lines which correspond to larger values of f (x;y). So if we wanted to maximise f (x;y) then we simply move the ruler down the page until we reach the \lowest" point in the feasible region\|this point will then be the feasible point that maximises f (x;y). Similarly, if we wanted to minimise f (x;y) then the \highest" feasible point will give the minimum value of f (x;y). ¡ 55 y 20 15 10 5 f (x;y) = 20 ¡ 10 f (x;y) = ¡ f (x;y) = 0 f (x;y) = 10 f (x;y) = 20 x 5 10 15 20 Figure 4.2: The feasible region corresponding to the constraints x x represent various level lines of f (x;y). 20 with objective function f (x;y) = x y and x 0, ‚ 2y. The dashed lines 0, y ‚ ¡ • ‚ Since our feasible region is a polygon, these points will always lie on vertices in the feasible region. (NOTE: We could have inﬂnitely many solutions if the gradient of a constraint = the gradient of the level lines... should I mention this?). The fact that the value of our objective function along the line of the ruler increases as we move it down and decreases as we move it up depends on this particular example. Some other examples might have that the function increases as we move the ruler up and decreases as we move it down. It is a general property, though, of linear objective functions that they will consistently increase or decrease as we move the ruler up or down. Knowing which direction to move the ruler in order to maximise/minimise f (x;y) = ax + by is as simple as looking at the sign of b (i.e. \is b negative, positive or zero?"). If b is positive, then f (x;y) increases as we move the ruler up and f (x;y) decreases as we move the ruler down. The opposite happens for the case when b is negative: f (x;y) decreases as we move the ruler up and f (x;y) increases as we move the ruler down. If b = 0 then we need to look at the sign of a. If a is positive then f (x;y) increases as we move the ruler to the right and decreases if we move the ruler to the left. Once again, the opposite happens for a negative. If we look again at the objective function mentioned earlier, f (x;y) = x 2), then we should ﬂnd that f (x;y) increases as we move the ruler down the page since b = 2 < 0. This is exactly what we found happening in Figure 4.2. 2y (a = 1 and b = ¡ ¡ The main points about linear programming we have encountered so far are ¡ † † † † The feasible region is always a polygon. Solutions occur at vertices of the feasible region. Moving a ruler parallel to the level lines of the objective function up/down to the top/bottom of the feasible region shows us which of the vertices is the solution. The direction in which to move the ruler is determined by the sign of b and also possibly by the sign of a. 56 (NOTE: I would like to mention f to determine ‘the direction in which to move the ruler’. Even if I neglect the fact that students certainly know nothing about partial diﬁerentiation, I’m still not sure whether I can mention and work with vectors in the plane...) r These points are su–cient to determine a method for solving any linear program. If we wish to maximise the objective function f (x;y) then: 1. Find the gradient of the level lines of f (x;y) (this is always going to be a b as we saw in Equation 4.1) ¡ 2. Place your ruler on the xy plane, making a line with gradient units on the x-axis and a units on the y-axis) ¡ a b (i.e. b ¡ 3. The solution of the linear program is given by appropriately moving the ruler. Firstly we need to check whether b is negative, positive or zero. (a) If b > 0, move the ruler up the page, keeping the ruler parallel to the level lines all the time, until it touches the \highest" point in the feasible region. This point is then the solution. (b) If b < 0, move the ruler in the opposite direction to get the solution at the \lowest" point in the feasible region. (c) If b = 0, check the sign of a i. If a < 0 move the ruler to the \leftmost" feasible point. This point is then the solution. ii. If a > 0 move the ruler to the \rightmost" feasible point. This point is then the solution. (NOTE: Point 3 is essentially trying to work with ing what it is or that it exists!) f without actually know- r 4.2 Gradient The syllabus requires: † Investigate numerically the average gradient between two points on a curve and develop an intuitive underst
anding of the concept of the gradient of a curve at a point (NOTE: this is undefined if this should be numerical, or an intro to differentiation. it is best to have it spread over both) perhaps 4.3 Old Content (please delete when ﬂnished) 4.3.1 Problems We often have to solve problems in which there are several variables, which we can change to suit us. We can now develop a method of dealing with such problems. 57 Worked Example 1: Q: A farmer grows wheat and maize. He has 20 ﬂelds of available land on which he can plant crops. He must grow at least 5 ﬂelds of maize. Also he cannot grow more than twice as much maize as wheat. Draw a graph to show the feasible region showing the possible number of ﬂelds of wheat and maize the farmer can plant. What is the maximum number of ﬂelds of wheat the farmer can plant? A: Step 1: Analyse the problem and assign the variables x and y. Let x be the number of ﬂelds of wheat the farmer plants. Let y be the number of ﬂelds of maize the farmer plants. Step 2: Write down the inequalities which are the restrictions on x and y. x + y 20 (the farmer only has 20 ﬂelds) (4.2) (at least 5 ﬂelds of maize must be planted) (4.3) (it is not possible to have a negative number of ﬂelds of wheat)(4.4) (4.5) (the farmer cannot plant more than twice as much maize as wheat) • y x 5 0 ‚ ‚ 2x y • Remember that every piece of information you are given is important, so check that you have not left out an inequality. Also note that often variables cannot be negative, which give further inequalities as in the case of x 0. Step 3: Solve for y in terms of x where possible. ‚ y y x y x + 20 • ¡ 5 0 ‚ ‚ 2x • (4.6) (4.7) (4.8) (4.9) Step 4: Plot a graph and ﬂnd the feasible region. 24 21 18 15 12 12 15 18 21 24 Figure 4.3: Graph of TODO Step 5: Answer the original question. 58 We need to ﬂnd the maximum number of ﬂelds of wheat which can be planted. This is the maximum x value which is in the feasible region. This occurs at the point (15,5). Thus the maximum x value is 15. Remembering to give the answer in terms of the original question: The farmer can plant a maximum of 15 ﬂelds of wheat. 4.3.2 Maximising or Minimising the Objective Function The objective function is a function of x and y. We are usually told to maximise or minimise this function. Worked Example 2: Q: Consider the same situation as in worked example 1. The farmer can make a proﬂt of R100 on every ﬂeld of wheat and R200 on every ﬂeld of maize that he grows. How many ﬂelds of wheat and maize must the farmer plant to maximise his proﬂt and what is this maximum proﬂt? Steps 1 - 4 are as in worked example 1. A: Step 5: Deﬂne the objective function. The objective function, in this case, is the proﬂt in terms of the number of ﬂelds of wheat and maize (the variables x and y). This is given by Step 6: Solve for y. P = 100x + 200y y = 1 2 ¡ x + P 200 (4.10) (4.11) Step 7: Maximise/minimise the objective function. In this case we need to maximise the objective function which is the proﬂt P. The greater P the larger the y intercept of the straight line of y as a function 1 2 (line A is an example of of x. However, the slope of the line will always be such a line). ¡ Now to maximise P we need the y-intercept to be as large as possible, but the line must still pass through the feasible region. Thus take a ruler and move it parallel to line A (keeping the slope the same). Move the ruler outwards until it is at the edge of the feasible region. This is line B, which is the line of maximum P . The point on the feasible region through which this line passes (in this case ( 20 3 and y = 40 3 ; 40 3 ). The proﬂt can be calculated from the objective function as P = R3333. 3 )) is the point giving this proﬂt (so x = 20 3 = 13 1 3 = 6 2 Step 8: Give the answer in terms of the question. For a maximum proﬂt of R3333, the farmer must plant 6 2 3 ﬂelds of wheat and 13 1 3 ﬂelds of maize. (NOTE: Further examples need to be included here, particularly add an example which uses discreet variables.) Worked Example 3: Q: A delivery company delivers wood to client A and bricks to client B. The company has a total of 5 trucks. A truck cannot travel more than 8 hours per day and it takes 4 hours make the trip to and back from client A and 2 hours for 59 20 15 10 5 A B 5 10 15 20 25 30 Figure 4.4: Graph of TODO client B. To honour an agreement with client B, at least 2 truck loads of bricks must be delivered per day. Also client A needs no more than 9 truck loads of wood per day. The delivery company makes a proﬂt of R100 per truck load of wood and R150 per truck load of bricks delivered. How many truck loads of wood and bricks should be delivered per day so as to maxmise the proﬂt? What is this maximum proﬂt? Note: Client A is in the opposite direction to client B, so each truck can only deliver a full truck load to A or B (a truck cannot take half a load to A and the other half to B). A: Step 1: Let x be the number of truck loads of wood the company delivers to client A per day. Let y be the number of truck loads of bricks the company delivers to client B per day. Step 2: Firstly, the total number of hours of delivery time available is 5 8 hours = 40 hours, since there are 5 trucks, which cannot be driven more than 8 hour per day. Delivery to client A takes 2 hours and delivery to client B takes 4 hours. Therefore £ The other inequalities are 2x + 4y 40 • (4.12) y x x;y 2 9 0 ‚ • ‚ (the company must delivered at least 2 truck loads of bricks per day) (4.13) (4.14) (client A needs no more than 9 truck loads of wood per day) (4.15) (we cannot have a negative number of truck loads) (4.16) 60 Furthermore, we know that x and y must be integers (in other words we cannot have a fractional truck load). These are therefore called discreet variables. Step 3: y = y ‚ x • x;y + 10 x 2 ¡ 2 9 0 ‚ (4.17) (4.18) (4.19) (4.20) Step 4: We now plot the constraints and the feasible region (see the graph at the end). The region enclosed by the constraints is the shaded region, but since the variables x and y can only take on positive integer values, the feasible region actually consists of the collection of dots showing the integer values in the shaded region. Step 5: The objective function is the proﬂt (in Rands), which is P = 100x + 150y (4.21) Step 6: Solving for y gives y = 2 3 ¡ x + P 150 (4.22) Step 7: We need to maximise the proﬂt P and therefore we need to maximise the y-intercept of the previously deﬂned straight line. Line A shows an arbitrary 2 straight line with slope 3 , which is drawn, for convenience, with intercepts x = 9 and y = 6. If a ruler is moved outwards parallel to this line (i.e. keeping the slope ﬂxed) to the edge of the feasible region, we obtain line B, which passes through the point (8;6).(NOTE: RULERS??? is this really the best way to do this? can we please have some equations of lines!) ¡ Therefore the maximum proﬂt occurs when x = 8 and y = 6. This proﬂt (in Rands) is P = 100x + 150y = 100(8) + 150(6) = 1700 (4.23) (4.24) (4.25) Note: We cannot use the point (9; 5 1 edge of the shaded region, because this is 5 1 a half a truck load). 2 ), which is actually the point at the 2 is not an integer (we cannot have Step 8: The maximum proﬂt is R1700 per day, which is obtained when the company delivers 8 truck loads of wood to client A and 6 truck loads of bricks to client B per day. 61 B 10 8 6 4 2 A 6 2 4 8 10 12 14 16 18 Figure 4.5: Graph of TODO Essay 1 : Diﬁerentiation in the Financial World Author: Fernando Durrell I lived in Cape Town (South Africa) all my life. I attended Thomas Wildschudtt I then attended St. Owen’s Senior School in Junior and Senior Primary Schools. Retreat up to half way through Grade 11 at which point I left for St. Joseph’s Senior School in Rondebosch (since St. Owen’s closed permanently at the end of my Grade 11 year). It was always one of my ambitions to attend the University of Cape Town (UCT) because it is a prestigious university. I applied to study medicine at UCT, but was not accepted, and so I enrolled for a science degree at UCT and have never regretted it. (I can stand only so much visible blood.) I wasn’t sure about what I wanted to do with my life so I enrolled for Mathematics, Applied Mathematics, Chemistry and Physics in my ﬂrst year at university. By the end of my ﬂrst year at UCT, I wanted to continue with the Mathematics stream. I completed by Bachelor of Science (BSc) degree with majors (main subjects) Mathematics and Applied Mathematics and then completed my BSc (Honours) degree in Applied I completed by Master of Science degree in Financial Mathematics Mathematics. and am currently registered for the degree of Doctor of Philosophy in Mathematics. Diﬁerentiation in the Financial World Most of us don’t really think about saving our money to buy something in the future - we have to spend it now! Our parents unfortunately (and one day when we’re older most probably we too) have to save for the future: for that possible time when they don’t have a job; they may save to purchase furniture for the house; or a present for your birthday. The most important thing adults save for is retirement - this is when they decide that they want to stop working. Their kids may not be able to care for them because they too may have families or may not have jobs themselves. So, adults have to save money while they are 62 working to support them when they retire. The amount of money they receive after they retire is called their pension - so adults save so that they can receive a pension when they retire. Suppose James is paying R100 every month toward his pension. When James retires, he wants every month to receive a bit more than the R100 he contributed toward his pension (while he was working). If he doesn’t get a bit more than R100 pension every month (when he retires), then he may as well save his money under his bed until he retires. Now, there are many adults like James who are saving for their pension. To whom do all these adults pay t
heir monthly pension savings? They pay their monthly pensions savings to a pension fund. Suppose there are ten million adults paying R100 every month to a pension fund. That means that each month the pension fund receives R100x10 000 000 = R1b (i.e. one billion rand) in total each month. Now, each adult, like James, will want to receive a monthly pension which is greater than R100 when they retire. So, the pension fund must ensure that, when a pension fund contributor retires, he/she receives more than R100 pension every month. There are many pension funds in the world, so, if the pension fund James is saving with is going to give him a R110 monthly pension and another pension fund is going to give him a R120 monthly pension, then he is going to save with the latter pension fund. So, pension funds can’t give pensioners too little pension. In fact, they have to give pensioners as big a pension as possible. Now, the pension fund can’t just put the monthly R1bn in the bank and let it earn interest and divide this amongst all pensioners. The government takes a lot of the interest earned by pension funds as tax. So, the pension funds have to make more money, and so they turn to the stock market. But the thing about the stock market is that one can lose a lot of money very quickly if one is not careful. The advantage about putting one’s money in the bank is that, when you come back the next day, your money will still be there. If you invest in the stock market today and you come back tomorrow, then you could have lost a substantial amount of money, but you could also have made a lot of money. So, for the pension fund, depositing the monthly R1b with the bank is appealing, but so is the stock market. The pension has to ﬂnd the best combination of the two (the bank and the stock market). That involves, ﬂnding out how much money to deposit with the bank and how much to invest in the stock market so that the pension fund makes as much money as possible. To solve this problem, involves diﬁerentiation, which is the topic of the next chapter. This is just one way in which diﬁerentiation is used in the ﬂnancial world. 63 Chapter 5 Diﬁerentiation 5.1 Limit and Derivative Calculus is fundamentally diﬁerent from the mathematics that you have studied previously. Calculus is more dynamic and less static. It is concerned with change and motion. It deals with quantities that approach other quantities. For that reason it may be useful to have an overview of the subject before beginning its In this section we give a glimpse of some of the main ideas intensive study. of calculus by showing how limits arise when we attempt to solve a variety of problems. 5.1.1 Gradients and limits A traditional slingshot is essentially a rock on the end of a string, which you rotate around in a circular motion and then release. When you release the string, in which direction will the rock travel? Many people mistakenly believe that the rock will follow a curved path. Newton’s First Law of Motion tells us that the path is straight. In fact, the rock follows a path along the tangent line to the circle, at the point of release. If we wanted to determine the path followed by the rock, we could do so, as tangent lines to circles are relatively easy to ﬂnd. Recall, from elementary geometry that a tangent line to a circle is a line that intersects the circle in exactly one point. In this chapter we will be concerned with tangent lines to a variety of functions, as the tangent line gives us the slope of a function at a point. Now let us consider the problem of trying to ﬂnd the equation of the tangent line t to a curve with equation y = f (x) at a given point P . 64 P f(x) Since we know that the point P lies on the tangent line, we can ﬂnd out the equation of t if we know its slope m. The problem is that we need two points to compute the slope and we only have one, namely P on t. To get around the problem we ﬂrst ﬂnd an approximation to m by taking a nearby point Q on the curve and computing the slope mP Q of the secant line P Q. f(x) P x Q a 65 From the ﬂgure we see that mP Q = f (x) x f (a) a ¡ ¡ (5.1) Now imagine that Q moves along the curve toward P .The secant line approaches the tangent line as its limiting position. This means that the slope mP Q of the secant line becomes closer and closer to the slope m of the tangent line as Q approaches P . We write m = lim P ! Q mP Q and we say that m is the limit of mP Q as Q approaches P along the curve. Since x approaches a as Q approaches P , we could also use Equation (5.1) to write ¡ ¡ The tangent problem has given rise to the branch of calculus called diﬁerential m = lim a ! (5.2) x f (x) x f (a) a calculus. 5.1.2 Diﬁerentiating f (x) = xn The central concept of diﬁerential calculus is the derivative. After learning how to calculate derivatives, we use them to solve problems involving rates of change. Deﬂnition: The derivative of a function f at a number a, denoted by f 0(a), is if this limit exists. f 0(a) = lim 0 ! h f (a + h) h ¡ f (a) (5.3) Let us use this deﬂnition to calculate the derivative of f (x) = x2, where n is a positive integer. f (x) ¡ f (x + h) h (x + h)2 h x2 + 2xh + h2 x2 ¡ x2 ¡ h 2xh + h2 h 2x + h f 0(x) = lim 0 ! h = lim 0 h ! = lim 0 h ! = lim 0 h ! = lim 0 h ! = 2x You should repeat this calculation for f (x) = x3 and (if you haven’t spotted a pattern yet!) for f (x) = x4. Then see if you can generalise what you are seeing to write down a formula for f 0(x) where f (x) = xn. (This isn’t a valid 66 mathematical way of arriving at a formula, but if you want to prove the general case you need to use the binomial theorem, which is outside the scope of your syllabus.) Hopefully you calculated the derivative of f (x) = x3 to be 3x2, and shortly after that spotted the pattern for powers of x: d dx (xn) = nxn ¡ 1 5.1.3 Other notations If we use the traditional notation y = f (x) to indicate that the dependent variable is y and the independent variable is x, then some common alternative notations for the derivative are as follows: f 0(x) = y0 = dy dx = df dx = d dx f (x) = Df (x) = Dxf (x) The symbols D and d=dx are called diﬁerential operators because they indicate the operation of diﬁerentiation, which is the process of calculating a derivative. It is very important that you learn to identify these diﬁerent ways of denoting the derivative, and that you are consistent in your usage of them when answering questions. Note Though we choose to use a fractional form of representation, dy IS NOT a fraction, i.e. dy with respect to x. Thus, dp is the \operator", operating on some function of x. dx is a limit and dx. dy dx means y diﬁerentiated dx does not mean dy dx means p diﬁerentiated with respect to x. The ‘ d dx ’ ¥ The syllabus requires: † † (grade 12) understand the limit concept in the context of approximating the rate of change or gradient of a function at a point (grade 12) establish derivatives of f (x) = xn from 1st principles and then generalise to obtain the derivative of f (x) = b f (x) = x3 f (x) = x2 1 x f (x) = 5.2 Rules of Diﬁerentiation In order to avoid diﬁerentiating functions from ﬂrst principles, we can establish certain rules. Rule 1 If f is a constant function, f (x) = c, then f 0(x) = 0 67 Rule 1 may also be written as d dx c = 0 (5.4) This result is geometrically evident if one considers the graph of a constant function. This is an horizontal line, which has slope 0. Rule 2: The Power Rule If f (x) = xn, where n is an integer, then f 0(x) = nxn ¡ 1 The Power Rule may also be written as d dx (xn) = nxn ¡ 1 (5.5) (5.6) This rule applies when n is a negative number. For example, the derivative of f (x) = 1 x is f 0(x) = x¡ 2, remembering that 1 x = x¡ 1. ¡ Rule 3: Linearity of Diﬁerentiation If c is a constant and both f and g are diﬁerentiable, then d dx (cf ) = c df dx df dx + d dx (f + g) = 5.2.1 Summary d dx c = 0 d dx (xn) = nxn ¡ 1 d dx (cf ) = c df dx dg dx (5.7) (5.8) d dx (f + g) = df dx + dg dx 5.3 Using Diﬁerentiation with Graphs The syllabus requires: † † (grade 12) find equation of a tangent to a graph (grade 12) sketch graph of a cubic function using diff to determine stationary points and their nature. x-axis intercept use factor theorem to determine 5.3.1 Finding Tangent Lines In section 5.1.1 we saw that ﬂnding the tangent to a function is the same as ﬂnding its slope at a particular point. The slope of a function at a point is just its derivative. If we want to ﬂnd a general formula for a tangent to a function, we diﬁerentiate the function. To ﬂnd the slope of the tangent at a particular point, we substitute that point’s x value into the function’s derivative. This will give us a single value, which is the slope of a straight line. We’ll look at one of these problems in the Worked Examples (section 5.4). 68 5.3.2 Curve Sketching Suppose we are given that f (x) = ax3 + bx2 + cx + d and we are asked to sketch the graph of this function. We will use our newfound knowledge of diﬁerentiation to solve this problem. There are FIVE steps to be followed: 1. If a > 0, then the graph is increasing from left to right, and has a maximum and then a minimum. As x increases, so does f (x). If a < 0, then the graph decreasing is from left to right, and has ﬂrst a minimum and then a maximum. f (x) decreases as x increases. 2. Determine the value of the y-intercept by substituting x = 0 into f (x) 3. Determine the x-intercepts by factorising ax3 + bx2 + cx + d = 0 and solving for x. First try to eliminate constant common factors, and to group like terms together so that the expression is expressed as economically as possible. Use the factor theorem if necessary. 4. Find the turning points of the function by working out the derivative df dx and setting it to zero, and solving for x. 5. Determine the y-coordinates of the turning points by substituting the x values obtained in the previous step, into the expression for f (x). 6. Step 6 of 5, Draw a neat sketch. The syllabus requ
ires: (grade 12) use the differentiation rules † Dx[f (x) g(x)] = Dx[f (x)] § § Dx[k:f (x)] = k:Dx[f (x)] Dx[g(x)] 5.4 Worked Examples Worked Example 2 : Finding derivatives from ﬂrst prin- ciples Question: Find the derivative of the function f (x) = x2 a. ¡ 8x + 9 at the number Answer: Step 1 : Write out the deﬂnition From deﬂnition (5.3) we have f 0(a) = lim 0 ! h f (a + h) h ¡ f (a) 69 Step 2 : Fill in the function f (x) and multiply out f 0(a) = lim 0 ! h = lim 0 h ! [(a + h)2 ¡ 8(a + h) + 9] h [a2 ¡ ¡ 8a + 9] a2 + 2ah + h2 8a ¡ 8h + 9 ¡ h a2 + 8a 9 ¡ ¡ Step 3 : Simplify f 0(a) = lim 0 ! h 2ah + h2 h h(2a + h h 8h 8) ¡ ¡ = lim 0 h ! = 2a 8 ¡ And you’re done! Worked Example 3 : Finding and using derivatives from ﬂrst principles Question: If f (x) = 4x + 2x2, ﬂnd f 0(x) from ﬂrst principles and hence calculate f 0(2). Answer: Step 1 : Write out deﬂnition (5.3) and ﬂll in f (x) f 0(x) = lim ! h 0 • = lim h ! 0 • f (x + h) h ¡ f (x) ‚ 4(x + h) + 2(x + h)2 h (4x + 2x2) ¡ ‚ Step 2 : Multiply out and simplify 4h + 4xh + 2h2 h ‚ [4 + 4x + 2h] h 0 • f 0(x) = lim ! = lim 0 h ! = 4 + 4x Step 3 : Substitute the value of x into f 0(x) Since f 0(x) = 4 + 4x then f 0(2) = 4 + 4(2) = 12 70 Worked Example 4 : Using Notation and Rules of Dif- ferentiation Question: Diﬁerentiate the following using the Rules of Diﬁerentiation listed above: a) y = t4 e) Du(um) c) h(x) = x6+x4 b)y = x1000 d) d dr (5r3) Answer: a) Step 1 : Write out the Power Rule, equation (5.5) d dt (tn) = ntn ¡ 1 Step 2 : In this case n = 4 so... y0 = 4t3 ¡ 1 Step 3 : Simplify dy dt = 4t3 b) Step 1 : Write out the Power Rule, equation (5.5) d dx (xn) = nxn ¡ 1 Step 2 : In this case n = 1000 so... y0 = 1000x999 ¡ 1 Step 3 : Simplify y0 = 1000x999 c) Step 1 : Write out the Second Linearity Rule, equation (5.8) dx (f + g) = df d dx + dg dx Step 2 : Write out the Power Rule, equation (5.5) d dx (xn) = nxn ¡ 1 Step 3 : Identify f and g, and diﬁerentiate them separately using the Power Rule f (x) = x6 so f 0(x) = 6x6 1 = 6x5 ¡ g(x) = x4 so g0(x) = 4x4 ¡ 1 = 4x3 Step 4 : Add the derivatives of f and g h0(x) = 6x5 + 4x3 d) 71 Step 1 : Write out the First Linearity Rule, equation (5.7) dx (cf ) = c df d dx Step 2 : Write out the Power Rule, equation (5.5) d dr (rn) = nrn ¡ 1 Step 3 : In this case n = 3 and c = 5 so... d 1 dr (5r3) = 5 3r3 ¡ £ Step 4 : Simplify d dr (5r3) = 15r2 e) Step 1 : Write out the Power Rule, equation (5.5) Du(un) = nun ¡ 1 Step 2 : In this case n = m so... Du(um) = mum ¡ 1 which cannot be simpliﬂed further Worked Example 5 : Finding tangent lines Question: Find the slope of the tangent to the graph of y(x) = 3x2 + 4x + 1 at x = 5. Answer: Step 1 : Diﬁerentiate y to get a general equation for the tangent to the graph y0(x) = 6x + 4 Step 2 : Substitute the value x = 5 into the tangent equation just calculated y0(5) = 6 5 + 4 = 34 £ So the slope of the tangent line to y(x) at x = 5 is 34. Worked Example 6 : Drawing graphs Question: Draw the graph of f (x) = x3 + 3x2. Answer: Step 1 : Basic shape of graph a is positive so from left to right, the graph has ﬂrst a maximum and then a minimum 72 Step 2 : y intercept y = x3 + 3x2 therefore y(0) = 0. Step 3 : x intercepts x3 + 3x2 = 0 x2(x + 3) = 0 x = 0 or x = 3 ¡ Step 4 : Turning points dy dx = 3x2 + 6x set this to zero 0 = 3x2 + 6x 0 = 3x(x + 2) x = 0 or x = 2 ¡ Step 5 : y-coordinate of the turning points y(0) = 0 and y( Local max at ( ¡ 2) = ( 2)2 = 4 2)3 + 3( ¡ 2; 4) and local min at (0; 0) ¡ ¡ Step 6 : Draw a neat sketch (-2; 4) (-3; 0) (0; 0) 73 5.5 Exercises 1. Draw the graph of y = x2 + x 6 for to this curve at x = 3, x = 1 and x = gradient of the curve at each of these points. ¡ x 6. Draw the tangents 5 ¡ 2, and hence ﬂnd a value for the ¡ • • 2. Draw the graph of y = x2 4x ¡ 4 x for 0 hence ﬂnd a value for the gradient of the curve at each of these points. 6. Draw tangents to the curve at x = 4, x = 3 and x = 2 and • • 3. Diﬁerentiate each of the following from ﬂrst principles to ﬂnd dy dx (a) y = 5x (b) y = 9x + 5 (c) y = 3x2 (d) y = x3 (e) y = x2 + 3x x2 + 7 ¡ (f) y = 5x (g) y = 1 x (h) y = 1 x2 2x2 ﬂnd f 0(x) from ﬂrst principles and hence evaluate f 0(4) ¡ 3 ﬂnd f 0(x) from ﬂrst principles and hence evaluate ¡ 2x ﬂnd f 0(x) from ﬂrst principles and hence evaluate 4. If f (x) = 3x 1) and f 0( ¡ 5. If f (x) = 2x2 + 5x 2) 1) and f 0( f 0( ¡ 6. If f (x) = x3 ¡ ¡ f 0(1),f 0(0) and f 0( 1) ¡ 1. Diﬁerentiate the following functions with respect to x: (a) x5 (b) x3 (c) 12x2 (d) 5x4 (e) 3x2 (f) 7 (g) x5=3 (h) x3=4 (i) x2=5 (j) 8x1=4 (k) px (l) px3 74 (m) 2=x (n) 3=x2 2. Find the gradient function dy dx for each of the following: 4 ¡ (a) y = x2 + 7x 7x2 (b) y = x (c) y = x3 + 7x2 (d) y = 3x2 + 7x ¡ (e) y = (x + 3)(x 1) (f) y = (2x + 3)(x + 2. Find the gradient of the following lines at the points indicated: ¡ (a) y = x2 + 4x at (0;0) x2 at (1;4) (b) y = 5x (c) y = 3x3 (d) y = 5x + x3 at ( (e) y = 3x + 1 (f) y = 2x2 1; ¡ x at (1;4) x + 4 ¡ 2x at (2;20) 6) ¡ x at (2;8) ¡ 4. Find the coordinates of the point(s) on the following lines where the gra- dient is given: (a) y = x2, gradient 8 (b) y = x2, gradient (c) y = x2 8 ¡ 4x + 5, gradient 2 x2, gradient 3 ¡ (d) y = 5x ¡ (e) y = x4 + 2, gradient (f) y = x3 + x2 4 ¡ x + 1, gradient 0 ¡ 5. If f (x) = x3 + 4x ﬂnd (a) f (1) (b) f 0(x) (c) f 0(1) (d) f "(x) (e) f "(1) 75 Chapter 6 Geometry (NOTE: we need motivation and history of geometry here. real world examples (and obscure ﬂgures) and some interesting facts (this is rich... e.g. architecture, like carpentry). What is a degree... why computer graphics, manufacturing... is it out of 360... etc.) 6.1 Polygons The syllabus requires: † † † † develop conjectures related to triangles, quadrilaterals and other polygons. using any logical method attempt to justify, explain or prove these conjectures define various polygons (isosceles, equilateral, right angled triangles, trapezium, isosceles trapezium, kite, parallelogram, rectangle, rhombus, square and the regular polygons) can tell when polygons are similar. similar equilateral triangles are the line drawn parallel to one side of a triangle divides the other 2 sides proportionally A polygon is a shape or ﬂgure with many straight sides. A polygon has interior angles. These are the angles that are inside the polygon. The number of sides of a polygon equals the number of interior angles. If a polygon has equal length sides and equal interior angles then the polygon is called a regular polygon. (NOTE: the language used in this notation section sometimes aims too high; words like \denote" and \line segment" may be daunting when simpler words may be used. the audience is only 15 and all of this is new to them. everything needs explained in detail, using simple language... and diagrams help too.) We denote a line segment that extends between a point A and some point B by line AB. The length of this line is just AB. So if we say, AB = CD we mean that the length of the line segment from A to B is equal to the length of the line segment from C to D. 76 ~AB is the line segment with length AB and direction from point A to point B. Similarly, ~BA is the line segment with length AB = BA and direction from point B to point A. Suppose we have two line segments AB and BC that join at a point B. We denote the angle B between the line segments by ^B. A line of symmetry divides a shape in such a way that it appears the same on both sides of the line. For example, if you divide a square along its one diagonal then you divide it into two triangles that are exactly the same i.e. they ﬂt perfectly on each other when the square is folded along the diagonal. If a line AB bisects a line CD then AB divides CD into half. A stop sign is in the shape of an octagon, an eight-sided polygon. Some coins are heptagonal and hexagonal. In the UK there are two heptagonal coins. The honeycomb of a beehive consist of hexagonal cells. (NOTE: these are true examples, but maybe best left till the list of names of polygons. examples here should try to motivate the study of polygons... how can we actually use the study of polygons to enrich our lives.) 6.1.1 Triangles A triangle is a three-sided polygon. The sum of the angles of a triangle is 180–. The exterior angle of any corner of a triangle is equal to the sum of the two opposite interior angles (NOTE: a diagram for these 2 rules). We have the following triangles: Equilateral All 3 sides are equal and each angle is 60–. (NOTE: need an example diagram) Isosceles Two equal angles occur opposite two equal sides and vice versa. (NOTE: need an example diagram) Right-angled This triangle has a right angle. The side opposite this angle is called the hypotenuse. Pythagoras’s Theorem is often applied to this type of triangle (NOTE: the students have not been subjected to Pythagoras yet... so place in a reference to the relevant chapter/section.) (NOTE: need an example diagram) Scalene This is any other triangle where the sides have diﬁerent lengths and angles are diﬁerent sizes. (NOTE: this is not on the syllabus but its small, and need an example diagram) 77 6.1.2 Quadrilaterals Quadrilaterals are four-sided polygons. The basic quadrilaterals are the trapezium, parallelogram, rectangle, rhombus, square and kite,. Trapezium This quadrilateral has one pair of parallel opposite sides. It may also be called a trapezoid. If the other pair of opposite sides is also parallel then the trapezium is a parallelogram. Another type of trapezium is the isosceles trapezium, where one pair of opposite sides is parallel, the other pair of sides is equal and the angles at the ends of each parallel side are equal. An isosceles trapezium has one line of symmetry and its diagonals are equal in length. (NOTE: need an example diagram) Parallelogram A parallelogram is a special type of trapezium. It is a quadrilateral with two pairs of opposite sides equal. Squares, rectangles and rhombuses are parallelograms. We have the following properties of parallelograms. Both pairs of opposi
te sides are parallel. Both pairs of opposite sides are equal in length. (NOTE: what does equal mean? is it in length, or direction, or both? we must be more precise in our wording) Both pairs of opposite angles are equal. Both diagonals bisect each other (i.e. they cut each other in half). There are not always lines of symmetry. (NOTE: what is a line of symmetry? we haven’t mentioned them before here. is this really a property?) (NOTE: need an example diagram) Rectangle This is a parallelogram with 90– angles. † Both pairs of opposite sides are parallel. Both pairs of opposite sides are equal. All angles are equal to 90–. Both diagonals bisect each other. Diagonals are equal in length. There are two lines of symmetry. (NOTE: need an example diagram) Rhombus This is a parallelogram with adjacent sides equal. † Both pairs of opposite sides are parallel. All sides are equal in length. Both pairs of opposite angles equal. Both diagonals bisect each other at 90–. Diagonals of a rhombus bisect both pairs of opposite angles. There are two lines of symmetry. (NOTE: need an example diagram) Square † This is a rhombus with all four angles equal to 90– or a rectangle with adjacent sides equal. 78 In a square both pairs of opposite sides are parallel. All sides are equal in length. All angles are equal to 90–. Both diagonals bisect each other at rightangles. Diagonals are equal in length and bisect both pairs of opposite angles. There are four lines of symmetry. (NOTE: need an example diagram) Kite A kite is a parallelogram with two pairs of adjacent sides equal. † Other properties of a kite are that the two pairs of adjacent sides are equal. One pair of opposite angles are equal where the angles must be between unequal sides. One diagonal bisects the other diagonal and one diagonal bisects one pair of opposite angles. Diagonals intersect at right-angles. There is one line of symmetry. (NOTE: need an example diagram) 6.1.3 Other polygons There are many other polygons, some of which are given in the table below. (NOTE: need an example diagram) 6.1.4 Similarity of Polygons If two polygons are similar, one is an enlargement of the other. This means that the two polygons will have the same angles and their sides will be in the same proportion. (NOTE: expand this with quick examples.) We can use the symbol to mean is similar to. Two polygons are similar if and only if either or both of the following are » true: † Corresponding (NOTE: deﬂne corresponding) angles are equal. Corresponding sides are all in proportion. † For example, CA F D (NOTE: need diagram here) ABC 4 » 4 DEF if ^A = ^D;^B =^E;^C =^F and AB DE = BC EF = (NOTE: we need a lot more examples here, speciﬂcally that all equilateral triangles are similar.) 6.1.5 Midpoint Theorem (NOTE: this section could really do with some interesting facts and examples.) Line joining the midpoints of two sides of a triangle is parallel to the third side ABC with midpoints M and equal to half the length of the third side. Given of AB and N of AC. 4 (NOTE: although proofs are nice... it is often more important to push the result itself, and why it is useful. if a proof is confusing, the student will simply skip the section because it is too hard... even though application may be easy.) Prove that M N is parallel to BC and M N = 1 2 BC. (NOTE: diagram ’join the dots style’ is in comments in the source ﬂle.) Extend MN by its own length to a point P. Join AP, CP and MC. M N = N P by construction. AN = N C given N 79 midpoint of AC. So APCM is a parm diags bisect each other. So CP = M A opp sides of parm equal. But BM = AM given M midpoint of AB. So CP = M B. Now CP M B AB is a line segment. So BMPC is a parm 1 pair of opp sides equal and parallel. So MN parallel to BC opp sides of parm parallel. Now M N = N P by construction. So M N = 1 M P . But M P = BC opp sides of parm equal. So M N = 1 AB APCM is a parm So CP BC 2 ⁄ k k 2 ⁄ 6.1.6 Extra Angles of regular polygons We have a formula to calculate the size of the interior angle of a regular polygon. ^A = n 2 ¡ n £ 180– (6.1) where n is the number of sides and ^A is any angle. Areas of Polygons k 2 £ Area of triangle: 1 £ (sum of perpendicular height perpendicular height sides) base Area of trapezium: 1 2 £ £ Area of parallelogram and rhombus: base Area of rectangle: length Area of square: length of side (NOTE: everything from here on in Extra is probably acceptable syllabus material, but it is here for now so i can see what needs to be brought back in to the main text. the theorems are not on the syllabus, but we should maybe include them since they use basic geometry techniques... but do not call them theorems, rather use them as in-line examples or worked examples.) perpendicular height length of side breadth £ £ £ Parallelograms 4 4 ABC and alternate angles =; AD alternate angles =; AB ADC: 1. \DAC = \ACB To show that a quadrilateral is a parallelogram, show any one of the ﬂrst four properties or that one pair of opposite sides are equal and parallel. Theorem 1: Given a parallelogram ABCD (with both pairs of opposite sides parallel), prove that the opposite sides and angles are equal. (ﬂgure 4 here) Proof: Join AC. In k BC 2. \BAC = \ACD CD 3. AC = AC (AAS) So AB = CD and DA = BC ABC common sides So · 4 corresponding sides in congruent triangles ^B = ^D; ^A = ^C corresponding angles in congruent triangles Hence opposite sides equal and opposite angles equal. Theorem 2: Given parallelogram ABCD with AC and BD joined and denote their intersection by O. Prove that AC and BD bisect each other. (ﬂgure 5 here) Proof: In alternate CD 3. angles =; AB AB = CD (AAS) So AO = OC; BO = OD corresp sides in congruent triangles Hence AC and BD bisect each other. CD 2. \ABD = \BDC k opposite sides of parm equal So COD: 1. \BAC = \ACD alternate angles =; AB AOB and CDO CDA ABO · 4 4 4 4 4 k k 80 Rectangles To prove that a quadrilateral is a rectangle you can ﬂrst prove that it is a parallelogram and then prove that it has a right-angle. Or you can directly prove that it has four right-angles. Theorem : Given rectangle ABCD prove ACD and that diagonals are equal in length. common sides (RHS) So 4 corresp sides in congruent triangles Hence diagonals are equal in (ﬂgure 6 here) Proof: opposite sides equal 2. DC = DC all angles equal So BCD: 1. AD = BC 4 are equal 3. \D = \C AC = BD length. BCD ADC · 4 In 4 Rhombuses AOB and AOD: 1. AB = AD To prove that a quadrilateral is a rhombus you can ﬂrst prove that it is a parallelogram. Then you can prove: all four sides equal, diagonals intersect at right-angles or diagonals bisect corner angles. Theorem : Given rhombus ABCD with diagonals intersecting at point O. Prove that the diagonals intersect at right-angles and that they bisect the corner angles. (ﬂgure 7 here) Proof: In all sides of rhombus equal 2. AO = AO 4 4 diags bisect each other So common sides are equal 3. OB = OD · (SSS) So \AOB = \AOD corresp angles in congruent triangles 4 But BD is a straight line So \AOB = \AOD = 90– sum of angles is 180– So AC and BD intersect therefore diagonals intersect Now \BAO = \DAO \ABO = corresp angles in congruent triangles Similarly, \CBO \BCO = \CDO = \ADO AOD ) \DCO Hence diagonals also bisect the corner angles. COB DOC AOB BOC · 4 · 4 COD AOD AOB ) ) 4 4 · 4 4 4 Squares To prove that a quadrilateral is a square you can prove that it is a rhombus and then prove that it has four right-angles or equal diagonals. You can also prove that it is a rectangle and then prove all four sides equal, diagonals intersect at right-angles or diagonals bisect corner angles. 6.2 Solids The syllabus requires: † † † † † † analyse, describe and represent the properties and relationships of geometric solids by calculating surface area, volume and the effect on these by scaling one or more dimension by k. estimate volume of everyday objects solids to consider: sphere, hemisphere, combinations with cylinders (grade 12) solids to consider: pyramid right circular cone, tetrahedron, classify geometric solids in various ways (including regular polyhedra) investigate the effect of a plane cutting the regular polyhedra in various ways 81 (grade 12) plane cutting right circular cone † 6.3 Coordinates The syllabus requires: † use coordinate systems to represent geometric figures and derive for any 2 points a formula for distance between points (NOTE: call it the metric), gradient of line between points and the coordinates of the midpoint of the line joining points (NOTE: SH: i assume they expect this to all be done in E2 or E3. seems oblivious to the non-triviality of doing this on any old surface) the syllabus author 6.4 Transformations The syllabus requires: † † † † † † † † generalise the effect of the following rigid transformations to translations, reflections in x,y, and x = y a point: recognise when an object is similar to another object under some transformation. conjecture and prove such similarities rotation of a point through 1800 vertices of a polygon after enlargement by factor k vertices of a polygon after shearing (base on x axis, opposite side parallel) emphasise that rigid transformations (trans, ref, rot, glide ref) preserve shape and size. size and shearing preserves area enlargement preserves shape but not (grade 12) generalise the effect on the point of stretch by k (NOTE: first i have ever heard of stretching a point) and rotation about the origin by an angle ﬁ0 (grade 12) identify and classify geometric border patterns and tessellations in terms of line symmetry, glide reflection symmetry, rotational symmetry and point symmetry 6.4.1 Shifting, Reecting, Stretching and Shrinking Graphs: Shifting Graphs Let us assume that we know some function f (x). What would happen if we a), where a is some positive constant? Well, deﬂned the function y = f (x ¡ the value of y at x is really just the value of the function f at x a moved to the point x. In other words, this deﬂnes the graph we would get if we shifted ¡ 82 the function f (
x) by a to the right. mathematical term... can we call it translation instead?) (NOTE: Shifting is not really a good Similarly, the function y = f (x + a), is the result of shifting the function by a to the left (the function value at x + a is moved to x). f (x) f (x a) ¡ a f (x + a) f (x) a Figure 6.1: Graph of a function with the translations x x Note that this is just a simple shift either left or right of the entire graph. x + a and x ! ! a. ¡ Now let us look at the function y b = f (x), where b is some positive constant, which is the result of replacing y by y b in the function y = f (x). This gives us that y = f (x) + b. The function is thus shifted upwards by the constant b. ¡ ¡ We can also replace y by y + b, which again just results in movement in the opposite direction. We can see this because y + b = f (x) implies that y = f (x) b, which shows that f (x) has been shifted downwards by b. ¡ f (x) + b b f (x) f (x) b f (x) b ¡ Figure 6.2: Graph of sin(x) with the translations y y ! that this is just a simple shift either up or down of the entire graph. y + b and b ! b. Note ¡ Now, what about relations in general? If we take a relation, which depends b (where a and b are positive a and y by y on x and y, and replaced x by x ¡ ¡ 83 ¡ a;y constants), then what would happen? Well, the relation value at the point (x b) would be moved to the point (x;y). Therefore the graph of the relation would be shifted right by a and upwards by b. Similarly, if we replaced x by x + a and y by y + b, then the function would be moved left by a and downwards by b. ¡ Reections x), where f (x) is a known function. This takes Now consider deﬂning y = f ( ¡ the function value at the point x to the point x. In other words, the function values on one side of the y-axis are moved to the other side of the y-axis. Thus the function is reected about the y-axis. ¡ Alternatively we can look at the function y = f (x). This is the same as ¡ saying y = f (x), which reects the function about the x-axis (every positive function value is changed to the corresponding negative function value and vice versa). ¡ f (x) f ( x) ¡ f (x) f (x) ¡ Figure 6.3: Graph of a function with the reections x ! f (x). Note that these are just reections in the vertical and horizontal axes, ! ¡ x and f (x) ¡ respectively. As before, we can generalise these idea to deal with relations. In all cases, x, then the relation will be reected about the y-axis and if ¡ y then there will be a reection about the x-axis. if we change x to we replace y by ¡ Note: We say that f (x) is symmetric about the y axis if f (x) = f ( x) (in other words, the function and its reection about the y-axis are the same). f (x), then we say that f (x) is symmetric about the x-axis. Similarly, if f (x) = ¡ ¡ 6.5 Stretching and Shrinking Graphs (NOTE: The ﬂgures are correct, but i think the ﬂgures are negating the truth.) We shall now look at what happens to f (x) if we consider the function y = f (ax), where a is a positive constant and a > 1. The point ax on the x-axis is further 84 ). Now the function value at ax is moved x from the y-axis than x (since j j to x, so the function is moved towards the x-axis by a factor of a. Thus the eﬁect is to shrink f (x) horizontally by a factor of a. ax j j > The function y = f ( x closer to the y-axis than x (as x so the function is stretched horizontally by a factor of a. a ) has the opposite eﬁect. The point x ). The function value at on the x-axis is a is moved to f (x) f (ax) f ( x a ) f (x) ax x x x a Figure 6.4: Graph of a function with the rescaling x that these are just a shrinking and stretching in the horizontal axis. ax and x ! ! x a . Note (NOTE: this is correct. vertical and horizontal rescalings are diﬁerent... i think the author got confused and thought the same thing happened in each. this needs ﬂxed.) Replacing y by by, where b is a positive constant and b > 1, gives by = f (x) and thus y = 1 b f (x). The function value at any point x is reduced by a factor of b. Therefore the graph shrinks vertically by a factor of b. b to give y b = f (x), we obtain the function y = bf (x). At each x value the function value is increased by a factor of b so the function is stretched vertically. Similarly, if we replace y with y Again these result can be used to deal with relations as well. If x and y are 85 1 b f (x) f (x) f (x) bf (x) Figure 6.5: Graph of a function with the rescaling f (x) ! bf (x). Note that these are just a shrinking and stretching in the vertical axis. 1 b f (x) and f (x) ! replaced by ax and by in any relation, the eﬁect is to shrink the graph of this relation by a factor of a horizontally and by a factor of b vertically. Similarly, a and y to y changing x to x b causes the relation to be stretched horizontally and vertically by factors of a and b respectively. 6.6 Mixed Problems If we perform many of these transformations on a given function, then we must combine the diﬁerent eﬁects. However, it is very important that we eﬁect these changes in the right order. Here are some examples of mixed problems. 6.7 Equation of a Line The syllabus requires: 86 derive formula for the equation of a line when given 2 points derive formula for the line parallel to a given line and passing through a point derive formula for the inclination of a line † † † 6.8 Circles The syllabus requires: † † † † (grade 12) tangent is perpendicular to the radius (grade 12) the line from the centre of a circle perpendicular to a chord bisects the chord (grade 12) angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc at the circle (grade 12) the opposite angle of a cyclic quadrilateral are supplementary the tangent chord theorem (NOTE: really... word for word) thats what it says, 6.8.1 Circles & Semi-circles Circles: A circle centered at the origin with radius r is described by the relation x2 + y2 = r2 (6.2) r) We can see that a circle is not a function, since both (0;r) and (0; satisfy the relation (in other words, the line x = 0 will always cut the circle at two points). ¡ Now, since x2 = r2 y2 and y2 is never negative, it follows that x2 and thus y2 = r2 ¡ r ¡ • x2 and therefore • x ¡ r ¡ • y • r. Therefore the domain of the relation is [ r. Thus the range of the relation is [ r2 r;r]. Similarly, r;r]. • ¡ ¡ Semi-Circles: The equation for a circle x2 + y2 = r2 can also be written as y = §p r2 x2 ¡ (6.3) Now let us we consider the positive and negative square roots separately. These describe semi-circles on either side of the x-axis. Thus the equations for two types of semi-circles are as follows: y = r2 x2 and y = r2 x2 (6.4) p The domain of each of these semi-circles is [ the ﬂrst semi-circle) and [ r;0] (for the second semi-circle). ¡ ¡p r;r] and the range is [0;r] (for ¡ ¡ ¡ 87 Note: These semi-circles are functions, since there is only one y value corresponding to each x value. 6.9 Locus The syllabus requires: † (grade 12) derive the equation of the locus of all points; equidistant from a given point, equidistant from 2 given points, equidistant from a given point and a line parallel to the x or y axis 6.10 Other Geometries The syllabus requires: basic knowledge of spherical geometry, taxicab geometry and fractals † 6.11 Unsorted 6.11.1 Fundamental vocabulary terms Measuring angles The magnitude of an angle does not depend on the length of its sides; it only depends on the relative direction of the two sides. E.g. the adjacent edges of a postcard are at an angle of 90– with respect to each other. But so is the Empire State Building in New York City with respect to Fifth Ave. and 34th Street (NOTE: its SA... maybe this should be the baobab tree with respect to the Limpopo river (i kid... i kid)). So we can’t use notions of length to measure angles. How do we measure angles then? We begin by ﬂnding a way to enumerate angles. What is the smallest angle you can draw? Two lines subtending almost no angle. Two coincident line segments pointing in the same direction subtend an angle of 0–, e.g. , lines AB and AC in the ﬂgure below. Now if we keep one line ﬂxed and move the other while still pivoted at the common vertex, we can obtain any other angle. Two line segments with a common vertex and facing in opposite direction are said to form an angle of 180–, e.g. , XY and XZ in the ﬂgure below. ZXY is a straight line. 0– A B Z C 180– X Y The choice of a measure of 180– for the angle subtended by the segments of a straight line is a matter of historical convention. Once we have decided what the measure of an angle formed by a straight line is, we have also ﬂxed the measure of all the other angles. This is because we would like the angles to obey some desirable properties. E.g. if we have an angle a– between two lines AB and AC, and another angle b– between AB and AD, we would like 88 the angle between AC and AD to be (a + b)–. This makes the measurement of angles intuitive and conforms to our notion of measurement of length, weight, etc. Similarly, a line segment that deﬂnes a direction exactly half way between AB and AC should create an angle of (a=2)–. D B B D b– a– A C A C a– (a=2)– An angle of 90– is termed a right angle. A right angle is half the measure of the angle subtended by a straight line (180–). An angle twice the measure of a straight line is 360–. An angle measuring 360– looks identical to an angle of 0–, except for the labelling. All angles after 360– also look like we have seen them before. Angles that measure more than 360– are largely for mathematical convenience to maintain continuity in our enumeration of angles. C Right Angle 90– A B 360– A B C We deﬂne some other terms at this point. These are simply labels for angles in particular ranges. Acute angle: An angle 0– and < 90–. ‚ Obtuse angle: An angle > 90– and < 180–. Straight angle: An angle measuring 180–. Reex angle: An angle > 180– and < 360–. † † † † C acute C A B A obtuse C B reex A B Once we can number or measure angles, we can al
so start comparing them. E.g. all right angles are 90–, hence equal. An obtuse angle is larger than an acute angle, etc. An alternative measure of angles is used on a compass. E.g. if North(N) is 0–, North-East(NE) is 45–, NNE is 22:5–, etc. 89 Special Angle Pairs In the previous section, we classiﬂed angles based on measurement. In this section we’ll examine some interesting properties of angles formed by a pair of intersecting lines. First we consider a single straight line, AB. There’s a point on the line AB called X. The measure of angle AXB is 180– as deﬂned in the previous section. Now let us draw another straight line intersecting the ﬂrst. Without loss of generality, let the point of intersection be X. We call the four angles formed with X as the vertex a, b, c and d. At this point, we introduce some deﬂnitions for convenience Vertical angles Deﬂnition: The angles formed by two intersecting straight lines that share a vertex but do not share any sides are called vertical angles. E.g. , a and c in the ﬂgure above are vertical angles. b and d are also vertical angles. Adjacent Angles Deﬂnition: Two angles that share a common vertex and a common side are called adjacent angles. E.g. , (a;b) and (c;d) are adjacent angles. Linear pairs Deﬂnition: The adjacent angles formed by two intersecting straight lines are said to form a linear pair. E.g. (a;b), (b;c), (c;d) and (d;a) all form linear pairs. Since the non common sides of a linear pair are part of the same straight line, the total angle formed by the linear pair is 180– by deﬂnition. E.g. a + b = 180–, etc. What can we say about the vertical angles? Looking at ﬂgure above, it seems like the vertical angles are equal to one another. We can prove the following result. Theorem: The vertical angles formed by intersection of two straight lines are equal. Proof: Since a and b form a linear pair, a + b = 180–. Similiarly, b and c form a linear pair, so, b + c = 180–. Thus, a + b = b + c. Since the angle b contributes equally to both sides of the equation, it can be cancelled out leaving, a = c. The proof for the pair (b;d) is identical. 90 This proof conﬂrms our intuition that vertical angles are equal in magnitude. This result will later be used when proving properties for parallel lines. We end this section with some more deﬂnitions. Supplementary and Complementary pairs Deﬂnition: Two angles are called supplementary if their sum equals 180–. E.g. angles that constitute a linear pair are supplementary. Deﬂnition: Two angles are called complementary if their sum equals 90–. Note that in order to be labelled supplementary or complementary, the two angles being considered need not be adjacent. E.g. x and y in the ﬂgure below are supplementary, but they are not adjacent and thus do not form a linear pair. C D x y x + y = 180– A B E F 6.11.2 Parallel lines intersected by transversal lines Two lines are said to intersect if there is a point that lies on both lines. Informally, two lines intersect if they meet at some point when extended indeﬂnitely in either direction. E.g. at a tra–c intersection, two or more streets intersect; the middle of the intersection is the common point between the streets. It is possible that two lines that lie on the same plane never intersect even when extended to inﬂnity in either direction. Such lines are termed parallel lines. E.g. the tracks of a straight railway line are parallel lines. We wouldn’t want the tracks to intersect as that would be catastrophic for the train! A section of the Australian National Railways Trans-Australian line is perhaps one of the longest pairs of man-made parallel lines. Longest Railroad Straight (Source: www.guinnessworldrecords.com) The Australian National Railways Trans-Australian line over the Nullarbor Plain, is 478 km. (297 miles) dead straight, from Mile 496, between Nurina and Loongana, Western Australia, to Mile 793, between Ooldea and Watson, South Australia. A transversal of two or more lines is a line that intersects these lines. E.g. in the ﬂgure below, AB and CD are two lines and EF is a transversal. We are interested in the properties of the angles formed by these intersecting lines, so we’ll introduce some deﬂnitions for various angle pairs. 91 Deﬂnitions: † † † † † Interior angles: When two lines are intersected by a transversal, the angles that lie between the two lines are called interior angles. E.g. in the ﬂgure above, 1, 2, 3 and 4 are interior angles. Exterior angles: When two lines are intersected by a transversal, the angles formed that lie outside the two lines are called exterior angles. E.g. 5, 6, 7 and 8 are exterior angles. Alternate interior angles: When two lines are intersected by a transversal, the interior angles that lie on opposite sides of the transversal are termed alternate interior angles. E.g. in the above example, 1 and 3 are are a pair of alternate interior angles. 2 and 4 are also alternate interior angles. Interior angles on the same side: As the name suggests, these are interior angles that lie on the same side of the transversal. E.g. (1,4) and (2,3). Corresponding angles: The angles on the same side of the transversal and the same side of the two lines are called corresponding angles. E.g. (1;5), (4;8) and (3;7), etc. , are pairs of corresponding angles. In order to prove relationships between the angles deﬂned above, we will assume the following postulate regarding parallel lines. Euclid’s Parallel Line Postulate: Postulate: If a straight line falling on two straight lines makes the two interior angles on the same side less than two right angles (180–), the two straight lines, if produced indeﬂnitely, meet on that side on which the angles are less than two right angles. The above is one of the fundamental postulates of Euclidean geometry and has no proof based on the other postulates. Now we’ll use the above postulate to prove some other properties. Theorem 1: If two parallel lines are intersected by a transversal, the sum of interior angles on the same side of the transversal is two right angles (180–). Proof: Consider parallel lines AB and CD intersected by the transversal EF in the ﬂgure above. Suppose that the sum of the interior angles is less than 180– on one side of the transveral, e.g. 1 + 4 < 180–. Then Euclid’s 92 ¡ ¡ ¡ (1) and (3) = 180– (4). So (2+3) = 360– Parallel Line Postulate implies that the AB and CD meet on that side of the transversal and are not parallel. This contradicts the assumption that the lines are parallel. Now suppose that the sum of the interior angles 1 and 4 is greater than (1+4). 180–. Now, (2) = 180– Since (1 + 4) > 180–, (2 + 3) < 180–. Thus the parallel line postulate implies that the lines will meet on that side of the transveral and are not parallel. Thus both pairs of interior angles on the same side need to sum up to 180– for the lines to be parallel. Theorem 2: If two parallel lines are intersected by a transversal, the alternate interior angles are equal. Proof: In the ﬂgure above, using Theorem 1, (1 + 4) = 180– Also, since AB is a straight line, 1 and 4 are supplementary. (4 + 3) = 180– Thus, 1 = 3. Similarly, 2 = 4. Theorem 3: If two parallel lines are intersected by a transversal, the corresponding angles are equal. Proof: Again using Theorem 1, in the ﬂgure above, (1 + 4) = 180– Also, since EF is a straight line, (4 + 5) = 180– So 1 = 5, etc. Theorem 4: The sum of the three angles in a triangle is 180. Proof: Consider triangle ABC shown in the ﬂgure below. The three angles are denoted 1, 2 and 3. We have to show that 1 + 2 + 3 = 180–. Consider a straight line DE through point A that is parallel to BC. We denote the angles between DE and the sides of the triangle as 4 and 5 Since DE is a straight line, 1 + 4 + 5 = 180–. Now DE is parallel to BC, and 2 and 4 are alternate interior angles (AB is the transversal). So 2 = 4. Similarly, 3 = 5. So substituting these in the ﬂrst equation, 1 + 2 + 3 = 180–. The following theorems help in determining when two lines are parallel to each other. Theorem 5: If two lines are intersected by a transversal such that any pair of interior angles on the same side is supplementary, then the two 93 lines are parallel. Proof: We’ll prove that if two lines are not parallel, the interior angles on the same side are not supplementary. We’ll prove this by contradiction. Assume that two non-parallel distinct lines are intersected by a transversal such that interior angles 1 and 4 are supplementary. 1 + 4 = 180– (Eq. (i)) Since the lines are not parallel, they have to intersect at some point Z. Since the two lines are distinct, they have to form a non-zero angle at their point of intersection. Z 9 X 1 4 Y XY Z is a triangle. So 1 + 4 + 9 = 180–, using Theorem 4. But using Eq. (i), 1 + 4 = 180–, so 9 = 0–. This contradicts the fact that distinct intersecting lines create a non-zero angle at their point of intersection. So our original assumption is not supportable and the interior angles 1 and 4 cannot be supplementary. Theorem 6: If two lines are intersected by a transversal such that a pair of alternate interior angles are equal, the lines are parallel. Proof: Left as an exercise. Theorem 7: If two lines are intersected by a transversal such that a pair of alternate corresponding angles are equal, the lines are parallel. Proof: Left as an exercise. Theorem 8: Prove that if a line AB is parallel to CD, and AB is parallel to EF, then CD is parallel to EF. Proof: Left as an exercise. (Hint: We can prove this in two steps: 1. Prove that if two lines are parallel, then a line that intersects one also intersects the other. 2. Use the equivalence of corresponding angles to get the result.) 94 Sides Name 5 6 7 8 10 15 pentagon hexagon heptagon octagon decagon pentadecagon Table 6.1: Table of some polygons and their number of sides. 95 Chapter 7 Trigonometry 7.1 Syllabus 7.1.1 Triangles The syllabus requires: { similarity of triangles is the basis of trig functions (NOTE: perhaps this is bes
t left in geometry) { solve problems in 2D by constructing and interpreting geometric and trig models including scale drawings, maps and building plans 7.1.2 Trigonometric Formul‰ The syllabus requires: { some history from various cultures { derive reduction formul‰ for trig ratios { recognise equivalence of trig expressions by reduction { solve 2D problems by establishing sin/cos/area rules (NOTE: perhaps just do the rules here, and do the problems in section 8.5) { use trig for height and distances (NOTE: SH isn’t this already done in 7.1.1?) 7.2 Radian and Degree Measure You should be familiar with the idea of measuring angles from geometry but have you ever stopped to think why there are 360 degrees in a circle? 96 The reason is purely historical 1. There are, in fact, many diﬁerent ways of measuring angles. The two most commonly used are degrees (the one you have been using up to now) and radians. arc le The radian measure of an angle is deﬂned as the ratio of the arc length subtending the angle to the radius of the circle. A = arclength radius (7.1) We know from geometry that the circumference of a circle is found using the equation c = 2…r. If we divide through by the radius of the circle, r, we ﬂnd that the radian angle subtended by the complete circumference, (or in other words the number of radians in a full circle) is 2…r r = 2…. This means that 2… radians is the same as 360–. With this in mind we can easily work out how to convert between degrees and radians. Deﬂnition: (rad) = (–) 2… 360 £ or (–) = (rad) 360 2… £ Using these formulae we can express common angles in radians. It is worth learning these as questions may be asked using either degrees, radians or a mixture of both. Degrees Radians 30– … 6 45– … 4 60– … 3 90– … 2 180– … 270– 3… 2 360– 2… 7.2.1 The unit of radians You may be wondering what the unit of radians is. The answer is that it doesn’t have one. This is because a radian is the ratio of two lengths: the arc length divided by the radius. Now, both of these will have the 1There are 360 degrees in a circle because the ancient Babylonians had a number system with base 60. A base is the number you count up to before you get an extra digit. The number system that we use everyday is called the decimal system (the base is 10), but computers use the binary system (the base is 2). 360 = 6 £ 60 so for them it make sense to have 360 degrees in a circle. 97 same unit, so when you divide them, the units simply disappear! This is what is known as a dimensionless quantity. Sometimes we write radians (or simply rad ) after the number to emphasise that we are using radians, but this is not necessary. In general, if an angle is expressed in terms of … it is meant to be in radians. Be careful though. If the question does not explicitly say whether the angle is measured in degrees or radians you need to use common sense to decide which to use. 7.3 Deﬂnition of the Trigonometric Functions 7.3.1 Trigonometry of a Right Angled Triangle Consider a right-angled triangle. We deﬂne b a c sin = cos = tan = a b c b a c (7.2) (7.3) (7.4) These are abbreviations for sine, cosine and tangent. These functions, known as trigonometric functions, relate the lengths of the sides of a triangle to its interior angles. How to remember the deﬂnitions Diﬁerent people have diﬁerent ways of remembering these ratios. One way involves deﬂning opposite to be side of the triangle opposite to the angle, hypotenuse to be the side opposite to the right-angle (just like we use the term in geometry) and adjacent to be the side next to the angle, which is not the hypotenuse. This is illustrated in the following picture, where we 98 show the adjacent, opposite and hypotenuse for the angle . e t i s o p p o hypotenuse adjacent So, using these deﬂnitions we have: sin = cos = tan = opposite hypotenuse adjacent hypotenuse opposite adjacent (7.5) (7.6) (7.7) There is a mnemonic to remember these: Sine S O Opposite H Hypotenuse C Cos A Adjacent H Hypotenuse T Tan O Opposite A Adjacent Another mnemonic that is perhaps easier to remember goes as follows: Silly Old Hens Sin = Cackle And Howl Cos = Till Old Age Tan = Opposite Hypotenuse Adjacent Hypotenuse Opposite Adjacent CAUTION! The deﬂnitions of opposite, adjacent and hypotenuse only make sense when you are working with right-angled triangles! Always check to make sure your triangle has a right angle before you use them, otherwise you will get the wrong answer. We will ﬂnd ways of working with the trigonometry of non right-angled triangles later in the chapter. By using the appropriate triangles it is possible to work out the following values of the sine, cosine and tangent functions for a number of common angles. 99 cos sin tan 0– 1 0 0 30– p3 2 1 2 1 p3 45– 1 p2 1 p2 1 60– 1 2 p3 2 p3 90– 0 1 ¡ 180– 1 ¡ 0 0 These values are useful to remember as they often occur in questions. They are also a good way of helping us to visualise the graphs of the sine, cosine and tangent functions. 7.3.2 Trigonometric Graphs Sine and Cosine Graphs Let us look back at our values for sin { 0– 0 30– 1 2 45– 1 p2 60– p3 2 90– 1 180– 0 sin As you can see, the function sin has a value of 0 at = 0–. Its value then smoothly increases until = 90– when its value is 1. We then know that it later decreases to 0 when = 180–. Putting all this together we can start to picture the full extent of the sine graph. The sine graph is shown in ﬂgure 7.1. 360 ¡ 180 ¡ 1 1 ¡ 180 360 Figure 7.1: The sine graph. Let us now look back at the values of cosine{ cos 0– 1 30– p3 2 45– 1 p2 60– 1 2 90– 0 180– 1 ¡ If you look carefully you will notice that the cosine of an angle is the same as the sine of the angle 90– . Take for example, ¡ cos 60– = 1 2 = sin 30– = sin (90– 60–) ¡ 100 This tells us that in order to create the cosine graph all we need to do is to shift the sine graph 90– to the left2. The cosine graph is shown in ﬂgure 7.2. 360 ¡ 180 ¡ 90– shift 180 360 1 1 ¡ Figure 7.2: The cosine graph (in black) with the sine graph (in gray). Tangent graph Now that we have the sine and cosine graphs there is an easy way to visualise the tangent graph. Let us look back at our deﬂnitions of sin and cos in a right angled triangle. sin cos = opposite hypotenuse adjacent hypotenuse = opposite adjacent = tan This is the ﬂrst of an important set of equations called trigonometric identities. An identity is an equation which holds true for any value which is put into it. In this case we have shown that tan = sin cos for any value of . So we know that for values of for which sin = 0, we must also have tan = 0. Also, if cos = 0 our value of tan = 0 is undeﬂned as we cannot divide by 0. The complete graph3 is shown in ﬂgure 7.3. 2You may have noticed that the transformation we are using is in fact a translation of 90– followed by a reection in the y axis due to a negative sign in front of the . However, because cosine is an even function (i.e. symmetric about the y axis) this reection doesn’t really matter! 3The dotted lines in the tangent graph are known as asymptotes and the graph is said to display asymptotic behaviour. This means that as approaches 90–, tan approaches inﬂnity. In other words, there is no deﬂned value of the function at the asymptote values. Another 1 graph which displays asymptotic behaviour is y = x whose asymptotes are the x and y axes themselves. 101 360 ¡ 180 ¡ 4 2 2 ¡ 4 ¡ 180 360 Figure 7.3: The tangent graph. 7.3.3 Secant, Cosecant, Cotangent and their graphs In the sections that follow it will often be useful to deﬂne the reciprocal functions of sine, cosine and tangent. We shall deﬂne them as follows{ csc = sec = cot = 1 sin 1 cos 1 tan = = = hypotenuse opposite hypotenuse adjacent adjacent opposite (7.8) (7.9) (7.10) j j j cos j and The graphs of these functions are shown in ﬂgures 7.4{7.6. There are a number of points worth noting about these graphs. Firstly, since sin j and are always less than or equal to 1 their reciprocal functions j must always be greater than or equal to 1. Secondly csc sec j notice that the sectant graph can be obtained from the cosecant graph by performing a 90– shift, just like we did with sine and cosine. Notice also that these graphs have asymptotes whenever their reciprocal function is 0. One important feature of all these trigonometric functions is that they are periodic with a period of 360–. This is most easily understood by looking back at a circle. j 102 360 ¡ 180 ¡ ¡ 180 360 Figure 7.4: The cosecant graph. 360 ¡ 180 ¡ ¡ 180 360 Figure 7.5: The sectant graph. 103 360 ¡ 180 ¡ 4 2 2 ¡ 4 ¡ 180 360 Figure 7.6: The cotangent graph. Imagine that we are measuring the angle on this circle. Now let us add 360– to our angle so that our line sweeps all the way around the circle and ends up back where it started as indicated in the diagram. There is no way of knowing whether we have swept around the circle in this way as everything ends up exactly where it started. In other words, if we add 360– to an angle we eﬁectively have the same angle we started with. Since our diagram is the same after the rotation the values of our trigonometric functions also remain unchanged. This is the reason that all trigonometric functions have a period of 360– { adding 360– degrees to an angle does nothing more than sweep it all the way round a circle back to where it 360–. began, so all of our functions must have the same value for and § 7.3.4 Inverse trigonometric functions Like all functions the trigonometric functions have inverses. These funcopposite hypotenuse in the case of inverse sine) and give tions take in ratios (such as out the angle it corresponds to. However, due to the periodicity of the 104 trigonometric functions there are many possible angles for one ratio. For example, both sin 30– and sin 150– give values of 1 2 can be either 30–, 150– or any one of an inﬂnite amount of other possibilities. 1 and Two notations are commonly used for the inverse functions { sin¡ 1 and arc versions can be used for a
ny of the six arcsin . Both the ¡ trigonometric functions. However, you must be careful not to confuse the reciprocal functions (csc, sec and cot) with the inverse functions (arcsin, arccos and arctan). They are diﬁerent functions with diﬁerent meanings and will give diﬁerent answers4. 2 so sin¡ 1 1 7.4 Trigonometric Rules and Identities 7.4.1 Translation and Reection We found earlier that all trigonometric functions are periodic, with a period of 360–. We can express this more formally by writing{ sin ( § 360–) = sin This identity states that the sine of an angle is unchanged if we add or subtract 360–. Another way to think of this is as a translation of the sine graph by 360– to the right or left. 360– shift 1 360 ¡ 180 ¡ 1 ¡ 180 360 As you can see, if we shift the whole graph by 360– left or right it will end up back on top of itself. The sine of an angle is therefore completely unchanged. Identities of this form can be very useful. We shall consider a few such identities here using the ideas of chapter (NOTE: Add in correct ref for transformations in geometry chapter). n First let us consider reecting the sin graph in the x and y axes. We know that if we reect in the x axis we will get the graph of sin . Figure 7.7 shows the original sine graph (gray) and its reection in the x axis. Just by looking at the graph we can see that reecting sine in the y axis would give the same result as reecting in the x axis did5. Mathematically, a ¡ 4Remember { you can have inverse reciprocal functions such as arccsc and sec¡1. 5If you have a mirror you can check this by putting it along the y axis. 105 360 ¡ 180 ¡ 1 1 ¡ 180 360 Figure 7.7: The sine graph reected in the x axis. reection in the y axis gives us the function sin ( second identity{ ¡ ). This gives us a sin ( ) = ¡ sin ¡ We could also obtain the black function in ﬂgure 7.7 by translating the gray sine graph 180– to the right or left. This tells us that{ sin ( ) = ¡ sin = sin( 180–) § ¡ Let us now look at the cosine function. Again we can use the fact that cosine is periodic with period 360– to give{ cos ( § 360–) = cos This time our reections are a little more complicated. Firstly let’s reect the cosine function in the y axis to generate cos ( ). Since the cosine graph is symmetric about the y axis the reection will not change our graph. This tells us that { ¡ cos ( ) = cos ¡ Reecting in the x axis we obtain ﬂgure 7.8. The ﬂrst point of interest 360 ¡ 180 ¡ 1 1 ¡ 180 360 Figure 7.8: The cosine graph reected in the x axis. is that unlike with sine we get diﬁerent graphs by reecting in the x and 106 y axes. We can however still consider the black graph as a translation of 180– to the left or right. As before this gives us an identity{ cos = cos( 180–) § ¡ One ﬂnal and very important class of translation identity are those that convert sine into cosine and vice versa. We have already seen one of these when we looked at the graphs of sine and cosine{ cos = sin (90– ) ¡ This can be written in many ways by using the identities we have already proven. One more sensible version is{ cos ( ¡ 90–) = sin It is now easier to see that the sine graph is just the cosine graph moved 90– to the right. We shall prove that these two identities are the same in the worked example at the end of this section. There are many more identities such as these for sine and cosine as well as tangent and the reciprocal functions. One way to ﬂnd such identities is to uses the addition and subtraction formulae which we will derive in section 7.4.6. Always remember to check if your expressions can be simpliﬂed using these identities. 7.4.2 Pythagorean Identities Consider a right angled triangle{ r x (0;0) y Let us use our deﬂnitions of sine and cosine in a right{angled triangle on the angle . sin = opposite hypotenuse = y r and cos = adjacent hypotenuse = x r y = r sin and x = r cos ) Using Pythagorases theorem we can say that{ (7.11) (7.12) (7.13) x2 + y2 = r2 107 Substituting in our expressions for x and y{ (r cos )2 + (r sin )2 = r2 r2 cos2 + r2 sin2 = r2 ) If we divide both sides through by r2 we arrive at{ cos2 + sin2 = 1 This is known as a Pythagorean identity. We can express the identity in terms of the other trigonometric functions by dividing through by either sin2 or cos2 on both sides{ sin2 sin2 sin2 cos2 + + cos2 sin2 cos2 cos2 = = 1 sin2 ) 1 cos2 ) 1 + cot2 = csc2 tan2 + 1 = sec2 These 3 versions of the Pythagorean Identity can be used in any question. They do not need to be used in a triangle they will work for any angle in any situation. Deﬂnition: Pythagorean Identities{ cos2 + sin2 = 1 1 + cot2 = csc2 tan2 + 1 = sec2 7.4.3 Sine Rule So far we have only dealt with the trigonometry of right angled triangles where we are able to use our deﬂnitions of sin , cos and tan . There are some rules which we can derive that hold true for any triangle. One of these is the sine rule. Consider a scalene triangle (i.e. one with all sides diﬁerent lengths and all angles diﬁerent){ a B b A p c 108 Note that the order in which the angles of the triangle are labelled is important. The three sides are labelled a, b, and c in any order. Then we deﬂne angle A as the angle opposite side a and so on. We can split it into two right-angled triangles by choosing a perpendicular, p, to one side (in this case side c) which passes through the opposite vertex, as shown. You may want to draw a few triangles and convince yourself that we can always do this regardless of the shape of the triangle. Now, since we have right angled triangles we can use our old deﬂnition of sin = opposite hypotenuse to ﬂnd{ sin A = p b and sin B = p a or, rearranging slightly{ p = b sin A and p = a sin B If we now set these two equations for p equal to each other (to eliminate the p) and rearrange again (by dividing through both sides by ab) we get the following{ p = b sin A = a sin B sin A a = sin B b ) (7.14) We could have chosen our perpendicular to go through any side of our triangle so let us repeat the proof with the perpendicular through a. b C p0 c a B Now we have{ p0 = b sin C = c sin B sin B b = sin C c ) (7.15) Check this result for yourself to make sure you understand where it came from! Putting equations 7.14 and 7.15 together we obtain the sine rule{ Deﬂnition: Sine Rule { sin A a = sin B b = sin C c 109 Many books now tell us that we have not proved the sine rule fully. Consider the following triangle{ b p a A B c This type of triangle which has all of its angles smaller than 90– is called an acute triangle. We have already proved the sine rule for acute triangles. If we ‘fold’ this acute triangle along the perpendicular we obtain the following{ p b a A A A0 B c This kind of triangle, which contains an angle greater than 90–, is called an obtuse triangle. Notice that we now have no way of ﬂnding a perpendicular through side c which goes through a vertex of the new triangle. Instead, the perpendicular must be drawn outside of the triangle. This is the reason that some books say that our proof is incomplete { the perpendicular which we used for acute triangles does not always exist in obtuse ones! We can either repeat our proof from the start for obtuse triangles(which is messy!) or we can use our knowledge of trigonometry to prove these other books wrong. Notice that the lengths of a, b and p have not changed, neither has the value of angle B. The only change is that we have replaced the old angle, A, with a new angle, A0. If we look at A and A0 we can see that they lie on a straight line so from our knowledge of geometry we can say that A + A0 = 180– or alternatively that A0 = 180– A. We have already shown that{ ¡ sin (180 ¡ ) = sin 110 so we can say that{ sin A0 = sin (180 A) = sin A ¡ We now know that although angle A has changed, its sine has stayed the same. Since we are only interested in the sine of the angle our equation must still be valid! Any obtuse triangle can be formed by ‘folding’ an acute triangle in this way so the sine rule must be true for any triangle we choose. 7.4.4 Cosine Rule Let us return to our scalene triangle We have deﬂned a perpendicular, p, just as we did before. This perpendicular divides side c into two parts. One part lies between vertex A and the perpendicular. We shall call its length x. The other part between the perpendicular and vertex B must, therefore, have a length of c x. Using the usual deﬂnitions of sine and cosine on the left had section of the triangle we ﬂnd{ ¡ sin A = p b p = b sin A and and cos A = x b x = b cos A ) Now we will use Pythagorases theorem on the right hand side of the triangle. a2 = p2 + (c = p2 + c2 x)2 2cx + x2 ¡ ¡ Substituting in p = b sin A and x = b cos A a2 = b2 sin2 A + c2 = b2(sin2 A + cos2 A) + c2 ¡ 2cb cos A + b2 cos2 A 2bc cos A ¡ (7.16) (7.17) (7.18) (7.19) Using the Pythagorean identity sin2 A + cos2 A = 1 we get the cosine rule. 111 Deﬂnition: Cosine rule { a2 = b2 + c2 2bc cos A ¡ 7.4.5 Area Rule One last triangle rule is an extension of the area formula for a triangle. height base From geometry{ area = 1 2 £ base £ height where height is measured perpendicular to the base. From trigonometry{ b height c sin = opposite adjacent = height b height = b sin ) Calling the base c we can write{ area = 1 2 £ base £ height = 1 2 bc sin where must be the angle between sides b and c. 112 Deﬂnition: Area rule { area = 1 2 bc sin 7.4.6 Addition and Subtraction Formulae Let us return to our scalene triangle complete with a perpendicular as we had before. ` p 1 x h ` ¡ 90– y The perpendicular, p, divides the top angle into two pieces, and `. We can use the fact that the internal angles of a triangle sum to 180– to ﬂnd that the bottom right angle must be 90– `. If we deﬂne one side of the triangle to be of legnth 1 for simplicity we can easily see from the left hand triangle that{ ¡ sin = cos = = sin p = cos From the right hand triangle we can see that{ cos ` = sin = p h ) y h ) h = p cos ` y = h sin ` = p sin ` cos ` Substituting in our expression for p we ge
t{ y = cos sin ` cos ` (7.20) (7.21) (7.22) (7.23) Putting these expressions togeher we ﬂnd that the base of the triangle, x + y, has a length{ x + y = sin + cos sin ` cos ` 113 + ` 1 90– sin + cos sin ` cos ` h ` ¡ We can now use the sine rule on the two angles that we know{ sin( + `) sin + cos sin ` cos ` = sin(90– 1 `) ¡ As we learnt in section 7.4.1, sin (90– cross multiplying we get{ ¡ ) = cos . Using this identity and sin( + `) sin + cos sin ` cos ` = cos ` 1 sin ( + `) = sin cos ` + cos sin ` ) (7.24) (7.25) This is the sine addition formula. If we take this formula and replace ) (remember, we can only do this with identities as they must with (90– be true for any value of their variables) we get{ ¡ sin (90– + `) = sin(90– ¡ As before we can use the identities ) cos ` + cos(90– ) sin ` ¡ ¡ ¡ ¡ Using these identities we can obtain the cosine subtraction formula{ sin (90– cos (90– `) = cos ` `) = sin ` (7.26) (7.27) sin (90– ( ¡ ¡ `)) = cos( ¡ `) = sin(90– ) cos ` + cos(90– ¡ = cos cos ` + sin sin ` ¡ ) sin ` (7.28) (7.29) We are missing two identities, the sine subtraction and cosine addition ` in the sine formulae. To obtain the sine subtraction we replace ` with addition formula. Since we know that sin( `) = cos ` from section 7.4.1 we ﬂnd that ¡ sin ` and cos( `) = ¡ ¡ ¡ sin ( ¡ `) = sin cos( ¡ = sin cos ` `) + cos sin( cos sin ` ¡ `) ¡ (7.30) (7.31) Proof of the cosine addition formula can be done in the same way, starting with the subtraction formula. cos ( + `) = cos cos( `) + sin sin( `) ¡ = cos cos ` ¡ sin sin ` ¡ (7.32) (7.33) 114 Similar equations can be found for the tangemt function. Proof of these is left to the reader6. Deﬂnition: Addition and Subtraction Formulae{ ¡ ¡ sin ( + `) = sin cos ` + cos sin ` cos sin ` sin ( `) = sin cos ` sin sin ` cos ( + `) = cos cos ` ¡ cos ( ¡ tan ( + `) = `) = cos cos ` + sin sin ` tan ` + tan tan tan ` 1 ¡ tan ` tan ¡ 1 + tan tan ` `) = tan ( ¡ (7.34) (7.35) (7.36) (7.37) (7.38) (7.39) 7.4.7 Double and Triple Angle Formulae Let us remind ourselves of the addition formulae for sine and cosine{ cos ( + `) = cos cos ` sin sin ` sin ( + `) = sin cos ` + cos sin ` ¡ (7.40) (7.41) If we set ` to be equal to we get the following equations{ cos ( + ) = cos (2) = cos cos sin2 (7.42) sin ( + ) = sin (2) = sin cos + cos sin = 2 cos sin (7.43) sin sin = cos2 ¡ ¡ These are known as the double angle formulae. By using the phythagorean identity cos2 + sin2 = 1 we can substitute into the cosine double angle formula for cos2 or sin2 to get diﬁerent forms. We can do the same for the tangent function. Deﬂnition: Double angle formulae{ cos (2) = cos2 sin2 ¡ = 2 cos2 1 ¡ 2 sin2 = 1 sin (2) = 2 cos sin ¡ tan (2) = 2 tan tan2 1 ¡ (7.44) (7.45) (7.46) (7.47) (7.48) 6HINT{Remember, tan = sin cos so you can use the sine and cosine addition formulae to ﬂnd the tangent one. To get it into the same form as in the deﬂnition you need to divide everything through by a factor. The fact there is a 1 in the denominator should give you a clue as to what the factor is! Once you have the addition forumula you need to remember that tan(¡) = ¡ tan to ﬂnd the subtraction formula. 115 Now that we have the double angle formulae it is easy to ﬂnd higher order multiple angle formulae. We shall derive the cosine triple angle formulae here. We start by taking the cosine addition faormula and setting ` = 2{ cos ( + 2) = cos (3) = cos cos (2) sin sin (2) ¡ We now substitute in the double angle formulae for cos (2) and sin (2). We have a choice of forms for the cos (2) formula. We shall choose cos (2) = cos2 sin2 . ¡ cos (3) = cos cos (2) sin sin (2) ¡ sin2 ) 3 sin2 cos ¡ ¡ = cos (cos2 = cos3 ¡ sin (2 cos sin ) (7.49) (7.50) (7.51) The corresponding sine triple angle formula is{ sin (3) = 3 cos2 sin sin3 ¡ 7.4.8 Half Angle Formulae We can rearrange the double angle formulae to ﬂnd the half angle formulae. We shall start by rearranging the cosine double angle formula of the form cos (2) = 2 cos2 1. ¡ 2 cos2 ) ¡ 1 = cos (2) 2 cos2 = 1 + cos (2) 1 + cos (2) 2 1 + cos (2) 2 cos2 = cos = §r ) ) (7.52) (7.53) (7.54) (7.55) Another way to write this is to halve both of the angles (we can do this because it is an identity, so must be valid for any angle) { cos 2 = §r 1 + cos 2 Using the same method to rearrange the identity cos (2) = 1 obtain{ ¡ 2 sin2 we §r We can now use the ratio identity to ﬂnd the tangent half angle formula{ sin = 2 1 ¡ cos 2 tan 2 = sin 2 cos 2 = § q q 1 cos 2 ¡ 1+cos 2 = §r 1 cos 1 + cos ¡ As with all identities the half angle formulae can be expressed in a number of ways. Some of these will be proven in the worked example and more 116 given in the summary of identities at the end of the chapter. Deﬂnition: Half Angle Formulae cos sin tan 2 2 2 = = = §r §r §r 1 1 + cos 2 cos 2 1 cos 1 + cos ¡ ¡ (7.56) (7.57) (7.58) ‘Product to Sum’ and ‘Sum to Product’ Identi- 7.4.9 ties For completeness we include a brief comment on the ‘product to sum’ and ‘sum to product’ identities. They can be derived from the addition and subtraction formulae. We shall only derive one of each type here as their derivations are broadly similar. We start by proving a product to sum identity. This identity us an experssion linking the product of two cosine functions (cos cos `) to a sum of cosine functions. The derivation is as follows{ cos ( + `) + cos ( ¡ `) = (cos cos ` = 2 cos cos ` ¡ (7.59) sin sin `) + (cos cos ` + sin sin `) (7.60) cos cos ` = ) 1 2 [cos ( + `) + cos ( `)] ¡ Sum to product identities are messier to prove. Here we prove the identity linking the sum of two cosines by exchange of variables. We substitute `0 into the product to sum identity above (the = 0 + `0 and ` = 0 primes just prevent us getting confused, we shall drop them later). ¡ (7.61) (7.62) 2 cos(0 + `0) cos(0 ¡ `0) = cos ((0 + `0) + (0 = cos(20) + cos(2`0) ¡ `0)) + cos ((0 + `0) (0 ¡ (7.63) `0)) ¡ (7.64) As always with identities we can divide all our variables by 2 for convenience (since it must be true for any angle) and drop out the primes to give{ cos + cos ` = 2 cos 0 + `0 2 ¶ cos 0 `0 ¡ 2 ¶ 7.4.10 Solving Trigonometric Identities A standard type of question in an exam is of the form \show that sin(2) tan = 2 cos2 ". As well as being important in examinations being able to prove 117 38:7– 100m Figure 7.9: Determining the height of a building using trigonometry. identities is a key mathematical skill. Most identities can be proven by using the standard identities we have already learnt earlier in this section. There are three ways that the two sides of an identity can diﬁer{ 1. The functions are diﬁerent e.g. sin cot = cos 2. The operations are diﬁerent e.g. sin cos3 = sin cos (1 3. The angles are diﬁerent e.g. sin = sin(2) 2 cos sin2 ) ¡ Of course, real identities (and even the examples above) contain a mixture of these three diﬁerences, but they can be solved by dealing with each of the diﬁerences, one at a time. 7.5 Application of Trigonometry Trigonometry is very important in many areas of every day life. In this section we shall learn to use trigonometry to solve problems which would otherwise require very complicated solutions. 7.5.1 Height and Depth One simple task is to ﬂnd the height of a building using trigonometry. We could just use tape measure lowered from the roof but this is impractical (and dangerous) for tall buildings. It is much more sensible to measure a distance along the ground and use trigonometry to ﬂnd the height of the building. Figure 7.9 shows a building whose height we do not know. We have walked 100m away from the building and measured the angle up to top. This angle is found to be 38:7–. We call this angle the angle of elevation. As you can see from ﬂgure 7.9 we now have a right angled triangle, one side of which 118 A 127– B 255– C Figure 7.10: Two lighthouses, A and B, and a boat, C. is the height of the building, which also includes our 100m distance and the angle of elevation. Using the standard deﬂnition of tangent{ tan 38:7– = = opposite adjacent height 100 ) height = 100 = 80m £ tan 38:7– (7.65) (7.66) (7.67) (7.68) 7.5.2 Maps and Plans Maps and plans are usually scale drawings. This means that they are an enlagement (usually with a negative scale factor so that they are smaller than the original) so all angles are unchanged. We can use this to make use of maps and plans by adding information from the real world. Let us imagine that there is a coastline with two lighthouses, one either side of a beach. This is shown in ﬂgure 7.10. The two lighthouses are 0:67km apart and one is exactly due east of the other. Let us suppose that no boat may get closer that 200m from the lighthouses in case it runs aground. How can the lighthouses tell how close the boat is? Both lighhouses take bearings to the boat (remember { a bearing is an angle measured clockwise from north). These bearings are shown on the map in ﬂgure 7.10. We can see that the two lighthouses and the boat form a triangle. Since we know the distance between the lighthouses and we have two angles we can use trigonometry to ﬂnd the remaining two sides of the triangle, the distance of the boat from the two lighthouses. Figure 7.11 shows this triangle more clearly. We need to know the legnths of the two sides AC and BC. We can choose to use either sine or cosine rule to ﬂnd our missing legnths. We shall use both here. Using the sine rule{ sin 119 (7.69) A 0:67km B 37– 15– 128– C Figure 7.11: Two lighthouses, A and B, and a boat, C. 7.6 Trigonometric Equations Trigonomeric equations often look very simple. Consider solving the equation sin = 0:7. We can take the inverse sine of both sides to ﬂnd that 1(0:7) = 1(0:7). If we put this into a calculator we ﬂnd that sin¡ = sin¡ 44:42–. This is true, however, it does not tell the whole story. As you 360 ¡ 180 ¡ 1 1 ¡ 180 360 Figure 7.12: The sine graph. The dotted line represents sin = 0:7. ¡ can see from ﬂgure 7.12, there are four possible angles with a sine of 0:7 between 360– and 360–. If we were to
extend the range of the sine graph to inﬂnity we would in fact see that there are an inﬂnite number of solutions to this equation! This di–culty (which is caused by the periodicity of the sine function) makes solving trigonometric equations much harder than they may seem to be. Any problem on trigonometric equations will require two pieces of information to solve. The ﬂrst is the equation itself and the second is the range in which your answers must lie. The hard part is making sure you ﬂnd all of the possible answers within the range. Your calculator will always give 90– and 90– for you the smallest answer (i.e. the one that lies between tangent and sine and one between 0– and 180– for cosine). Bearing this in mind we can already solve trigonometric equations within these ranges. ¡ 120 Worked Example 7 : Question: Find the values of x for which sin 3x = 0:5 if it is given that 0 < x < 90–. Answer: Because we are told that x is an acute angle, we can simply apply an inverse trigonometric function to both sides. sin x = 0:5 x = arcsin 0:5 x = 30– ) ) (7.70) (7.71) (7.72) (7.73) We can, of course, solve trigonometric equations in any range by drawing the graph. Worked Example 8 : Question: For what values of x does sin x = 0:5, when x < 360–? Answer: 360– < ¡ Step 1 : Draw the graph We take look at the graph of sin x = 0:5 on the interval [-360, 360]. We want to know when the y value of the graph is 0.5, so we draw in a line at y=0.5. 360 ¡ 180 ¡ 1 1 ¡ 180 360 Step 2 : Notice that this line touches the graph four times. This means that there are four solutions to the equation. Step 3 : Read oﬁ those x values from the graph as x = and 150–. 210–,30– 330–, ¡ ¡ 121 1 0 1 ¡ 1st 2nd 3rd 4th 90 180 270 360 – 180 +VE +VE -VE -VE – 90 2nd +VE 1st +VE 3rd -VE 4th -VE – 270 – 0 /360 – Figure 7.13: The graph and unit circle showing the sign of the sine function. 1 90 180 270 360 360 ¡ 270 ¡ 180 ¡ 90 ¡ 1 ¡ This method can be time consuming and inexact. We shall now look at how to solve these problems algebraically. 7.6.1 Solution using CAST diagrams The Sign of the Trigonometric Function The ﬂrst step to ﬂnding the trigonometry of any angle is to determine the sign of the ratio for a given angle. We shall do this for the sine function ﬂrst. In ﬂgure 7.13 we have split the sine graph unto four quadrants, each 90– wide. We call them quadrants because they correspond to the four quadrants of the unit circle. We notice from ﬂgure 7.13 that the sine graph is positive in the 1st and 2nd quadrants and negative in the 3rd and 4th. Figure 7.14 shows similar graphs for cosine and tangent. All of this can be summed up in two ways. Table 7.1 shows which trrigonometric fuctions are positive and which are negative in each quadrant. A more convenient way of writing this is to note that all fuctions are positive in the 1st quadrant, only sine is positive in the 2nd, only tangent in the 3rd and only 122 1 0 1 ¡ 1st 2nd 3rd 4th 90 180 270 360 +VE -VE -VE +VE ¡ 1st 2nd 3rd 4th 90 180 270 360 +VE -VE +VE -VE Figure 7.14: Graphs showing the sign of the cosine and tangent functions. 2nd 1st sin +VE +VE -VE cos +VE -VE +VE tan +VE 3rd 4th -VE -VE -VE +VE -VE Table 7.1: The signs of the three basic trigonometric functions in each quadrant. 123 cosine in the 4th. We express this using the CAST digram (ﬂgure 7.15). This diagram is known as a CAST diagram as the letter, taken anticlock- 180– 90– S T A C 270– 0–/360– S T A C Figure 7.15: The two forms of the CAST diagram. wise from the bottom right, read C-A-S-T. The letter in each quadrant tells us which trigonometric functions are positive in that quadrant. The ‘A’ in the 1st quadrant stands for all (meaning sine, cosine and tangent are all positive in this quadrant). ‘S’, ‘C’ and ‘T’ ,of course, stand for sine, cosine and tangent. The diagram is shown in two forms. The version on the left shows the CAST diagram including the unit circle. This version is useful for equations which lie in large or negative ranges. The simpler version on the right is useful for ranges between 0– and 360–. Magnitude of the trigonometric functions Now that we know which quadrants our solutions lie in we need to know which angles in these quadrants satisfy our equation. Calculators give us the smallest possible answer (sometimes negative) which satisﬂes the equation. For example, if we wish to solve sin = 0:3 we can apply the inverse sine function to both sides of the equation to ﬂnd{ = arcsin 0:3 = 17:46– However, we know that this is just one of inﬂnitely many possible answers. We get the rest of the answers by ﬂnding relationships between this small angle, , and answers in other quadrants. To do this we need to condider the modulus7 of the sine graph. = As you can see in ﬂgure 7.16 there is a solution to the equation 0:3 in every quadrant. The 1st quadrant solution is of course 17:46– as our calculator told us. The 2nd quadrant solution can be seen to be 180– . Another way to see this is to look at the identity sin ¡ j j 7This means we plot only the magnitude of the function. This is the same as reecting sin = sin(180– ) ¡ negative sections in the x axis. 124 1 0.3 0 0 1st 2nd 3rd 4th 90 180 270 360 Figure 7.16: The modulus of the sine graph. proved in section 7.4. Using the same logic the 3rd quadrant solution can be seen to be (180– + ) and the 4th quadrant solution (360– ). It is now left to the reader to show, using similar graphs for cosine and tangent, that these relationships are true for all three of the trigonometric functions. These rules can be expressed in a simpler way. If we deﬂne the solution lying between 0– and 90– as `{ ¡ Deﬂnition: { If we are in the 1st or 3rd quadrants our solution is the lower boundary of the quadrant plus `. { If we are in the 2nd or 4th quadrants our solution is the upper boundary of the quadrant minus `. 7.6.2 Solution Using Periodicity ¡ Up until now we have only solved trigonometric equations where the argument (the bit after the function, e.g. the in cos or the (2x 7) in tan(2x 7)) has been . If there is anything more complicated than this we need to be a little more careful. 360–. We want soLet us try to solve tan(2x) = 2:5 in the range 0– lutions for positive tangent so using our CAST diagram we know to look in the 1st and 3rd quadrants. Our calulator tells us that arctan(2:5) = 68:2–. This is our ﬂrst quadrant solution for 2x. Our 3rd quadrant lies between 180– and 270– so our solution is 180– + 68:2–. Putting this together{ ¡ • • x 2x = 68:2– x = 34:1– ) or or 248:2– 124:1– 125 x • • Notice that we did not divide by the 2 until we had found our answers. Now try to put x = 214:1 into the equation. This gives 2x = 428:2 and we ﬂnd that tan(428:2) = 2:5! This solution, x = 214:1, lies within the range 0– 360– so we should have included it in our answer. Why did we not ﬂnd this solution before? The answer is that when we halved our solutions for 2x to ﬂnd x we also 360– so, after halved our range. We looked for solutions for 0– • halving, our ﬂnal answer gave us solutions in the range 0– 180–. There are two ways of dealing with this. We could redo the problem looking the the range 0– 720–. This will work but there is a simpler method. We know that all the trigonometric functions are periodic with a period of 360–. This means we can add (or subtract) a factor of 360n– (where n is an integer) our solution to ﬂnd another equally valid solution. Let us try this with tan(2x) = 2:5. If n = 0 we regain our original answers{ 2x 2x • • • • • x 2x = 68:2– or 248:2– Adding 360– (n = 1) to our solutions for 2x we ﬂnd the next two solutions{ 2x = 68:2– + 360– or = 428:2– ) or x = 214:1– ) 248:2– + 360– or 608:2– 304:1– 7.6.3 Linear Triginometric Equations Just like with regular equation solving without trigonometric functions the equations can become a lot more complicated. You should solve these just like normal equations and once you have a signal trigonometric ratio isolated, then you follow the strategy outlined in the previous section.(ADD AN EXAMPLE HERE) 7.6.4 Quadratic and Higher Order Trigonometric Equations The simplest quadratic trigonometric equation is of the form{ sin2 x 2 = 1:5 ¡ ¡ This type of equation can be easily solved by rearranging to get a more familiar linear equation{ sin2 x = 0:5 sin x = ) p0:5 § (7.74) (7.75) This gives two linear trigonometric equations. The solutions to either of these equations will satisfy the original quadratic. (ADD AN EXAMPLE HERE) The next level of complexity comes when you need to solve a trinomial 126 which contains trig functions. Here you can make you life a lot easier if you use temporary variables. Consider solving{ tan2 (2x + 1) + 3 tan (2x + 1) + 2 = 0 Here you should notice that tan(2x+1) occurs twice in the equation, hence we let y = tan(2x + 1) and rewrite: y2 + 3y + 2 = 0 That should look rather more familiar so that you can immediately write down the factorised form and the solutions: (y + 1)(y + 2) = 0 ¡ Next one just substitutes back for the temporary variable: ) y = 1 OR y = 2 ¡ tan (2x + 1) = 1 ¡ or tan (2x + 1) = 2 ¡ And then we are left with two linear trigonometric equations. Be careful: sometimes one of the two solutions will be outside the range of the In that case you need to discard that solution. trigonometric function. For example sonsicer the same equation with cosines instead of tangents{ cos2 (2x + 1) + 3 cos (2x + 1) + 2 = 0 Using the same method we ﬂnd that{ cos (2x + 1) = 1 ¡ or cos (2x + 1) = 2 ¡ 1 and The second solution cannot be valid as cosine must lie between 1. We must, therefore, reject the second equation. Only solutions to the ﬂrst equation will be valid. ¡ 7.6.5 More Complex Trigonometric Equations Here are two examples on the level of the hardest trig equations you are likely to encounter. They require using everything that you have learnt in this chapter. If you can solve these, you should be able to solve anything! (ADD AN EXAMPLE HERE) 127 7.7 Summary of the Trigonomertic Rules and Ident
ities Pythagorean Identities Reciprocal Identities Ratio Identities cos2 + sin2 = 1 1 + cot2 = csc2 tan2 + 1 = sec2 csc = 1 sin sec = 1 cos tan = 1 cot tan = sin cos cot = cos sin Odd/Even Identities Periodicity Identities Cofunction Identities ) = sin sin( ¡ ¡ cos( ) = cos ¡ tan ) = tan( ¡ ¡ cot ) = cot( ¡ ¡ ) = csc( csc ¡ ¡ ) = sec sec( ¡ sin( cos( tan( cot( csc( sec( § § § § § § 360–) = sin 360–) = cos 180–) = tan 180–) = cot 360–) = csc 360–) = sec sin(90– cos(90– tan(90– cot(90– csc(90– sec(90– ¡ ¡ ¡ ¡ ¡ ¡ ) = cos ) = sin ) = cot ) = tan ) = sec ) = csc Double Angle Identities Addition/Subtraction Identities Half Angle Identities sin(2) = 2 sin cos sin ( + `) = sin cos ` + cos sin ` sin ( ¡ `) = sin cos ` cos sin ` ¡ cos (2) = cos2 ¡ cos (2) = 2 cos2 cos (2) = 1 sin2 1 ¡ 2 sin2 ¡ tan (2) = 2 tan tan2 1 ¡ cos ( + `) = cos cos ` cos ( sin sin ` `) = cos cos ` + sin sin ` ¡ ¡ tan ( + `) = tan `+tan tan tan ` tan tan ( ¡ 1+tan tan ` 1 `) = tan ` ¡ ¡ sin 2 = cos 2 = 1 cos 2 ¡ 1+cos 2 §q §q tan 2 = 1 cos 1+cos ¡ §q Sine Rule Area Rule Cosine Rule sin A a = sin B b = sin C c Area = 1 Area = 1 Area = 1 2 bc cos A 2 ac cos B 2 ab cos C a2 = b2 + c2 b2 = a2 + c2 c2 = a2 + b2 2bc cos A 2ac cos B 2ab cos C ¡ ¡ ¡ 128 cos cos ` = 1 Product to Sum Identities 2 [cos( + `) + cos( `) 2 [cos( 2 [sin( + `) + sin( ¡ ¡ sin sin ` = 1 sin cos ` = 1 cos( + `)] `)] ¡ `)] sin + sin ` = 2 sin ¡ sin ¡ sin ` = 2 cos cos + cos ` = 2 cos Sum to Product Identities ‡ ‡ · +` 2 +` 2 +` 2 · +` 2 · cos sin ` ¡ 2 · ` · ` ¡ 2 ‡ ‡ cos ‡ sin ¡ sin ¡ cos ¡ cos ` = 129 Chapter 8 Solving Equations 8.1 Linear Equations The syllabus requires: { solve linear equations 8.1.1 Introduction Let’s imagine you have a friend called Joseph. He picked up your test results from the Biology class and now he refuses to tell what you scored, or what he scored! Obviously you are trying everything to get him to tell you, and he decides to tease you and makes you work it out for yourself. He says the following: \I have 2 marks more than you and the sum of both our marks is equal to 14. How much did we get?" Now if the numbers are simple like in the example, you might be able to work it out in your head. Can you? But to make it easier, you can use a linear equation! This is how it works: We use a placeholder for your amount and that placeholder is x. So: Y ou = x Then we need a placeholder for Joseph Joseph = y BUT the trick is that we have some information about Joseph’s mark, which is that Joseph has 2 more than you. We need to use that, so how 130 about: Joseph = you + 2 Or Joseph = (x + 2) Or y = (x + 2) Now we need to use the last bit of information we have and that is: Y ou + Joseph = 14 Or using placeholders x + y = 14 Or substituting y x + (x + 2) = 14 What we have here is the actual linear equation. You already know what an equation is but what does linear mean? Linear means the highest power of the unknown variable, usually called x, is one. 8.1.2 Solving Linear equations - the basics To ﬂnd out what your test result is we need to now simplify this equation until we only have the x on the one side of the equal sign and a value on the other side. There are a few rules on how to simplify these equations to get a value for x. They can be organized into 3 groups. Once we have worked through them and we are sure about them, then we can attempt to ﬂnd out what the answer to our problem is. So here they are: Rule one - Addition or subtraction You are allowed to subtract or add any amount as long as you do it on both sides of the equal sign: Example 18.1) (8.2) (8.3) ) ) x + 5 x = ¡ 11 ¡ 131 Example 2 15 Rule two - Multiply or divide The same principle applies for multiplication and division: Example 1: Example 2: 2x = 9 2x = 20 4 = 5 4 £ ) ) Rule three - Fractions (8.4) (8.5) (8.6) (8.7) (8.8) (8.9) (8.10) (8.11) (8.12) If x is multiplied by a fraction we need to divide both sides of the equal sign with that fraction to get x alone. We do that by ipping the fraction around and then multiplying both sides with it Example 1: )x( 2 3 14 3 x = ) = 7( 2 3 ) ) ) (8.13) (8.14) (8.15) These are the basic rules to apply when simplifying a linear equation. But most linear equations will require a few combinations of these before x is 132 sitting alone on the one side of the equal sign. That means we might have to make use of the rules above a number of times one after the other. Let’s do a few examples where we will use multiple steps to solve the equation: 8.1.3 Solving linear equations - Combining the basics in a few steps TIP: Start with eliminating the terms without x, that way you avoid having to calculate too many fractions Example 1: Example 2 = 62 7 + 5x = 62 7 + 5x ¡ 5x = 55 5x 55 5 5 x = 11 = 55 = 5x + 3 4 3 4 = 5x + 3 4 ¡ 3 4 ) = 5x = 5x 55 54( ¡ 1 4 217 4 217 4 £ 1 5 = 5x 1 5 £ Doing that is the the same as dividing by 5 217 20 ) = x (8.16) (8.17) (8.18) (8.19) (8.20) (8.21) (8.22) (8.23) (8.24) (8.25) (8.26) TIP: Start by moving all the terms with x to the one side and all the terms without x to the opposite side of the equal sign. Remember we can do that by changing the sign of the term 133 Example 3: 5x = 3x + 45 ) ) ) ) ) ) ) ) Example 4: Example 5: 3x = 45 5x ¡ 2x = 45 2x 45 2 2 1 2 x = 22 = 23x 12 = 6 + 2x 2x 12 = 6 2x = 6 + 12 ¡ ¡ 23x 23x ¡ ¡ 21x = 18 x = 21 18 12 6x + 34x = 2x ¡ 6x + 34x = 2x 6x + 34x ¡ 2x = 24 64 ¡ 24 ¡ 64 ¡ 24 ¡ 64 ¡ ¡ ¡ 12 12 ) ¡ ) ¡ ¡ 100 ) ) ) 26x = x = x = ¡ ¡ ¡ 100 26 50 13 (8.27) (8.28) (8.29) (8.30) (8.31) (8.32) (8.33) (8.34) (8.35) (8.36) (8.37) (8.38) (8.39) (8.40) (8.41) (8.42) We simpliﬂed the answer - but this is not necessarily a required step TIP: If there are parentheses (brackets) in the equation, start by removing them - multiply with the coe–cient Example 6(3x 4) = 8 ¡ 9x + 12 = 8 9x = 8 12 9x = 4 9 ¡ 4 ¡ (8.43) (8.44) (8.45) (8.46) (8.47) see the term is now positive - do you remember why? 134 Example 7: lets start with the parentheses - don’t forget the minus! 6x + 3x = 4 5(2x 3) ¡ ¡ next we move the like terms to their own sides 6x + 3x = 4 10x + 15 ¡ ) 6x + 3x + 10x = 4 + 15 19x = 19 x = 1 ) ) ) Example 8: (8.48) (8.49) (8.50) (8.51) (8.52) 8(3x 14) 34 = 2(4x 22) 5(3 + 2x) (8.53) ¡ Looks like a big one? Lets take it step by step ¡ ¡ ¡ 24x 112 ¡ ¡ ) 34 = 8x 44 ¡ ¡ 15 ¡ 10x (8.54) that’s all the brackets gone ) ) 24x 24x ¡ ¡ 34 112 ¡ ¡ 8x + 10x = and now solve! 8x + 10x = 15 44 15 + 112 + 34 ¡ ¡ 44 ¡ ¡ ) ) 26x = 87 x = 87 26 (8.55) (8.56) (8.57) (8.58) And that is it for our examples. This covers all the types of linear equations you can be expected to solve. It’s the best to always keep your priority of steps in mind and then just simply do them one by one. If you are unsure about your answer you can just substitute it into your original equation and see if you get the same value for both sides: Example : 5(x 3) = 5 ¡ 15 = 5 5x 5x = 5 + 15 ¡ ) ) ) ) 5x = 20 x = 4 135 (8.59) (8.60) (8.61) (8.62) (8.63) Test: 3) = 5 5(4 ¡ 5(1) = 5 5 = 5 ) ) (8.64) (8.65) (8.66) and there we see it works! Now lets get back to our original problem of your test results! The linear equation was: x + (x + 2) = 14 x + x + 2 = 14 2 x + x = 14 ¡ 2x = 12 x = 6 ) ) ) ) (8.67) (8.68) (8.69) (8.70) (8.71) You scored 6 for your Biology test and Joseph scored 6 +2 = 8! 8.2 Quadratic Equations The syllabus requires: { solve quadratic equations by factorisation, completing the square and quadratic formula { identify ‘‘not real’’ numbers and how they occur. (see 2.1) 8.2.1 The Quadratic Function (NOTE: these notes have just been copied and pasted from the older structure notes and have not been written to the syllabus. it needs a serious edit and the worked example methodology needs redone (we now do inline examples and analogies with worked examples and exercises at the end of the chapter.)) A quadratic or parabolic function is a function of the form f (x) = ax2 + bx + c. (NOTE: very quick notes by sam for simple quadratics... this would be best as a decision tree. make the student appreciate that its basically trial and error to get the answer, but you can do some detective work ﬂrst to eliminate most possibilities { write the problem in the form ax2 + bx + c = 0 (with a positive) { write down two brackets, with an x in each, leaving room for a num- ber on each side ( x )( x ) (8.72) 136 { write out your options (in a table at the side) for multiplying two numbers together to give a. these numbers should go in front of the xs in your brackets. { if c is positive, then the other two numbers you need are either both positive or both negative. they are both negative if b is negative, and both positive id b is positive. if c is negative, it means only one of your numbers is negative, the other one beng positive. { your two numbers should multiply to give c, so write out your options if each number is multiplied by the (in a table oﬁ to the side). number in front of the x in the other bracket, then added together it gives you b. so try diﬁerent combinations of the numbers you have written). if it doesn’t work, go back to the 3rd step and try a diﬁerent combination of numbers to give you a. { once you get an answer, multiply out your brackets again just to make sure it works (sanity check). damn thats long winded!) Worked Example: Q: Draw a graph of the quadratic function y = x2 A: First let us set up a table of x and y values: 6. x ¡ ¡ x : y : -5 24 -4 14 -3 6 -2 0 -1 -4 0 -6 1 -6 2 -4 3 0 4 6 5 14 The graph of this function is shown in ﬂgure 8.1. Notice that the function 3). This shows that the x-intercepts can also be written as y = (x + 2)(x (where y = 0) are x = 2 and x = 3, which agrees with the graph. The y-intercept (where x = 0) is at y = ¡ ¡ 6. ¡ 8.2.2 Writing a quadratic function in the form f (x) = a(x p)2 + q. ¡ Consider the general form of the quadratic function y = ax2 + bx + c. Now adding and subtracting the same factor b2 4a from this expression does not change anything. Therefore y = ax2 + bx + b2 4a ¡ b2 4a + c Taking out a factor of a
then gives y = a(x2 + b a + ( b 2a )2) + c b2 4a ¡ (8.73) (8.74) 137 25 20 15 10 Figure 8.1: Graph of y = x2 x 6 ¡ ¡ The expression in brackets is then a perfect square so that y = a(x + ( b 2a ))2 + c ¡ = a(x ( ¡ ¡ b 2a ))2 + (c b2 4a ¡ b2 4a ) which can be written in the form y = a(x p)2 + q ¡ (8.75) (8.76) (8.77) where p = b 2a and q = c b2 4a . ¡ ¡ ¡ p)2 is a perfect square and therefore always positive, (x p)2 is Since (x at a minimum of 0 when x = p. This means that y is minimum (if a0) or maximum (if a < 0) when x = p and y = q. This point (p;q) is therefore called the turning point. ¡ In Now notice that the quadratic function is symmetric about x = p. v) = av2 + q for any real number v. This other words f (p + v) = f (p means that the part of the quadratic to the right of the vertical line x = p looks like the part to the left of x = p ipped about this line. Therefore we call the line x = p the axis of symmetry of the parabola. ¡ Worked Example: Q: Consider the quadratic function f (x) = into the form f (x) = a(x axis of symmetry. Plot a graph of f (x) showing all the intercepts. 5. Put this function p)2 + q and thus ﬂnd the turning point and ¡ ¡ ¡ x2 +6x 138 f (p v) ¡ v v f (p + v) x = p Figure 8.2: Graph of TODO (NOTE: original BMP (qf-gen1.bmp) missing, i made this up on the y) A: First we shall write the quadratic in the form f (x) = a(x f (x x2 + 6x (x2 (x2 (x2 (x2 ¡ ¡ ¡ ¡ 6x + 5) 6x + 9 6x + 9) + 4 3)2 + 4 ¡ 9 + 5) p)2 + q. ¡ (8.78) (8.79) (8.80) (8.81) (8.82) ¡ Therefore the turning point is (3,4) and the axis of symmetry is x = 3 (in other words p = 3 and q = 4). ¡ Now to plot the graph we need to know the intercepts. The y-intercept 5. The x-intercepts can be found by solving the equation is y = f (0) = f (x) = 0 as follows: ¡ x2 + 6x 5 = 0 ¡ ¡ 6x + 5 = 0 ¡ x2 (x 1)(x x = 1 or x = 5 ¡ ¡ 5) = 0 ) ) ) (8.83) (8.84) (8.85) (8.86) Note that the technique of writing x2 5) is called factorisation. We shall learn more about this in the following chapter. For now just check that this is true by multiplying out the brackets. 6x + 5 = (x 1)(x ¡ ¡ ¡ Thus the graph of the quadratic function f (x) = x2 + 6x ¡ 5 is ¡ 8.2.3 What is a Quadratic Equation? An equation of the form ax2 + bx + c = 0 is called a quadratic equation. Solving this equation for x is the same as ﬂnding the roots (x-intercepts) of the quadratic function f (x) = ax2 + bx + c. 139 4 3 2 1 x1 1 2 3 4 x2 5 Figure 8.3: Graph of TODO (NOTE: original BMP (qf-eg2.bmp) missing, i made this up on the y) 8.2.4 Factorisation We have already seen examples of solving for the roots of a quadratic function by writing this function as the multiple of two brackets. For example, x2 2 or x = 3. This is called factorising the quadratic function. 3) = 0 means that either x = 6 = (x + 2)(x ¡ ¡ ¡ ¡ x Knowing how to factorise a quadratic takes some practice, but here some general ideas which are useful. { First divide the entire equation by any common factor of the coe–cients, so as to obtain an equation of the form ax2 + bx + c = 0 where a, b and c have no common factors. For example, 2x2 + 4x + 2 = 0 can be written as x2 + 2x + 1 = 0 by dividing by 2. { Now, if ax2 + bx + c = (rx + s)(ux + v), then sv = c and ru = a. Therefore, by ﬂnding all the factors of a and c, one can try all the combinations and see if there is one which gives the correct result for b = su + rv. { Once writing the equation in the form (rx + s)(ux + v) = 0, it then follows that the two solutions are x = s r and x = u v . ¡ ¡ Worked Examples: Example 1: Q: Solve the equation x2 + 3x 4. ¡ A: Since a = 1, if this equation can be factorised, it must have the form x2 + 3x 4 = (x + s)(x + v) = x2 + (s + v)x + sv (8.87) ¡ 4, we know that s = 1, v = 4 or s = 1, 4 (excluding the options which just involve interchanging s and v, Now, as sv = v = which makes no diﬁerence to the ﬂnal answer). 2, v = 2 or s = ¡ ¡ ¡ ¡ 140 Also s + v = 3, so s = quadratic equation can be written as ¡ 1 and v = 4 is the correct combination. Thus the x2 + 3x 4 = (x ¡ ¡ 1)(x + 4) = 0 (8.88) Therefore the solutions are x = 1 and x = the roots of the quadratic function f (x) = 4. Example 2: Q: Find ¡ 2x2 + 4x ¡ ¡ 2. 2x2 + 4x 2 = 0. ¡ 2. This gives A: We must ﬂnd the solutions to the equation f (x) = ¡ First we divide both sides of the equation be a factor of the equation ¡ Now, let us assume that x2 ¡ 2x + 1 = 0 (8.89) x2 ¡ 2x + 1 = (x + s)(x + v) = x2 + (s + v)x + sv (8.90) Then sv = 1 and therefore either s = v = 1 or s = v = s + v = 2, it follows that s = v = 1 and thus 1. Since ¡ ¡ ¡ ¡ x2 ¡ 2x + 1 = (x ¡ 1) = (x ¡ ¡ 1)2 = 0 (8.91) The only solution is therefore x = 1. Example 3: Q: Solve the equation 2x2 12 = 0. 5x A: This equation has no common factors, but still has a = 2. Therefore, we must look for a factorisation in the form ¡ 1)(x 2x2 5x 12 = (2x + s)(x + v) = 2x2 + (s + 2v)x + sv (8.92) ¡ ¡ We see that sv = considered below. Note: ¡ 12 and s + 2v = 5. All the options for s and v are Since we now have the factor of 2x in the ﬂrst ¡ s 2 -2 3 -3 4 -4 6 -6 v -6 6 -4 4 -3 3 -2 2 s + 2v -10 10 -5 5 -2 2 2 -2 bracket, it does make a diﬁerence, in this case, whether we interchange 6, v = 2 give the s and v values. For example, s = 2, v = ¡ diﬁerent solutions. We must therefore consider both options. 6 and s = ¡ We can see that the combination s = 3 and v = Therefore one can check that 4 gives s + 2v = ¡ 5. ¡ 2x2 ¡ 5x + 12 = (2x + 3)(x 4) = 0 (8.93) Therefore the solutions are x = ¡ ¡ 3 2 and x = 4. 141 8.2.5 Completing the Square It is not always possible to factorize a quadratic function. We shall now derive a general formula, which gives the solutions to any quadratic equation. Consider a general quadratic equation ax2 + bx + c = 0. Adding and subtracting b2 equation. Thus 4a from the left-hand side does not change the ax2 + bx + b2 4a ¡ b2 4a + c = 0 (8.94) Taking out a factor of a from the 1st 3 terms gives a(x2 + b a a(x2 + ) b2 4a2 ) ¡ b2 4a + c = 0 + ( b 2a )2) = b2 4a ¡ c + b a (8.95) (8.96) We can now see that the term in brackets is the perfect square (x + b and therefore 2a )2 a(x + )2 = b 2a b2 4a ¡ c (8.97) Now dividing by a and taking the square root of either side gives the expression x + b 2a = §r b2 4a2 ¡ c a Finally, solving for x implies that x = b 2a § r b2 4a2 ¡ c a ¡ = b 2a § r ¡ b2 4ac ¡ 4a2 and taking the square root of 4a2 to obtain 2a gives b x = ¡ § 4ac pb2 2a ¡ (8.98) (8.99) (8.100) These are the solutions to the quadratic equation. Notice that there are two solutions in general, but these may not always exists (depending on the sign of the expression b2 4ac under the square root). ¡ Worked Examples: Example 1: Q: Solve for the roots of the function f (x) = 2x2 + 3x 7. ¡ A: One should ﬂrst try to factorise this expression, but in this case it turns out that this is not possible. Therefore we must make use of the general formula as follows: 142 4(2)( 7) ¡ b x = ¡ § (3) = ¡ 4ac pb2 2a ¡ § p (3)2 2(2) ¡ 3 = ¡ 3 = ¡ p56 § 4 2p14 § 4 (8.101) (8.102) (8.103) (8.104) Therefore the two roots of the quadratic function are x = ¡ 3+2p14 4 and 3 2p14 ¡ 4 ¡ . Example 2: Q: Solve for the solutions to the quadratic equation x2 5x + 8. ¡ A: Again it is not possible to factorise this equation. The general formula shows that b x = ¡ § 4ac pb2 2a ¡ 5)2 ( ¡ 2(1) p ( = ¡ ¡ 5(1)(8) ¡ (8.105) (8.106) (8.107) (8.108) Since the expression under the square root is negative these are not real solutions (p 7 is not a real number). Therefore there are no real solutions to the quadratic equation x2 5x + 8. This means that the quadratic function f (x) = x2 5x + 8 has no x-intercepts, but the entire function ¡ lies above the x-axis. ¡ ¡ Note to self: maybe add quadratic example about distance, velocity and acceleration ... object falling under action of gravity (giving formula for distance as a function of time)?. 8.2.6 Theory of Quadratic Equations What is the Discriminant of a Quadratic Equation? Consider a general quadratic function of the form f (x) = ax2 + bx + c. The discriminant is deﬂned as ¢ = b2 4ac. This is the expression under the square root in the formula for the roots of this function. We have already seen that whether the roots exist or not depends on whether this factor ¢ is negative or positive. ¡ 143 The Nature of the Roots Real Roots: Consider ¢ case there are solutions to the equation f (x) = 0 given by the formula 0 for some quadratic function f (x) = ax2 + bx + c. In this ‚ b x = ¡ § pb2 2a 4ac b = ¡ ¡ p¢ § 2a (8.109) since the square roots exists (the expression under the square root is nonnegative.) These are the roots of the function f (x). There are various possibilities: Equal Roots: If ¢ = 0, then the roots are equal and, from the formula, these are given by x = b 2a ¡ (8.110) Unequal Roots: There will be 2 unequal roots if ¢ > 0. The roots of f (x) are rational if ¢ is a perfect square (a number which is the square of a rational number), since, in this case, p¢ is rational. Otherwise, if ¢ is not a perfect square, then the roots are irrational. Imaginary Roots: If ¢ < 0, then the solution to f (x) = ax2 + bx + c = 0 contains the square root of a negative number and therefore there are no real solutions. We therefore say that the roots of f (x) are imaginary (the function f (x) does not intersect the x-axis). Summary of Cases: { Real Roots (¢ 0) ‚ Equal Roots (¢ = 0) Unequal Roots (¢ > 0) ⁄ ⁄ Rational Roots (¢ a perfect square) Irrational Roots (¢ not a perfect square) ¢ ¢ { Imaginary Roots (¢ < 0) Note to self: maybe add pictures showing these cases graphically? 144 Worked Examples: Example 1: Q: Consider the function f (x) = 2x2 + 5x equation f (x) = 0, discuss the nature of the roots of f (x). A: We need to calculate and classify ¢ = b2 for the roots. ¡ ¡ 11. Without solving the 4ac according to the cases ¢ = (5)2 ¡ 4(2)( 11) = 25 + 88 = 113 ¡ (8.111) Now ¢ is positive, so the roots are real and unequal. Also, since 113 is not a perfect square, the roots are irrational. Example 2: Q: Consider
the quadratic function f (x) = x2 + bx + (2b some constant. Classify the roots of this function as far as possible. ¡ 5), where b is A: Let us calculate the discriminant ¢ = b2 4(1)(2b 5) = b2 8b + 20 (8.112) ¡ We shall now use a useful trick, which is to write the above expression as a perfect square plus a number. ¡ ¡ ¢ = b2 8b + 20 = (b2 8b + 16) + 4 = (b 4)2 + 4 (8.113) ¡ 0 because this is a perfect square. Therefore we know that ¡ ¡ Now (b ¢ ¡ 4 > 0. 4)2 ‚ ‚ We can thus say that f (x) has real unequal roots. We do not know whether ¢ is a perfect square, since we do not know that value of the constant b, and therefore we cannot say whether the roots are rational or irrational. 8.3 Cubic Equations The syllabus requires: { (grade 12) solve cubic equations using factor theorem ‘‘and other techniques’’ (NOTE: SH: i want to hit whoever wrote this syllabus) 8.4 Exponential Equations The syllabus requires: { solve exponential equations { (grade 12) switch between log and exp form of an equation 145 8.5 Trigonometric Equations The syllabus requires: { solve trig equations (NOTE: this implies that this section comes after trig, which is also probably after geometry) 8.6 Simultaneous Equations The syllabus requires: { solve simultaneous equations algebraically and graphically 8.7 Inequalities The syllabus requires: { solve linear inequalities in 1 and 2 variables and illustrate graphically { solve quadratic inequalities in 1 variable and illustrate graphically 8.7.1 Linear Inequalities Let us say that we are given a general inequality as follows: 0; ‚ ax + by + c ax + by + c > 0; ax + by + c < 0 (8.114) Now there are many possible values of x and y, for which this may be true (these will depend on the values of the constants a, b and c). The set of all the (x;y) values which satisfy this inequality is called the solution set. ax + by + c or • 0 We shall now see how to draw the solution set on a graph. Let’s consider the following example. Worked Example 1: Q: Find the solution set of the inequality 2x + y ¡ A: First we solve for y by writing the inequality as 3 0. ‚ y 2x + 3 ‚ ¡ (8.115) 2x+3 is a straight line and the points (x;y), which Now the function y = satisfy the inequality are therefore all the points above the line. These can be drawn as ¡ The shaded section on the graph, which shows the solution set, is called the feasible region. Now sometimes x and y must satisfy more than one inequality. In this case, we consider each inequality separately and then the feasible region is where the feasible regions of each inequality overlap. 146 3 2 1 3 2 1 2 3 4 Figure 8.4: Graph of y = 2x + 3. Points satisfying y ¡ 2x + 3 are shaded ‚ ¡ Worked Example 2: Q: Graphically represent the solution set for the following inequalities: ¡ 2x + 10 • ¡ ¡ A: Solving for y gives 1 y y < 2x + 5 ‚ y x + 10 • ¡ (8.116) (8.117) (8.118) (8.119) (8.120) (8.121) Now we draw the solution set to each of the inequalities separately and ﬂnd the region where these overlap as shown below. We draw the line y = 2x + 5 as a dashed line because the inequality is < and not (the line is not included in the feasible region). • 8.7.2 What is a Quadratic Inequality A quadratic inequality is an inequality of the form ax2 + bx + c > 0, ax2 + bx + c 0, ax2 + bx + c < 0 or ax2 + bx + c 0. ‚ • 8.7.3 Solving Quadratic Inequalities Solving a quadratic inequality corresponds to working out in what region a quadratic function lies above or below the x-axis. Here are some examples showing how this is done. 147 10 10 Figure 8.5: Graph of TODO Worked Examples: Example 1: Q: Find all the solutions to the inequality x2 A: Consider the function f (x) = x2 f (x) ‚ 5x + 6. We need to ﬂnd out where 0; in other words, where the function f (x) lies above/on the x-axis. 5x + 6 ¡ ¡ 0. We shall ﬂrst work out where f (x) intersects the x-axis by solving the equation which can be factorised to give x2 ¡ 5x + 6 = 0 (x ¡ 3)(x ¡ 2) = 0 (8.122) (8.123) The x-intercepts are therefore x1 = 2 and x2 = 3. We can see from ﬂgure 8.6 that f (x) is above/on the x-axis when x or x 2. 3 ‚ Therefore the solution to the quadratic inequality is or in interval notation ( [3; ;2] ). x : x f ‚ 3 or x 2 g • ¡1 [ 1 Note: The x-intercepts are included in this solution, since the f (x) inequality includes the solution f (x) = 0. 0 • ‚ • Example 2: Q: Solve the quadratic inequality Let f (x) = A: quadratic equation ¡ x2 ¡ x2 ¡ ¡ 3x + 5 > 0. 3x + 5. The x-intercepts are solutions to the 148 8 6 4 2 x1 2 x2 3 1 Figure 8.6: Graph of f (x) = x2 5x + 6 ¡ x2 ¡ x2 + 3x 3x + 8.124) (8.125) which has solutions (using the formula for the roots of a quadratic function (NOTE: reference that equation)) given by 4(1)( 5) ¡ 3 x = ¡ § (3)2 ¡ 2(1) p p29 3 = ¡ § 2 3 x1 = ¡ x2 = ¡ p29 ¡ 2 3 + p29 2 (8.126) (8.127) (8.128) (8.129) The graph of f (x) is shown in ﬂgure 8.7. The points x1 and x2 (where the function f (x) cuts the x axis) are labelled. 3 Now f (x) > 0 (the function is above the x-axis) when ¡ p29 ¡ 2 < x < 3+p29 2 ¡ . Therefore the solution to the inequality is 3 or in interval notation ( ¡ 3+p29 2 Note: The x-intercepts are not included in the solution because the > sign has been used and therefore f (x) = 0 does not deﬂne a solution to the inequality. < x < ¡ p29 ¡ 2 , g ; ¡ ). 3 x : ¡ f 3+p29 2 p29 ¡ 2 Example 3: Q: Solve the inequality 4x2 4x + 1 0. • ¡ 149 ! ! 6 4 2 x1 4 ¡ 3 ¡ 2 ¡ 1 ¡ x2 1 Figure 8.7: Graph of f (x) = x2 ¡ ¡ 3x + 5 1:5 1:0 0:5 0:5 1:0 Figure 8.8: Graph of f (x) = 4x2 4x + 1 ¡ A: Let f (x) = 4x2 f (x) = (2x ¡ 4x + 1. Factorising this quadratic function gives 1)2, which shows that f (x) = 0 only when x = 1 2 . The function f (x) lies below/on the x-axis only at the x-intercept. Therefore the only solution to the inequality is x = 1 2 . ¡ 8.8 Intersections The syllabus requires: { find solutions of 2 lines and interpret the common solution as the intersection 150 " " " # # { find solutions of linear and quadratic, interpret the common solution(s) as intersections 151 Chapter 9 Working with Data 9.1 Statistics The syllabus requires: { organise univariate numerical data to determine measure of central tendency; mean, median, mode and when each is appropriate. measure of dispersion; ranger, percentiles, quantities, interquartile, semi-interquartile range { represent data effectively, choosing from; bar and compound bar graphs, histograms, frequency polygons, pie charts, dot plots, line and broken line graphs, stem and leaf plots, box and whisker diagrams { (grade 12) variance, standard deviation { (grade 12) draw a suitable random sample from a populations, and understand the importance of sample size in predicting the mean and standard deviation { (grade 12) identifies data which is normally distributed about a mean by investigating appropriate histograms and frequency polygons 9.2 Function Fitting The syllabus requires: { represent bivariate data as a scatter plot and suggest what function (linear, quadratic, exp) would best describe it { tell the difference between symmetric and skewed data and make relevant deductions { (grade 12) use appropriate technology to calculate linear regression line which best fits a given set of bivariate data 152 9.3 Probability The syllabus requires: { Venn diagrams to solve probability problems. must know P (s) = 1qquadP (AorB) = P (A) + P (B) P (AandB) ¡ { identify dependent and independent events and calculate the prob of 2 independent events occurring by applying the product rule for independent events P (AandB) = P (A):P (B) { identify mutually exclusive events. calculate prob of the events occuring by applying additive rule for mutually exclusive events P (AorB) = P (A) + P (B) { identify complementary events P (notA) = 1 { use prob models for comparing experimental results with theory; P (A) ¡ need many trials to get comparable results... coin example flipping a { comparing experimental results with each other { potential sources of bias, error in measurement, potential uses and misuses of stats and charts (NOTE: if SA people could get some popular TV adverts as examples, that would be good) { converts this theory into a project (NOTE: i don’t know how much of this project stuff we should do in the book. just ignore its existence) possibly 9.4 Permutations and Tree Diagrams The syllabus requires: { tree diagrams and other methods of listing all options to generalise counting principle (successive choices) { calculate the probability of compound events which are not independent { assess the odds in a variety of games of chance, lotteries, raffles { (grade 12) use investigate and solve problems involving the number of arrangements (permutations) of a number of discrete objects (when order matters) m! (m different items), m items selected from n { (grade 12) investigate and solve problems involving the number of possible solutions when order is not important (combinations) of m items from n where all are different or distinguishable { (grade 12) uses permutations and combinations to correctly calculate the probability of specified events occurring { determines the odds of various games of chance and the probability of events which depend on combinations and permutations 153 9.5 Finance The syllabus requires: { use simple and compound interest for relevant problems (hire purchase and inflation) { effective and nominal interest { understand fluctuating foreign exchange rates and their effect on local prices, travelling prices, imports and exports { solve straight line (simple) depreciation and depreciation on a reducing budget (compound depreciation) { (grade 12) apply geometric series to solve problems (future values of annuities, bond repayments, sinking fund contributions including the difference in time taken to pay when the monthly payment is changed) { (grade 12) critically analyse investment and loan options and make informed decisions to the best options (pyramid and micro lenders schemes) 9.6 Worked Examples TODO 9.7 Exercises TODO 154 Part II Old
Maths 155 Chapter 10 Worked Examples 10.1 Exponential Numbers (NOTE: All of these worked examples need to be updated to use the FHSST internal environments and also to use the new rules (i changed the originals). They are for the Exponential Numbers section) Worked Example 9 : Manipulating Exponential Num- bers Question: Simplify the expression 42:33 63 2 = 22 and 6 = 2:3 it follows that A: Noting that 4 = 2 £ 42:33 63 = = (22)2:32 (2:3)3 24:33 23:33 3 ¡ = 24 = 21 = 2 (10.1) (10.2) (10.3) (10.4) (10.5) Example 2: Q: Simplify ( 5 A: First note that 20 = 2 2 )2:20. 2 £ £ 5 = 22:5. Therefore 156 ( 5 2 )2:20 = 52 22 :22:5 = 52+1 = 53 = 5 £ = 125 5 5 £ (10.6) (10.7) (10.8) (10.9) (10.10) Example 3: Q: Solve for the variable x in the equation 2x+1 = 2x + 8. A: ¡ 2x+1 = 2x + 8 2x = 8 2x:2 2x(2 1) = 8 ¡ 2x = 8 = 2 x = 3 £ 2 (10.11) (10.12) (10.13) (10.14) (10.15) 2 = 23 £ ) ) ) ) It is also possible to talk about zero, negative and even fractional exponents. We shall assume that laws 1{5 are also true in these cases. This gives us 3 more laws. a0 = a1 1 = ¡ a1 a1 = a a = 1 Law 7 Since n = 0 ¡ ¡ 1 an n, it follows from laws 2 and 6 that n = a¡ a¡ n = a0 ¡ n = a0 an = 1 an (10.16) (10.17) This deﬂnes what is meant by a negative exponent. Note to self: add aside lay-out to following paragraph Aside: A fraction to the power of a negative exponent is the same as the inverse fraction to the power of the corresponding positive exponent. Therefore ( a b )¡ n = 1 a b = a an bn = ( b a )n (10.18) 157 Example 1: Q: Simplify the expression ( 1 A: 2 )¡ 3. ( 1 2 )¡ 3 = 1 ( 1 2 )3 = 1 1 23 = 23 = 8 (10.19) Example 2: Q: Simplify (27) 1 3 : A: Now 27 = 3 £ q 3 £ 2 9 . 3 = 33 and 9 = 3 3 = 32, so £ (27) 1 3 : q 2 9 = (33) 1 3 : 2 r 32 2 32 ) 1 2 = 33: 1 3 :( 1 2 1 2 = 31:2 (32) 3:p2 3 = p2 = Example 3: Q: Simplify (an + an) A: 1 2 (an + an) 1 2 = (2an) 1 2 1 2 1 2 :(an) = (2) = p2 an: 1 = p2 a n 2 2 (10.20) (10.21) (10.22) (10.23) (10.24) (10.25) (10.26) (10.27) (10.28) Example 4: Q: Find a solution to the equation ax b are constants. A: m n ¡ b = 0, where a and ) ) ) ) ) b = 0 m n ax ax ¡ m n = b b a xm = (x m n = x m n )n = ( x = (xm) 1 m = (( )n b a b )nn ( r 158 (10.29) (10.30) (10.31) (10.32) (10.33) (10.34) Therefore x = m 0. q ( b a )n is a solution to the equation ax m n b = ¡ (NOTE: surds) Worked Example: Q: Simplify the expression p32 2 . A: p32 2 = p32 p4 32 4 = r = p8 = p4:2 = p4:p2 = 2p2 (10.35) (10.36) (10.37) (10.38) (10.39) (10.40) Worked Example: Q: Which of the numbers 3p100 and p20 is bigger? (You may not use a calculator to answer this question.) A: The two numbers must ﬂrst be converted into like surds. Since we have a cube root and a square root, we must ﬂrst ﬂnd the lowest common multiple of 2 and 3 which is 6. We then convert each of the surds into 6th roots as follows: 3p100 = 3 q p1002 = 3:2p1002 = 6p10000 (10.41) p20 = 2 3p203 = 2:3p203 = 6p8000 q (10.42) Now, since 100000 is bigger than 8000, it follows that 6p10000 is bigger than 6p8000. Therefore 3p100 is bigger than p20. Worked Examples: Example 1: Q: Rationalize the denominator of the fraction 1 p6 . A: The denominator can be changed into the rational number 6 by multiplying the numerator and denominator by p6. Therefore 1 p6 = 1 p6 £ p6 p6 = p6 6 159 (10.43) Example 2: Q: Rationalize the denominator of p2 (p3+p8) . A: Multiplying the numerator and denominator by (p3 p8) gives ¡ p2 (p3+p8) = = = = = (p3 (p3 p8) p8) ¡ ¡ p2 (p3 + p8) £ p8) p2(p3 3 p2p3 ¡ 8 ¡ p2p8 ¡ 5 ¡ p16 ¡ 5 p6 ¡ 5 p6 ¡ 4 (10.44) (10.45) (10.46) (10.47) (10.48) 1.1.4 d) Equations Here we shall solve some equations involving surds. Worked Examples: Example 1: Q: Find a solution to the following equation x + 3 ¡ p6x + 13 = 0 (10.49) A: First the square root must be moved to the right-hand side of the equation, and everything else to the left. x + 3 = p6x + 13 (10.50) Now we square both sides of the equation. (x + 3)2 = 6x + 13 x2 + 6x + 9 = 6x + 13 x2 = 4 x = 2 or x = 2 ¡ ) ) ) (10.51) (10.52) (10.53) (10.54) When we square both sides of the equation, it is possible that we introduce extra solutions, which may not actually satisfy the original equation. This is the reason one should always check the answers by substituting these into the original equation. x = 2: 160 p6x + 13 = 2 + 3 6(2) + 13 = 5 p25 = 5 ¡ ¡ 5 = 0 (10.552: x+3 p6x + 13 = 2+3 6( 2) + 13 = ¡ ¡ ¡p Therefore, in this case, both the solutions x = 2 and x = to the original equation ¡ ¡ 1 ¡ p1 = 1 1 = 0 (10.56) ¡ 2 are solutions ¡ Note to self: Maybe the following example should only be included after the chapter on quadratics? But need non-trivial example where not all solutions valid. Example 2: Q: Solve the equation px + 7 = x ¡ 1 ¡ (10.57) As before, we ﬂrst move the surd to the right-hand side and the other terms to the left. x + 1 = px + 7 (10.58) Squaring both sides of the equation gives (x + 1)2 = x + 7 x2 + 2x + 1 = x + 7 x2 + x ¡ (x + 3)(x 2 or x = 2 ¡ ) ) ) ) (10.59) (10.60) (10.61) (10.62) (10.63) Again we must check these answers: x = -3: px + : 3 + 7 = p ¡ p4 = 10.64) ¡ px + 7 = 2 x ¡ p2 + 7 = 2 p9 = 2 3 = 1 ¡ ¡ ¡ ¡ (10.65) Therefore x = only solution. 3 does not satisfy the original equation, so x = 2 is the ¡ 161 6 Scientiﬂc Notation Example 1: The speed of light in a vacuum c is c = 2:9979 = 2:9979 £ £ 108 m s 100000000 m s (10.66) (10.67) (10.68) = 299790000 m s Example 2: The approximate radius of the hydrogen atom (called the Bohr radius a0) is a0 = 5:3 = 5:3 = 5:3 £ £ £ 11 m 10¡ 1 1011 m 1 10000000000 0:00000000001 m m = 5:3 = 0:000000000053 m £ (10.69) (10.70) (10.71) (10.72) (10.73) Note to self: check the above number ... I think a0 = 0:53”A ? ... Example 3: Q: The universe is known to be 13.7 billion years old. Convert this number into scientiﬂc notation. A: The age of the universe is 13700000000 years (since 1 billion is a thousand million or 1000000000). The decimal point must move 10 places to the left to convert this number into 1.37. Since 13700000000 is much bigger than 1.37, the power 10m must make the number 1.37 bigger and therefore m must be positive (so that 10m is greater than 1). Therefore m = 10 (the decimal point moves 10 places). Therefore 13700000000 = 1:37 1:37 1010 years. £ 1010. The age of the universe is thus £ Example 4: Q: The charge on an electron is e = 0:0000000000000000001602 C. What is this constant in scientiﬂc notation. A: The decimal point must be moved 19 places to the right to change this number into 1.602 (so a = 1:602). Now e is much smaller than 1 and thus 162 the exponent m must be negative, so that 10m is less than 1. Therefore m = 19. ¡ Therefore e = 1:602 10¡ 19 C. £ Note: As a general rule, if the decimal point must move to the left then the exponent m is positive and if the decimal point moves to the right then m is negative. patterns sequences Worked Example 10 : Calculating the nth Term of a Sequence ¡ is given by 2 + 4(n Question: Check that the formula for the nth term of the sequence 1), and calculate 2;6;10;14;18;::: g f the thousandth term. Answer: The sequence we are given starts at two, and each term is equal to the previous term plus four. We can check whether the formula is valid by going through the ﬂrst few terms, and seeing whether the terms in the sequence correspond to the If we substitute n = 1 into the terms given by the formula. formula, we should get the ﬂrst term of the sequence, and this is indeed the case: 2 = 2 + 4(1 1). If we substitute n = 2, we ¡ get the second term, namely 6. We can continue in this way, substituting n = 3, n = 4, and so on - each time we get the expected value back. This means that the formula is indeed valid. To calculate the thousandth term, we must substitute 1000 into the formula. When we do so we get 2 + 4(1000 1) = 2 + 4(999) = 3998. ¡ Worked Example 11 : Calculating the Sequence When Given the Formula ¡ Question:The formula for the nth term of a sequence is given 1).Write down the ﬂrst four terms of the sequence, and by 2(2n describe the pattern that you observe. Answer:This question is similar to the one in the previous example. To get the ﬂrst term of the sequence, we must substitute 1 into the formula. Doing so gives us 2(21 1) = 2(20) = 2(1) = 2. To ﬂnd the second term we must substitute two into the sequence: 2(22 1) = 2(2) = 4. For the third and fourth terms we substitute three and four respectively, and so we ﬂnd that the ﬂrst four terms of the sequence are 2;4;8;16;:::. >From these four terms we can see that every term in the sequence is equal to the previous term multiplied by two. ¡ ¡ 163 Worked Example 12 : Calculating the Sequence When Given the Formula Question:The ﬂrst term of a sequence is 3, and the formula for the (n + 1)th term is given by an+1 = an + 2n. Write down the ﬂrst four terms of the sequence. Answer:We start by calculating the second term, since we already know the ﬂrst. To do this we need to substitute n = 1 into the formula, since the formula is for the (n + 1)th term, NOT for the nth term, as was previously the case. So, substituting n = 1, we get that a1+1 = a1 + 2(1) = 3 + 2 = 5: To get the next term, we must substitute n = 2: a2+1 = a2 + 2(2) = 5 + 4 = 9. Lastly, we calculate that a4 = 9 + 2(3) = 15. So the ﬂrst four terms of the sequence are 3;5;9;15;:::. Worked Example 13 : Checking That a Given Se- quence is Arithmetic ¡ Question:Check that the sequence given by the formula an = 1) is an arithmetic sequence, and ﬂnd d for this se3 + 4(n quence. Answer:We must check to see that the diﬁerence between successive terms is a constant. There are two ways of doing this: we could write out the ﬂrst few terms of the sequence and check that they are evenly spread - i.e. they diﬁer by a constant amount, or we could do the calculation in general, using the formula directly. We will do the example using both these approaches alternately. ¡ ¡ 3 = 11 7 = 15 11 = 19 { First approach: It is easy to calculate the ﬂrst few terms using the formula. They are given by 3;7;11;15;19;:::. We can
see that the diﬁerence between successive terms is always 4, since 7 15 = 4, so the sequence is indeed an arithmetic sequence, and d = 4 { Second approach: We know that the formula for the nth term is given by an = 3 + 4(n 1). From this it should be clear that the (n + 1)th term is given by an+1 = 3 + 4((n + 1) If we work out the diﬁerence between successive terms, we get that an+1 ¡ 1) = 4n we got using the previous method. ¡ 4n + 4 = 4 = d, which is the same answer that 1) = 3 + 4n. an = 3 + 4n 4(n ¡ ¡ ¡ ¡ ¡ ¡ ¡ 3 Worked Example 14 : Calculating the Formula for the nth term of a Sequence Question:Find a formula for the nth term of the sequence 6; 17; 28; 39;:::. Which term in the sequence equals 688? 164 Answer:First we must check that the given sequence is in fact an arithmetical one, otherwise we can’t use the formula. This is easily seen, since the diﬁerence between successive terms is 11. Now we must use (??). We know that a1 = 6 and that If we substitute these into the formula, we see that d = 11. an = 6 + 11(n 1). Lastly, we must ﬂnd which term equals 688. Notice that in the ﬂrst worked example we were given an n (namely 1000), and asked to calculate an. Now we are given an an and asked to calculate n. We can do this easily by rearranging the formula: ¡ an = 688 = 6 + 11(n 6 = 11(n 6 ¡ 688 688 ¡ 11 = n 1) 1) ¡ ¡ 1 ¡ n = 6 688 ¡ 11 + 1 = 63 We conclude that the 63rd term will equal 688. geometric Worked Example 15 : Checking That a Given Se- quence is Geometric ¡ 1) is a geometric sequence, and ﬂnd r for this sequence. Question:Check that the sequence given by the formula an = 2(3n Answer:We must check to see that the ratio between successive terms is a constant. As in example four, there are two ways of doing this: we could write out the ﬂrst few terms of the sequence and check that successive terms diﬁer by a common factor, or we could do the calculation in general, using the formula directly. We will do the example using both these approaches alternately. { First approach: It is easy to calculate the ﬂrst few terms using the formula. They are given by 2;6;18;54;162;:::. We can see that the ratio between successive terms is always 3, since 6 6 = 54 54 = 3, so the sequence is indeed an arithmetic sequence, and r = 3 18 = 162 2 = 18 { Second approach: We know that the formula for the nth 1). From this it should be clear term is given by an = 2(3n ¡ that the (n + 1)th term is given by an+1 = 2(3n). If we work out the ratio between successive terms, we get that an+1 an = 2(3 n+1 = 3, which is the same answer that we got using the previous method. 2(3n¡1) = 3n ¡ n ) Worked Example 16 : Calculating the Formula for the nth term of a Sequence 165 Question:Find a formula for the nth term of the sequence 2;4;8;16;:::. Which term in the sequence equals 8192? Answer:First we must check that the given sequence is in fact a geometric one, otherwise we can’t use the formula. This is easily seen, since the ratio between successive terms is 2. Now we must use (??). We know that a1 = 2 and that r = 2. If 1). we substitute these into the formula, we see that an = 2(2n Lastly, we must ﬂnd which term equals 8192. We can do this easily by rearranging the formula: ¡ a n = 688 (688 688 = 6 + 11(n = = ¡ 6)=11 + 1 = 6 ¡ 6)=11 11(n n ¡ 63 ¡ 1 1) ¡ 1) n = (688 ¡ We conclude that the 13th term will equal 8192. 10.2 series Worked Example 17 : Calculating Sn Question:Calculate S4 for the series 2 + 4 + 8 + 16 + 32 + :::. Answer:Recall that S4 is the sum of the ﬂrst four terms of the series. This is given by 2 + 4 + 8 + 16 = 30. Worked Example 18 : Calculating a Series Question:Calculate ﬂrst ﬂve terms of the series which corresponds to the sequence an = 2 + 2(n Answer:First we calculate the ﬂrst ﬂve terms of the sequence. They are 2,4,6,8,10. To get the corresponding series, we simply need to put addition signs in between the terms: 2+4+6+8+10. 1). ¡ Worked Example 19 : Checking Sn for a given series 2 [4 + 2(n Question:For the series 2+4+6+8+:::, check that the formula for the sum of the ﬂrst n terms is given by Sn = n 1)] Answer:Let us start by writing out the ﬂrst few terms of Sn. S1 equals the ﬂrst term of the series, so S1 = 2. S2 is the sum of the ﬂrst two terms, so S2 = 2 + 4 = 6. S3 is the sum of the ﬂrst three terms, namely 12. Continuing in this fashion, we can see that the ﬂrst few terms of Sn are 2;6;12;20;28;:::.What we need to determine is whether these correspond to the given formula for Sn, and indeed they do. We must simply note that when we substitute 1 into n 1)], we get 2, when we substitute 2, we get 6, when we substitute 3, we get 12, then 20, then 28, 2 [4 + 2(n ¡ ¡ 166 and so on. This means that the formula does indeed give us the sum up to n terms. Worked Example 20 : Calculating the Sequence that Corresponds to a Given Sn ¡ 4n. Find the sequence which corresponds to this series. Question:For a certain series, Sn is given by the formula Sn = 2n2 Answer:Let us start by writing out the ﬂrst few terms of Sn: 2;0;6;16;30;:::. Think carefully about what this means - S1 = ¡ 2 means that the ﬂrst term of the series is -2, S2 = 0 means ¡ that the sum of the ﬂrst two terms is 0. Therefore the second term must be 2, since when we add 2 to -2 we get 0. S3 = 6 means that the sum of the ﬂrst three terms of the series is 6, so, by similar reasoning, the third term must be 6. S4 = 16 means that the fourth term must be 10. Now one should start to see the pattern - the ﬂrst few terms of the sequence which corresponds to this series are 2;2;6;10;:::, so we are dealing ¡ with an arithmetical sequence that has a common diﬁerence of 4. Worked Example 21 : Calculating Sn for a Given Series Question:Calculate the value of the series 1 + 4 + 7 + 10 + 13 + ::: + 46. Answer:We wish to ﬂnd S16 (since 46 is the 16th term of the series). Of course we can do this with a calculator, but there is a much quicker way. S16 = 1 + 4 + ::: + 43 + 46 S16 = 46 + 43 + ::: + 4 + 1 2S16 = 47 + 47 + ::: + 47 + 47 2S16 = 16 47 = 752 £ S16 = 752 2 = 376 Worked Example 22 : Using the Formula for Sn Question:Find the sum of all the integers between 1 and 100, i.e. ﬂnd 1+2+3+...+99+100. Answer:Since we are dealing with an arithmetic series,we can use the formula for Sn that we have derived. In order to use it we need to know which values to put in for a1, d, and n. It should Since the series starts at 1, we know that a1 = 1. 167 also be clear that d = 1 (since the common diﬁerence between successive terms is 1), and that n = 100 (since we are summing up to the hundredth term). Now it is just a question of plugging these values into the formula. S100 = n S100 = 100 S100 = 50(101) = 5050 2 [2a1 + (n ¡ 2 [2 + (100 1)d] 1)] ¡ Worked Example 23 : Calculating Sn for a Geomet- ric Series in General Question:Calculate a formula for Sn for any geometric series, given r and a1. Answer:This example is slightly diﬁerent from the case for arithmetic series. Try writing out the calculation for yourself to make sure that you understand all the steps. r £ Sn = a1 + a1r + a1r2 + ::: + a1rn Sn = a1r + a1r2 + ::: + a1rn ¡ 2 + a1rn 2 + a1rn ¡ 1 + a1rn 1 rSn ¡ Sn = Sn(r ¡ ¡ ¡ ¡ a1 + 0 + 0::: + 0 + a1rn 1) = a1rn a1 ¡ a1(rn r ¡ = 1, otherwise we would have This formula is only valid when r 0 in the denominator. When r = 1, Sn = a1 +a1 +:::+a1 = na1. (10.74) Sn = ¡ 1 1) Worked Example 24 : Using the Formula for Sn Question:What is 2 + 4 + 8 + 16 + ::: + 32768? Answer:We are dealing with a geometric series, so we have to use equation (10.74). In order to use it we need to know which values to put in for a1, r, and n. Since the series starts at 2, we know that a1 = 2. It should also be clear that d = 2 (since the common ratio between successive terms is 2). It is not so clear what n should be, but we can work it out using the equation for the nth term of a geometric series. 32768 = 2(2n 1 an = a1rn ¡ 1) = 2n ¡ log 32768 = n log 2 log 32768 log 2 = 15 n = 168 6 Now that we have values for a1,n, and r, we can use the formula to work out S15. n Sn = a1(r r ¡ S15 = 2(215 1) 1) ¡ 1 ¡ 1 S15 = 2(215 ¡ 2 ¡ 1) = 65534 Worked Example 25 : Sigma Notation Question:Calculate the values of the following expressions: { { { { 7 k=2 k + 1 4 t=2 t2 4 t=2 2t 4 k=1 3 P P P Answer: P { We have to sum the expression k +1 from k = 2 until k = 7: (1+1)+(2+1)+(3+1)+(4+1)+(5+1)+(6+1)+(7+1) = 35. 4 t=2 t2 = 22 + 32 + 42 = 4 + 9 + 16 = 29 4 t=2 2t = 22 + 23 + 24 = 4 + 8 + 16 = 28 4 k= = 12 { { { P P P Worked Example 26 : Converging or diverging Question:Which of the following inﬂnite series do you think will converge, and which do you think will diverge: { { { { P P P 1k=1 k 1 1k=1 k 1 1k=1 k2 1k=1( ¡ 1)(k + 1) P Answer:We don’t really have any systematic way of working this out yet, but we can easily guess by using our calculators. { Working out the ﬂrst few terms of the sum, we get S1 = 1, S2 = 1 + 2 = 3, S3 = 1 + 2 + 3 = 6, S4 = 1 + 2 + 3 + 4 = 10, S5 = 1 + 2 + 3 + 4 + 5 = 15. Clearly this is getting larger and larger, and it would be reasonable to guess that if we kept on adding terms, we would not get a ﬂnite number. So this series diverges. { As before, we can work out the ﬂrst few terms by hand: 6 , S4 = 2 1 S1 = 1, S2 = 1 + 1 12 , S5 = 2 17 60 , and, using a calculator, S20 = 3:6. This is still not absolutely clear, so now we can write a computer program It turns out that to work out even higher values of Sn. 2 , S3 = 169 S100 = 5:19, S1000 = 7:49, and S1000000 = 14:4. This series is in fact divergent. Even though it grows at a very slow rate, it never stops getting larger when we add more terms. { As we mentioned in the text above, this series never gets larger than 1, no matter how many terms we care to add. In fact, the more terms we add, the closer the series gets to 1 (try it on your calculator), so we say that the series converges to 1. 1 = 0, S3 = 1 { This example seems a bit strange at ﬂrst. Let us try to write out a few terms of Sn to give us an idea of what is happening: S1 = 1, S2 = 1 1+
1 = 1, S4 = 0, S5 = 1. It seems that the values we get are oscillating between 0 and 1. Remember we said that a series converges if we get closer and closer to some number when we add more terms - and that is clearly not what is happening here. That means that this series is divergent, even though it never gets larger than 1. ¡ ¡ Worked Example 27 : Using the formula for S 1 Question:Calculate the sum of the inﬂnite series Answer:We are asked to determine the sum of the inﬂnite series , which is 1 and 1, so we can use the formula. The formula between ¡ = a gives S 1 ¡ P 8 + :::. The common ratio is clearly 1 r = 1 = 2. 1 ¡ 1k=1( 1 2 )k 1 2 1 1 ¡ 10.3 functions Q: State the domain and range of the function y = x2 notation and interval notation. ¡ 4 in set-builder A: There is nothing to stop x from taking on the value of any real number, but, since x2 cannot be negative, we see that y 4. Thus the domain of the function is ‚ ¡ x : x † R g f or ) ; ( 1 ¡1 and the range is given by y : y f ‚ 4 and y † R g or [ ¡ 4; ) 1 (10.75) (10.76) Worked Example: Q: Plot a graph of the function f (x) = A: The x-intercept is x + 1. ¡ 170 x + 1 = 0 x = 1 ¡ ) The y-intercept is ¡ Therefore the graph is as follows: y = (0) + 1 = 1 (10.77) (10.78) (10.79 Figure 10.1: Graph of f (x) = 1 x ¡ Now we have seen that, in general, the constant b is the y-intercept, but what is a? The bigger a the faster the y-values change when the x values change. Therefore a is called the slope and shows how steep the straight line is. Worked Example: Example 1: Q: We are given a straight line graph f (x) = ax + b and two diﬁerent = x2. What is the slope of the line? points (x1;y1) and (x2;y2), where x1 6 A: Now we need to calculate the slope a and we know that y1 = ax1 + b and y2 = ax2 + b, since (x1;y1) and (x2;y2) are points on the straight line. This means that y2 ¡ y1 = (ax1 + b) = ax2 ¡ = a(x2 ¡ ¡ ax1 + b x1) (ax2 + b) b ¡ (10.80) (10.81) (10.82) 171 $ $ Therefore the slope describes the change in y (sometimes called ¢y = x1) between any two y2 ¡ diﬁerent points on the line, i.e. y1) divided by the change in x (¢x = x2 ¡ a = ¢y y2 ¡ y1 x2 ¡ x1 ¢x = 0) because one cannot x1 6 We have used the fact that x1 6 = x2 (i.e. x2 ¡ = y2 gives an inﬂnite slope divide by zero. (The case of x1 = x2 but y1 6 and this describes a vertical line at constant x.)(NOTE: This section needs to work in the relevance of ﬂgure 10.2 more) (10.83) = y2 y1 ¢y ¢x x1 x2 Figure 10.2: Graph showing ¢x and ¢y for a line of the form y = ax + b Note: A straight line with a positive slope (a > 0) increases from left to right and a straight line with a negative slope (a < 0) increases from right to left. a > 0 a < 0 Figure 10.3: A straight line with positive slope (a > 0) and a straight line with negative slope (a < 0) Example 2: Q: Consider the straight line shown in the graph below: 172 % % Figure 10.4: Graph of straight line with y0 = 2 and x0 = 5 ¡ Find the equation of the straight line describing this graph. A: We need to ﬂnd f (x) = ax + b, so we need the a and b values for this line. We see that the y-intercept is y = 2. Since we have two points (the y-intercept (0,-2) and the x-intercept (5,0)), we can ﬂnd the slope using the equation from the previous example. Therefore 2 so b = ¡ ¡ a = y2 ¡ x2 ¡ y1 x1 0 = ¡ 5 2) ( ¡ 0 ¡ = 2 5 (10.84) So the equation of the straight line is f (x) = 2 5 x 2. ¡ Now, for a general parabola of the form f (x) = ax2 + c where a is positive, the term ax2 is always positive so the function is at a minimum when x = 0. Therefore the arms of the parabola point upwards. Otherwise, if a is negative, then ax2 is always negative and thus the function is maximum at x = 0. This means that the arms of the parabola must point downwards. Worked Examples: Example 1: Q: Find the intercepts and thus plot a graph of the function f (x) = x2 + 4. ¡ A: The y-intercept is and the x-intercepts are y = (0)2 + 4 = 4 ¡ (10.85) 173 & a > 0 a < 0 Figure 10.5: A parabola of the form y = ax2 + c with positive curvature (a > 0) and a parabola with negative curvature (a < 0)(NOTE: is curvature the right word to use here?) x2 + 4 = 0 x2 = 4 2 x = § ¡ ) ) (10.86) (10.87) (10.88) Since a = 1 is negative, we know that the arms of the parabola must point downwards. We can now use the three intercepts to draw the graph of this function Figure 10.6: Graph of f (x) = x2 + 4 ¡ 174 ’ ’ ’ Example 2: Q: Plot a graph of the parabola f (x) = x2 + 1. A: The y-intercept is y = (0)2 + 1 = 1 (10.89) and the x-intercepts are x2 + 1 = 0 x2 = 1 ¡ ) (10.90) (10.91) (10.92) This is not possible if x is a real number, so there are no x-intercepts. The parabola must therefore be entirely above the x-axis. This agrees with the fact that we know the arms of the parabola point upwards because a = 1 is positive. Thus the graph is as follows: 5 4 3 2 1 0 Figure 10.7: Graph of the parabola f (x) = x2 + 1 1 ¡ 1 175 ( Example 2: Q: Plot a graph of the hyperbola xy = ¡ A: A table of x and y values is as follows: 1. x : y = 1 x : -4 1 4 -2 1 2 -4 1 2 -2 1 -1 2 - 1 2 4 - 1 4 This gives the graph shown below Figure 10.8: Graph of the hyperbola xy = 1 ¡ We can see that a hyperbola has no x or y-intercepts. However, there are two general forms for hyperbolae, depending on whether a is positive or negative. Both types of hyperbola are symmetric about the lines y = x and y = x (in other words, the part of the hyperbola on one side of the line is just the reection of the part on the other side). ¡ In the ﬂrst case (a > 0), the hyperbola intersects line y = x at two points. At these intersections xy = a and y = x (10.93) which implies that 176 ) ) ) ) ) ) ) ) ) ) a > 0 a < 0 Figure 10.9: A hyperbola of the form y = a x with positive coe–cient (a > 0) and straight line y = x. A hyperbola with negative curvature (a < 0) and straight line y = x. ¡ x(x) = x2 = a pa x = § ) (10.94) (10.95) and, as y = x, it follows that the points of intersection are ( and (pa;pa). ¡ pa; ¡ pa) In the second case (a < 0), the hyperbola intersects the line y = two intersection points occur when x. The ¡ xy = a and y = x ¡ and therefore x) = x2 = a ¡ x( ¡ x2 = x = a ¡ p § a ¡ ) ) and, since y = pa). ¡ x, the two intersection points are ( (10.96) (10.97) (10.98) (10.99) pa;pa) and (pa; ¡ ¡ Worked Examples: Example 1: Q: Draw a graph of the hyperbola xy = 9. A: First we note that, since a = 9 is positive, the hyperbola must be in the top right and bottom left quadrants. The points at which it intersects the line y = x are (-3,-3) and (3,3). It is also clear that the points (-1,-9), (-9,-1), (1,9) and (9,1) are part of the hyperbola. Thus the graph is 177 Figure 10.10: Graph of the hyperbola xy = 9 1:5 1:0 0:5 0:5 1:0 1:5 1:5 ¡ 1:0 ¡ 0:5 ¡ 0:5 ¡ 1:0 ¡ 1:5 ¡ 1 2 ; 1 2 ) and Figure 10.11: Graph of a hyperbolic function turning at the points ( ( 1 2 ; 1 2 ) ¡ ¡ Example 2: Q: The graph of a hyperbolic function is shown below. 178 * * * * * * + + What is the function deﬂning this hyperbola? A: Since the hyperbola is in the top left and bottom right quadrants, we known that a < 0. Now the point ( 2 ) lies on the hyperbola, so 2 ; 1 1 ¡ a = xy = ( ¡ 1 2 )( 1 2 ) = 1 4 ¡ (10.100) which is negative as we originally worked out. Therefore the hyperbolic function is f (x) = 1 4x . ¡ Worked Example: Q: Plot graph of the relation x2 + y2 = 4. A: This is the equation of a circle centered at the origin with radius 2. The graph is as follows: 2 2 ¡ 2 ¡ 2 Figure 10.12: Graph of the circle x2 + y2 = 4 Worked Example: Q: Draw the semi-circles described by the equations y = 9 x2 and y = ¡ Also give the domains and ranges of these semi-circles. p ¡ x2 9 ¡p (10.101) A: These equations describe semi-circles with radius 3. The ﬂrst semicircle lies above the x-axis (since the positive square root is being considered, the y values are all positive) and the second semi-circle is below the x-axis (the y values are all negative because the equation involves the negative root). The graphs of these semi-circles are thus as follows: From the above graphs we can see that, for the semi-circle y = p9 the domain is [ x2, 3;3] and the range is [0;3] and, for the semi-circle y = ¡ p9 ¡ ¡ ¡ x2, the domain is [ ¡ 3;3] and the range is [-3,0]. 179 3 3 ¡ 3 3 ¡ 0 3 3 ¡ Figure 10.13: Semi-circles of radius 3, centered at the origin. The equation for the left semi-circle is y = p9 x2 whereas the right semi-circle is governed by y = p9 x2 ¡ ¡ ¡ The equation x2 + y2 = r2 can also be used to solve for x which gives x = r2 y2 (10.102) §p Again the positive and negative roots each describe a semi-circle, but in this case, on either side of the y-axis. Therefore the equations for two other types of semi-circles are ¡ x = r2 p y2 ¡ and x = r2 ¡p y2 ¡ (10.103) The domains of these semi-circles are [0;r] (in the ﬂrst case) and [ (in the second case). In both cases, the range is [ r;r]. r;0] ¡ ¡ Note: Again y = ¡ Thus these semi-circles are not functions. r and y = r corresponds to the same x = 0 value. 180 Worked Example: Q: Plot the graphs of the following relations 1 ¡p p State the domains and ranges of these relations. x = x = and ¡ 1 y2 y2 ¡ (10.104) A: The above equations describe semi-circles of radius 1 on either side of the y axis (in the ﬂrst case x is always positive and in the second x is always negative). The graphs of these relations are as follows ¡ Figure 10.14: Semi-circles of radius 1 on either side of the y axis. In the ﬂrst y2 and in the second case x is always case x is always positive as x = y2. (NOTE: This entire chapter needs a rethink on negative as x = 1 how it references graphs... instead of non-descriptive sentences like \in the ﬂrst p case", we need to use 1 p ¡ ¡ ¡ ref a lot more) n For the ﬂrst semi-circle, the domain is [0;1] and the range is [ the case of the second semi-circle, the domain is [ [ ¡ 1;1]. ¡ 1;1] and in ¡ 1;0] and the range is Worked Example: Q: Plot a graph of the function f (x) = x + 2 j j ¡ 5. A: Let us ﬂrst work out a table of x and f (x) values as fol
lows: x: f (x) -8 1 -7 0 -6 -1 -5 -2 -4 -3 -3 -4 -2 -5 -1 -4 0 -3 1 -2 2 - The graph of the function f (x) is thus shown below. 181 Figure 10.15: Graph of the function f (x) = x + 2 j j ¡ 5 Worked Examples: Example 1: Q: Consider the function y = 2 4. Plot a graph of this absolute x ¡ j value function, showing all the intercepts, the turning point and the axis of symmetry. 1 j ¡ 4) A: We can see that, since b = 1 and c = and the axis of symmetry is x = 1. Also, because a = 2 is positive, the absolute value function will be V-shaped. 4, the turning point is (1; ¡ ¡ Now let us ﬂnd the y-intercept. At 0(1) 4 = ¡ 2 (10.105) ¡ Since the turning point is below the x-axis and the graph points upwards, we suspect that this function does have x-intercepts. Let us try to ﬂnd these intercepts, which are the x-intercepts of the two straight lines y = 2(x 4 and y = 2( x + 1) 1) 4. ¡ ¡ ¡ ¡ y = 2(x 2x ¡ 6 = 0 1) ¡ 2x = 6 x = 3 4 = 0 ¡ (10.106) (10.107) (10.108) (10.109) (10.110) ) ) ) and 182 , , , , , , , , , , , , , , , , , y = 2( 2x x + 1) ¡ 2 = 0 ) ¡ ¡ 2x = 10.111) (10.112) (10.113) (10.114) (10.115) Therefore the graph of the absolute values function is as follows Figure 10.16: Graph of the absolute values function y = 2 j 4 1 j ¡ ¡ Example 2: Q: Plot a graph of the function f (x) = turning point and axis of symmetry. x+3 1 showing the intercepts, ¡j j¡ A: The turning point is of this function is ( symmetry is given by x = like an upside down V. 3. As a = ¡ ¡ 1) and the axis of 1 is negative the graph is shaped ¡ ¡ 3; At x = 0, the y-intercept is ¡j j ¡ 1 = 1 = f (0) = (0) + 3 3 j ¡ ¡j Now this function points downwards and obtains its maximum at the turning point ( 1 is negative). Since the function is never greater than -1, it cannot become positive and therefore cannot cross the x-axis. Thus this absolute value function has no x-intercepts. 1). This point is below the x-axis (since y = (10.116; The graph of the function is shown below. 183 ¡ Figure 10.17: Graph of the function f (x) = x + 3 1 ¡j j ¡ Worked Example: Q: What the does the relation x2 graph of this relation. ¡ 2x + y2 + 4y = 4 describe? Plot a ¡ A: We can see that it is not going to be easy to solve to x and y, but let us try another trick, which is to write the relation as the sum of perfect squares. We know that x2 ¡ 2x = (x2 ¡ 2x + 1) and also that y2 + 4y = (y2 + 4y + 4) 1 = (x 1)2 1 ¡ ¡ 4 = (y + 2)2 4 ¡ ¡ ¡ Therefore the relation can be written as follows: and thus 1)2 (x ¡ ¡ 1 + (y + 2)2 4 = 4 ¡ ¡ 1)2 + (y + 2)2 = 1 (x ¡ (10.117) (10.118) (10.119) (10.120) But this is just the equation for a circle centered at the origin with radius 1 (x2 + y2 = 1), where x and y have been replaced by x 1 and y + 2. So we can see that this is a circle which has been moved 1 to the right and 2 downwards. Therefore this relation describes a circle centered at the point (1,-2) with radius 1 (as shown in the diagram below). ¡ Note: point (a;b) is In general, the equation for a circle of radius r centered at the a)2 + (y (x ¡ ¡ b)2 = r2 (10.121) 184 . . 1 2 (1; 2) ¡ 1 ¡ 2 ¡ 3 ¡ Figure 10.18: Graph of the circle (x 1)2 + (y + 2)2 = 1 ¡ Worked Example: Q: Show that the equation for an ellipse centered at the origin can be derived from that of a circle centered at the origin by performing suitable transformations. Draw a graph of the resulting ellipse. A: We shall start with the equation of a circle centered at the origin, which is given by x2 + y2 = r2, and replace x by r a x and y by r b y. We can see that the equation for the circle becomes y)2 = r2 r b b2 = r2 x)2 + ( r ( a r2 x2 a2 + r2 y2 y2 x2 b2 = 1 a2 + ) ) (10.122) (10.123) (10.124) which is the equation for an ellipse centred at the origin with y-intercepts (0; b) and (0;b) and x-intercepts ( a;0) and (a;0). ¡ ¡ The graph of this ellipse is as follows: a b b a Figure 10.19: Graph of an ellipse centered at the origin with y-intercepts (0; a;0) and (a;0) and (0;b) and x-intercepts ( ¡ b) ¡ 185 / 0 0 0 0 10.3.1 Worked Examples: Example 1: Q: Consider the parabolic function f (x) = x2. What is the equation for the function which can be obtained by stretching f (x) vertically by a factor of a, then shifting this function to the right by p and upwards by q? A: We start with the function y = x2 and must be very careful to apply these changes in the right order. First we stretch f (x) vertically by a factor of a. This means that we must change y to y a , which gives = x2 y a y = ax2 ) (10.125) (10.126) Now we shift y = ax2 to the right by p. In other words, we change x to x p, which gives ¡ y = a(x p)2 ¡ (10.127) p)2 upwards by q. Therefore we must replace y Finally we shift y = a(x by y ¡ q. This means that ¡ q = a(x y y = a(x ¡ p)2 ¡ p)2 + q ¡ ) (10.128) (10.129) which is one way of writing the general formula for a quadratic function (see section 1.4). Example 2: Let f (x) = x + 6. What is the function which is the result of Q: reecting f (x) about the y-axis and then shrinking this function by a factor of 2 horizontally and then vertically? Plot graphs of the initial and ﬂnal functions. There are other ways of getting this ﬂnal function from f (x) by performing only two changes. Give an example of such a method. A: We start with the initial function y = x + 6. If we reect this about the x-axis, we must change x to x and this gives ¡ ¡ Now we must shrink this function by a factor of 2 horizontally. Thus changing x to 2x result in the equation y = x + 6 (10.130) ¡ If we now shrink this by a factor of 2 vertically (changing y to 2y) we get y = 2x + 6 (10.131) 186 2y = y = 2x + 6 ¡ x + 3 ¡ ) (10.132) (10.133) The graphs of the initial function y = x + 6 and the ﬂnal function y = x + 3 are shown below Figure 10.20: Graph of the initial function y = x + 6 (left) and the ﬂnal function y = x + 3 (right) ¡ Finally we must look for two transformations which give this same result. We suspect that there must be a reection involved, because the ﬂnal funcx and the initial function contains x (also the above graph tion involves shows that the straight lines are perpendicular). Therefore, as before, we start by reecting the function f (x) about the y-axis. This gives ¡ y = x + 6 ¡ (10.134) Now we see that to get the ﬂnal equation y = x + 3 we must also add -3 to this equation. This is the same as replacing x by x + 3, so let us try shifting the equation by 3 to the left. This gives the equation ¡ ¡ ¡ Therefore, two changes which give the same ﬂnal function are a reection about the y-axis followed by a shift by 3 to left. ) y = y = (x + 3) + 6 x + 3 (10.135) (10.136) 187 1 2 2 Appendix A GNU Free Documentation License 2000,2001,2002 Free Software Foundation, Inc. Version 1.2, November 2002 Copyright c 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. 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n obtuse angle. A straight angle, previously defined as the union of opposite rays, is an angle whose degree measure is 180. In the figure above, DEF is a straight angle. Thus, h , the sides of DEF, are opposite rays mDEF 180. Note that EF and form a straight line. h ED and EXAMPLE 1 D is a point in the interior of ABC, mABD 15, and mDBC 90. a. Find mABC. b. Name an acute angle. A D c. Name a right angle. d. Name an obtuse angle. B C 14365C01.pgs 7/9/07 4:37 PM Page 18 18 Essentials of Geometry Solution a. mABC mABD mDBC 15 90 105 b. ABD is an acute angle. c. DBC is a right angle. d. ABC is an obtuse angle. A D B C Answers a. mABC 105 b. ABD c. DBC d. ABC Exercises Writing About Mathematics 1. Explain the difference between a half-line and a ray. h PS 2. Point R is between points P and S. Are h PR and the same ray, opposite rays, or neither the same nor opposite rays? Explain your answer. Developing Skills 3. For the figure shown: a. Name two rays. b. Name the vertex of the angle. c. Name the sides of the angle. d. Name the angle in four ways. 4. a. Name the vertex of BAD in the figure. F y E D b. Name the sides of BAD. c. Name all the angles with A as vertex. h AB d. Name the angle whose sides are e. Name the ray that is a side of both BAD and BAC. f. Name two angles in whose interior regions point R lies. and h . AC g. Name the angle in whose exterior region point S lies. A h AB h AC h. Are i. Is BAC a straight angle? Explain your answer. opposite rays? Explain your answer. and C S D R B 14365C01.pgs 7/9/07 4:37 PM Page 19 5. For the figure shown: a. Name angle 1 in four other ways. b. Name angle 2 in two other ways. c. Name angle 3 in two other ways. d. Name the point of intersection of AC and . BD e. Name two pairs of opposite rays. f. Name two straight angles each of which has its vertex at E. g. Name two angles whose sum is ABC. More Angle Definitions 19 D 1 2 E C 3 B A 1-6 MORE ANGLE DEFINITIONS Congruent Angles DEFINITION Congruent angles are angles that have the same measure. C F B A E D In the figures, ABC and DEF have the same measure. This is written as mABC mDEF. Since congruent angles are angles that have the same measure, we may also say that ABC and DEF are congruent angles, symbolized as ABC DEF. We may use either notation. ABC DEF mABC mDEF The angles are congruent. The measures of the angles are equal. Note: We do not write ABC DEF. Two angles are equal only if they name the union of the same rays. For example, ABC CBA. When two angles are known to be congruent, we may mark each one with the same number of arcs, as shown in the figures above, where each angle is marked with a single arc. 14365C01.pgs 7/9/07 4:37 PM Page 20 20 Essentials of Geometry Bisector of an Angle DEFINITION A bisector of an angle is a ray whose endpoint is the vertex of the angle, and that divides that angle into two congruent angles OC For example, if mAOC mCOB. If 1 m/COB 5 1 2m/AOB , 2mCOB. is the bisector of AOB, then AOC COB and , mAOB 2mAOC, and mAOB bisects AOB, we may say that h OC 2m/AOB m/AOC g AB , ACB is a straight angle, and mACB = 180. If Let C be any point on h is the bisector of ACB, each of the two congruent angles, ACD and CD DCB, has degree measure 90 and is a right angle. Any line, ray, or line segment at C. that is a subset of , and contains C, is said to be perpendicular to g AB h CD For example, CD is perpendicular to g . AB DEFINITION Perpendicular lines are two lines that intersect to form right angles. Since rays and line segments are contained in lines, any rays or line seg- ments that intersect to form right angles are also perpendicular. DEFINITION The distance from a point to a line is the length of the perpendicular from the point to the line. P A P is a point not on g AB , and PR ⊥ g AB . The segment PR is called the per- B pendicular from P to . The point R at which the perpendicular meets the line g AB g AB is PR (the is called the foot of the perpendicular. The distance from P to R length of PR ). Adding and Subtracting Angles T P S R DEFINITION If point P is a point in the interior of RST and RST is not a straight angle, or if P is any point not on straight angle RST, then RST is the sum of two angles, RSP and PST. 14365C01.pgs 7/9/07 4:37 PM Page 21 More Angle Definitions 21 h SP In the figure, separates RST into two angles, RSP and PST, such that the following relations are true for the measures of the angles: mRST mRSP mPST mRSP mRST mPST mPST mRST mRSP EXAMPLE 1 h CD In the figure, bisects ACB. a. Name 1 in two other ways. b. Write a conclusion that states that two angles are con- gruent. c. Write a conclusion that states that two angles have C equal measures. Solution a. ACD and DCA. Answer b. ACD BCD or 2 1. Answer c. mACD mBCD or m2 m1. Answer P R S T B 2 1 D A Exercises Writing About Mathematics 1. Tanya said that if g RST separates PSQ into two congruent angles, then g RST is the bisector of PSQ. Do you agree with Tanya? Explain why or why not. 2. If an obtuse angle is bisected, what kind of an angle is each of the two angles formed? Developing Skills In 3–10, if an angle contains the given number of degrees, a. state whether the angle is an acute angle, a right angle, an obtuse angle, or a straight angle, b. find the measure of each of the angles formed by the bisector of the given angle. 3. 24 7. 82 4. 98 8. 180 5. 126 9. 57 6. 90 10. 3 14365C01.pgs 7/9/07 4:37 PM Page 22 22 Essentials of Geometry In 11–13, find the measure of each of the following: 11. of a right angle 1 2 12. of a right angle 1 3 13. of a straight angle 4 5 In 14 and 15, in each case, use the information to: a. Draw a diagram. b. Write a conclusion that states the congruence of two angles. c. Write a conclusion that states the equality of the measures of two angles. h CD h AC 14. 15. bisects ACB. is the bisector of DAB. 16. If a straight angle is bisected, what types of angles are formed? 17. If a right angle is bisected, what types of angles are formed? In 18–20, complete each statement, which refers to the figure shown. 18. mLMN mLMP m______ 19. mLMP mLMN m______ 20. mLMN m______ mNMP In 21–24, use the figure to answer each question. 21. mABE mEBC m______ 22. mBEC mCED m______ 23. mADC mCDE m______ 24. mAEC mAEB m______ N P M L D C E A B 25. Every horizontal line is perpendicular to every vertical line. Draw, using pencil and paper or geometry software, a horizontal line ACB and a vertical line DCF. a. Name four congruent angles in your diagram. b. What is the measure of each of the angles that you named in part a? c. Name two rays that bisect ACB. d. Name two rays that bisect DCF. 26. and Applying Skills h h PQ PR what is the value of x? h BD sented by 5a 6, what is mABC? are opposite rays and 27. h PS bisects QPR. If mQPS is represented by 4x 30, bisects ABC. If mABD can be represented by 3a 10 and mDBC can be repre- 14365C01.pgs 7/9/07 4:37 PM Page 23 Triangles 23 h RT h BD 28. 29. bisects QRS. If mQRS 10x and mSRT 3x 30, what is mQRS? bisects ABC and h MP bisects LMN. If mCBD = mPMN, is ABC LMN? Justify your answer. Hands-On Activity For this activity, use a sheet of wax paper, patty paper, or other paper thin enough to see through. You may work alone or with a partner. 1. Draw a straight angle, ABC. 2. Fold the paper through the vertex of the angle so that the two opposite rays correspond, and crease the paper. Label two points, D and E, on the crease with B between D and E. 3. Measure each angle formed by the crease and one of the opposite rays that form the straight angle. What is true about these angles? 4. What is true about DE and AC ? 1-7 TRIANGLES DEFINITION A polygon is a closed figure in a plane that is the union of line segments such that the segments intersect only at their endpoints and no segments sharing a common endpoint are collinear. A polygon consists of three or more line segments, each of which is a side of the polygon. For example, a triangle has three sides, a quadrilateral has four sides, a pentagon has five sides, and so on. In the definition of a polygon, we used the word closed. We understand by the word closed that, if we start at any point on the figure and trace along the sides, we will arrive back at the starting point. In the diagram, we see a figure that is not closed and is, therefore, not a polygon. DEFINITION A triangle is a polygon that has exactly three sides. The polygon shown is triangle ABC, written as ABC. In ABC, each of the points A, B, and C is a vertex of the triangle. are the AB sides of the triangle. , and CA BC , C A B 14365C01.pgs 7/9/07 4:37 PM Page 24 24 Essentials of Geometry Included Sides and Included Angles of a Triangle If a line segment is the side of a triangle, the endpoints of that segment are the vertices of two angles. For example, in ABC, the endpoints of are the vertices of A and B. We say that the side, , is included between the angles, A and B. In ABC: AB AB C 1. AB 2. BC 3. CA is included between A and B. is included between B and C. is included between C and A. A B In a similar way, two sides of a triangle are subsets of the rays of an angle, and we say that the angle is included between those sides. In ABC: 1. A is included between 2. B is included between 3. C is included between AC and AB . BA CA and BC . and CB . Opposite Sides and Opposite Angles in a Triangle For each side of a triangle, there is one vertex of the triangle that is not an endpoint of that side. For example, in ABC, C is not an endpoint of . We say that . AB AB is opposite A Similarly, and A is opposite is the side opposite C and that C is the angle opposite side AC is opposite B and B is opposite ; also AB AC BC BC . The length of a side of a triangle may be represented by the lowercase form of the letter naming the opposite vertex. For example, in ABC, BC a, CA b, and AB c. Classifying Triangles According to Sides Scalene Triangle Z Isosceles Triangle T Equilateral Triangle N X Y R S L M 14365C01.pgs 7/9/07 4:37 PM Page 25 Triangles 25 DEFINITION A scalene
triangle is a triangle that has no congruent sides. An isosceles triangle is a triangle that has two congruent sides. An equilateral triangle is a triangle that has three congruent sides. Parts of an Isosceles Triangle In isosceles triangle RST, the two congru, are called the legs ent sides, and TS of the triangle. The third side, , is called the base. TR RS The angle formed by the two congruent sides of the isosceles triangle, T, is called the vertex angle of the isosceles triangle. The vertex angle is the angle opposite the base. T vertex angle leg l e g base angle base angle R base S The angles whose vertices are the endpoints of the base of the triangle, R and S, are called the base angles of the isosceles triangle. The base angles are opposite the legs. Classifying Triangles According to Angles Acute Triangle B A C DEFINITION G H Right Triangle Obtuse Triangle M L K L An acute triangle is a triangle that has three acute angles. A right triangle is a triangle that has a right angle. An obtuse triangle is a triangle that has an obtuse angle. In each of these triangles, two of the angles may be congruent. In an acute triangle, all of the angles may be congruent. DEFINITION An equiangular triangle is a triangle that has three congruent angles. 14365C01.pgs 7/9/07 4:37 PM Page 26 26 Essentials of Geometry Right Triangles In right triangle GHL, the two sides of the triangle , are called the that form the right angle, GH legs of the right triangle. The third side of the triangle, , the side opposite the right angle, is called the hypotenuse. and HL GL G g e l hyp oten use H leg L EXAMPLE 1 E In DEF, mE 90 and EF ED. D F a. Classify the triangle according The triangle is an isosceles triangle. to sides. Answers b. Classify the triangle according The triangle is a right triangle. to angles. c. What sides are the legs of the triangle? The legs of the triangle are and . ED EF d. What side is opposite the right FD is opposite the right angle. angle? e. What angle is opposite ? EF Angle D is opposite EF . f. What angle is included between Angle F is between EF and . FD EF and ? FD Using Diagrams in Geometry In geometry, diagrams are used to display and relate information. Diagrams are composed of line segments that are parallel or that intersect to form angles. From a diagram, we may assume that: • A line segment is a part of a straight line. • The point at which two segments intersect is a point on each segment. • Points on a line segment are between the endpoints of that segment. • Points on a line are collinear. • A ray in the interior of an angle with its endpoint at the vertex of the angle separates the angle into two adjacent angles. From a diagram, we may NOT assume that: • The measure of one segment is greater than, or is equal to, or is less than that of another segment. • A point is a midpoint of a line segment. 14365C01.pgs 7/9/07 4:37 PM Page 27 Triangles 27 • The measure of one angle is greater than, is equal to, or is less than that of another angle. • Lines are perpendicular or that angles are right angles (unless indicated with an angle bracket). • A given triangle is isosceles or equilateral (unless indicated with tick marks). • A given quadrilateral is a parallelogram, rectangle, square, rhombus, or trapezoid. EXAMPLE 2 Tell whether or not each statement may be assumed from the given diagram. g PR (1) is a straight line. Yes (2) S, T, and R are collinear. (3) PSR is a right angle. (4) QTR is isosceles. No (5) QRW is adjacent to WRU. Yes Yes No P V Q W S RT U (6) Q is between V and T. (7) PQR and VQT intersect. (8) R is the midpoint of . TU Yes Yes No (9) QT PS No (10) mQRW mVQP No (11) PQTS is a trapezoid. No Exercises Writing About Mathematics 1. Is the statement “A triangle consists of three line segments” a good definition? Justify your answer. 2. Explain the difference between the legs of an isosceles triangle and the legs of a right triangle. Developing Skills In 3 and 4, name the legs and the hypotenuse in each right triangle shown. 3. C 4. L K A B J 14365C01.pgs 7/9/07 4:37 PM Page 28 28 Essentials of Geometry In 5 and 6, name the legs, the base, the vertex angle, and the base angles in each isosceles triangle shown. 5. N 6. T L M R S In 7–10, sketch, using pencil and paper or geometry software, each of the following. Mark congruent sides and right angles with the appropriate symbols. 7. An acute triangle that is isosceles 8. An obtuse triangle that is isosceles 9. A right triangle that is isosceles 11. The vertex angle of an isosceles triangle is ABC. Name the base angles of this triangle. 12. In DEF, 13. In RST, S is included between which two sides? is included between which two angles? 10. An obtuse triangle that is scalene DE 14. a. Name three things that can be assumed from the diagram to the right. b. Name three things that can not be assumed from the diagram to the right. Applying Skills D G A F C E B 15. The degree measures of the acute angles of an obtuse triangle can be represented by 5a 12 and 3a 2. If the sum of the measures of the acute angles is 82, find the measure of each of the acute angles. 16. The measures of the sides of an equilateral triangle are represented by x + 12, 3x 8, and 2x 2. What is the measure of each side of the triangle? 17. The lengths of the sides of an isosceles triangle are represented by 2x 5, 2x 6, and 3x 3. What are the lengths of the sides of the triangle? (Hint: the length of a line segment is a positive quantity.) 18. The measures of the sides of an isosceles triangle are represented by x 5, 3x 13, and 4x 11. What are the measures of each side of the triangle? Two answers are possible. 14365C01.pgs 7/9/07 4:37 PM Page 29 Chapter Summary 29 CHAPTER SUMMARY Undefined Terms • set, point, line, plane Definitions to Know • A collinear set of points is a set of points all of which lie on the same straight line. • A noncollinear set of points is a set of three or more points that do not all lie on the same straight line. • The distance between two points on the real number line is the absolute value of the difference of the coordinates of the two points. • B is between A and C if and only if A, B, and C are distinct collinear points and AB BC AC. • A line segment, or a segment, is a set of points consisting of two points on a line, called endpoints, and all of the points on the line between the endpoints. • The length or measure of a line segment is the distance between its end- points. • Congruent segments are segments that have the same measure. • The midpoint of a line segment is a point of that line segment that divides the segment into two congruent segments. • The bisector of a line segment is any line, or subset of a line, that intersects the segment at its midpoint. • A line segment, RS , is the sum of two line segments, RP and PS , if P is between R and S. • Two points, A and B, are on one side of a point P if A, B, and P are collinear and P is not between A and B. • A half-line is a set of points on one side of a point. • A ray is a part of a line that consists of a point on the line, called an end- point, and all the points on one side of the endpoint. • Opposite rays are two rays of the same line with a common endpoint and no other point in common. • An angle is a set of points that is the union of two rays having the same endpoint. • A straight angle is an angle that is the union of opposite rays and whose degree measure is 180. • An acute angle is an angle whose degree measure is greater than 0 and less than 90. • A right angle is an angle whose degree measure is 90. • An obtuse angle is an angle whose degree measure is greater than 90 and less than 180. 14365C01.pgs 7/9/07 4:37 PM Page 30 30 Essentials of Geometry P R S T Properties of the Real Number System • Congruent angles are angles that have the same measure. • A bisector of an angle is a ray whose endpoint is the vertex of the angle, and that divides that angle into two congruent angles. • Perpendicular lines are two lines that intersect to form right angles. • The distance from a point to a line is the length of the perpendicular from the point to the line. • If point P is a point in the interior of RST and RST is not a straight angle, or if P is any point not on straight angle RST, then RST is the sum of two angles, RSP and PST. • A polygon is a closed figure in a plane that is the union of line segments such that the segments intersect only at their endpoints and no segments sharing a common endpoint are collinear. • A triangle is a polygon that has exactly three sides. • A scalene triangle is a triangle that has no congruent sides. • An isosceles triangle is a triangle that has two congruent sides. • An equilateral triangle is a triangle that has three congruent sides. • An acute triangle is a triangle that has three acute angles. • A right triangle is a triangle that has a right angle. • An obtuse triangle is a triangle that has an obtuse angle. • An equiangular triangle is a triangle that has three congruent angles. Addition Multiplication Closure Commutative Property a b b a Associative Property a b is a real number. a b is a real number. a b b a (a b) c a (b c) a (b c) (a b) c a 0 a and 0 a a a 1 a and 1 a a a 0, a (a) 0 a ? and ab ac a (b c) 1 a 5 1 Distributive Property a(b c) ab ac Identity Property Inverse Property Multiplication Property ab 0 if and only if a 0 or b 0 of Zero VOCABULARY 1-1 Undefined term • Set • Point • Line • Straight line • Plane 1-2 Number line • Coordinate • Graph • Numerical operation • Closure property of addition • Closure property of multiplication • Commutative 14365C01.pgs 7/9/07 4:37 PM Page 31 Review Exercises 31 property of addition • Commutative property of multiplication • Associative property of addition • Associative property of multiplication • Additive identity • Multiplicative identity • Additive inverses • Multiplicative inverses • Distributive property • Multiplication property of zero • 1-3 Definition • Collinear set of points •
Noncollinear set of points • Distance between two points on the real number line • Betweenness • Line segment • Segment • Length (Measure) of a line segment • Congruent segments 1-4 Midpoint of a line segment • Bisector of a line segment • Sum of two line segments 1-5 Half-line • Ray • Endpoint • Opposite rays • Angle • Sides of an angle • Vertex • Straight angle • Interior of an angle • Exterior of an angle • Degree • Acute angle • Right angle • Obtuse angle • Straight angle 1-6 Congruent angles • Bisector of an angle • Perpendicular lines • Distance from a point to a line • Foot • Sum of two angles 1-7 Polygon • Side • Triangle • Included side • Included angle • Opposite side • Opposite angle • Scalene triangle • Isosceles triangle • Equilateral triangle • Legs • Base • Vertex angle • Base angles • Acute triangle • Right triangle • Obtuse triangle • Equiangular triangle • Hypotenuse REVIEW EXERCISES 1. Name four undefined terms. 2. Explain why “A line is like the edge of a ruler” is not a good definition. In 3–10, write the term that is being defined. 3. The set of points all of which lie on a line. 4. The absolute value of the difference of the coordinates of two points. 5. A polygon that has exactly three sides. 6. Any line or subset of a line that intersects a line segment at its midpoint. 7. Two rays of the same line with a common endpoint and no other points in common. 8. Angles that have the same measure. 9. A triangle that has two congruent sides. 10. A set of points consisting of two points on a line and all points on the line between these two points. 11. In right triangle LMN, mM 90. Which side of the triangle is the hypotenuse? 14365C01.pgs 7/9/07 4:37 PM Page 32 32 Essentials of Geometry 12. In isosceles triangle RST, RS ST. a. Which side of the triangle is the base? b. Which angle is the vertex angle? 13. Points D, E, and F lie on a line. The coordinate of D is 3, the coordinate of E is 1, and the coordinate of F is 9. a. Find DE, EF, and DF. b. Find the coordinate of M, the midpoint of DF . g AB c. intersects EF at C. If g AB is a bisector of EF , what is the coordinate of C? 14. Explain why a line segment has only one midpoint but can have many bisectors. In 15–18, use the figure shown. In polygon ABCD, the midpoint of BD . AC and BD intersect at E, 15. Name two straight angles. 16. What angle is the sum of ADE and EDC? 17. Name two congruent segments. 18. What line segment is the sum of AE and EC ? B A E C In 19–22, use the figure shown. AB ' BC and h BD bisects ABC. 19. Name two congruent angles. 20. What is the measure of ABD? 21. Name a pair of angles the sum of whose measures is 180. 22. Does AB BC AC? Justify your answer. D A B D C Exploration Euclidean geometry, which we have been studying in this chapter, focuses on the plane. Non-Euclidian geometry focuses on surfaces other than the plane. For instance, spherical geometry focuses on the sphere. In this Exploration, you will draw on a spherical object, such as a grapefruit or a Styrofoam ball, to relate terms from Euclidean geometry to spherical geometry. 14365C01.pgs 7/9/07 4:37 PM Page 33 Review Exercises 33 1. Draw a point on your sphere. How does this point compare to a point in Euclidean geometry? 2. The shortest distance between two points is called a geodesic. In Euclidean geometry, a line is a geodesic. Draw a second point on your sphere. Connect the points with a geodesic and extend it as far as possible. How does this geodesic compare to a line in Euclidean geometry? 3. Draw a second pair of points and a second geodesic joining them on your sphere. The intersection of the geodesics forms angles. How do these angles compare to an angle in Euclidean geometry? 4. Draw a third pair of points and a third geodesic joining them on your sphere. This will form triangles, which by definition are polygons with exactly three sides. How does a triangle on a sphere compare to a triangle in Euclidean geometry? Consider both sides and angles. 14365C02.pgs 7/9/07 4:40 PM Page 34 CHAPTER 2 CHAPTER TABLE OF CONTENTS 2-1 Sentences, Statements, and Truth Values 2-2 Conjunctions 2-3 Disjunctions 2-4 Conditionals 2-5 Inverses, Converses, and Contrapositives 2-6 Biconditionals 2-7 The Laws of Logic 2-8 Drawing Conclusions Chapter Summary Vocabulary Review Exercises Cumulative Review 34 LOGIC Mathematicians throughout the centuries have used logic as the foundation of their understanding of the relationships among mathematical ideas. In the late 17th century, Gottfried Leibniz (1646–1716) organized logical discussion into a systematic form, but he was ahead of the mathematical thinking of his time and the value of his work on logic was not recognized. It was not until the 19th century that George Boole (1815–1864), the son of an English shopkeeper, developed logic in a mathematical context, representing sentences with symbols and carefully organizing the possible relationships among those sentences. Boole corresponded with Augustus DeMorgan (1806–1871) with whom he shared his work on logic. Two important relationships of logic are known today as DeMorgan’s Laws. not (p and q) (not p) or (not q) not (p or q) (not p) and (not q) Boolean algebra is key in the development of com- puter science and circuit design. 14365C02.pgs 7/9/07 4:40 PM Page 35 2-1 SENTENCES, STATEMENTS, AND TRUTH VALUES Sentences, Statements, and Truth Values 35 Logic is the science of reasoning. The principles of logic allow us to determine if a statement is true, false, or uncertain on the basis of the truth of related statements. We solve problems and draw conclusions by reasoning from what we know to be true. All reasoning, whether in mathematics or in everyday living, is based on the ways in which we put sentences together. Sentences and Their Truth Values When we can determine that a statement is true or that it is false, that statement is said to have a truth value. In this chapter we will study the ways in which statements with known truth values can be combined by the laws of logic to determine the truth value of other statements. In the study of logic, we use simple declarative statements that state a fact. That fact may be either true or false. We call these statements mathematical sentences. For example: 1. Congruent angles are angles that True mathematical sentence have the same measure. 2. 17 5 12 3. The Brooklyn Bridge is in New York. 4. 17 3 42 True mathematical sentence True mathematical sentence False mathematical sentence 5. The Brooklyn Bridge is in California. False mathematical sentence Nonmathematical Sentences and Phrases Sentences that do not state a fact, such as questions, commands, or exclamations, are not sentences that we use in the study of logic. For example: 1. Did you have soccer practice today? This is not a mathematical sentence because it asks a question. 2. Go to your room. This is not a mathematical sentence because it gives a command. An incomplete sentence or part of a sentence, called a phrase, is not a math- ematical sentence and usually has no truth value. For example: 1. Every parallelogram This is not a mathematical sentence. 2. 19 2 This is not a mathematical sentence. 14365C02.pgs 7/9/07 4:40 PM Page 36 36 Logic Some sentences are true for some persons and false for others. For example: 1. I enjoy reading historical novels. 2. Summer is the most pleasant season. 3. Basketball is my favorite sport. Conclusions based on sentences such as these do not have the same truth value for all persons. We will not use sentences such as these in the study of logic. Open Sentences In the study of algebra, we worked with open sentences, that is, sentences that contain a variable. The truth value of the open sentence depended on the value of the variable. For example, the open sentence x 2 5 is true when x 3 and false for all other values of x. In some sentences, a pronoun, such as he, she, or it, acts like a variable and the name that replaces the pronoun determines the truth value of the sentence. 1. x 2 8 Open sentence: the variable is x. 2. He broke my piggybank. Open sentence: the variable is he. 3. Jenny found it behind the sofa. Open sentence: the variable is it. In previous courses, we learned that the domain or replacement set is the set of all elements that are possible replacements for the variable. The element or elements from the domain that make the open sentence true is the solution set or truth set. For instance: Open sentence: 14 x 9 Variable: x Domain: {1, 2, 3, 4, 5} Solution set: {5} When x 5, then 14 5 9 is a true sentence. The method we use for sentences in mathematics is the same method we apply to sentences in ordinary conversation. Of course, we would not use a domain like {1, 2, 3, 4} for the open sentence “It is the third month of the year.” Common sense tells us to use a domain consisting of the names of months. The following example compares this open sentence with the algebraic sentence used above. Open sentences, variables, domains, and solution sets behave in exactly the same way. Open sentence: It is the third month of the year. Variable: It Domain: {Names of months} Solution set: {March} When “It” is replaced by “March,” then “March is the third month of the year” is a true sentence. 14365C02.pgs 7/9/07 4:40 PM Page 37 Sentences, Statements, and Truth Values 37 Sometimes a solution set contains more than one element. If Elaine has two brothers, Ken and Kurt, then for her, the sentence “He is my brother” has the solution set {Ken, Kurt}. Here the domain is the set of boys’ names. Some people have no brothers. For them, the solution set for the open sentence “He is my brother” is the empty set, ∅ or { }. Identify each of the following as a true sentence, a false sentence, an open sentence, or not a mathematical sentence at all. Answers a. Football is a water sport. False sentence b. Football is a team sport. True sentence c. He is a football player. Open sentence: the variable is he. d. Do you like football? Not a math
ematical sentence: this is a question. e. Read this book. f. 3x 7 11 g. 3x 7 Not a mathematical sentence: this is a command. Open sentence: the variable is x. Not a mathematical sentence: this is a phrase or a binomial. EXAMPLE 1 EXAMPLE 2 Use the replacement set 2, tence “It is an irrational number.” , 2.5, U 2p 3 2 2 " V to find the truth set of the open sen- Solution Both 2 and 2.5 are rational numbers. 2 Both p and rational number and an irrational number is an irrational number, are irrational. are irrational numbers. Since the product or quotient of a and 2p 3 " 2 2 " Answer 2p 3 , 2 U 2 V " Statements and Symbols A sentence that can be judged to be true or false is called a statement or a closed sentence. In a statement, there are no variables. A closed sentence is said to have a truth value. The truth values, true and false, are indicated by the symbols T and F. 14365C02.pgs 7/9/07 4:40 PM Page 38 38 Logic Negations In the study of logic, you will learn how to make new statements based upon statements that you already know. One of the simplest forms of this type of reasoning is negating a statement. The negation of a statement always has the opposite truth value of the given or original statement and is usually formed by adding the word not to the given statement. For example: 1. Statement: Neil Armstrong walked on the moon. Negation: Neil Armstrong did not walk on the moon. 2. Statement: A duck is a mammal. Negation: A duck is not a mammal. (True) (False) (False) (True) There are other ways to insert the word not into a statement to form its negation. One method starts the negation with the phrase “It is not true that . . .” For example: 3. Statement: A carpenter works with wood. Negation: Negation: A carpenter does not work with wood. It is not true that a carpenter works with wood. (True) (False) (False) Both negations express the same false statement. Logic Symbols The basic element of logic is a simple declarative sentence. We represent this basic element by a single, lowercase letter. Although any letter can be used to represent a sentence, the use of p, q, r, and s are the most common. For example, p might represent “Neil Armstrong walked on the moon.” The symbol that is used to represent the negation of a statement is the symbol placed before the letter that represents the given statement. Thus, if p represents “Neil Armstrong walked on the moon,” then p represents “Neil Armstrong did not walk on the moon.” The symbol p is read “not p.” Symbol Statement in Words Truth Value p p q q There are 7 days in a week. There are not 7 days in a week. 8 9 16 8 9 16 True False False True When p is true, then its negation p is false. When q is false, then its nega- tion q is true. A statement and its negation have opposite truth values. 14365C02.pgs 7/9/07 4:40 PM Page 39 Sentences, Statements, and Truth Values 39 It is possible to use more than one negation in a sentence. Each time another negation is included, the truth value of the statement will change. For example: Symbol r r (r) Statement in Words Truth Value A dime is worth 10 cents. A dime is not worth 10 cents. It is not true that a dime is not worth 10 cents. True False True We do not usually use sentences like the third. Note that just as in the set of real numbers, (a) a, (r) always has the same truth value as r. We can use r in place of (r). Therefore, we can negate a sentence that contains the word not by omitting that word. r: A dime is not worth 10 cents. (r): A dime is worth 10 cents. The relationship between a statement p and its negation p can be summarized in the table at the right. When p is true, p is false. When p is false, p is true. p T F p F T In this example, symbols are used to represent statements. The truth value of each statement is given. k: Oatmeal is a cereal. m: Massachusetts is a city. (True) (False) For each sentence given in symbolic form: a. Write a complete sentence in words to show what the symbols represent. b. Tell whether the statement is true or false. (1) k (2) m Answers a. Oatmeal is not a cereal. a. Massachusetts is not a city. b. False b. True EXAMPLE 3 Exercises Writing About Mathematics 1. Explain the difference between the use of the term “sentence” in the study of grammar and in the study of logic. 14365C02.pgs 7/9/07 4:40 PM Page 40 40 Logic 2. a. Give an example of a statement that is true on some days and false on others. b. Give an example of a statement that is true for some people and false for others. c. Give an example of a statement that is true in some parts of the world and false in others. Developing Skills In 3–10, tell whether or not each of the following is a mathematical sentence. 3. Thanksgiving is on the fourth Thursday 4. Albuquerque is a city in New Mexico. in November. 5. Where did you go? 7. Be quiet. 9. y 7 3y 4 6. Twenty sit-ups, 4 times a week 8. If Patrick leaps 10. Tie your shoe. In 11–18, all of the sentences are open sentences. Find the variable in each sentence. 11. She is tall. 13. 2y 17 12. We can vote at age 18. 14. 14x 8 9 15. This country has the third largest population. 16. He hit a home run in the World Series. 17. It is my favorite food. 18. It is a fraction. In 19–26: a. Tell whether each sentence is true, false, or open. b. If the sentence is an open sentence, identify the variable. 19. The Statue of Liberty was given to the United States by France. 20. They gained custody of the Panama Canal on December 31, 1999. 21. Tallahassee is a city in Montana. 23. 6x 4 16 25. 6(2) 4 16 22. A pentagon is a five-sided polygon. 24. 6(10) 4 16 26. 23 32 In 27–31, find the truth set for each open sentence using the replacement set {Nevada, Illinois, Massachusetts, Alaska, New York}. 27. Its capital is Albany. 28. It does not border on or touch an ocean. 29. It is on the east coast of the United States. 30. It is one of the states of the United States of America. 31. It is one of the last two states admitted to the United States of America. 14365C02.pgs 7/9/07 4:40 PM Page 41 Sentences, Statements, and Truth Values 41 In 32–39, use the domain {square, triangle, rectangle, parallelogram, rhombus, trapezoid} to find the truth set for each open sentence. 32. It has three and only three sides. 33. It has exactly six sides. 34. It has fewer than four sides. 35. It contains only right angles. 36. It has four sides that are all equal in measure. 37. It has two pairs of opposite sides that are parallel. 38. It has exactly one pair of opposite sides 39. It has interior angles with measures that are parallel. whose sum is 360 degrees. In 40–47, write the negation of each sentence. 40. The school has an auditorium. 41. A stop sign is painted red. 42. The measure of an obtuse angle is greater than 90°. 44. Michigan is not a city. 46. 3 4 5 6 43. There are 1,760 yards in a mile. 45. 14 2 16 12 47. Today is not Wednesday. In 48–56, for each given sentence: a. Write the sentence in symbolic form using the symbols shown below. b. Tell whether the sentence is true, false, or open. Let p represent “A snake is a reptile.” Let q represent “A frog is a snake.” Let r represent “Her snake is green.” 48. A snake is a reptile. 49. A snake is not a reptile. 50. A frog is a snake. 51. A frog is not a snake. 52. Her snake is green. 53. Her snake is not green. 54. It is not true that a frog is a snake. 55. It is not the case that a snake is not a reptile. 56. It is not the case that a frog is not a snake. In 57–64, the symbols represent sentences. p: Summer follows spring. q: August is a summer month. r: A year has 12 months. s: She likes spring. For each sentence given in symbolic form: a. Write a complete sentence in words to show what the symbols represent. b. Tell whether the sentence is true, false, or open. 57. p 61. (p) 58. q 62. (q) 59. r 63. (r) 60. s 64. (s) 14365C02.pgs 7/9/07 4:40 PM Page 42 42 Logic 2-2 CONJUNCTIONS We have identified simple sentences that have a truth value. Often we wish to use a connecting word to form a compound sentence. In mathematics, sentences formed by connectives are also called compound sentences or compound statements. One of the simplest compound statements that can be formed uses the connective and. In logic, a conjunction is a compound statement formed by combining two simple statements using the word and. Each of the simple statements is called a conjunct. When p and q represent simple statements, the conjunction p and q is written in symbols as p ∧ q. For example: p: A dog is an animal. q: A sparrow is a bird. p ∧ q: A dog is an animal and a sparrow is a bird. This compound sentence is true because both parts are true. “A dog is an animal is true and “A sparrow is a bird” is true. When one or both parts of a conjunction are false, the conjunction is false. For example, • “A dog is an animal and a sparrow is not a bird” is false because “A spar- row is not a bird” is the negation of a true statement and is false. • “A dog is not an animal and a sparrow is a bird” is false because “A dog is not an animal” is the negation of a true statement and is false. • “A dog is not an animal and a sparrow is not a bird” is false because both “A dog is not an animal” and “A sparrow is not a bird” are false. We can draw a diagram, called a tree diagram, to show all possible combinations of the truth values of p and q that are combined to make the compound statement p ∧ q. p is true p is false q is true p is true and q is true. q is false p is true and q is false. q is true p is false and q is true. q is false p is false and q is false. These four possible combinations of the truth values of p and q can be displayed in a chart called a truth table. The truth table can be used to show the possible truth values of a compound statement that is made up of two simple statements. For instance, write a truth table for p ∧ q. 14365C02.pgs 7/9/07 4:40 PM Page 43 STEP 1. In the first column, we list the truth values of p. For each possible truth value of p, there are two possible truth v
alues for q. Therefore, we list T twice and F twice. STEP 2. In the second column, we list the truth values of q. In the two rows in which p is true, list q as true in one and false in the other. In the two rows in which p is false, list q as true in one and false in the other. STEP 3. In the last column, list the truth values of the conjunction, p ∧ q. The conjunction is true only when both p and q are true. The conjunction is false when one or both conjuncts are false. Conjunctions 43 The conjunction, p and q, is true only when both parts are true: p must be true and q must be true. For example, let p represent “It is spring,” and let q represent “It is March.” CASE 1 Both p and q are true. On March 30, “It is spring” is true and “It is March” is true. Therefore, “It is spring and it is March” is true. CASE 2 p is true and q is false. On April 30, “It is spring” is true and “It is March” is false. Therefore, “It is spring and it is March” is false. CASE 3 p is false and q is true. On March 10, “It is spring” is false and “It is March” is true. Therefore, “It is spring and it is March” is false. CASE 4 Both p and q are false. On February 28, “It is spring” is false and “It is March” is false. Therefore, “It is spring and it is March” is false. 14365C02.pgs 7/9/07 4:40 PM Page 44 44 Logic A compound sentence may contain both negations and conjunctions at the same time. For example: Let p represent “Ten is divisible by 2.” Let q represent “Ten is divisible by 3.” Then p ∧ q represents “Ten is divisible by 2 and ten is not divisible by 3.” Here p is true, q is false, and q is true. Then for p ∧ q, both parts are true so the conjunction is true. This can be summarized in the following table EXAMPLE 1 Let p represent “Albany is the capital of New York State.” (True) Let q represent “Philadelphia is the capital of Pennsylvania.” (False) For each given sentence: a. Write the sentence in symbolic b. Tell whether the statement is form. true or false. (1) Albany is the capital of New York State and Philadelphia is the capital of Pennsylvania. (2) Albany is the capital of New York State and Philadelphia is not the capital of Pennsylvania. (3) Albany is not the capital of New York State and Philadelphia is the capital of Pennsylvania. (4) Albany is not the capital of New York State and Philadelphia is not the capital of Pennsylvania. (5) It is not true that Albany is the capital of New York State and Philadelphia is the capital of Pennsylvania. Solution (1) The statement is a conjunction. Since p is true and q is false, the statement is false. (2) The statement is a conjunction. Since p is true and q is true, the statement is true Answers: a. p ∧ q b. False Answers: a. p ∧ q b. True 14365C02.pgs 7/9/07 4:40 PM Page 45 Conjunctions 45 (3) The statement is a conjunction. Since p is false and q is false, the statement is false. (4) The statement is a conjunction. Since p is false and q is true, the statement is false Answers: a. p ∧ q b. False Answers: a. p ∧ q b. False (5) The phrase “It is not true that” applies to the entire conjunction. Since p is true and q is false, (p ∧ q) is false and the negation of (p ∧ q) is true. p T q F p ∧ q (p ∧ q) F T Answers: a. (p ∧ q) b. True EXAMPLE 2 Solution Use the domain {1, 2, 3, 4} to find the truth set for the open sentence (x 3) ∧ (x is a prime) Let x 1 Let x 2 Let x 3 Let x 4 (1 3) ∧ (1 is a prime) (2 3) ∧ (2 is a prime) (3 3) ∧ (3 is a prime) (4 3) ∧ (4 is a prime) T ∧ F False T ∧ T True F ∧ T False F ∧ F False A conjunction is true only when both simple sentences are true. This condition is met here when x 2. Thus, the truth set or solution set is {2}. Answer {2} EXAMPLE 3 Three sentences are written below. The truth values are given for the first two sentences. Determine whether the third sentence is true, is false, or has an uncertain truth value. Today is Friday and I have soccer practice. Today is Friday. I have soccer practice. (False) (True) (?) Solution Since the conjunction is false, at least one of the conjuncts must be false. But “Today is Friday” is true. Therefore, “I have soccer practice” must be false. Answer “I have soccer practice” is false. 14365C02.pgs 7/9/07 4:40 PM Page 46 46 Logic EXAMPLE 4 Three sentences are written below. The truth values are given for the first two sentences. Determine whether the third sentence is true, is false, or has an uncertain truth value. Today is Monday and the sun is shining. Today is Monday. The sun is shining. (False) (False) (?) Solution (1) Use symbols to represent the sentences. Indicate their truth values. p: Today is Monday. q: The sun is shining. p ∧ q: Today is Monday and the sun is shining. (False) (?) (False) (2) Construct a truth for the conjunction. Study the truth values of p and p ∧ q. Since p ∧ q is false, the last three rows apply. Since p is false, the choices are narrowed to the last two rows. Therefore q could be either true or false Answer The truth value of “The sun is shining” is uncertain. Exercises Writing About Mathematics 1. Is the negation of a conjunction, (p ∧ q), the same as p ∧ q? Justify your answer. 2. What must be the truth values of p, q, and r in order for (p ∧ q) ∧ r to be true? Explain your answer. Developing Skills In 3–12, write each sentence in symbolic form, using the given symbols. Let p represent “It is hot.” Let q represent “It is raining.” Let r represent “The sky is cloudy.” 3. It is hot and it is raining. 4. It is hot and the sky is cloudy. 5. It is not hot. 6. It is not hot and the sky is cloudy. 7. It is raining and the sky is not cloudy. 8. It is not hot and it is not raining. 14365C02.pgs 7/9/07 4:40 PM Page 47 9. The sky is not cloudy and it is not hot. 10. The sky is not cloudy and it is hot. 11. It is not the case that it is hot and it is raining. 12. It is not the case that it is raining and it is not hot. In 13–20, using the truth value for each given statement, tell if the conjunction is true or false. Conjunctions 47 A piano is a percussion instrument. A piano has 88 keys. A flute is a percussion instrument. A trumpet is a brass instrument. (True) (True) (False) (True) 13. A flute is a percussion instrument and a piano is a percussion instrument. 14. A flute is a percussion instrument and a trumpet is a brass instrument. 15. A piano has 88 keys and is a percussion instrument. 16. A piano has 88 keys and a trumpet is a brass instrument. 17. A piano is not a percussion instrument and a piano does not have 88 keys. 18. A trumpet is not a brass instrument and a piano is a percussion instrument. 19. A flute is not a percussion instrument and a trumpet is a brass instrument. 20. It is not true that a piano is a percussion instrument and has 88 keys. In 21–28, complete each sentence with “true” or “false” to make a correct statement. 21. When p is true and q is true, then p ∧ q is _______. 22. When p is false, then p ∧ q is _______. 23. If p is true, or q is true, but not both, then p ∧ q is _______. 24. When p ∧ q is true, then p is ______ and q is ______. 25. When p ∧ q is true, then p is ______ and q is ______. 26. When p ∧ q is true, then p is ______ and q is ______. 27. When p is false and q is true, then (p ∧ q) is _______. 28. If both p and q are false, then p ∧ q is _______. Applying Skills In 29–36, three sentences are written. The truth values are given for the first two sentences. Determine whether the third sentence is true, is false, or has an uncertain truth value. 29. It is noon and I get lunch. (True) 30. I have the hiccups and I drink some It is noon. (True) I get lunch. (?) water. (False) I have the hiccups. (True) I drink some water. (?) 14365C02.pgs 7/9/07 4:40 PM Page 48 48 Logic 31. I have the hiccups and I drink some 32. I play tennis and Anna plays golf. water. (False) I have the hiccups. (False) I drink some water. (?) (False) I play tennis. (True) Anna plays golf. (?) 33. Pam sees a movie and Pam loves going 34. Jon and Edith like to eat ice cream. to the theater. (True) Pam sees a movie. (True) Pam loves going to the theater. (?) (False) Jon likes to eat ice cream. (True) Edith likes to eat ice cream. (?) 35. Jordan builds model trains and model 36. Bethany likes to chat and surf on the planes. (False) Jordan builds model trains. (False) Jordan builds model planes. (?) internet. (False) Bethany likes to surf on the internet. (True) Bethany likes to chat on the internet. (?) In 37 and 38, a compound sentence is given using a conjunction. Use the truth value of the compound sentence to determine whether each sentence that follows is true or false. 37. In winter I wear a hat and scarf. (True) 38. I do not practice and I know that I a. In winter I wear a hat. b. In winter I wear a scarf. c. In winter I do not wear a hat. should. (True) a. I do not practice. b. I know that I should practice. c. I practice. 2-3 DISJUNCTIONS In logic, a disjunction is a compound statement formed by combining two simple statements using the word or. Each of the simple statements is called a disjunct. When p and q represent simple statements, the disjunction p or q is written in symbols as p ∨ q. For example: p: Andy rides his bicycle to school. q: Andy walks to school. p ∨ q: Andy rides his bicycle to school or Andy walks to school. In this example, when is the disjunction p ∨ q true and when is it false? 1. On Monday, Andy rode his bicycle part of the way to school when he met a friend. Then he walked the rest of the way to school with his friend. Here p is true and q is true. The disjunction p ∨ q is true. 2. On Tuesday, Andy rode his bicycle to school and did not walk to school. Here p is true and q is false. The disjunction p ∨ q is true. 3. On Wednesday, Andy did not ride his bicycle to school and walked to school. Here p is false and q is true. The disjunction p ∨ q is true. 14365C02.pgs 7/9/07 4:40 PM Page 49 Disjunctions 49 4. On Thursday, it rained so Andy’s father drove him to school. Andy did not ride his bicycle to school and did not walk to school. Here p is
false and q is false. The disjunction p ∨ q is false. The disjunction p or q is true when any part of the compound sentence is true: p is true, q is true, or both p and q are true. The only case in which the disjunction p or qis false is when both p and q are false. The truth values of the disjunction p ∨ q are summarized in the truth table to the right. The possible combinations of the truth values of p and of q, shown in the first two columns, are the same as those used in the truth table of the conjunction. The third column gives the truth values for the disjunction, p ∨ q EXAMPLE 1 Use the following statements: Let k represent “Kurt plays baseball.” Let a represent “Alicia plays baseball.” Let n represent “Nathan plays soccer.” Write each given sentence in symbolic form. a. Kurt or Alicia play baseball. b. Kurt plays baseball or Nathan plays soccer. c. Alicia plays baseball or Alicia does not play baseball. d. It is not true that Kurt or Alicia play baseball. e. Either Kurt does not play baseball or Alicia does not play baseball. f. It’s not the case that Alicia or Kurt play baseball. Answers k ∨ a) k ∨ a (a ∨ k) EXAMPLE 2 Symbols are used to represent three statements. For each statement, the truth value is noted. k: “Every line segment has a midpoint.” m: “A line has a midpoint.” q: “A ray has one endpoint.” (True) (False) (True) For each sentence given in symbolic form: a. Write a complete sentence in words to show what the symbols represent. b. Tell whether the statement is true or false. 14365C02.pgs 7/9/07 4:40 PM Page 50 50 Logic (1) k ∨ q (2) k ∨ m Answers a. Every line segment has a midpoint or every ray has one endpoint. b. T ∨ T is a true disjunction. a. Every line segment has a midpoint or a line has a midpoint. b. T ∨ F is a true disjunction. (3) m ∨ q a. A line has a midpoint or a ray does not have one endpoint. b. F ∨ T F ∨ F, a false disjunction. (4) (m ∨ q) a. It is not the case that a line has a midpoint or a ray has one endpoint. b. (F ∨ T) T, a false disjunction. EXAMPLE 3 Find the solution set of each of the following if the domain is the set of positive integers less than 8. a. (x 4) ∨ (x 3) b. (x 3) ∨ (x is odd) c. (x 5) ∧ (x 3) Solution The domain is {1, 2, 3, 4, 5, 6, 7}. a. The solution set of x 4 is {1, 2, 3} and the solution set of x 3 is {4, 5, 6, 7}. The solution set of the disjunction (x 4) ∨ (x 3) includes all the numbers that make x 4 true together with all the numbers that make x 3 true. Answer {1, 2, 3, 4, 5, 6, 7} Note: The solution set of the disjunction (x 4) ∨ (x 3) is the union of the solution sets of the individual parts: {1, 2, 3} {4, 5, 6, 7} {1, 2, 3, 4, 5, 6, 7}. b. The solution set of (x 3) is {4, 5, 6, 7} and the solution set of (x is odd) is {1, 3, 5, 7}. The solution set of the disjunction (x 3) ∨ (x is odd) includes all the numbers that make either x 3 true or x is odd true. Answer {1, 3, 4, 5, 6, 7} Note: The solution set of the disjunction (x 3) ∨ (x is odd) is the union of the solution sets: {4, 5, 6, 7} {1, 3, 5, 7} {1, 3, 4, 5, 6, 7}. c. The solution set of (x 5) is {6, 7} and the solution set of (x 3) is {1, 2}. The solution set of (x 5) ∧ (x 3) is {6, 7} {1, 2} or the empty set, . Answer 14365C02.pgs 7/9/07 4:40 PM Page 51 Disjunctions 51 Two Uses of the Word Or When we use the word or to mean that one or both of the simple sentences are true, we call this the inclusive or. The truth table we have just shown uses truth values for the inclusive or. Sometimes, however, the word or is used in a different way, as in “He is in grade 9 or he is in grade 10.” Here it is not possible for both simple sentences to be true at the same time. When we use the word or to mean that one and only one of the simple sentences is true, we call this the exclusive or. The truth table for the exclusive or will be different from the table shown for disjunction. In the exclusive or, the disjunction p or qwill be true when p is true, or when q is true, but not both. In everyday conversation, it is often evident from the context which of these uses of or is intended. In legal documents or when ambiguity can cause difficulties, the inclusive or is sometimes written as and/or. We will use only the inclusive or in this book. Whenever we speak of disjunction, p or q will be true when p is true, when q is true, when both p and q are true. Exercises Writing About Mathematics 1. Explain the relationship between the truth set of the negation of a statement and the com- plement of a set. 2. Explain the difference between the inclusive or and the exclusive or. Developing Skills In 3–12, for each given statement: a. Write the statement in symbolic form, using the symbols given below. b. Tell whether the statement is true or false. Let c represent “A gram is 100 centigrams.” Let m represent “A gram is 1,000 milligrams.” Let k represent “A kilogram is 1,000 grams.” Let l represent “A gram is a measure of length.” (True) (True) (True) (False) 3. A gram is 1,000 milligrams or a kilogram is 1,000 grams. 4. A gram is 100 centigrams or a gram is a measure of length. 5. A gram is 100 centigrams or 1,000 milligrams. 6. A kilogram is not 1,000 grams or a gram is not 100 centigrams. 7. A gram is a measure of length or a kilogram is 1,000 grams. 14365C02.pgs 7/9/07 4:40 PM Page 52 52 Logic 8. A gram is a measure of length and a gram is 100 centigrams. 9. It is not the case that a gram is 100 centigrams or 1,000 milligrams. 10. It is false that a kilogram is not 1,000 grams or a gram is a measure of length. 11. A gram is 100 centigrams and a kilogram is 1,000 grams. 12. A gram is not 100 centigrams or is not 1,000 milligrams, and a gram is a measure of length. In 13–20, symbols are assigned to represent sentences. Let b represent “Breakfast is a meal.” Let s represent “Spring is a season.” Let h represent “Halloween is a season.” For each sentence given in symbolic form: a. Write a complete sentence in words to show what the symbols represent. b. Tell whether the sentence is true or false. 13. s ∨ h 17. b ∨ s 14. b ∧ s 18. (s ∧ h) 15. s ∨ h 19. (b ∨ s) 16. b ∧ h 20. b ∧ s In 21–27, complete each sentence with the words “true” or “false” to make a correct statement. 21. When p is true, then p ∨ q is ______. 22. When q is true, then p ∨ q is ______. 23. When p is false and q is false, then p ∨ q is _______. 24. When p ∨ q is false, then p is _______ and q is _______. 25. When p ∨ q is false, then p is _______ and q is _______. 26. When p is false and q is true, then (p ∨ q) is _______. 27. When p is false and q is true, then p ∨ q is ______. Applying Skills In 28–32, three sentences are written. The truth values are given for the first two sentences. Determine whether the third sentence is true, is false, or has an uncertain truth value. 28. May is the first month of the year. (False) January is the first month of the year. (True) May is the first month of the year or January is the first month of the year. (?) 29. I will study more or I will fail the course. (True) I will fail the course. (False) I will study more. (?) 30. Jen likes to play baseball and Mason likes to play baseball. (False) Mason likes to play baseball. (True) Jen likes to play baseball. (?) 14365C02.pgs 7/9/07 4:40 PM Page 53 Conditionals 53 31. Nicolette is my friend or Michelle is my friend. (True) Nicolette is my friend. (True) Michelle is my friend. (?) 32. I practice the cello on Monday or I practice the piano on Monday. (True) I do not practice the piano on Monday. (False) I practice the cello on Monday. (?) 2-4 CONDITIONALS A sentence such as “If I have finished my homework, then I will go to the movies” is frequently used in daily conversation. This statement is made up of two simple statements: p: I have finished my homework. q: I will go to the movies. The remaining words, if . . . then, are the connectives. In English, this sentence is called a complex sentence. In mathematics, however, all sentences formed using connectives are called compound sentences or compound statements. In logic, a conditional is a compound statement formed by using the words if . . . then to combine two simple statements. When p and q represent simple statements, the conditional if p then q is written in symbols as p → q. The symbol p → q can also be read as “p implies q” or as “p only if q.” Here is another example: p: It is January. q: It is winter. p → q: If it is January, then it is winter. or It is January implies that it is winter. or It is January only if it is winter. Certainly we would agree that the compound sentence if p then q is true for this example: “If it is January, then it is winter.” However, if we reverse the order of the simple sentences to form the new conditional if q then p, we will get a sentence with a different meaning: q → p: If it is winter, then it is January. When it is winter, it does not necessarily mean that it is January. It may be February, the last days of December, or the first days of March. Changing the order in which we connect two simple statements in conditional does not always give a conditional that has the same truth value as the original. 14365C02.pgs 7/9/07 4:40 PM Page 54 54 Logic Parts of a Conditional The parts of the conditional if p then q can be identified by name: p is the hypothesis, which is sometimes referred to as the premise or the antecedent. It is an assertion or a sentence that begins an argument. The hypothesis usually follows the word if. q is the conclusion, which is sometimes referred to as the consequent. It is the part of a sentence that closes an argument. The conclusion usually follows the word then. There are different ways to write the conditional. When the conditional uses the word if, the hypothesis always follows if. When the conditional uses the word implies, the hypothesis always comes before implies. When the conditional uses the words only if, the conclusion follows the words only if. p → q: If it is January, then it is winter. d d hypothesis conclusion p → q: It is J
anuary implies that it is winter. d d hypothesis conclusion p → q: It is January only if it is winter. d d hypothesis conclusion All three sentences say the same thing. We are able to draw a conclusion about the season when we know that the month is January. Although the word order of a conditional may vary, the hypothesis is always written first when using symbols. Truth Values for the Conditional p → q In order to determine the truth value of a conditional, we will consider the statement “If you get an A in Geometry, then I will buy you a new graphing calculator.” Let p represent the hypothesis, and let q represent the conclusion. p: You get an A in Geometry. q: I will buy you a new graphing calculator. Determine the truth values of the conditional by considering all possible combinations of the truth values for p and q. CASE 1 You get an A in Geometry. (p is true.) I buy you a new graphing calculator. (q is true.) We both keep our ends of the agreement. The conditional statement is true. p T q T p → q T 14365C02.pgs 7/9/07 4:40 PM Page 55 CASE 2 You get an A in Geometry. (p is true.) I do not buy you a new graphing calculator. (q is false.) I broke the agreement because you got an A in Geometry but I did not buy you a new graphing calculator. The conditional statement is false. Conditionals 55 p T q F p → q F CASE 3 You do not get an A in Geometry. (p is false.) I buy you a new graphing calculator. (q is true.) You did not get an A but I bought you a new graphing calculator anyway. Perhaps I felt that a new calculator would help you to get an A next time. I did not break my promise. My promise only said what I would do if you did get an A. The conditional statement is true. p → q T T p q F CASE 4 You do not get an A in Geometry. (p is false.) I do not buy you a new graphing calculator. (q is false.) Since you did not get an A, I do not have to keep our agreement. The conditional statement is true. p F q F p → q T Case 2 tells us that the conditional is false only when the hypothesis is true and the conclusion is false. If a conditional is thought of as an “agreement” or a “promise,” this corresponds to the case when the agreement is broken. Cases 3 and 4 tell us that when the hypothesis is false, the conclusion may or may not be true. In other words, if you do not get an A in Geometry, I may or may not buy you a new graphing calculator. These four cases can be summarized as follows: A conditional is false when a true hypothesis leads to a false conclusion. In all other cases, the conditional is true Hidden Conditionals Often the words “if . . . then” may not appear in a statement that does suggest a conditional. Instead, the expressions “when” or “in order that” may suggest that the statement is a conditional. For example: 1. “When I finish my homework I will go to the movies.” p → q: If I finish my homework, then I will go to the movies. 14365C02.pgs 7/9/07 4:40 PM Page 56 56 Logic 2. “In order to succeed, you must work hard” becomes p → q: If you want to succeed, then you must work hard. 3. “2x 10; therefore x 5” becomes p → q: If 2x 10, then x 5. EXAMPLE 1 For each given sentence: a. Identify the hypothesis p. b. Identify the conclusion q. (1) If Mrs. Shusda teaches our class, then we will learn. (2) The assignment will be completed if I work at it every day. (3) The task is easy when we all work together and do our best. Solution (1) If Mrs. Shusda teaches our class, then we will learn. i p q e a. p: Mrs. Shusda teaches our class. b. q: We will learn. (2) The assignment will be completed if I work at it every day. i f q p a. p: I work at it every day. b. q: The assignment will be completed. (3) Hidden Conditional: If we all work together and do our best, then the task is easy. i p e q a. p: We all work together and we do our best. b. q: the task is easy. Note: In (3), the hypothesis is a conjunction. If we let r represent “We all work together” and s represent “We do our best,” then the conditional “If we all work together and we do our best, then the task is easy” can be symbolized as (r ∧ s) → q. EXAMPLE 2 Identify the truth value to be assigned to each conditional statement. (1) If 4 4 8, then 2(4) 8. (2) If 2 is a prime number, then 2 is odd. 14365C02.pgs 7/9/07 4:40 PM Page 57 Conditionals 57 (3) If 12 is a multiple of 9, then 12 is a multiple of 3. (4) If 2 3 then 2 3 is a positive integer. Solution (1) The hypothesis p is “4 4 8,” which is true. The conclusion q is “2(4) 8,” which is true. The conditional p → q is true. Answer (2) The hypothesis p is “2 is a prime number,” which is true. The conclusion q is “2 is odd,” which is false. The conditional p → q is false. Answer (3) The hypothesis p is “12 is a multiple of 9,” which is false. The conclusion q is “12 is a multiple of 3,” which is true. The conditional p → q is true. Answer (4) The hypothesis p is “2 3,” which is false. The conclusion q is “2 3 is a positive integer,” which is false. The conditional p → q is true. Answer EXAMPLE 3 For each given statement: a. Write the statement in symbolic form using the symbols given below. b. Tell whether the statement is true or false. Let m represent “Monday is the first day of the week.” Let w represent “There are 52 weeks in a year.” Let h represent “An hour has 75 minutes.” (True) (True) (False) (1) If Monday is the first day of the week, then there are 52 weeks in a year. (2) If there are 52 weeks in a year, then an hour has 75 minutes. (3) If there are not 52 weeks in a year then Monday is the first day of the week. (4) If Monday is the first day of the week and there are 52 weeks in a year, then an hour has 75 minutes. Answers a. m → w b. T → T is true. a. w → h b. T → F is false. a. ~w → m b. F → T is true. a. (m ∧ w) → h b. (T ∧ T) → F T → F is false. 14365C02.pgs 7/9/07 4:40 PM Page 58 58 Logic Exercises Writing About Mathematics 1. a. Show that the conditional “If x is divisible by 4, then x is divisible by 2” is true in each of the following cases: (1) x 8 (2) x 6 (3) x 7 b. Is it possible to find a value of x for which the hypothesis is true and the conclusion is false? Explain your answer. 2. For what truth values of p and q is the truth value of p → q the same as the truth value of q → p? Developing Skills In 3–10, for each given sentence: a. Identify the hypothesis p. b. Identify the conclusion q. 3. If a polygon is a square, then it has four right angles. 4. If it is noon, then it is time for lunch. 5. When you want help, ask a friend. 6. You will finish more quickly if you are not interrupted. 7. The perimeter of a square is 4s if the length of one side is s. 8. If many people work at a task, it will be completed quickly. 9. 2x 7 11 implies that x 2. 10. If you do not get enough sleep, you will not be alert. In 11–16, write each sentence in symbolic form, using the given symbols. p: The car has a flat tire. q: Danny has a spare tire. r: Danny will change the tire. 11. If the car has a flat tire, then Danny will change the tire. 12. If Danny has a spare tire, then Danny will change the tire. 13. If the car does not have a flat tire, then Danny will not change the tire. 14. Danny will not change the tire if Danny doesn’t have a spare tire. 15. The car has a flat tire if Danny has a spare tire. 16. Danny will change the tire if the car has a flat tire. 14365C02.pgs 7/9/07 4:40 PM Page 59 In 17–24, for each given statement: a. Write the statement in symbolic form, using the symbols given below. b. Tell whether the conditional statement is true or false, based upon the truth values given. Conditionals 59 b: The barbell is heavy. t: Kylie trains. l: Kylie lifts the barbell. (True) (False) (True) 17. If Kylie trains, then Kylie will lift the barbell. 18. If Kylie lifts the barbell, then Kylie has trained. 19. If Kylie lifts the barbell, the barbell is heavy. 20. Kylie lifts the barbell if the barbell is not heavy. 21. Kylie will not lift the barbell if Kylie does not train. 22. Kylie trains if the barbell is heavy. 23. If the barbell is not heavy and Kylie trains, then Kylie will lift the barbell. 24. If the barbell is heavy and Kylie does not train, then Kylie will not lift the barbell. In 25–31, find the truth value to be assigned to each conditional statement. 25. If 4 8 12, then 8 4 12. 27. If 1 1 1, then 1 1 1 1. 29. 6 6 66 if 7 7 76. 26. If 9 15, then 19 25. 28. 24 3 8 if 24 8 3. 30. 48 84 if 13 31. 31. If every rhombus is a polygon, then every polygon is a rhombus. In 32–39, symbols are assigned to represent sentences, and truth values are assigned to these sentences. Let j represent “July is a warm month.” Let d represent “I am busy every day.” Let g represent “I work in my garden.” Let f represent “I like flowers.” (True) (False) (True) (True) For each compound statement in symbolic form: a. Write a complete sentence in words to show what the symbols represent. b. Tell whether the compound statement is true or false. 32. j → g 36. (j ∧ f ) → d 33. d → g 37. (j ∧ g) → f 34. f → g 38. j → (d ∧ f ) 35. g → j 39. g → (j ∨ d) In 40–45 supply the word, phrase, or symbol that can be placed in the blank to make each resulting sentence true. 40. When p and q represent two simple sentences, the conditional if p then q is written symboli- cally as _______. 41. The conditional if q then p is written symbolically as _______. 14365C02.pgs 7/9/07 4:40 PM Page 60 60 Logic 42. The conditional p → q is false only when p is _______ and q is _______. 43. When the conclusion q is true, then p → q must be _______. 44. When the hypothesis p is false, then p → q must be _______. 45. If the hypothesis p is true and conditional p → q is true, then the conclusion q must be _______. Applying Skills In 46–50, three sentences are written in each case. The truth values are given for the first two sentences. Determine whether the third sentence is true, is false, or has an uncertain truth value. 46. If you read in dim light, then you can strain your eyes. (True) You read in dim light. (True) You can strain your
eyes. (?) 47. If the quadrilateral has four right angles, then the quadrilateral must be a square. (False) The quadrilateral has four right angles. (True) The quadrilateral must be a square. (?) 48. If n is an odd number, then 2n is an even number. (True) 2n is an even number. (True) n is an odd number. (?) 49. If the report is late, then you will not get an A. (True) The report is late. (False) You will not get an A. (?) 50. Area 1 2bh if the polygon is a triangle. (True) The polygon is a triangle. (True) Area (?) 1 2bh 2-5 INVERSES, CONVERSES, AND CONTRAPOSITIVES The conditional is the most frequently used statement in the construction of an argument or in the study of mathematics. We will use the conditional frequently in our study of geometry. In order to use the conditional statements correctly, we must understand their different forms and how their truth values are related. There are four conditionals that can be formed from two simple statements, p and q, and their negations. The conditional: The converse: p → q q → p The inverse: The contrapositive: p → q q → p 14365C02.pgs 7/9/07 4:40 PM Page 61 Inverses, Converses, and Contrapositives 61 The Inverse The inverse of a conditional statement is formed by negating the hypothesis and the conclusion. For example, the inverse of the statement “If today is Monday, then I have soccer practice” is “If today is not Monday, then I do not have soccer practice.” In symbols, the inverse of (p → q) is (p → q). The following examples compare the truth values of given conditionals and their inverses. 1. A true conditional can have a false inverse. Let p represent “A number is divisible by ten.” Let q represent “The number is divisible by five.” (p → q): (p → q): Conditional If a number is divisible by ten, then it is divisible by five. f i p q Inverse If a number is not divisible by ten, then it is not divisible by five. g i p q We can find the truth value of these two statements when the number is 15. p → q: If 15 is divisible by 10, then 15 is divisible by 5. p: 15 is divisible by 10. q: 15 is divisible by 5. p: 15 is not divisible by 10. q: 15 is not divisible by 5. False True F → T is T True False T → F is F p → q: If 15 is not divisible by 10, then 15 is not divisible by 5. In this case, the conditional and the inverse have opposite truth values. 2. A false conditional can have a true inverse. Let p represent “Two angles are congruent.” Let q represent “Two angles are both right angles.” Conditional (p → q): If two angles are congruent, g p then the two angles are both right angles. i q Inverse (p → q): If two angles are not congruent, i p then the two angles are not both right angles. i q We can find the truth value of these two statements with two angles A and B when mA 60 and mB 60. 14365C02.pgs 7/9/07 4:40 PM Page 62 62 Logic p: Angle A and B are congruent. q: Angle A and B are both right angles. p → q: If A and B are congruent, then A and B are both right angles. p: Angle A and B are not congruent. q: Angle A and B are not both right angles. p → q: If A and B are not congruent, then A and B are not both right angles. True False T → F is F False True F → T is T Again, the conditional and the inverse have opposite truth values. 3. A conditional and its inverse can have the same truth value. Let r represent “Twice Talia’s age is 10.” Let q represent “Talia is 5 years old.” Conditional If twice Talia’s age is 10, then Talia is 5 years old. f e r s Inverse If twice Talia’s age is not 10, then Talia is not 5 years old. g f r s (r → s): (r → s): When Talia is 5, r is true and s is true. The conditional is true. When Talia is 5, r is false and s is false. The inverse is true. Both the conditional and its inverse have the same truth value. When Talia is 6, r is false and s is false. The conditional is true. When Talia is 6, r is true and s is true. The conditional is true. Again, the conditional and its inverse have the same truth value. These three illustrations allow us to make the following conclusion: A conditional (p → q) and its inverse (p → q) may or may not have the same truth value. This conclusion can be shown in the truth table. Note that the conditional (p → q) and its inverse (p → q) have the same truth value when p and q have the same truth value. The conditional (p → q) and its inverse (p → q) have opposite truth values when p and q have opposite truth values. Conditional Inverse 14365C02.pgs 7/9/07 4:40 PM Page 63 Inverses, Converses, and Contrapositives 63 The Converse The converse of a conditional statement is formed by interchanging the hypothesis and conclusion. For example, the converse of the statement “If today is Monday, then I have soccer practice” is “If I have soccer practice, then today is Monday.” In symbols, the converse of (p → q) is (q → p). To compare the truth values of a conditional and its converse, we will con- sider some examples. 1. A true conditional can have a false converse. Let p represent “x is a prime.” Let q represent “x is odd.” Conditional (p → q): If x is a prime, then x is odd. d p q Converse (q → p): If x is odd, then x is a prime. d d d q p When x 9, p is false and q is true. Therefore, for this value of x, the condi- tional (p → q) is true and its converse (q → p) is false. In this example, the conditional is true and the converse is false. The condi- tional and its converse do not have the same truth value. 2. A false conditional can have a true converse. Let p represent “x is divisible by 2.” Let q represent “x is divisible by 6.” Conditional (p → q): If x is divisible by 2, then x is divisible by 6. e p Converse (q → p): If x is divisible by 6, then x is divisible by 2. q e e e q p When x 8, p is true and q is false. Therefore, for this value of x, the condi- tional (p → q) is false and its converse (q → p) is true. In this example, the conditional is false and the converse is true. Again, the conditional and its converse do not have the same truth value. 3. A conditional and its converse can have the same truth value. Let p represent “Today is Friday.” Let q represent “Tomorrow is Saturday.” Conditional (p → q): If today is Friday, then tomorrow is Saturday. e f Converse (q → p): If tomorrow is Saturday, then today is Friday. f e q p p q 14365C02.pgs 7/9/07 4:40 PM Page 64 64 Logic On Friday, p is true and q is true. Therefore, both (p → q) and (q → p) are true. On any other day of the week, p is false and q is false. Therefore, both (p → q) and (q → p) are true. A conditional and its converse may have the same truth value. These three examples allow us to make the following conclusion: A conditional (p → q) and its converse (q → p) may or may not have the same truth value. This conclusion can be shown in the truth table. Note that the conditional (p → q) and its converse (q → p) have the same truth value when p and q have the same truth value. The conditional (p → q) and its converse (q → p) have different truth values when p and q have different truth values. The Contrapositive Conditional p → q Converse We form the inverse of a conditional by negating both the hypothesis and the conclusion. We form the converse of a conditional by interchanging the hypothesis and the conclusion. We form the contrapositive of a conditional by doing both of these operations: we negate and interchange the hypothesis and conclusion. In symbols, the contrapositive of (p → q) is (q → p). 1. A true conditional can have a true contrapositive. Let p represent “Gary arrives late to class.” Let q represent “Gary is marked tardy.” Conditional If Gary arrives late to class, then Gary is marked tardy. f g p q (p → q): (q → p): Contrapositive If Gary is not marked tardy, i q then Gary does not arrive late to class. i p If Gary arrives late to class, p is true and q is true. Therefore, (p → q) is true. Also, if p and q are true, p is false and q is false, so (q → p) is true. If p is false, that is, if Gary does not arrive late to class, q is false and (p → q) is true. Similarly, if p and q are false, p is true and q is true, so (q → p) is true. 14365C02.pgs 7/9/07 4:40 PM Page 65 Inverses, Converses, and Contrapositives 65 2. A false conditional can have a false contrapositive. Conditional If x is an odd number, then x is a prime number. f f p q Contrapositive If x is not a prime number, then x is not an odd number. g i (p → q): (q → p): q p → q F → T is true T → F is false T → T is true F → F is true p q → p F → T is true T → F is false F → F is true T → T is true Let x 2 Let x 9 Let x 11 Let x 8 For each value of x, the conditional and its contrapositive have the same truth value. These illustrations allow us to make the following conclusion: A conditional (p → q) and its contrapositive (q → p) always have the same truth value: When a conditional is true, its contrapositive must be true. When a conditional is false, its contrapositive must be false. This conclusion can be shown in the following truth table. Conditional Contrapositive Logical Equivalents A conditional and its contrapositive are logical equivalents because they always have the same truth value. The inverse of (p → q) is (p → q). The contrapositive of (p → q) is formed by negating and interchanging the hypothesis and conclusion of (p → q). The negation of p is p and the negation of q is q. Therefore, the contrapositive of (p → q) is (q → p). For instance: 14365C02.pgs 7/9/07 4:40 PM Page 66 66 Logic Conditional If Jessica likes waffles, then Jessica eats waffles. f f (p → q): p q Inverse If Jessica does not like waffles, i (p → q): p then Jessica does not eat waffles. i q Contrapositive If Jessica eats waffles, then Jessica likes waffles. f f q p of the inverse (q → p): Notice, however, that the contrapositive of the inverse is the same as the converse of the original conditional. Thus, the inverse and the converse of (p → q) are contrapositives of each other. Since a conditional and its contrapositive always have the same truth value, the converse and the inverse always ha
ve the same truth value. This can be verified by constructing the following truth table EXAMPLE 1 Write the inverse, converse, and contrapositive of the given conditional: If today is Tuesday, then I play basketball. Solution Inverse: If today is not Tuesday, then I do not play basketball. Converse: If I play basketball, then today is Tuesday. Contrapositive: If I do not play basketball, then today is not Tuesday. EXAMPLE 2 Write the inverse, converse, and contrapositive of the given conditional: If a polygon is a square then it has four right angles. Solution Inverse: If a polygon is not a square, then it does not have four right angles. Converse: If a polygon has four right angles, then it is a square. Contrapositive: If a polygon does not have four right angles, then it is not a square. 14365C02.pgs 7/9/07 4:40 PM Page 67 Inverses, Converses, and Contrapositives 67 Write the inverse, converse, and contrapositive of the given conditional: If M is the midpoint of AB , then AM MB. Inverse: If M is not the midpoint of AB , then AM MB. Converse: If AM MB, then M is the midpoint of . AB Contrapositive: If AM MB, then M is not the midpoint of AB . EXAMPLE 3 Solution EXAMPLE 4 Given the true statement, “If the polygon is a rectangle, then it has four sides,” which statement must also be true? (1) If the polygon has four sides, then it is a rectangle. (2) If the polygon is not a rectangle, then it does not have four sides. (3) If the polygon does not have four sides, then it is not a rectangle. (4) If the polygon has four sides, then it is not a rectangle. Solution A conditional and its contrapositive always have the same truth value. The contrapositive of the given statement is “If a polygon does not have four sides then it is not a rectangle.” Answer (3) Exercises Writing About Mathematics 1. Samuel said that if you know that a conditional is true then you know that the converse of the conditional is true. Do you agree with Samuel? Explain why or why not. 2. Kate said that if you know the truth value of a conditional and of its converse then you know the truth value of the inverse and the contrapositive. Do you agree with Kate? Explain why or why not. Developing Skills In 3–6, for each statement, write in symbolic form: a. the inverse b. the converse c. the contrapositive. 3. p → q 4. t → w 5. m → p 6. p → q 14365C02.pgs 7/9/07 4:40 PM Page 68 68 Logic In 7–10: a. Write the inverse of each conditional statement in words. b. Give the truth value of the conditional. c. Give the truth value of the inverse. 7. If 6 3, then 6 3. 8. If a polygon is a parallelogram, then the polygon has two pairs of parallel sides. 9. If 3(3) 9, then 3(4) 12. 10. 1f 22 4, then 32 6. In 11–14, write the converse of each statement in words. 11. If you lower your cholesterol, then you eat Quirky oatmeal. 12. If you enter the Grand Prize drawing, then you will get rich. 13. If you use Shiny’s hair cream, then your hair will curl. 14. If you feed your pet Krazy Kibble, he will grow three inches. In 15–18: a. Write the converse of each conditional statement in words. b. Give the truth value of the conditional. c. Give the truth value of the converse. 15. If a number is even, then the number is exactly divisible by 2. 16. If 0.75 is an integer, then it is rational. 17. If 8 1 7, then 82 12 72. 18. If 4(5) 6 20 6, then 4(5) 6 14. In 19–23: a. Write the contrapositive of each statement in words. b. Give the truth value of the conditional. c. Give the truth value of the contrapositive. 19. If Rochester is a city, then Rochester is the capital of New York. 20. If two angles form a linear pair, then they are supplementary. 21. If 3 2 1, then 4 3 2. 22. If all angles of a triangle are equal in measure, then the triangle is equiangular. 23. If 0, then 1 2 1 2 is a counting number. In 24–28, write the numeral preceding the expression that best answers the question. 24. When p → q is true, which related conditional must be true? (1) q → p (2) p → q (3) p → q (4) q → p 25. Which is the contrapositive of “If March comes in like a lion, it goes out like a lamb”? (1) If March goes out like a lamb, then it comes in like a lion. (2) If March does not go out like a lamb, then it comes in like a lion. (3) If March does not go out like a lamb, then it does not come in like a lion. (4) March goes out like a lion if it comes in like a lamb. 14365C02.pgs 7/9/07 4:40 PM Page 69 Biconditionals 69 26. Which is the converse of “If a rectangular prism is a cube, then its surface area is 6s2”? (1) If a rectangular prism is not a cube, then its surface area is not 6s2. (2) If the surface area of a rectangular prism is 6s2, then it is a cube. (3) If the surface area of a rectangular prism is not 6s2, then it is not a cube. (4) If the surface area of a cube is 6s2, then it is a rectangular prism. 27. Which is the inverse of “If z 4, then 2z 9”? (1) If 2z 9, then z 4. (2) If 2z 9, then z 4. (3) If z 4, then 2z 9. (4) If z 4, then 2z 9. 28. Which is the contrapositive of “If y is greater than 3, then 2y 10y is not equal to 36”? (1) If 2y 10y is not equal to 36, then y is greater than 3. (2) If 2y 10y equals 36, then y is not greater than 3. (3) If 2y 10y is not equal to 36, then y is not greater than 3. (4) If 2y 10y equals 36, then y is greater than 3. Applying Skills In 29–34, assume that each conditional statement is true. Then: a. Write its converse in words and state whether the converse is always true, sometimes true, or never true. b. Write its inverse in words and state whether the inverse is always true, sometimes true, or never true. c. Write its contrapositive in words and state whether the contrapositive is always true, sometimes true, or never true. 29. If Derek lives in Las Vegas, then he lives in Nevada. 30. If a bin contains 3 red marbles and 3 blue marbles, then the probability of picking a red 1 marble from the bin is . 2 31. If a polygon has eight sides, then it is an octagon. 32. If a garden grows carrots, then it grows vegetables. 33. If the dimensions of a rectangle are 8 feet by 6 feet, then the area of the rectangle is 48 square feet. 34. If a number has 7 as a factor, then it is divisible by 7. 2-6 BICONDITIONALS A biconditional is the conjunction of a conditional and its converse. For the conditional (p → q), the converse is (q → p). The biconditional can be written as (p → q) ∧ (q → p) or in the shorter form p ↔ q, which is read p if and only if q. 14365C02.pgs 7/9/07 4:40 PM Page 70 70 Logic Recall that a conjunction is true only when both parts of the compound statement are true. Therefore, (p → q) ∧ (q → p) is true only when (p → q) is true and its converse (q → p) is true. In the last section you learned that a conditional (p → q) and its converse (q → p) are both true when p and q are both true or both false. This is shown in the table belowp → q) ∧ (q → p The biconditional p if and only if q is true when p and q are both true or both false. In other words, p ↔ q is true when p and q have the same truth value. When p and q have different truth values, the biconditional is false. Applications of the Biconditional There are many examples in which the biconditional is always true. Consider the following: 1. Every definition is a true biconditional. Every definition can be written in reverse order. Both of the following statements are true: • Congruent segments are segments that have the same measure. • Line segments that have the same measure are congruent. We can restate the definition as two true conditionals: • If two line segments are congruent, then they have the same measure. • If two line segments have the same measure, then they are congruent. Therefore, this definition can be restated as a true biconditional: • Two line segments are congruent if and only if they have the same measure. 2. Biconditionals are used to solve equations. We know that when we add the same number to both sides of an equation or when we multiply both sides of an equation by the same number, the derived equation has the same solution set as the given equation. That is, any number that makes the first equation true will make the derived equation true. 14365C02.pgs 7/9/07 4:40 PM Page 71 Biconditionals 71 For example: p: 3x 7 19 q: 3x 12 p → q: If 3x 7 19, then 3x 12. (7 was added to both sides of q → p: If 3x 12, then 3x 7 19. (7 was added to both sides of the equation.) the equation.) When x 4, both p and q are true and both p → q and q → p are true. When x 1 or when x equals any number other than 3, both p and q are false and both p → q and q → p are true. Therefore, the biconditional “3x 7 19 if and only if 3x 12” is true. The solution of an equation is a series of biconditionals: 3x 7 19 12 3x 4 x 3x 7 19 if and only if 3x 12. 3x 12 if and only if x 4. 3. A biconditional states that two logical forms are equivalent. Two logical forms that always have the same truth values are said to be equivalent. We have seen that a conditional and its contrapositive are logically equivalent and that the converse and inverse of a conditional are logically equivalent. There are many other statements that are logically equivalent. The table below shows that (p ∧ q) and p ∨ q are logically equivalent. We can write the true biconditional (p ∧ q) ↔ (p ∨ q). p ∧ q EXAMPLE 1 Determine the truth value to be assigned to the biconditional. Germany is a country in Europe if and only if Berlin is the capital of Germany. Solution “Germany is a country in Europe” is true. “Berlin is the capital of Germany” is true. Therefore, the biconditional “Germany is a country in Europe if and only if Berlin is the capital of Germany” is also true. Answer 14365C02.pgs 7/9/07 4:40 PM Page 72 72 Logic EXAMPLE 2 The statement “I go to basketball practice on Monday and Thursday” is true. Determine the truth value to be assigned to each statement. a. If today is Monday, then I go to basketball practice. b. If I go to basketball practice, then today is Monday. c. Today is Monday if and only if I go to basket
ball practice. Solution Let p: “Today is Monday,” and q: “I go to basketball practice.” a. We are asked to find the truth value of the following conditional: p → q: If today is Monday, then I go to basketball practice. On Monday, p is true and q is true. Therefore, p → q is true. On Thursday, p is false and q is true. Therefore, p → q is true. On every other day, p is false and q is false. Therefore, p → q is true. “If today is Monday, then I go to basketball practice” is always true. Answer b. We are asked to find the truth value of the following conditional: q → p: If I go to basketball practice then today is Monday. On Monday, p is true and q is true. Therefore, q → p is true. On Thursday, p is false and q is true. Therefore, q → p is false. On every other day, p is false and q is false. Therefore, q → p is true. “If I go to basketball practice, then Today is Monday” is sometimes true and sometimes false. Answer c. We are asked to find the truth value of the following biconditional: p ↔ q: Today is Monday if and only if I go to basketball practice. The conditionals p → q and q → p do not always have the same truth value. Therefore, the biconditional “Today is Monday if and only if I go to basketball practice” is not always true. We usually say that a statement that is not always true is false. Answer EXAMPLE 3 Determine the truth value of the biconditional. 3y 1 28 if and only if y 9. 14365C02.pgs 7/9/07 4:40 PM Page 73 Biconditionals 73 Solution When y = 9, 3y 1 28 is true and y 9 is true. Therefore, 3y 1 28 if and only if y 9 is true. When y 9, 3y 1 28 is false and y 9 is false. Therefore, 3y 1 28 if and only if y 9 is true. 3y 1 28 if and only if y 9 is always true. Answer Exercises Writing About Mathematics 1. Write the definition “A prime number is a whole number greater than 1 and has exactly two factors” as a biconditional. 2. Tiffany said that if the biconditional p ↔ q is false, then either p → q is true or q → p is true but both cannot be true. Do you agree with Tiffany? Explain why or why not. Developing Skills In 3–16, give the truth value of each biconditional. 3. y 7 30 if and only if y 23. 4. B is between A and C if and only if AB AC BC. 5. z 9 13 if and only if z 2 6. 6. A parallelogram is a rhombus if and only if the parallelogram has four sides of equal length. 7. A real number is positive if and only if it is greater than zero. 8. An angle is an acute angle if and only if its degree measure is less than 90. 9. An element belongs to the intersection of sets F and G if and only if it belongs to both F and G. 10. An integer is odd if and only if it is not divisible by 2. 11. Two angles have the same measure if and only if they are right angles. 12. I live in the United States if and only if I live in New York State. 13. A rational number has a multiplicative inverse if and only if it is not zero. 14. An angle is an acute angle if and only if it has a degree measure of 50. 15. x 5 if and only if x 3. 16. Today is Friday if and only if tomorrow is Saturday. 14365C02.pgs 7/9/07 4:40 PM Page 74 74 Logic Applying Skills 17. Let p represent “x is divisible by 2.” Let q represent “x is divisible by 3.” Let r represent “x is divisible by 6.” a. Write the biconditional (p ∧ q) ↔ r in words. b. Show that the biconditional is always true for the domain {2, 3, 5, 6, 8, 9, 11, 12}. c. Do you think that the biconditional is true for all counting numbers? Explain your answer. 18. A gasoline station displays a sign that reads “Open 24 hours a day, Monday through Friday.” a. On the basis of the information on the sign, is the conditional “The gasoline station is closed if it is Saturday or Sunday” true? b. On the basis of the information on the sign, is the conditional “If the gasoline station is closed, it is Saturday or Sunday” true? c. On the basis of the information on the sign, is the biconditional “The gasoline station is closed if and only if it is Saturday or Sunday” true? d. Marsha arrives at the gasoline station on Monday and finds the station closed. Does this contradict the information on the sign? e. Marsha arrives at the gasoline station on Saturday and finds the station open. Does this contradict the information on the sign? In 19–22, write a biconditional using the given conditionals and tell whether each biconditional is true or false. 19. If a triangle is isosceles, then it has two congruent sides. If a triangle has two congruent sides then it is isosceles. 20. If two angles are both right angles, then they are congruent. If two angles are congruent, then they are both right angles. 21. If today is Thursday, then tomorrow is not Saturday. If tomorrow is not Saturday, then today is Thursday. 22. If today is not Friday, then tomorrow is not Saturday. If tomorrow is not Saturday, then today is not Friday. 2-7 THE LAWS OF LOGIC We frequently want to combine known facts in order to establish the truth of related facts. To do this, we can look for patterns that are frequently used in drawing conclusions. These patterns are called the laws of logic. 14365C02.pgs 7/9/07 4:40 PM Page 75 The Laws of Logic 75 The Law of Detachment A valid argument uses a series of statements called premises that have known truth values to arrive at a conclusion. For example, Cynthia makes the following true statements to her parents: I want to play baseball. If I want to play baseball, then I need a glove. These are the premises of Cynthia’s argument. The conclusion that Cynthia wants her parents to make is that she needs a glove. Is this conclusion valid? Let p represent “I want to play baseball.” Let q represent “I need a glove.” Then p → q represents “If I want to play baseball, then I need a glove.” We know that the premises are true, that is, p is true and p → q is true. The only line of the truth table that satisfies both of these conditions is the first in which q is also true. Therefore, “I need a glove” is a true conclusion The example just given does not depend on the statement represented by p and q. The first line of the truth table tells us that whenever p → q is true and p is true, then q must be a true conclusion. This logical pattern is called the Law of Detachment: If a conditional (p → q) is true and the hypothesis (p) is true, then the con- clusion (q) is true. EXAMPLE 1 If the measure of an angle is greater than 0° and less than 90°, then the angle is an acute angle. Let “mA 40°, which is greater than 0° and less than 90°” be a true statement. Prove that A is an acute angle. Solution Let p represent “mA 40°, which is greater than 0° and less than 90°.” Let q represent “the angle is an acute angle.” Then p → q is true because it is a definition of an acute angle. Also, p is true because it is given. Then by the Law of Detachment, q is true. Answer “A is an acute angle” is true. 14365C02.pgs 7/9/07 4:40 PM Page 76 76 Logic The Law of Disjunctive Inference We know that a disjunction is true when one or both statements that make up the disjunction are true. The disjunction is false when both statements that make up the disjunction are false. For example, let p represent “A real number is rational” and q represent “A real number is irrational.” Then p ∨ q represents “A real number is rational or a real number is irrational,” a true statement. When the real number is p, then “A real number is rational” is false. Therefore “A real number is irrational” must be true. When the real number is 7, then “A real number is irrational” is false. Therefore “A real number is rational” must be true. The truth table shows us that when p is false and p ∨ q is true, only the third line of the table is satisfied. This line tells us that q is true. Also, when q is false and p ∨ q is true, only the second line of the table is satisfied. This line tells us that p is true. The example just given illustrates a logical pattern that does not depend on the statements represented by p and q. When a disjunction is true and one of the disjuncts is false, then the other disjunct must be true. This logical pattern is called the Law of Disjunctive Inference If a disjunction (p ∨ q) is true and the disjunct (p) is false, then the other disjunct (q) is true. If a disjunction (p ∨ q) is true and the disjunct (q) is false, then the other disjunct (p) is true. EXAMPLE 2 What conclusion can be drawn when the following statements are true? I will walk to school or I will ride to school with my friend. I do not walk to school. Solution Since “I do not walk to school” is true, “I walk to school” is false. The disjunc- tion “I will walk to school or I will ride to school with my friend” is true and one of the disjuncts, “I walk to school,” is false. By the Law of Disjunctive Inference, the other disjunct, “I ride to school with my friend,” must be true. Alternative Solution Let p represent “I walk to school.” Let q represent “I ride to school with my friend.” 14365C02.pgs 7/9/07 4:40 PM Page 77 The Laws of Logic 77 Make a truth table for the disjunction p ∨ q and eliminate the rows that do not apply. We know that p ∨ q is true. We also know that since p is true, p is false. (1) Eliminate the last row of truth values in which the disjunction is false. (2) Eliminate the first two rows of truth values in which p is true. (3) Only one case remains: q is true. Answer I ride to school with my friend Note: In most cases, more than one possible statement can be shown to be true. For example, the following are also true statements when the given statements are true: I do not walk to school and I ride to school with my friend. If I do not walk to school, then I ride to school with my friend. If I do not ride to school with my friend, then I walk to school. EXAMPLE 3 From the following true statements, is it possible to determine the truth value of the statement “I will go to the library”? If I have not finished my essay for English class, then I will go to the library. I have finished my essay for English class. Solution When “I have finished my essay f
or English class” is true, its negation, “I have not finished my essay for English class” is false. Therefore, the true conditional “If I have not finished my essay for English class, then I will go to the library” has a false hypothesis and the conclusion, “I will go to the library” can be either true or false. Alternative Solution Let p represent “I have not finished my essay for English class.” Let q represent “I will go to the library.” Make a truth table for the disjunction p → q and eliminate the rows that do not apply. We know that p → q is true. We also know that since p is true, p is false. 14365C02.pgs 7/9/07 4:40 PM Page 78 78 Logic (1) Eliminate the second row of truth values in which the conditional is false. (2) Eliminate the first row of truth values in which p is true. The second row in which p is true has already been eliminated. (3) Two rows remain, one in which q is true and the other in which q is false Answer “I will go to the library” could be either true or false. EXAMPLE 4 Draw a conclusion or conclusions base on the following true statements. If I am tired, then I will rest. I do not rest. Solution A conditional and its contrapositive are logically equivalent. Therefore, “If I do not rest, then I am not tired” is true. By the Law of Detachment, when the hypothesis of a true conditional is true, the conclusion must be true. Therefore, since “I do not rest” is true, “I am not tired” must be true. Alternative Solution Let p represent “I am tired.” Let q represent “I will rest.” Make a truth table for the disjunction p → q and eliminate the rows that do not apply. We know that p → q is true. We also know that since q is true, q is false. (1) Eliminate the second row of truth values in which the conditional is false. (2) Eliminate the first and third rows of truth values in which q is true. (3) Only one case remains: p is false. Since p is false, p is true. “I am not tired” is true Answer I am not tired. 14365C02.pgs 7/9/07 4:40 PM Page 79 The Laws of Logic 79 Exercises Writing About Mathematics 1. Clovis said that when p → q is false and q ∨ r is true, r must be true. Do you agree with Clovis? Explain why or why not. 2. Regina said when p ∨ q is true and q is true, then p ∧ q must be true. Do you agree with Regina? Explain why or why not. Developing Skills In 3–14, assume that the first two sentences in each group are true. Determine whether the third sentence is true, is false, or cannot be found to be true or false. Justify your answer. 3. I save up money or I do not go on the trip. 4. If I speed, then I get a ticket. I go on the trip. I save up money. 5. I like swimming or kayaking. I like kayaking. I like swimming. 7. x 18 if x 14. x 14 x 18 I speed. I get a ticket. 6. I like swimming or kayaking. I do not like swimming. I like kayaking. 8. I live in Pennsylvania if I live in Philadelphia. I do not live in Philadelphia. I live in Pennsylvania. 9. If I am late for dinner, then my dinner 10. If I am late for dinner, then my dinner will be cold. I am late for dinner. My dinner is cold. will be cold. I am not late for dinner. My dinner is not cold. 11. I will go to college if and only if I work this summer. I do not work this summer. I will go to college. 13. If I am late for dinner, then my dinner will be cold. My dinner is not cold. I am not late for dinner. 12. The average of two numbers is 20 if the numbers are 17 and 23. The average of two numbers is 20. The two numbers are 17 and 23. 14. If I do not do well in school, then I will not receive a good report card. I do well in school. I receive a good report card. Applying Skills In 15–27, assume that each given sentence is true. Write a conclusion using both premises, if possible. If no conclusion is possible, write “No conclusion.” Justify your answer. 15. If I play the trumpet, I take band. I play the trumpet. 16. 6 " 6 " is rational or irrational. is not rational. 14365C02.pgs 7/9/07 4:40 PM Page 80 80 Logic 17. If 2b 6 14, then 2b 8. If 2b 8, then b 4. 2b 6 14 19. If k is a prime, then k 8. k = 8 18. If it is 8:15 A.M., then it is morning. It is not morning. 20. x is even and a prime if and only if x 2. x 2 21. It is February or March, and it is not 22. If x is divisible by 4, then x is divisible summer. It is not March. by 2. x is divisible by 2. 23. On Saturdays, we go bowling or we 24. I study computer science, and wood shop fly kites. or welding. Last Saturday, we did not go bowling. I do not take woodshop. 25. Five is a prime if and only if five has 26. If x is an integer greater than 2 and x exactly two factors. Five is a prime. is a prime, then x is odd. x is not an odd integer. 27. If a ray bisects an angle, the ray divides the angle into two congruent angles. Ray DF does not divide angle CDE into two congruent angles. 2-8 DRAWING CONCLUSIONS Many important decisions as well as everyday choices are made by applying the principles of logic. Games and riddles also often depend on logic for their solution. EXAMPLE 1 The three statements given below are each true. What conclusion can be found to be true? 1. If Rachel joins the choir then Rachel likes to sing. 2. Rachel will join the choir or Rachel will play basketball. 3. Rachel does not like to sing. Solution A conditional and its contrapositive are logically equivalent. Therefore, “If Rachel does not like to sing, then Rachel will not join the choir” is true. By the Law of Detachment, when the hypothesis of a true conditional is true, the conclusion must be true. Therefore, since “Rachel does not like to sing” is true, “Rachel will not join the choir” must be true. Since “Rachel will not join the choir” is true, then its negation, “Rachel will join the choir,” must be false. 14365C02.pgs 7/9/07 4:40 PM Page 81 Drawing Conclusions 81 Since “Rachel will join the choir or Rachel will play basketball” is true and “Rachel will join the choir” is false, then by the Law of Disjunctive Inference, “Rachel will play basketball” must be true. Answer Rachel does not join the choir. Rachel will play basketball. Alternative Solution Let c represent “Rachel joins the choir,” s represent “Rachel likes to sing,” and b represent “Rachel will play basketball.” Write statements 1, 2, and 3 in symbols: 1. c → s 2. c ∨ b 3. s Using Statement 1 c → s is true, so s → c is true. (A conditional and its contrapositive always have the same truth value.) Using Statement 3 s is true and s → c is true, so c is true. (Law of Detachment) Also, c is false. Using Statement 2 c ∨ b is true and c is false, so b must be true. (Law of Disjunctive Inference) Answer Rachel does not join the choir. Rachel will play basketball. EXAMPLE 2 If Alice goes through the looking glass, then she will see Tweedledee. If Alice sees Tweedledee, then she will see the Cheshire Cat. Alice does not see the Cheshire Cat. Show that Alice does not go through the looking glass. Solution A conditional and its contrapositive are logically equivalent. Therefore, “If Alice does not see the Cheshire Cat, then Alice does not see Tweedledee” is true. By the Law of Detachment, when the hypothesis of a true conditional is true, the conclusion must be true. Therefore, since “Alice does not see the Cheshire Cat” is true, “Alice does not see Tweedledee” must be true. Again, since a conditional and its contrapositive are logically equivalent, “If Alice does not see Tweedledee, then Alice does not go through the looking 14365C02.pgs 7/9/07 4:40 PM Page 82 82 Logic glass” is also true. Applying the Law of Detachment, since “Alice does not see Tweedledee” is true, “Alice does not go through the looking glass” must be true. Alternative Solution Let a represent “Alice goes through the looking glass.” t represent “Alice sees Tweedledee.” c represent “Alice sees the Cheshire cat.” 1. a → t Then, in symbols, the given statements are: 3. c We would like to conclude that a is true. 2. t → c Using Statement 2 t → c is true, so c → t is true. (A conditional and its contrapositive always have the same truth value.) Using Statement 3 c is true and c → t is true, so t is true. (Law of Detachment) Using Statement 1 a → t is true, so t → a is true. (A conditional and its contrapositive always have the same truth value.) Therefore, by the Law of Detachment, since t is true and t → a is true, a is true. EXAMPLE 3 Three siblings, Ted, Bill, and Mary, each take a different course in one of three areas for their senior year: mathematics, art, and thermodynamics. The following statements about the siblings are known to be true. • Ted tutors his sibling taking the mathematics course. • The art student and Ted have an argument over last night’s basketball game. • Mary loves the drawing made by her sibling taking the art course. What course is each sibling taking? Solution Make a table listing each sibling and each subject. Use the statements to fill in the information about each sibling. 14365C02.pgs 7/9/07 4:40 PM Page 83 (1) Since Ted tutors his sibling taking the math course, Ted cannot be taking math. Place an ✗ in the table to show this. (2) Similarly, the second state- Drawing Conclusions 83 Math Art Thermodynamics ✗ Ted Bill Mary Ted ment shows that Ted cannot be taking art. Place an ✗ to indicate this. The only possibility left is for Ted to be taking thermodynamics. Place a to indicate this, and add ✗’s in the thermodynamics column to show that no one else is taking this course. Mary Bill ✗ Math Art Thermodynamics ✗ ✗ ✗ (3) The third statement shows that Mary is not taking art. Therefore, Mary is taking math. Since Ted is taking thermodynamics and Mary is taking math, Bill must be taking art. Math Art Thermodynamics Ted Bill Mary ✗ ✗ ✗ ✗ ✗ ✗ Answer Ted takes thermodynamics, Bill takes art, and Mary takes mathematics. Exercises Writing About Mathematics 1. Can (p ∨ q) ∧ r be true when p is false? If so, what are the truth values of q and of r? Justify your answer. 2. Can (p ∨ q) ∧ r be true when r is false? If so, what are the truth values of p and of q? Justify yo

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