C H A P T E R 9 Probability and Calculus

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© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. C H A P T E R 9 Probability and Calculus Se... ction 9.1 Discrete Probability ............................................................................549 Section 9.2 Continuous Random Variables ..........................................................555 Section 9.3 Expected Value and Variance............................................................560 Review Exercises ........................................................................................................569 Test Yourself .............................................................................................................578 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 549 C H A P T E R 9 Probability and Calculus Section 9.1 Discrete Probability 1. (a) S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} (b) A = {HHH, HHT, HTH, THH} (c) B = {HTT, THT, TTH, TTT} 2. (a) S = {HH, HT, T1, T 2, T3, T 4, T5, T6} (b) A = {T 4, T5, T6} (c) B = {HH} 3. (a) { } , , , , , , , , , , , , , , , , , , , , , , , , , , S III IIO IIU IOI IUI OII UII IOO IOU IUO IUU OIO OIU UIO UIU OOI OUI UOI UUI OOO OOU OUO UOO OUU UOU UUO UUU = (b) A = {III, IIO, IIU, IOI, IUI , OII , UII} (c) { } , , , , , , , , , , , , , , , , , , , B III IIO IIU IOI IUI OII UII IOU IUO IUU OIU UIO UIU OUI UOI UUI OUU UOU UUO UUU = Skills Warm Up 1. 1 2 2 9 3 9 1 6 2 9 9 9 9 9 1 x x x x + + = + + = = = 2. 1 5 1 1 1 3 12 8 12 24 8 10 3 2 24 24 24 24 24 24 24 23 24 24 24 23 24 1 x x x x x + + + + = + + + + = + = + = = 3. ( 1 ) ( 3 ) ( 8 ) ( 3 ) ( 1 ) 32 16 16 16 16 16 16 0 + 1 + 2 + 3 + 4 = = 2 4. ( 1 ) ( 2 ) ( 6 ) ( 2 ) ( 1 ) 24 12 12 12 12 12 12 0 + 1 + 2 + 3 + 4 = = 2 5. ( )2 (1 ) ( )2 (1 ) ( )2 (1 ) 1 1 1 4 2 4 4 4 2 0 − 1 + 1 − 1 + 2 − 1 = + = 6. ( )2 ( 1 ) ( )2 ( 2 ) ( )2 ( 6 ) ( )2 ( 2 ) ( )2 ( 1 ) 4 2 2 4 12 12 12 12 12 12 12 12 12 0 − 2 + 1 − 2 + 2 − 2 + 3 − 2 + 4 − 2 = + + + = 1 7. 3 8 = 0.375 = 37.5% 8. 9 11 ≈ 0.8182 = 81.82% 9. 13 24 ≈ 0.5417 = 54.17% 10. 112 256 = 0.4375 = 43.75% 550 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 4. (a) S = {SSSS, CCCC, MMMM, OOOO, SSSC, SSCS, SCSS, CSSS, …} The sample space S consists of all distinct orders of four methods of the types S, C, M, or O. (b) { } , , , , , , , , , , , , A MMMM MMMS MMMC MMMO MMSM MMCM MMOM MSMM MCMM MOMM SMMM CMMM OMMM = (c) Event B consists of all outcomes in S for which there is no more than one C. 5. (a) (b) 6. (a) (b) 7. (a) ( ) 12 9 21 7 72 72 72 24 P college = + = = (b) ( ) 7 17 24 24 P not college = 1 − = (c) ( ) 32 4 72 9 P girl, not college = = 8. (a) ( ) 26 1 52 2 P red = = (b) ( ) 4 1 52 13 P 5 = = (c) ( ) 6 3 52 26 P black face card = = 9. P(3) = 1 − (0.20 + 0.35 + 0.15 + 0.05) = 0.25 10. P(1) = 1 − (0.05 + 0.25 + 0.30 + 0.15 + 0.10) = 0.15 11. The table represents a probability distribution because each distinct value of x corresponds to a probability P(x) such that 0 ≤ P(x) ≤ 1, and the sum of all P(x) is 1. 12. The table does not represent a probability distribution because the sum of all values of P(x) does not equal 1: P(0) + P(1) + P(2) + P(3) + P(4) + P(5) = 0.05 + 0.30 + 0.10 + 0.40 + 0.15 + 0.20 = 1.2 13. The table does not represent a probability distribution because P(4) is not a probability satisfying the inequality 0 ≤ P(x) ≤ 1, and because the sum of all values of P(x) is not equal to 1. 14. The table represents a probability distribution because each distinct value of x corresponds to a probability P(x) such that 0 ≤ P(x) ≤ 1, and the sum of all values of P(x) is 1. 15. (a) ( ) ( ) ( ) ( ) 3 6 6 20 20 20 15 3 20 4 P 1 ≤ x ≤ 3 = P 1 + P 2 + P 3 = + + = = (b) ( ) ( ) ( ) ( ) 6 6 4 20 20 20 16 20 4 5 P x ≥ 2 = P 2 + P 3 + P 4 = + + = = Random variable 0 1 2 3 4 Frequency 1 4 6 4 1 Random variable 0 1 2 3 Frequency 1 3 3 1 Random variable, x 0 1 2 3 Probability, P(x) 18 3 8 3 8 18 0.1 3 0.3 0.4 0.2 0 1 2 P(x) x 2 30 4 30 6 30 8 30 x ) x P ) 0 1 2 3 4 5 0.1 3 4 0.3 0.2 0 1 2 x )x P ) Random variable, x 0 1 2 3 4 Probability, P(x) 1 16 4 16 6 16 4 16 1 16 Section 9.1 Discrete Probability 551 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 16. (a) ( ) ( ) ( ) ( ) 8 6 3 20 20 20 17 20 P x ≤ 2 = P 0 + P 1 + P 2 = + + = (b) ( ) ( ) 17 20 3 20 2 1 2 1 P x > = − P x ≤ = − = 17. (a) ( 3) (0) (1) (2) (3) 0.041 0.189 0.247 0.326 0.803 P x ≤ = P + P + P + P = + + + = (b) ( 3) (4) (5) 0.159 0.038 0.197 P x > = P + P = + = 18. (a) (1 2) (1) (2) 0.189 0.441 0.630 P ≤ x ≤ = P + P = + = (b) ( 2) (0) (1) 0.027 0.189 0.216 P x < = P + P = + = 19. (a) (b) (15 44) (15 24) (25 34) (35 44) 0.198 0.262 0.255 0.715 P − = P − + P − + P − = + + = (c) ( 35) (35 44) (45 54) (55 64) (65 and over) 0.255 0.195 0.069 0.017 0.536 P a ≥ = P − + P − + P − + P = + + + = (d) ( 24) (14 and under) (15 24) 0.004 0.198 0.202 P a ≤ = P + P − = + = 20. (a) (b) ( 2) (2) (3 or more) 0.167 0.092 0.259 P x ≥ = P + P = + = (c) ( 2) (0) (1) (2) 0.548 0.193 0.167 0.908 P x ≤ = P + P + P = + + = (d) ( 1) (1) (2) (3 or more) 0.193 0.167 0.092 0.452 P x ≥ = P + P + P = + + = 21. (a) S = {gggg, gggb, ggbg, gbgg, bggg, ggbb, gbbg, gbgb, bgbg, bbgg, bggb, gbbb, bgbb, bbgb, bbbg, bbbb} (b) x 0 1 2 3 4 P(x) 1 16 4 16 6 16 4 16 1 16 x 0 1 3 4 P(x) 2 8 20 4 20 0.4 0.3 0.2 0.1 0 1 2 3 4 5 x )x P ) x 0 1 3 0.1 0.2 0.3 0.4 0.5 P(x) 2 14 and under 15-24 25-34 35-44 45-54 55-64 65 and over 0.05 0.10 0.15 0.20 0.25 0.30 P(a) a x 0 1 3 or more P(x) 2 0.1 0.2 0.3 0.4 0.5 0.6 552 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (c) (d) 1 15 16 16 Probability of at least one boy 1 probability of all girls P 1 = − = − = 22. (a) {( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )} 1, 1 , 1, 2 , 1, 3 , 1, 4 , 1, 5 , 1, 6 , 2, 1 , 2, 2 , 2, 3 , 2, 4 , 2, 5 , 2, 6 , 3, 1 , 3, 2 , 3, 3 , 3, 4 , 3, 5 , 3, 6 , 4, 1 , 4, 2 , 4, 3 , 4, 4 , 4, 5 , 4, 6 , 5, 1 , 5, 2 , 5, 3 , 5, 4 , 5, 5 , 5, 6 , 6, 1 , 6, 2 , 6, 3 , 6, 4 , 6, 5 , 6, 6 S = (b) (c) (d) ( ) ( ) ( ) ( ) 3 2 1 6 1 36 36 36 36 6 P 10 ≤ x ≤ 12 = P 10 + P 11 + P 12 = + + = = = 0.16 23. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 3 8 3 1 48 16 16 16 16 16 16 1 3 8 3 1 1 7 16 16 16 16 16 16 8 1 2 3 4 5 3 1 3 2 3 3 3 4 3 5 3 14 0.875 0.9354 E x V x σ V x = + + + + = = = − + − + − + − + − = = = = = 24. ( ) ( ) 2 2 2 2 2 1 4 2 2 3 2 4 1 5 1 23 2.3 10 10 10 10 10 10 1 23 4 2 23 2 3 23 2 4 23 1 5 23 1 181 1.81 10 10 10 10 10 10 10 10 10 10 100 181 E x V x σ = ⎛ ⎞ + ⎛ ⎞ + ⎛ ⎞ + ⎛ ⎞ + ⎛ ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = ⎛ − ⎞ ⎛ ⎞ + ⎛ − ⎞ ⎛ ⎞ + ⎛ − ⎞ ⎛ ⎞ + ⎛ − ⎞ ⎛ ⎞ + ⎛ − ⎞ ⎛ ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = 181 1.345 100 10 = ≈ 25. ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 1 1 1 1 4 5 5 5 5 5 5 4 1 4 1 4 1 4 1 4 1 5 5 5 5 5 5 5 5 5 5 3 1 0 3 5 3 1 0 3 5 8.16 2.857 E x V x σ V x = − + − + + + = = − − + − − + − + − + − = = ≈ 3 4 2 16 4 16 6 16 0 1 2 x P(x) x 2 3 4 5 6 7 8 9 10 11 12 P(x) 1 36 1 18 1 12 19 5 36 1 6 5 36 19 1 12 1 18 1 36 x P(x) 2 4 6 8 10 12 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 Section 9.1 Discrete Probability 553 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 26. ( ) ( ) ( ) ( ) ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) 5000 0.008 2500 0.052 300 0.940 112 5000 112 0.008 2500 112 0.052 300 112 0.940 597,056 597,056 772.694 E x V x σ = − − + = = − − + − − + − = = ≈ 27. (a) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) 1 1 1 1 10 4 4 4 4 4 1 1 1 1 4 4 4 4 1 2 3 4 2.5 1 2.5 2 2.5 3 2.5 4 2.5 1.25 1.25 1.118 E x V x σ = + + + = = = − + − + − + − = = ≈ (b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) 1 2 3 4 3 2 1 80 16 16 16 16 16 16 16 16 1 2 3 4 3 2 1 16 16 16 16 16 16 16 5 2 2 3 4 5 6 7 8 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 2.5 2.5 1.581 E x V x σ = + + + + + + = = = − + − + − + − + − + − + − = = = ≈ 28. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) 1 4 6 4 1 16 16 16 16 16 1 4 6 4 1 16 16 16 16 16 0 1 2 3 4 2 0 2 1 2 2 2 3 2 4 2 1 1 1 E x V x σ = + + + + = = − + − + − + − + − = = ≈ 29. (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 10 0.25 15 0.30 20 0.25 30 0.15 40 0.05 18.50 thousand or 18,500 10 18.50 0.25 15 18.50 0.30 20 18.50 0.25 30 18.50 0.15 40 18.50 0.05 65.25 8.078 E x V x σ V x = + + + + = = − + − + − + − + − = = ≈ (b) Expected revenue: R = $4.95(18.50)(1000) = $91,575 30. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) 30 0.10 40 0.20 50 0.50 60 0.15 80 0.05 49 thousand dollars 30 49 0.10 40 49 0.20 50 49 0.50 60 49 0.15 80 49 0.05 119 119 10.909 or $10,909 E x V x σ = + + + + = = − + − + − + − + − = = ≈ 31. E(x) = 0(0.995) + 30,000(0.0036) + 60,000(0.0011) + 90,000(0.0003) = 201 Each customer should be charged $201. 32. E(x) = 0(0.936) + 1000(0.040) + 5000(0.020) + 10,000(0.004) = $180 33. (a) (b) x 0 1 2 3 4 P(x) 40 160 61 160 40 160 17 160 2 160 0 1 2 3 4 0.1 0.2 0.3 0.4 P(x) x 554 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (c) ( ) ( ) ( ) ( ) 61 40 17 160 160 160 118 59 160 80 1 3 1 2 3 0.7375 P ≤ x ≤ = P + P + P = + + = = = (d) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 40 61 40 17 2 5 160 160 160 160 160 4 5 40 5 61 5 40 5 17 5 2 39 4 160 4 160 4 160 4 160 4 160 40 39 40 0 1 2 3 4 0 1 2 3 4 0.975 0.975 0.9874 E x V x σ = + + + + = = − + − + − + − + − = = = = ≈ On average, you can expect the player to get 1.25 hits per game. The variance and standard deviation are measures of how spread out the data are. 34. Because News Station A said that there is a 40% chance of rain, there is a higher chance that it will rain on two or three of the days. So, graph (i) is for News Station A and graph (ii) is for News Station B. 35. ( ) ( 1 ) ( )(37 ) 2 ( ) ( 1 ) (37 ) 36 38 38 38 38 38 38 36 38 35 1 $0.05 or 36 0 1 $0.05 E x = + − = − ≈ − E x = + = − = − 36. ( ) ( 1 ) ( 1 ) ( 25 ) (2973) 3000 3000 3000 3000 E x = 2950 + 400 + 20 + 0 ≈ $1.28 Net gain = 1.28 − 2 = −$0.72 ALTERNATIVE METHOD: ( ) ( 1 ) ( 1 ) ( 25 ) ( )(2973) 3000 3000 3000 3000 E x = 2948 + 398 + 18 + −2 ≈ −$0.72 37. City 1: Expected value = 0.3(20) + 0.7(−4) = 3.2 million City 2: Expected value = 0.2(50) + 0.8(−9) = 2.8 million The company should open the store in City 1. 38. City 1: Expected value = 0.4(20) + 0.6(−4) = 5.6 City 2: Expected value = 0.25(50) + 0.75(−9) = 5.75 The company should open the store in City 2. 39. Answers will vary. x 2950 400 20 0 P(x) 1 3000 1 3000 25 3000 2973 3000 x 2948 398 18 −2 P(x) 1 3000 1 3000 25 3000 2973 3000 Section 9.2 Continuous Random Variables 555 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 9.2 Continuous Random Variables 1. f is a probability density function because ] 8 8 0 0 1 1 8 8 ∫ dx = x = 1 and ( ) 1 8 f x = ≥ 0 on [0, 8]. 2. f is not a probability density function because ] 4 4 0 0 1 1 4 5 5 5 ∫ dx = x = ≠ 1. 3. f is a probability density function because 2 4 4 0 0 4 14 1 8 8 2 x dx x x − ⎛ ⎞⎤ = ⎜ − ⎟⎥ = ⎝ ⎠⎥⎦ ∫ and ( ) 4 0 8 f x x − = ≥ on [0, 4]. 4. f is a probability density function because ( ) 0 18 f x = x ≥ on [0, 6] and 2 6 6 0 0 1. 18 36 x dx x ⎤ = = ⎥⎦ ∫ Skills Warm Up 1. The domain of the rational function f (x) 1 x = consists of all real numbers except x = 0.The value of 1 x is positive for all positive real numbers x. So, f (x) is continuous and nonnegative on [0, 1]. 2. The polynomial function f (x) = x2 − 1is continuous at every real number. Because f (0) = 02 − 1 = −1, however, f (x) is continuous but not nonnegative on [0, 1]. 3. The polynomial function f (x) = 3 − x is continuous at every real number, but f (5) = 3 − 5 = −2.So, f (x) is continuous but not nonnegative on [1, 5]. 4. The function ( ) x 1 x f x e e = − = is continuous at every real number and its value is positive on [0, 1].So, f (x) is continuous and nonnegative on [0, 1]. 5. ] 4 4 0 0 1 1 4 4 ∫ dx = x = 1 6. 2 2 1 1 2 2 1 2 1 2 2 1 4 1 3 4 1 4 x dx x dx x x − ⎛ ⎞ = ⎜ − ⎟ ⎝ ⎠ = ⎡ − ⎤ ⎢⎣ ⎥⎦ = − = ∫ ∫ 7. ( ) 3 3 0 0 3 0 3 3 lim 3 lim lim 1 1 t b t b t b b b b e dt e dt e e ∞ − − →∞ − →∞ − →∞ = = ⎡⎣− ⎤⎦ = − + = ∫ ∫ 0 0 8 0.25 0 0 4 1 0 0 4 1 0 0. 6 0 6 556 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 5. f (x) is not a probability density function because 1 7 7 3 0 0 1 7 ∫ e−x dx = ⎡⎣−e−x ⎤⎦ ≈ 0.349 ≠ 1 6. f is a probability density function because ( ) 1 6 6 f x = e−x > 0 on [0, ∞) and 6 6 0 0 1 6 lim 0 1 1. x x b b e dx e ∞ − − → ∞ ∫ = ⎡⎣− ⎤⎦ = + = 7. f is not a probability density function because ( ) 2 3 2 2 0 0 4 4 32 3 3 3 2 4 4 8 6.895 1. − x dx = − ⎡⎣ − x ⎤⎦ = − + ≈ ≠ ∫ 8. f is not a probability density function because ( ) ( ) ( ) 2 2 2 2 3 0 0 3 4 2 0 12 1 12 12 4 3 16 0 16 1 x x dx x x dx x x − = − = ⎡⎣ − ⎤⎦ = − − = − ≠ ∫ ∫ and because f (x) = 12x2 (1 − x) is negative on (1, 2). 9. f is a probability density function because ( ) 2 (3 ) 0 9 f x = x − x ≥ on [0, 3] and 2 3 3 3 2 0 0 2 2 2 1. 3 9 3 27 x x dx x x ⎛ ⎞ ⎡ ⎤ ⎜ − ⎟ = ⎢ − ⎥ = ⎝ ⎠ ⎣ ⎦ ∫ 10. f is a probability density function because f (x) = 0.4e−0.4x > 0 on [0, ∞) and 0.4 0.4 0 0 0.4 lim 0 1 1. x x b b e dx e ∞ − − → ∞ ∫ = − ⎦⎤ = + = 11. 2 4 4 1 1 15 1 2 2 2 15 kx dx kx k k ⎡ ⎤ = ⎢ ⎥ = = ⇒ = ⎣ ⎦ ∫ 12. 4 4 4 3 0 0 64 1 1 4 64 kx dx kx k k ⎤ = = = ⇒ = ⎥⎦ ∫ 13. ( ) 3 2 2 2 2 2 4 4 3 32 1 3 3 32 k x dx k x x k k − − ⎛ ⎞⎤ − = −⎥ ⎜ ⎟ ⎝ ⎠⎥⎦ = = ⇒ = ∫ ` 14. ( ) ( ) 1 11 2 3 2 0 0 1 3 2 5 2 0 1 2 2 3 5 4 1 15 15 4 k x x dx k x x dx k x x k k − = − = ⎡ − ⎤ ⎢⎣ ⎥⎦ = = ⇒ = ∫ ∫ 15. 2 2 0 0 1 2 lim 2 2 1 x x b b ke dx ke k k ∞ − − →∞ ∫ = ⎡⎣− ⎤⎦ = = ⇒ = 16. 1 1 b b a a k dx kx k k b a b a = ⎤ = = ⇒ = − − ⎥⎦ ∫ 0 0 3 0.2 0 0. 5 0 10 0 0 2 5 0 0 2 2 10 0 0 0.5 3 0 0 1 Section 9.2 Continuous Random Variables 557 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 17. 1 3 3 3 b b a a dx x b a ⎤ − = = ⎥⎦ ∫ (a) (0 2) 2 0 2 3 3 P x − < < = = (b) (1 2) 2 1 1 3 3 P x − < < = = (c) (1 3) 3 1 2 3 3 P x − < < = = (d) ( 2) 2 2 0 3 P x − = = = 18. 1 10 10 10 b b a a dx x b a ⎤ − = = ⎥⎦ ∫ (a) (0 6) 6 0 3 10 5 P x − < < = = (b) (4 6) 6 4 1 10 5 P x − < < = = (c) (8 10) 10 8 1 10 5 P x − < < = = (d) ( 2) (2 10) 10 2 4 10 5 P x P x − ≥ = < < = = 19. 2 2 2 50 100 100 b b a a x dx x b a ⎡ ⎤ − = ⎢ ⎥ = ⎣ ⎦ ∫ (a) (0 6) 36 0 9 100 25 P x − < < = = (b) (4 6) 36 16 1 100 5 P x − < < = = (c) (8 10) 100 64 9 100 25 P x − < < = = (d) ( 2) (2 10) 100 4 24 100 25 P x P x − ≥ = ≤ ≤ = = 20. 2 2 2 2 25 25 25 b b a a x dx x b a ⎡ ⎤ − = ⎢ ⎥ = ⎣ ⎦ ∫ (a) (0 3) 9 0 9 25 25 P x − < < = = (b) (1 3) 9 1 8 25 25 P x − < < = = (c) (3 5) 25 9 16 25 25 P x − < < = = (d) ( 1) (1 5) 25 1 24 25 25 P x P x − ≥ = < < = = 1 2 3 1 y x 2 3 10 10 1 x y x 4 8 12 1 5 2 5 y 1 2 3 4 5 1 x y 558 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 21. 3 3 2 3 2 1 16 16 3 8 b b a a x dx = ⎛⎜ ⎞⎟ x ⎤⎥ = ⎡⎣b b − a a ⎤⎦ ⎝ ⎠ ⎦ ∫ (a) (0 2) 2 0.354 4 P < x < = ≈ (b) (2 4) 1 2 0.646 4 P < x < = − ≈ (c) (1 3) 1(3 3 1) 0.525 8 P < x < = − ≈ (d) ( 3) 3 3 0.650 8 P x ≤ = ≈ 22. ( )2 ( ) 5 5 5 1 1 4 1 4 1 4 1 1 b b a a dx x x b a ⎤ ⎛ ⎞ = − ⎥ = − ⎜ − ⎟ + + + ⎝ ⎠ + ⎥⎦ ∫ (a) (0 2) 5 1 1 5 4 3 6 P < x < = − ⎛⎜ − ⎞⎟ = ⎝ ⎠ (b) (2 4) 5 1 1 1 4 5 3 6 P < x < = − ⎛⎜ − ⎞⎟ = ⎝ ⎠ (c) (1 3) 5 1 1 5 4 4 2 16 P < x < = − ⎛⎜ − ⎞⎟ = ⎝ ⎠ (d) ( 3) (0 3) 5 1 1 15 4 4 16 P x ≤ = P ≤ x ≤ = − ⎛⎜ − ⎞⎟ = ⎝ ⎠ 23. 1 3 3 3 3 3 b t t b a b a a e dt e e e − = − ⎤ = − − − ∫ ⎦ (a) P(t < 2) = e−0 3 − e−2 3 ≈ 0.4866 (b) P(t ≥ 2) = e−2 3 − 0 ≈ 0.5134 (c) P(1 < t < 4) = e−1 3 − e−4 3 ≈ 0.4529 (d) P(t = 3) = 0 24. ( ) 3 ( ) ( 3 3) 2 2 3 3 48 16 16 48 256 256 3 256 256 b b b a a a t t b a a b t dt t t ⎡ ⎤ − + − − = ⎢ − ⎥ = ⎡⎣ − ⎤⎦ = ⎣ ⎦ ∫ (a) ( ) ( ) (48 2) ( 64 8) 5 2 4 2 256 32 P t P t + − + < − = − < < − = = (b) ( ) ( ) ( ) ( ) 48 2 8 64 5 2 2 4 256 32 P t P t + − > = < < = = (c) ( ) ( ) ( ) 48 2 1 1 47 1 1 256 128 P t + − − − < < = = (d) ( 2) 1 ( 2) 1 5 27 32 32 P t > − = − P t < − = − = 25. ( ) 1 30 30 30 b b a a P a t b dt t b a ⎤ − < < = = = ⎥⎦ ∫ (a) (0 5) 5 0 1 30 6 P t − ≤ ≤ = = (b) (18 30) 30 18 2 30 5 P t − ≤ < = = 1 2 3 4 5 x y 2 16 4 16 6 16 x 1 1 2 3 4 5 y 2 3 1 3 1 1 2 3 4 t y t 1 4 −4 −2 2 4 y Section 9.2 Continuous Random Variables 559 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 26. ( ) (0.41 0.08 ) 0.41 0.04 2 (0.41 0.04 2 ) (0.41 0.04 2 ) b b a a P a < x < b = ∫ − x dx = ⎡⎣ x − x ⎤⎦ = b − b − a − a (a) ( ) ( ) ( )2 ( ) ( )2 P 0 ≤ x ≤ 3 = ⎡⎣0.41 3 − 0.04 3 ⎤⎦ − ⎡⎣0.41 0 − 0.04 0 ⎤⎦ = 0.87 (b) ( ) ( ) ( )2 ( ) ( )2 P 2 ≤ x ≤ 4 = ⎡⎣0.41 4 − 0.04 4 ⎤⎦ − ⎡⎣0.41 2 − 0.04 2 ⎤⎦ = 0.34 27. Using substitution, let u = 9 − t. Then t = 9 − u and dt = −du. ( ) ( ) ( ) ( ) 1 2 3 2 1 2 5 2 3 2 5 5 324 324 5 324 5 2 324 5 9 9 9 6 t t dt u u du u u du u u du − = − − = − = − ∫ ∫ ∫ ∫ 5 5 2( )5 2 ( )3 2 324 324 5 9 9 6 9 b b a a ∫ t − t dt = ⎡⎣ − t − − t ⎤⎦ (a) ( ) ( ) 3 0 1 3 P t < 3 = ∫ f t dt = 1 − 6 ≈ 0.1835 (b) ( ) ( ) 8 4 25 7 81 81 P 4 < t < 8 = ∫ f t dt = 5 − ≈ 0.6037 28. (a) Model A predicts a higher probability of selling between 0 and 2000 units, because the area of the region bounded by Model A is greater than that of Model B for 0 ≤ x ≤ 2. (b) Model B predicts a higher probability of selling between 4000 and 6000 units, because the area of the region bounded by Model B is greater than that of Model A for 4 ≤ x ≤ 6. 29. 1 3 3 3 3 3 b t t b a b a a ∫ e− dt = −e− ⎤⎦ = e− − e− (a) P(0 < t < 2) = e−0 3 − e−2 3 = 1 − e−2 3 ≈ 0.487 (b) P(2 < t < 4) = e−2 3 − e−4 3 ≈ 0.250 (c) P(t > 2) = 1 − P(0 < t < 2) = e−2 3 ≈ 0.513 30. 4 3 4 3 4 3 4 3 43 b t t b a b a a ∫ e− = −e− ⎤⎦ = e− − e− (a) P(0 < t < 1) = 1 − e−4 3 ≈ 0.736 (b) P(1 < t < 2) = e−4 3 − e−8 3 ≈ 0.194 (c) P(0 ≤ t ≤ 3) = 1 − e−4 ≈ 0.982 31. 1 5 5 5 5 5 b t t b a b a a ∫ e− dt = −e− ⎤⎦ = e− − e− (a) P(0 < t < 6) = 1 − e−6 5 ≈ 0.699 (b) P(2 < t < 6) = e−2 5 − e−6 5 ≈ 0.369 (c) ( 8) 8 5 lim b 5 8 5 0.202 b P t e− e− e− →∞ > = − = ≈ 32. 1 3.5 3.5 3.5 3.5 3.5 b t t b a b a a ∫ e− = −e− ⎤⎦ = e− − e− (a) P(0 < t < 1) = 1 − e−1 3.5 ≈ 0.249 (b) P(2 < t < 4) = e−2 3.5 − e−4 3.5 ≈ 0.246 (c) ( ) 3.5 5 3.5 3.5 5 3.5 5 3.5 5 5 lim lim 0 0.240 t b b b b P t e− e− e− e− e− →∞ →∞ ≥ = ⎡⎣− ⎤⎦ = − = − = ≈ 33. ( ) sin 1 cos 1 cos cos 30 15 2 15 2 15 15 b b a a P a x b x dx x b a π π ⎡ π ⎤ ⎛ π π ⎞ ≤ ≤ = = ⎢− ⎥ = − ⎜ − ⎟ ⎣ ⎦ ⎝ ⎠ ∫ (a) (0 10) 1 cos 10 cos 0 1 3 0.75 2 15 2 2 P x ⎛ π ⎞ ⎛ ⎞ ≤ ≤ = − ⎜ − ⎟ = − ⎜− ⎟ = ⎝ ⎠ ⎝ ⎠ There is a 75% probability of receiving up to 10 inches of rain. (b) (10 15) 1 cos 15 cos 10 1 1 0.25 2 15 15 2 2 P x ⎛ π π ⎞ ⎛ ⎞ ≤ ≤ = − ⎜ − ⎟ = − ⎜− ⎟ = ⎝ ⎠ ⎝ ⎠ There is a 25% probability of receiving between 10 and 15 inches of rain. (c) (0 5) 1 cos 5 cos 0 1 1 0.25 2 15 2 2 P x ⎛ π ⎞ ⎛ ⎞ ≤ < = − ⎜ − ⎟ = − ⎜− ⎟ = ⎝ ⎠ ⎝ ⎠ There is a 25% probability of receiving less than 5 inches of rain. (d) (12 15) 1 cos 15 cos 12 0.095 2 15 15 P x ⎛ π π ⎞ ≤ ≤ = − ⎜ − ⎟ ≈ ⎝ ⎠ There is about a 9.5% probability of receiving between 12 and 15 inches of rain. 560 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 34. 1 6 1 6 ( ) 36 6 6 b x x b a a ∫ xe− dx = − e− x + ⎤⎦ (a) ( ) ( ) ( ) 6 6 1 0 1 1 6 6 P x < 6 = − e−x x + 6 ⎤⎦ = − 12e− − 6 ≈ 0.264 The probability that weekly demand for a certain product will be less than 6 tons is 0.264 (b) ( ) ( ) 6 12 1 2 6 1 6 P 6 < x < 12 = − x−x x + 6 ⎤⎦ = 2e− − 3e− ≈ 0.330 The probability that weekly demand for a certain product will be more than 6 tons but less than 12 tons is 0.330. (c) P(x > 12) = 1 − P(x ≤ 12) = 1 − (1 − 3e−2 ) ≈ 0.406 The probability that weekly demand for a certain product will be at least 12 tons is 0.406. 35. 1 16 1 6 ( 6) 36 6 b b x x a a xe− dx e− x − ⎤ = + ⎥⎦ ∫ ( ) ( ) ( ) ( ) ( ) 6 0 6 6 6 0 0.90 1 6 0.90 6 1 6 6 0.90 6 1 6 0.10 6 6 0.60 x b b b b P x b e x e b b e b e − − − − < < = − + ⎤⎦ = − ⎡⎣ + − ⎤⎦ = − + = − + = Solving for b with a graphing utility or using a computer, b ≈ 23.34 tons. Section 9.3 Expected Value and Variance Skills Warm Up 1. ] 0 0 1 0.5 16 1 0.5 16 1 0.5 16 8 m m dx x m m = = = = ∫ 2. ( ) 3 0 3 0 3 0 3 3 3 1 0.5 3 0.5 0.5 1 0.5 0.5 0.5 ln 0.5 3 3 ln 0.5 2.0794 m t t m m m m m e dt e e e e e e m m − − − − − − = − ⎤⎦ = − − = − + = − = − = − = = − ≈ ∫ 3. ( ) 2 2 2 3 3 3 0 0 1 1 2 0 8 4 2 6 6 6 3 x dx = x ⎤ = − = = ⎥⎦ ∫ 4. ( ) ( ) 2 2 2 1 1 2 3 2 1 2 8 4 4 3 3 3 3 4 2 4 2 2 x x dx x x dx x x − = − = ⎣⎡ − ⎦⎤ = − = ∫ ∫ 5. ( ) 1 1 8 8 8 b b a a P a x b dx x b a ⎤ − ≤ ≤ = = = ⎥⎦ ∫ (a) ( 2) (0 2) 2 0 1 8 4 P x P x − ≤ = ≤ ≤ = = (b) (3 7) 7 3 1 8 2 P x − < < = = Section 9.3 Expected Value and Variance 561 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. (a) ( ) 2 3 3 0 0 1 3 3 6 2 b a μ x f x dx x dx x ⎛ ⎞ ⎤ = = ⎜ ⎟ = ⎥ = ⎝ ⎠ ⎦ ∫ ∫ (d) (b) ( ) 2 3 3 2 2 2 3 2 0 0 1 3 9 3 9 3 3 2 9 4 4 4 b a σ x f x dx μ x dx x ⎛ ⎞ ⎛ ⎞ ⎤ = − = ⎜ ⎟ − ⎜ ⎟ = ⎥ − = − = ⎝ ⎠ ⎝ ⎠ ⎦ ∫ ∫ (c) 3 3 4 2 σ = = 2. (a) ( ) 2 4 4 0 0 1 2 4 8 b a μ x f x dx x dx x ⎛ ⎞ ⎤ = = ⎜ ⎟ = ⎥ = ⎝ ⎠ ⎦ ∫ ∫ (d) (b) ( ) ( ) 3 4 2 2 2 4 2 2 0 0 1 2 4 16 12 4 4 12 3 3 3 b a σ x f x dx μ x dx x ⎛ ⎞ ⎤ = − = ⎜ ⎟ − = ⎥ − = − = ⎝ ⎠ ⎦ ∫ ∫ (c) 2 3 σ = 3. (a) ( ) 3 6 6 0 0 4 18 54 b a μ t f t dt t t dt t ⎛ ⎞ ⎤ = = ⎜ ⎟ = ⎥ = ⎝ ⎠ ⎦ ∫ ∫ (d) (b) ( ) 4 6 2 2 2 6 2 2 2 0 0 4 4 18 16 2 18 72 b a σ t f t dt μ t t dt t ⎛ ⎞ ⎤ = − = ⎜ ⎟ − = ⎥ − = − = ⎝ ⎠ ⎦ ∫ ∫ (c) σ = 2 4. (a) ( ) 3 4 4 0 0 8 8 24 3 b a μ x f x dx x x dx x ⎛ ⎞ ⎤ = = ⎜ ⎟ = ⎥ = ⎝ ⎠ ⎦ ∫ ∫ (d) (b) ( ) 2 4 4 2 2 2 4 2 0 0 8 64 8 8 3 32 9 9 b a σ x f x dx μ x x dx x ⎛ ⎞ ⎛ ⎞ ⎤ = − = ⎜ ⎟ − ⎜ ⎟ = ⎥ − = ⎝ ⎠ ⎝ ⎠ ⎦ ∫ ∫ (c) 8 2 2 9 3 σ = = 5. (a) ( ) 4 4 4 1 2 1 1 4 4 1 4 ln 4 ln 4 1.848 3 3 3 3 b a x f x dx x dx dx x x x μ = = ⎛⎜ ⎞⎟ = = ⎤⎥ = ≈ ⎝ ⎠ ⎦ ∫ ∫ ∫ (b) ( ) ( ) ( ) 2 4 2 2 2 4 2 2 2 1 2 1 4 4 ln 4 4 16 ln 4 4 16 ln 4 0.583 3 3 3 9 9 b a x f x dx x dx x x σ = − μ = ⎛⎜ ⎞⎟ − ⎡⎢ ⎤⎥ = ⎤⎥ − = − ≈ ⎝ ⎠ ⎣ ⎦ ⎦ ∫ ∫ (c) σ = 0.583 ≈ 0.76 Skills Warm Up —continued— 6. ( ) (6 6 2 ) 3 2 2 3 (3 2 2 3) (3 2 2 3) b b a a P a ≤ x ≤ b = ∫ x − x dx = ⎡⎣ x − x ⎤⎦ = b − b − a − a (a) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 2 1 3 2 3 1 2 2 2 2 2 P x ≤ = P 0 ≤ x ≤ = ⎡⎢3 − 2 ⎤⎥ − ⎡⎣3 0 − 2 0 ⎤⎦ = ⎣ ⎦ (b) ( ) ( ) ( ) ( ) ( ) 1 3 3 2 3 3 1 2 1 3 27 5 11 4 4 4 4 4 4 32 32 16 P ≤ x ≤ = ⎡⎢3 − 2 ⎤⎥ − ⎡⎢3 − 2 ⎤⎥ = − = ⎣ ⎦ ⎣ ⎦ 1 2 3 x y Mean 2 3 1 3 x 1 2 3 Mean y 4 1 4 1 1 2 3 4 5 6 t 2 3 1 3 Mean y x Mean y 1 2 3 4 0.5 1.0 562 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (d) 6. (a) ( ) 1 1 3 2 1 5 2 7 2 0 0 0 5 5 5 2 5 0.714 2 2 2 7 7 b a μ x f x dx x x dx x dx x ⎛ ⎞ ⎛ ⎞ ⎤ = = ⎜ ⎟ = = ⎜ ⎟ ⎥ = ≈ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ ∫ ∫ (b) ( ) 2 1 2 2 2 1 2 3 2 1 7 2 9 2 0 0 0 5 5 5 25 5 2 25 5 25 20 2 7 2 49 2 9 49 9 49 441 b a σ = x f x dx − μ = x ⎛⎜ x ⎞⎟ dx − ⎛⎜ ⎞⎟ = x dx − = ⎛⎜ ⎞⎟ x ⎤⎥ − = − = ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ ∫ ∫ (c) 2 5 21 σ = (d) 7. (a) ( ) 2 2 2 3 0 0 1 1 1 1 2 2 2 6 3 b a μ = x f x dx = x⎛⎜ − x⎞⎟ dx = x − x ⎤⎥ = ⎝ ⎠ ⎦ ∫ ∫ (b) ( ) 2 2 2 2 2 2 2 3 4 0 0 1 1 2 1 1 4 2 2 3 3 8 9 9 b a σ = x f x dx − μ = x ⎛⎜ − x⎞⎟ dx − ⎛⎜ ⎞⎟ = x − x ⎤⎥ − = ⎝ ⎠ ⎝ ⎠ ⎦ ∫ ∫ (c) 2 2 9 3 σ = = (d) 8. (a) ( ) 4 4 2 3 0 0 1 1 1 1 4 2 8 4 24 3 b a μ = x f x dx = x⎛⎜ − x⎞⎟ dx = x − x ⎤⎥ = ⎝ ⎠ ⎦ ∫ ∫ (d) (b) ( ) 2 2 2 2 4 2 0 4 3 4 0 1 1 4 2 8 3 1 1 16 8 6 32 9 9 b a x f x dx x x dx x x σ = − μ = ⎛⎜ − ⎞⎟ − ⎛⎜ ⎞⎟ ⎝ ⎠ ⎝ ⎠ = − ⎤ − = ⎥⎦ ∫ ∫ (c) 8 2 2 9 3 σ = = x 1 2 1 3 4 Mean y 1 3 1 2 x Mean y 1 2 2 y x Mean x Mean y 1 2 3 4 1 Section 9.3 Expected Value and Variance 563 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 9. (a) ( ) ( ) 1 0 1 3 4 0 6 1 2 3 1 2 2 b a x f x dx x x x dx x x μ = = ⎡⎣ − ⎤⎦ = − ⎤ = ⎥⎦ ∫ ∫ (d) (b) ( ) ( ) 2 2 2 2 1 2 0 1 4 5 0 6 1 1 2 3 6 1 2 5 4 1 20 b a x f x dx x x x dx x x σ = − μ = ⎡⎣ − ⎤⎦ − ⎛⎜ ⎞⎟ ⎝ ⎠ = − ⎤ − ⎥⎦ = ∫ ∫ (c) 1 1 5 20 2 5 10 σ = = = 10. (a) ( ) ( ) 4 0 4 3 4 0 3 4 32 1 3 2 8 128 b a x f x dx x x x dx x x μ = = ⎡⎢ − ⎤⎥ ⎣ ⎦ = − ⎤ = ⎥⎦ ∫ ∫ (d) (b) ( ) ( ) ( ) 2 2 2 4 2 2 0 4 4 5 0 3 4 2 32 3 3 4 32 160 4 5 b a x f x dx x x x dx x x σ = − μ = ⎡⎢ − ⎤⎥ − ⎣ ⎦ = − ⎤ − ⎥⎦ = ∫ ∫ (c) 4 2 2 5 5 5 5 σ = = = 11. (a) ( ) ( ) ( ) ( ) 4 4 3 2 4 0 0 0 3 3 1 8 16 16 40 5 4 4 4 3 8 1.6 b a μ = ∫ x f x dx = ∫ x − x dx = ∫ x − x dx = − − x x + ⎤⎦ = = (b) ( ) ( ) ( ) ( ) ( ) 2 2 2 4 2 2 4 2 0 0 3 2 2 4 0 3 8 3 64 16 5 16 25 1 64 128 64 192 280 25 35 25 175 4 4 4 15 48 128 1.097 b a x f x dx x x dx x x dx x x x σ = − μ = − − = − − = − − + + ⎤ − = − = ≈ ⎦ ∫ ∫ ∫ (c) σ ≈ 1.047 (d) 12. (a) ( ) ( ) ( ) 9 0 9 9 3 2 0 0 1 9 18 1 9 1 9 6 18 45 18 3.6 5 b a x f x dx x x dx x xdx x x μ = = ⎛⎜ − ⎞⎟ ⎝ ⎠ = − = − − + ⎤⎥⎦ = = ∫ ∫ ∫ 1 2 1 1 2 y x Mean x Mean y 1 1 2 3 4 x 1 2 3 4 Mean 1 2 1 4 y 564 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (b) ( ) ( ) ( ) 2 2 2 2 9 2 0 9 3 2 2 0 1 9 18 18 5 1 9 5 36 216 324 315 25 972 175 b a x f x dx x x dx x x x σ = − μ = ⎛⎜ − ⎞⎟ − ⎛⎜ ⎞⎟ ⎝ ⎠ ⎝ ⎠ = − − + + ⎤ − ⎥⎦ = ∫ ∫ (d) (c) 972 18 21 175 35 σ = = 13. Mean 1 (0 11) 5.5 2 = + = 0 Median : 1 0.5 11 0.5 11 5.5 m m dx m m = = = = ∫ 14. 1 ( ) 2 Mean = 0 + 20 = 10 0 Median : 0.05 0.5 0.05 0.5 10 m m dx m m = = = = ∫ 15. ( )( ) 1 2 0 1 6 Mean = ∫ x 4 1 − 2x dx = ( ) 0 2 0 2 1 2 1 2 1 2 Median : 41 2 4 4 4 4 0.1465 m m m xdx x x m m m = − = − ⎤⎦ = − = ⇒ ≈ ∫ ( 1 ) 2 m ≈ 0.8536 is not in the interval ⎡⎣0, ⎤⎦. 16. 1 0 Mean 4 2 4 3 3 9 = x⎛⎜ − x⎞⎟ dx = ⎝ ⎠ ∫ 0 2 0 2 Median : 4 2 1 3 3 2 4 1 3 3 2 4 1 0.4189 3 3 2 m m m x dx x x m m m = ⎛ − ⎞ = ⎜ ⎟ ⎝ ⎠ ⎤ − = ⎥⎦ − = ⇒ ≈ ∫ 17. 9 ( ) 9 0 0 Mean 1 9 9 9 t e t dt t e t ∞ = ∞ ⎛ − ⎞ = − + − ⎤ = ⎜ ⎟ ⎥ ⎝ ⎠ ⎦ ∫ 9 9 9 0 0 Median 1 1 1 9 2 m = m e−t dt = −e−t ⎤ = − em = ⎥⎦ ∫ 9 1 ln 1 2 9 2 9 ln 1 6.238 2 e m m m − = ⇒ − = = − ≈ 18. 2 5 ( ) 2 5 0 0 Mean 2 1 2 5 5 2 5 2 t e t dt t e t ∞ = ∞ ⎛ − ⎞ = − + − ⎤ ⎜ ⎟ ⎥ ⎝ ⎠ ⎦ = ∫ 2 5 2 5 0 0 2 5 2 5 Median 2 5 1 1 2 1 2 ln 1 2 5 2 5 ln 1 5 ln 2 1.733 2 2 2 m m t t m m e dt e e e m m − − − − = = − ⎤⎥⎦ = − = = ⇒ − = = − = ≈ ∫ 19. ( ) 1 , 10 f x = [0, 10] is a uniform probability density function. Mean: 0 10 5 2 2 a + b + = = ( )2 ( )2 10 0 100 25 Variance: 12 12 12 3 b − a − = = = Standard deviation: 10 0 2.887 12 12 b − a − = ≈ 20. ( ) 1, [0, 6] 6 f x = is a uniform probability density function. Mean: 0 6 3 2 2 a + b + = = Variance: ( ) ( )2 6 0 36 3 12 12 12 b − a − = = = Standard deviation: 6 0 6 3 1.732 12 12 12 b − a − = = = ≈ Mean x y 1 2 3 4 5 6 7 8 9 0.5 Section 9.3 Expected Value and Variance 565 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 21. ( ) 1 8, 8 f x = e−x [0, ∞) is an exponential probability density function with 1. 8 a = Mean: 1 8 a = 2 Variance: 1 64 a = Standard deviation: 1 8 a = 22. ( ) 5 5 3, 3 f x = e− x [0, ∞) is an exponential probability density function with 5. 3 a = Mean: 1 1 3 a 5 3 5 = = 2 ( )2 Variance: 1 1 9 a 5 3 25 = = Standard deviation: 1 1 3 a 5 3 5 = = 23. ( ) 1 ( 100)2 242 , 11 2 f x e x π = − − (−∞, ∞) is a normal probability density function with μ = 100 and σ = 11. Mean: μ = 100 Variance: σ 2 = 121 Standard deviation: σ = 11 24. ( ) 1 ( 30)2 72, 6 2 f x e x π = − − (−∞, ∞) is a normal probability density function. Mean: μ = 30 Variance: σ 2 = 36 Standard deviation: σ = 6 25. Mean = 0 Standard deviation = 1 P(0 ≤ x ≤ 0.85) ≈ 0.3023 26. Mean = 0 Standard deviation = 1 P(−1.21 ≤ x ≤ 1.21) ≈ 0.7737 27. Mean = 6 Standard deviation = 6 P(x ≥ 2.23) ≈ 0.6896 28. 43 Mean = 43 Standard deviation = P(x ≥ 0.27) ≈ 0.8167 29. Mean = 8 Standard deviation = 2 P(3 ≤ x ≤ 13) ≈ 0.9876 30. Mean = 2 Standard deviation = 1.5 P(−2.5 ≤ x ≤ 2.5) ≈ 0.6292 31. μ = 50, σ = 10 (a) ( ) ( 50)2 2(102 ) 55 55 1 10 2 0.3085 x P x e dx π ∞ − − > = ≈ ∫ (b) ( ) ( 50)2 2(102 ) 60 60 1 10 2 0.1587 x P x e dx π ∞ − − > = ≈ ∫ (c) ( ) 60 60 1 ( 50)2 2(102 ) 10 2 0.8413 x P x e dx π − − −∞ < = ≈ ∫ (d) ( ) 55 ( 50)2 2(102 ) 30 30 55 1 10 2 0.6687 x P x e dx π − − < < = ≈ ∫ 32. μ = 70, σ = 14 (a) ( ) ( 70)2 2(142 ) 65 65 1 14 2 0.6395 x P x e dx π ∞ − − > = ≈ ∫ (b) ( ) 98 98 1 ( 70)2 2(142 ) 14 2 0.9772 x P x e dx π − − −∞ < = ≈ ∫ (c) ( ) 49 49 1 ( 70)2 2(142 ) 14 2 0.0668 x P x e dx π − − −∞ < = ≈ ∫ (d) ( ) 75 ( 70)2 2(142 ) 56 56 75 1 14 2 0.4809 x P x e dx π − − < < = ≈ ∫ 566 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 33. (a) Since the arrival time t (in minutes) is uniformly distributed between 10:00 A.M. and 10:10 A.M., let ( ) 1 10 f t = over [0, 10], where t = 0 corresponds to 10:00 A.M. (b) Mean 10 5 2 = = The mean is 10:05 A.M. Standard deviation 10 2.887 min 12 = ≈ (c) 10 3 1 1 1 7 3 0.30 10 10 10 − ∫ dx = − = = 34. (a) Let ( ) 1 5 f t = over [0, 5], where t = 0 represents 10:00 A.M. and t = 5 represents 10:05 A.M. (b) Mean: 5 2 μ = or 2.5 minutes after 10:00 A.M. Standard deviation: 5 5 3 12 6 1.443 min σ = = ≈ (c) ( ) 5 5 3 3 3 5 1 1 2 0.40 5 5 5 P t dt t ⎤ < < = = = = ⎥⎦ ∫ 35. (a) Because μ = 2, ( ) 1 2 2 f t = e−t , 0 ≤ t < ∞. (b) ( ) 1 2 2 1 1 2 0 0 1 2 P 0 < t < 1 = ∫ e−t dt = −e−t ⎤⎦ = 1 − e− ≈ 0.3935 36. (a) Because ( ) 1 4 4 μ = 4, f t = e−t , 0 ≤ t < ∞. (b) ( ) 3 4 4 3 1 4 3 4 1 1 1 4 P 1 < t < 3 = ∫ e−t dt = −e−t ⎤⎦ = e− − e− ≈ 0.3064 37. (a) Because μ = 5, ( ) 1 5 5 f t = e−t , 0 ≤ t < ∞. (b) ( ) ( ) 10 5 5 10 2 0 0 1 5 P μ − σ < t < μ + σ = P 0 < t < 10 = ∫ e−t dt = −e−t ⎤⎦ = 1 − e− ≈ 0.865 = 86.5% 38. (a) Because μ = 15, ( ) 1 15 15 f t = e−t , 0 ≤ t < ∞. (b) ( ) ( ) 30 15 15 30 2 0 0 1 15 P μ − σ < t < μ + σ = P 0 < t < 30 = ∫ e−t dt = −e−t ⎤⎦ = 1 − e− ≈ 0.865 = 86.5% 39. (a) ( ) ( ) 3 7 7 7 2 2 0 0 0 6 7 6 7 6 7 1 343 343 343 2 3 μ x x dx x x dx x x ⎡ ⎤ = − = − = ⎢ − ⎥ = ⎣ ⎦ ∫ ∫ ( ) ( ) ( ) 4 7 2 7 2 2 7 2 3 3 0 0 0 6 7 1 6 7 1 6 7 1 7 1 5 343 343 343 3 4 2 2 σ x x dx x x dx x x ⎡ ⎤ = − − = − − = ⎢ − ⎥ − = − = ⎣ ⎦ ∫ ∫ 5 10 1.581 2 2 σ = = ≈ (b) ( ) ( ) ( )( ) 3 2 3 2 2 0 0 0 2 3 2 6 7 6 7 6 7 6 7 1 343 343 343 2 3 343 2 3 2 7 343 0 3.54 28 98 2 3 12 3.5 or 7 147 4 m m x x dx m x x dx x x m m m m m m m m m ⎡ ⎤ ⎡ ⎤ − = − = ⎢ − ⎥ = ⎢ − ⎥ = ⎣ ⎦ ⎣ ⎦ − = ⇒ = − − − ± = = ∫ ∫ In the interval [0, 7], m = 3.5. (c) ( ) ( ) ( ) 3 2.5811 2.5811 2 0.5811 0.5811 1 10 1 10 0.5811 2.581 2 2 6 7 6 7 0.286 343 343 2 3 P x P x P x x x dx x x μ σ μ σ − − ⎛ ⎞ − < < + = ⎜⎜ − < < + ⎟⎟ ≈ − < < ⎝ ⎠ ⎡ ⎤ = − = ⎢ − ⎥ ≈ ⎣ ⎦ ∫ Section 9.3 Expected Value and Variance 567 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 40. (a) ( )( ) ( ) 4 10 10 10 3 2 3 2 2 2 2 3 2 10 3 12 20 3 4 10 6 256 256 256 4 μ x x x dx x x x dx x x x ⎡ ⎤ = − − = − + − = ⎢− + − ⎥ = ⎣ ⎦ ∫ ∫ ( )( ) ( ) ( ) 5 3 10 2 10 2 2 10 4 3 2 4 2 2 2 3 2 10 6 3 12 20 36 3 3 20 256 256 256 5 3 196 36 16 5 5 σ x x x dx x x x dx x x x ⎡ ⎤ = − − − = − + − − = ⎢− + − ⎥ ⎣ ⎦ = − = ∫ ∫ 16 4 1.789 5 5 σ = = ≈ (b) ( )( ) ( ) 3 2 2 2 2 2 3 2 3 2 10 3 12 20 3 6 20 256 256 256 3 3 6 20 3 56 1 256 3 256 3 2 m m x x dx m x x dx x x x m m m ⎡ ⎤ − − = − + − = ⎢− + − ⎥ ⎣ ⎦ ⎛ ⎞ ⎛ ⎞ = ⎜− + − ⎟ − ⎜− ⎟ = ⎝ ⎠ ⎝ ⎠ ∫ ∫ So, 3 2 3 2 18 60 56 128 0 18 60 72. m m m m m m − + − + = = − + + By the Rational Zero Theorem or a graphing utility, we have m = 6. (c) ( ) ( )( ) 3 7.7889 7.7889 2 4.2111 4.2111 6 4 6 4 4.2111 7.7889 5 5 3 2 10 3 6 20 0.626 256 256 3 P x P x x x dx x x x ⎛ ⎞ ⎜ − < < + ⎟ ≈ < < ⎝ ⎠ ⎡ ⎤ = − − = ⎢− + − ⎥ ≈ ⎣ ⎦ ∫ 41. ( ) 2 1, 0 3 12 f x x x + = ≤ ≤ 3 0 3 2 3 0 Expected value 2 1 12 1 2 12 3 2 15 8 x x dx x x ⎛ + ⎞ = ⎜ ⎟ ⎝ ⎠ ⎡ ⎤ = ⎢ + ⎥ ⎣ ⎦ = ∫ 42. ( ) 0.28 0 0 0.28 0.28 0.28 0.5 1 0.5 0.5 1 ln 0.5 0.28 2.4755 m m x m m f x dx e dx e e m − − − = = − + = = = − ≈ ∫ ∫ 43. (a) 2 3 ( ) 0 2 3 3 0 1 Use integration by parts. 9 lim 3 2 1 6 9 3 x b x x b x e dx x e x e μ ∞ − − − → ∞ = ⎡ ⎛ ⎞ ⎤ = − ⎢ − ⎜− − ⎟ ⎥ = ⎣ ⎝ ⎠ ⎦ ∫ ( ) ( ) ( ) ( ( ) ) 2 3 3 2 0 3 3 2 3 0 0 1 6 Use integration by parts. 9 lim 9 1 36 3 9 0 96 36 18 Use part a . x b x x b x e dx xe xe dx σ ∞ − − ∞ − →∞ = − ⎡ ⎤ = ⎢− ⎥ + − ⎣ ⎦ = + − = ∫ ∫ σ = 18 = 3 2 ≈ 4.243 (b) ( ) ( ) ( ) 4 4 3 3 0 0 4 1 4 1 1 1 1 3 0.615 9 3 P x > = − P x < = − xe−x dx = − ⎡− e−x x + ⎤ ≈ ⎢⎣ ⎥⎦ ∫ 568 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 44. Draw a vertical line to form two regions with equal areas. 45. (a) 340 400 2.5 24 x μ σ − − = = − Your battery fell short of the expected life by 2.5 standard deviations. (b) ( ) ( ) ( ) ( ) 400 2 1152 400 2 1152 340 1 24 2 340 1 24 2 0.9938 x x f x e P x e dx π π − − ∞ − − = > = ≈ ∫ Note: 1 − P(0 < x < 340) ≈ 0.9938 46. (a) 174 150 3 1.5 16 2 x μ σ − − = = = Your score exceeded the national mean by 1.5 standard deviations. (b) ( ) ( ) ( ) ( ) 150 2 512 174 150 2 512 0 1 16 2 174 1 16 2 0.9332 x x f x e P x e dx π π − − − − = > = ≈ ∫ 47. u = 4.5, σ = 0.5 (a) ( ) ( ) ( ) ( ) 5 4.5 2 2 0.5 2 4 5 2 4.5 2 4 4 5 1 0.5 2 2 0.6827 2 x x P x e dx e dx π π − − − − < < = = ≈ ∫ ∫ (b) ( ) ( ) ( ) ( ) 3 4.5 2 2 0.5 2 0 3 2 4.5 2 0 3 1 0.5 2 2 2 0.0013 0.13% x x P x e dx e dx π π − − − − < = = ≈ = ∫ ∫ No, only about 0.13% of the batteries will last less than 3 years. 48. μ = 14.50, σ = 1.50 (a) ( ) ( ) ( ) ( ) ( ) 14 14.50 2 2 1.5 2 14 2 14.50 2 9 11 11 11 14 1 2 1.50 2 3 2 0.3596 Using Simpson’s Rule with 10 35.96% P x e x dx e x dx n π π < < = − − = − − ≈ = = ∫ ∫ (b) ( ) 2( 14.50)2 9 16 16 2 0.1587 3 2 P x e x dx π > = ∫ ∞ − − ≈ No, only about 15.87% will be paid more than $16 per hour. 49. ( ) 1 ( 266)2 2(16)2 1 ( 266)2 512 16 2 40.106 f x e x e x π = − − = − − (a) (b) ( ) ( ) 280 240 240 280 0.757 or 76% P ≤ x ≤ = f x dx ≈ ∫ (c) ( ) ( ) 280 P x 280 f x dx 0.191 or 19.1% ∞ > = ∫ ≈ 50. (a) ( ) 1 ( 21.1)2 2(5.1)2 5.1 2 x f x e π − − = (b) ( ) 36 ( 21.1)2 2(5.1)2 24 24 36 1 5.1 2 0.2831 or 28.31% P x e x dx π ≤ ≤ = − − ≈ ∫ (c) ( ) ( 21.1)2 2(5.1)2 26 26 1 5.1 2 0.1683 or 16.83% x P x e dx π ∞ − − > = ≈ ∫ Probability Distance from nucleus Median 0.03 −0.01 0 400 0 5 40 0.1 Review Exercises for Chapter 9 569 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 51. (a) ( ) 1 ( 21.0)2 2(5.6)2 5.6 2 f x e x π = − − (b) ( ) 30 ( 21.0)2 2(5.6)2 25 25 30 1 0.1835 or 18.35% 5.6 2 P x e x dx π ≤ ≤ = ∫ − − ≈ (c) ( ) 18 ( 21.0)2 2(5.6)2 0 18 1 0.2960 or 29.60% 5.6 2 P x e x dx π < = ∫ − − ≈ Review Exercises for Chapter 9 1. (a) S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} (b) A = {9, 10, 11, 12} (c) B = {1, 3, 5, 7, 9, 11} 2. (a) S = {c, a, l, s, u} (b) A = {a, u} (c) B = {c} 3. (a) S = {000, 001, 002, 003,…, 997, 998, 999} The sample space consists of all distinct orders of three digits from 0 to 9. (b) A = {100, 101, 102, 103,…, 198, 199} (c) B = {050,100,150, 200, 250, 300, 350,…, 900, 950} 4. (a) {2 3 , 10 J Q K A 3 ,10 J Q K A 3 ,10 J Q K A 3 ,10 J Q K A S = ♣, ♣, ♣, ♣, ♣, ♣, ♣, 2♠ ♠, ♠, ♠, ♠, ♠, ♠, 2♥ ♥, ♥, ♥, ♥, ♥, ♥, 2♦ ♦, ♦, ♦, ♦, ♦, ♦} … … … … The sample space consists of 52 cards, 2 through A (13 cards), of 4 suits. (b) A = {2♥, 3♥,…, K♥, A♥, 2♦, 3♦,…, K♦, A♦} The event consists of 26 cards, 2 through A (13 cards) of hearts and diamonds. (c) B = {5♣, 5♠} 5. 6. 0 8 34 0.1 x 0 1 2 3 n(x) 1 3 3 1 4 3 2 1 0 1 2 3 x n(x) x 0 1 2 3 4 1 2 3 4 5 6 n(x) x 0 1 2 3 4 n(x) 1 4 6 4 1 570 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 7. 8. 9. The table does not represent a probability distribution because the sum of the probabilities does not equal 1. 10. 11. (a) ( ) ( ) ( ) ( ) 7 5 3 15 5 18 18 18 18 6 P 2 ≤ x ≤ 4 = P 2 + P 3 + P 4 = + + = = (b) ( ) ( ) ( ) ( ) 5 3 2 10 5 18 18 18 18 9 P x ≥ 3 = P 3 + P 4 + P 5 = + + = = 12. (a) ( ) 1 2 3 11 11 11 P x < 0 = + = (b) ( ) 3 1 4 11 11 11 P x > 1 = + = 13. ( ) ( ) ( )2 ( )2 ( )2 ( )2 ( )2 0 1 1 3 2 2 3 3 4 1 2 10 10 10 10 10 0 2 1 1 2 3 2 2 2 3 2 3 4 2 1 7 10 10 10 10 10 5 7 35 5 5 E x V x σ = ⎛ ⎞ + ⎛ ⎞ + ⎛ ⎞ + ⎛ ⎞ + ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = − ⎛ ⎞ + − ⎛ ⎞ + − ⎛ ⎞ + − ⎛ ⎞ + − ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = = 14. ( ) ( ) 2 2 2 2 2 1 1 2 1 3 2 4 3 5 1 13 8 8 8 8 8 4 1 13 1 2 13 1 3 13 2 4 13 3 5 13 1 23 4 8 4 8 4 8 4 8 4 8 16 23 23 16 4 E x V x σ = ⎛ ⎞ + ⎛ ⎞ + ⎛ ⎞ + ⎛ ⎞ + ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = ⎛ − ⎞ ⎛ ⎞ + ⎛ − ⎞ ⎛ ⎞ + ⎛ − ⎞ ⎛ ⎞ + ⎛ − ⎞ ⎛ ⎞ + ⎛ − ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = = 15. ( ) ( ) ( ) ( ) ( ) ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) 0 0.006 1 0.240 2 0.614 3 0.140 1.888 0 1.888 0.006 1 1.888 0.240 2 1.888 0.614 3 1.888 0.140 0.391456 0.391456 0.6257 E x V x σ = + + + = = − + − + − + − = = ≈ 16. ( ) ( ) ( ) ( ) ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) 0 0.310 1 0.685 2 0.005 0.695 0 0.695 0.310 1 0.695 0.685 2 0.695 0.005 0.221975 0.221975 0.4711 E x V x σ = + + = = − + − + − = = ≈ x 0 1 2 3 ( ) P x 18 3 8 3 8 18 x 0 1 2 3 4 P(x) 1 16 1 4 3 8 1 4 1 16 1 25 3 25 1 5 7 25 x )x P ) 0 1 2 3 4 5 1 2 3 4 5 x 6 18 4 18 2 18 ) x) P x −2 1 3 P(x) 4 11 5 Review Exercises for Chapter 9 571 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 17. (a) E(x) = 10(0.10) + 15(0.20) + 20(0.50) + 30(0.15) + 40(0.05) = 20.5 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) 10 20.5 0.10 15 20.5 0.2 20 20.5 0.5 30 20.5 0.15 40 20.5 0.05 49.75 V x = − + − + − + − + − = σ = 49.75 ≈ 7.05 (b) R = 20.5(1000)(3.95) = $80,975 18. ( ) ( 1 ) ( 1 ) ( 50 ) ( )(1948 ) 2000 2000 2000 2000 E x = 2995 + 995 + 15 + −5 = −$2.50 Expected net gain: −$2.50 19. E(x) = 0(0.10) + 1(0.28) + 2(0.39) + 3(0.17) + 4(0.04) + 5(0.02) = 1.83 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) 2 2 2 2 2 2 0 1.83 0.10 1 1.83 0.28 2 1.83 0.39 3 1.83 0.17 4 1.83 0.04 5 1.83 0.02 1.1611 V x = − + − + − + − + − + − ≈ σ = V(x) ≈ 1.0775 20. E(x) = 0(0.12) + 1(0.31) + 2(0.43) + 3(0.12) + 4(0.02) = 1.61 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) V x = 0 − 1.61 0.12 + 1 − 1.61 0.31 + 2 − 1.61 0.43 + 3 − 1.61 0.12 + 4 − 1.61 0.02 = 0.8379 σ = V(x) ≈ 0.9154 21. f is a probability density function because 12 12 0 0 1 1 12 12 ∫ dx = x⎤⎦ = 1and ( ) 1 12 f x = ≥ 0 on [0, 12]. 22. f is not a probability density function because 8 8 1 1 1 1 7 8 8 8 ∫ dx = x⎤⎦ = ≠ 1. 23. f is not a probability density function because ( ) 1( ) 4 f x = 3 − x is negative on (3, 4]. 24. f is a probability density function because ( ) 3 2 (2 ) 0 4 f x = x − x ≥ on [0, 2] and ( ) ( ) 2 2 2 3 0 0 3 4 2 0 3 2 3 2 4 4 3 2 1. 4 3 4 x x dx x x dx x x − = 2 − ⎡ ⎤ = ⎢ − ⎥ = ⎣ ⎦ ∫ ∫ x $2995 $995 $15 −$5 P(x) 1 2000 1 2000 50 2000 1948 2000 0 0 12 2 12 0 0 8 0.25 −0.5 0 4 1 1 0 2 −1 572 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 25. f is a probability density function because 9 9 1 1 1 12 1 4 4 dx x x = ⎡ ⎤ = ∫ ⎣ ⎦ and ( ) 1 0 4 f x x = ≥ on [1, 9]. 26. f is not a probability density function because f (x) = 8.75x3 2 (1 − x) is negative on (1, 2]. 27. f (x) is not a probability density function because 8 8 8 8 0 0 1 8 ∫ e−x dx = ⎡⎣−e−x ⎤⎦ ≈ 0.632 ≠ 1. 28. f (x) is a probability density function. 5 5 0 0 1 5 lim 1 x x b b e dx e ∞ − − →∞ ∫ = ⎡⎣− ⎤⎦ = and ( ) 1 5 5 f x = e− x ≥ 0 on [0, ∞). 29. (a) ( ) 5 5 0 0 1 1 5 16 16 16 P 0 < x < 5 = ∫ dx = x⎤⎦ = (b) ( ) 13 13 12 12 1 1 1 16 16 16 P 12 < x < 13 = ∫ dx = x⎤⎦ = (c) ( ) 16 16 5 5 1 1 11 16 16 16 P x ≥ 5 = ∫ dx = x⎤⎦ = (d) ( ) 12 12 8 8 1 1 1 16 16 4 P 8 < x < 12 = ∫ dx = x⎤⎦ = 30. (a) ( ) 2 3 3 0 0 0 3 9 32 64 64 P x x dx x ⎤ < < = = = ⎥⎦ ∫ (b) ( ) 2 8 8 7 7 7 8 15 32 64 64 P x x dx x ⎤ < < = = = ⎥⎦ ∫ (c) ( ) 2 6 6 4 4 4 6 5 32 64 16 P x x dx x ⎤ < < = = = ⎥⎦ ∫ (d) ( ) 2 8 8 6 6 6 7 32 64 16 P x x dx x ⎤ ≥ = = = ⎥⎦ ∫ 31. (a) ( ) ( ) 2 2 2 0 0 0 2 1 10 1 10 9 50 50 2 25 P x x dx x x ⎡ ⎛ ⎞⎤ < < = − = ⎢ ⎜ − ⎟⎥ = ⎢⎣ ⎝ ⎠⎥⎦ ∫ (b) ( ) ( ) 2 10 10 7 7 7 1 10 1 10 9 50 50 2 100 P x x dx x x ⎡ ⎛ ⎞⎤ ≥ = − = ⎢ ⎜ − ⎟⎥ = ⎢⎣ ⎝ ⎠⎥⎦ ∫ (c) ( ) ( ) 2 5 5 0 0 5 1 10 1 10 3 50 50 2 4 P x x dx x x ⎡ ⎛ ⎞⎤ ≤ = − = ⎢ ⎜ − ⎟⎥ = ⎢⎣ ⎝ ⎠⎥⎦ ∫ (d) ( ) ( ) 2 9 9 8 8 8 9 1 10 1 10 3 50 50 2 100 P x x dx x x ⎡ ⎛ ⎞⎤ < < = − = ⎢ ⎜ − ⎟⎥ = ⎢⎣ ⎝ ⎠⎥⎦ ∫ 0 1 9 1 0 0 2 2 0 0 8 0.15 0 0 10 0.3 4 8 12 16 y x 2 16 y x 1 8 1 4 2 4 6 8 2 4 6 8 10 0.1 0.2 y x Review Exercises for Chapter 9 573 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 32. (a) ( ) ( ) 3 2 2 2 3 3 2 1 9 1 9 25 36 36 3 27 P x x dx x x − − ⎡ ⎛ ⎞⎤ < = − = ⎢ ⎜ − ⎟⎥ = ⎢⎣ ⎝ ⎠⎥⎦ ∫ (b) ( ) ( ) 3 3 3 2 2 2 2 1 9 1 9 25 36 36 3 27 P x x dx x x − − ⎡ ⎛ ⎞⎤ > − = − = ⎢ ⎜ − ⎟⎥ = ⎢⎣ ⎝ ⎠⎥⎦ ∫ (c) ( ) ( ) 3 2 2 2 1 1 1 2 1 9 1 9 2 36 36 3 3 P x x dx x x − − ⎡ ⎛ ⎞⎤ − < < = − = ⎢ ⎜ − ⎟⎥ = ⎢⎣ ⎝ ⎠⎥⎦ ∫ (d) ( ) ( ) 3 3 3 2 2 2 2 1 9 1 9 2 36 36 3 27 P x x dx x x ⎡ ⎛ ⎞⎤ > = − = ⎢ ⎜ − ⎟⎥ = ⎢⎣ ⎝ ⎠⎥⎦ ∫ 33. (a) ( ) 1 2 1 2 0 2 0 0 1 2 2 2 2 1 1 3 P x dx x x ⎛ < < ⎞ = = ⎡− ⎤ = ⎜ ⎟ ⎢ + ⎥ ⎝ ⎠ + ⎣ ⎦ ∫ (b) ( ) 3 4 3 4 1 4 2 1 4 1 3 2 2 16 4 4 1 1 35 P x dx x x ⎛ < ⎞ = = ⎡− ⎤ = ⎜ ⎟ ⎢ + ⎥ ⎝ ⎠ + ⎣ ⎦ ∫ (c) ( ) 1 1 1 2 2 1 2 1 2 2 1 2 1 1 3 P x dx x x ⎛ ≥ ⎞ = = ⎡− ⎤ = ⎜ ⎟ ⎢ + ⎥ ⎝ ⎠ + ⎣ ⎦ ∫ (d) ( ) 3 10 3 10 1 10 2 1 10 1 3 2 2 40 10 10 1 1 143 P x dxx x ⎛ < < ⎞ = = ⎡− ⎤ = ⎜ ⎟ ⎢ + ⎥ ⎝ ⎠ + ⎣ ⎦ ∫ 34. (a) ( ) 9 9 3 2 4 4 4 9 3 1 19 128 64 64 P < x < = x dx = ⎡⎢ x ⎤⎥ = ⎣ ⎦ ∫ (b) ( ) 16 16 3 2 4 4 4 16 3 1 7 128 64 8 P < x < = x dx = ⎡⎢ x ⎤⎥ = ⎣ ⎦ ∫ (c) ( ) 9 9 3 2 0 0 9 3 1 27 128 64 64 P x < = x dx = ⎡⎢ x ⎤⎥ = ⎣ ⎦ ∫ (d) ( ) 12 12 3 2 0 0 0 12 3 1 3 3 0.6495 128 64 8 P < x < = x dx = ⎡⎢ x ⎤⎥ = ≈ ⎣ ⎦ ∫ 35. (a) ( ) 5 12 0 12 5 0 5 12 1 12 0 5 1 0.341 t t P t e dt e e − − − < < = = ⎡⎣− ⎤⎦ = − ≈ ∫ (b) ( ) 12 12 9 12 12 9 3 4 1 1 12 9 12 0.104 t t P t e dt e e e − − − < < = = ⎡⎣− ⎤⎦ = − ≈ ∫ 36. (a) ( ) 6 5 6 5 5 6 1 4 4 1 4 2 2 1 0.207 2 2 P t dt t t < < = − = ⎡ − ⎤ ⎢⎣ ⎥⎦ = − ≈ ∫ (b) ( ) 13 8 13 8 8 13 1 4 4 1 4 2 3 1 1 0.5 2 2 P t dt t t < < = − = ⎡ − ⎤ ⎢⎣ ⎥⎦ = − = = ∫ x −2 −1 1 2 3 1 16 3 16 y 0.5 1 1 2 y x x 1 20 1 10 y 4 8 12 16 574 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 37. (a) 7 7 2 0 0 1 1 7 7 14 2 μ = x⎛⎜ ⎞⎟ dx = ⎡⎢ x ⎥⎤ = ⎝ ⎠ ⎣ ⎦ ∫ (b) ( ) 2 7 0 7 1 49 2 7 12 V x = ⎛⎜ x − ⎞⎟ ⎛⎜ ⎞⎟ dx = ⎝ ⎠ ⎝ ⎠ ∫ (c) 49 7 7 3 12 2 3 6 σ = V x = = = (d) 38. (a) 3 12 12 0 0 8 72 216 μ x x dx x ⎛ ⎞ ⎡ ⎤ = ⎜ ⎟ = ⎢ ⎥ = ⎝ ⎠ ⎣ ⎦ ∫ (b) ( ) ( ) 12 2 0 8 8 72 V x = x − ⎛ x ⎞ dx = ⎜ ⎟ ⎝ ⎠ ∫ (c) σ = V (x) = 8 = 2 2 (d) 39. (a) 3 3 1 2 1 3 3 ln 3 ln 3 1.648 2 2 2 x dx x x μ = ⎛⎜ ⎞⎟ = ⎡⎢ ⎤⎥ = ≈ ⎝ ⎠ ⎣ ⎦ ∫ (b) ( ) 2 2 3 1 2 3 ln 3 3 3 3 ln 3 0.284 2 2 2 V x x dx x = ⎛ − ⎞ ⎛ ⎞ = − ⎛ ⎞ ≈ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ∫ (c) ( ) 2 3 3 ln 3 0.533 2 σ = V x = − ⎛⎜ ⎞⎟ ≈ ⎝ ⎠ (d) 40. (a) 3 2 8 8 0 0 8 8 32 96 8 3 μ x x dx x x ⎛ − ⎞ ⎡ ⎤ = ⎜ ⎟ = ⎢− + ⎥ = ⎝ ⎠ ⎣ ⎦ ∫ (b) ( ) 2 8 0 8 8 32 3 32 9 V x x x dx ⎛ ⎞ ⎛ − ⎞ = ⎜ − ⎟ ⎜ ⎟ = ⎝ ⎠ ⎝ ⎠ ∫ (c) ( ) 32 4 2 9 3 σ = V x = = (d) 41. (a) ( ) 4 3 3 3 0 0 2 3 2 3 9 18 9 2 μ x x x dx x x ⎛ ⎞ ⎡ ⎤ = ⎜ − ⎟ = ⎢− + ⎥ = ⎝ ⎠ ⎣ ⎦ ∫ (b) ( ) ( ) 2 3 0 3 2 3 9 2 9 20 V x = ⎛⎜ x − ⎞⎟ ⎛⎜ x − x ⎞⎟ dx = ⎝ ⎠ ⎝ ⎠ ∫ (c) ( ) 9 3 3 5 20 2 5 10 σ = V x = = = (d) 1 2 3 4 5 6 7 y x 2 7 Mean y x 1 6 1 12 Mean 2 4 6 8 10 12 1 2 3 1 2 y x Mean y x Mean 1 2 3 4 5 6 7 8 1 8 1 2 0.1 0.2 0.3 0.4 0.5 y x Mean Review Exercises for Chapter 9 575 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 42. (a) 4 4 5 2 0 0 3 3 12 16 40 5 μ = x⎛⎜ x ⎞⎟ dx = ⎡⎢ x ⎤⎥ = ⎝ ⎠ ⎣ ⎦ ∫ (b) ( ) 2 4 0 12 3 192 5 16 175 V x = ⎛⎜ x − ⎞⎟ ⎛⎜ x ⎞⎟ dx = ⎝ ⎠ ⎝ ⎠ ∫ (c) ( ) 192 8 21 175 35 σ = V x = = (d) 43. 0 0 1 1 1 14 14 14 1 1 14 2 7 m m dx x m m m = ⎡⎣ ⎤⎦ = = = ∫ 44. 2 0 0 2 2 2 3 3 1 2 2 2 3 1 2 2 3 1 1 2 2 2 3 1 0 3 9 4 3 5 2 2 3 5 0.3820 2 m m x dx x x m m m m m m m m ⎛ − ⎞ = ⎡ − ⎤ ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ = − − = − + = ± − ± = = − = ≈ ∫ 45. 4 0 4 1 2 1 2 0.25 1 2.7726 m x m e dx e m − − = − = ≈ ∫ 46. 5 6 5 6 0 5 1 6 2 1 0.8318 ∫ m e− x dx = − e− m = ⇒ m ≈ 47. ( ) 1 2 f x = is a uniform probability density function. ( ) ( ) ( ) 2 2 0 2 1 2 2 2 0 1 12 12 3 2 0 2 3 12 12 12 3 a b b a V x b a μ σ + + = = = − − = = = − − = = = = 48. ( ) 1 9 f x = is a uniform probability density function. ( ) ( ) ( ) 2 2 0 9 9 2 2 2 9 0 27 12 12 4 9 0 9 3 3 12 12 12 2 a b b a V x b a μ σ + + = = = − − = = = − − = = = = 49. ( ) 1 6 6 f x = e− x is an exponential probability density function with 1. 6 a = ( ) ( )2 1 6 1 6 1 36 1 6 36 6 V x μ σ = = = = = = 50. ( ) 4 4 5 5 f x = e− x is an exponential probability density function with 4. 5 a = ( ) ( )2 1 5 4 5 4 1 25 4 5 16 25 5 16 4 V x μ σ = = = = = = 51. ( ) 1 ( 16)2 18 3 2 f x e x π = − − is a normal probability density function, with mean 16 and standard deviation 3. ( ) 2 ( )2 16 3 9 3 V x μ σ σ = = = = = 52. ( ) 1 ( 40)2 50 5 2 f x e x π = − − is a normal probability density function, with mean 40 and standard deviation 5. ( ) 2 ( )2 40 5 25 5 V x μ σ σ = = = = = y x Mean 1 2 3 4 3 8 3 16 576 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 53. (a) ( ) 1 , [0, 20], 20 f t = where t = 0 corresponds to 7:00 A.M. (b) ( ) ( ) 2 20 20 0 0 20 2 0 A.M. 1 10 or 7:10 20 40 10 1 100 20 3 100 10 10 3 5.8 min 3 3 3 t dt t V x t dt μ σ ⎛ ⎞ ⎡ ⎤ = ⎜ ⎟ = ⎢ ⎥ = ⎝ ⎠ ⎣ ⎦ = − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = = = ≈ ∫ ∫ (c) 20 20 4 4 1 1 1 1 1 4 1 0.2 20 20 5 5 − dt = − ⎡⎢ t⎤⎥ = − = = ⎣ ⎦ ∫ 54. (a) ( ) 1 , [0, 15], 15 f t = where t = 0 corresponds to 7:00 A.M. (b) ( ) 2 15 15 0 0 2 15 0 A.M. 1 15 7.5 or 7:07:30 15 30 2 15 1 75 2 15 4 75 5 3 min 4 2 t dt t V x t dt μ σ ⎛ ⎞ ⎡ ⎤ = ⎜ ⎟ = ⎢ ⎥ = = ⎝ ⎠ ⎣ ⎦ = ⎛ − ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = = ∫ ∫ (c) 15 15 4 4 1 1 1 1 1 11 4 0.267 15 15 15 15 − dt = − ⎡⎢ t⎤⎥ = − = ≈ ⎣ ⎦ ∫ 55. (a) ( ) 1 15 15 f t = e−t , 0 ≤ t < ∞ (b) ( ) 10 15 0 15 10 0 2 3 1 15 0 10 1 0.4866 t t P t e dt e e − − − < < = = ⎡⎣− ⎤⎦ = − ≈ ∫ 56. (a) ( ) 1 350 350 f t = e−t , 0 ≤ t < ∞ (b) ( ) 500 350 400 350 500 400 8 7 10 7 1 350 400 500 0.0793 t t P t e dt e e e − − − − < < = = ⎡⎣− ⎤⎦ = − ≈ ∫ 57. (a) ( ) ( ) ( ) ( ) 4 3 8 8 0 0 8 2 0 3 8 3 256 1024 32 4 3 8 16 256 5 16 4 5 5 5 x x x dx x x V x x x x dx μ σ ⎛ ⎞ ⎡ ⎤ = ⎜ − ⎟ = ⎢− + ⎥ ⎝ ⎠ ⎣ ⎦ = − ⎛ − ⎞ = ⎜ ⎟ ⎝ ⎠ = = ∫ ∫ (b) ( ) 2 3 2 3 0 0 Median 3 8 3 3 256 64 256 64 256 m m x x dx x x m m ⎛ ⎞ ⎡ ⎤ = ⎜ − ⎟ = ⎢ − ⎥ = − ⎝ ⎠ ⎣ ⎦ ∫ ( )( ) 2 3 3 2 2 3 1 64 256 2 12 128 0 4 8 32 0 4 or 4 4 3 m m m m m m m m m − = − + = − − − = = = ± In the interval [0, 8], m = 4. Review Exercises for Chapter 9 577 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (c) ( ) ( ) ( ) 2 2 5.7889 5.7889 2.2111 2.2111 4 4 5 4 4 5 5 5 2.2111 5.7889 3 8 3 0.6261 or 62.61% 256 64 256 P x P x P x x x dx x x μ σ μ σ ⎛ ⎞ − < < + = ⎜⎜ − < < + ⎟⎟ ⎝ ⎠ ≈ < ⎛ ⎞ ⎡ ⎤ = ⎜ − ⎟ = ⎢ − ⎥ ≈ ⎝ ⎠ ⎣ ⎦ ∫ 58. (a) ( ) ( ) ( ) 3 4 5 5 0 0 2 5 0 6 5 2 3 5 125 25 250 2 5 6 5 5 2 125 4 5 5 4 2 x x x dx x x V x x x x dx μ σ ⎛ ⎞ ⎡ ⎤ = ⎜ − ⎟ = ⎢ − ⎥ = ⎝ ⎠ ⎣ ⎦ = ⎛ − ⎞ ⎛ − ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = = ∫ ∫ (b) ( ) 2 3 2 3 0 0 Median 6 5 3 2 3 2 125 25 125 25 125 m m x x dx x x m m ⎛ ⎞ ⎡ ⎤ = ⎜ − ⎟ = ⎢ − ⎥ = − ⎝ ⎠ ⎣ ⎦ ∫ ( )( ) 2 3 3 2 2 3 2 1 25 125 2 4 30 125 0 2 5 2 10 25 0 5 or 5 5 3 2 2 m m m m m m m m m − = − + = − − − = ± = = In the interval [0, 5], 5. 2 m = (c) ( ) ( ) ( ) 2 3 3..6180 3.6180 1.3820 1.3820 5 5 5 5 1.3820 3.6180 2 2 2 2 6 5 3 2 0.6261 or 62.61% 125 25 125 P x P x P x x x dx x x μ σ μ σ ⎛ ⎞ − < < + = ⎜⎜ − < < + ⎟⎟ ≈ < < ⎝ ⎠ ⎛ ⎞ ⎡ ⎤ = ⎜ − ⎟ = ⎢ − ⎥ ≈ ⎝ ⎠ ⎣ ⎦ ∫ 59. ( ) 1 ( 168)2 2(25)2 1 ( 168)2 1250 25 2 25 2 f x e x e x π π = − − = − − (a) (b) ( ) ( ) 105 70 70 105 0.0058 or 0.58% P < x < = f x dx ≈ ∫ (c) ( ) ( ) ( ) 120 0 120 1 0 120 1 1 0.0274 0.9726 or 97.26% P x P x f x dx > = − < < = − ≈ − = ∫ or ( ) ( ) 120 P x 120 f x dx 0.9726 or 97.26% ∞ > = ∫ ≈ 60. ( ) ( ) ( ) ( ) 42 2 2 3 2 42 2 18 1 3 2 1 3 2 x x f x e e π π − − − − = = (a) (b) ( ) ( ) 45 40 40 45 0.5889 or 58.89% P < x < = f x dx ≈ ∫ (c) ( ) ( ) 50 0 0 50 0.9962 or 99.62% P < x < = f x dx ≈ ∫ 0 90 250 0.025 0 30 54 0.2 578 Chapter 9 Probability and Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 9 Test Yourself 1. (a) { , , , , , , , , , , , , , , , } S TTTT TTTF TTFT TTFF TFTT TFTF TFFT TFFF FTTT FTTF FTFT FTFF FFTT FFTF FFFT FFFF = (b) (c) 2. ( ) 20 5 P red and not a face card = 52 = 13 3. (a) ( ) ( ) ( ) 3 7 5 16 16 8 P x < 3 = P 1 + P 2 = + = (b) ( ) ( ) ( ) 1 5 3 16 16 8 P x ≥ 3 = P 3 + P 4 = + = 4. (a) P(7 ≤ x ≤ 10) = P(7) + P(8) + P(9) + P(10) = 0.21 + 0.13 + 0.19 + 0.42 = 0.95 (b) P(x > 8) = P(9) + P(10) + P(11) = 0.19 + 0.42 + 0.05 = 0.66 5. ( ) ( 2 ) ( 1 ) ( 4 ) ( 3 ) 9 10 10 10 10 5 E x = 0 + 1 + 2 + 3 = = 1.8 ( ) ( )2 ( 2 ) ( )2 ( 1 ) ( )2 ( 4 ) ( )2 ( 3 ) 10 10 10 10 V x = 0 − 1.8 + 1 − 1.8 + 2 − 1.8 + 3 − 1.8 = 1.16 σ = V(x) ≈ 1.077 6. E(x) = −2(0.141) + (−1)(0.305) + 0(0.257) + 1(0.063) + 2(0.234) = −0.056 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( ) ( )2 ( )2 ( ) 2 0.056 0.141 1 0.056 0.305 0 0.056 0.257 1 0.056 0.063 2 0.056 0.234 1.864864 V x = − + + − + + + + + + + = σ = V(x) ≈ 1.366 7. f (x) is not a probability density function because 8 8 0 0 1 1 1 1. 16 16 2 dx = ⎡⎢ x⎤⎥ = ≠ ⎣ ⎦ ∫ 8. f is a probability density function because 2 1 1 1 1 3 1 1. 6 2 12 x dx x x − − − ⎡ ⎤ = ⎢ − ⎥ = ⎣ ⎦ ∫ Random variable, x 0 1 2 3 4 Frequency of x, n(x) 1 4 6 4 1 Random variable, x 0 1 2 3 4 Probability, P(x) 1 16 4 16 6 16 4 16 1 16 1 2 3 4 P(x) x 2 16 4 16 6 16 8 16 0 0 8 0.125 0 −1 1 1 7 8 9 10 11 P(x) x 0.1 0.2 0.3 0.4 Chapter 9 Test Yourself 579 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 9. f (x) is a probability density function. 3 4 3 4 0 8 34 e x dx e x 1 ∞ − = ⎡− − ⎤ ∞ = ∫ ⎣ ⎦ and f (x) ≥ 0 over the interval [0, ∞). 10. 2 1 2 1 2 1 2 9 9 9 9 b b a a x dx = ⎡ x ⎤ = b − a ⎢⎣ ⎥⎦ ∫ (a) (0 1) 1(1)2 1(0)2 1 9 9 9 P ≤ x ≤ = − = (b) (2 3) 1(3)2 1(2)2 5 9 9 9 P ≤ x ≤ = − = 11. 4( 3) (4 4 3) 2 2 4 (2 2 4 ) (2 2 4 ) b b b a a a ∫ x − x dx = ∫ x − x dx = ⎡⎣ x − x ⎤⎦ = b − b − a − a (a) P(0 < x < 0.5) = ⎡⎣2(0.5)2 − 0.54⎤⎦ − ⎡⎣2(0)2 − 04⎤⎦ = 0.4375 (b) P(0.25 ≤ x < 1) = ⎡⎣2(1)2 − 14⎤⎦ − ⎡⎣2(0.25)2 − 0.254⎤⎦ ≈ 0.879 12. 2 2 ( 2 2 ) 2 b b x x b a a a xe dx e e e − − − − ⎤ = − = − − ⎥⎦ ∫ (a) P(x < 1) = −(e−12 − e0 2 ) = 1 − e−1 ≈ 0.632 (b) ( ) 2 2 ( 2 1) 1 1 1 1 2 lim lim 0.368 b x x b b b P x xe dx e e e e ∞ − − − − − →∞ →∞ ≥ = = ⎡− ⎤ = − + = ≈ ∫ ⎢⎣ ⎥⎦ 13. ( ) 1 , [0, 14] 14 f x = 0 14 7 2 2 μ a b + + = = = ( ) ( ) ( ) 2 2 14 0 49 12 12 3 b a V x − − = = = 14 4.041 12 12 σ b a − = = ≈ 14. ( ) [ ] 2 32 f x = 3x − x , 0, 1 ( ) ( ) 1 2 1 2 3 0 0 3 4 1 0 3 3 2 2 3 5 8 8 x 3x x dx 3x x dx x x μ = − = − = ⎡⎣ − ⎤⎦ = ∫ ∫ ( ) ( ) ( ) ( ) 1 2 2 2 0 1 3 4 0 4 5 1 0 3 5 2 8 3 25 2 64 3 3 25 4 10 64 9 25 20 64 19 320 3 3 V x x x x dx x x dx x x = − − = − − = ⎡⎣ − ⎤⎦ − = − = ∫ ∫ σ = V(x) ≈ 0.244 15. f (x) = e−x , [0, ∞); f is an exponential probability density function with a = 1. 1 1 1 a 1 μ = = = ( ) 2 2 1 1 1 1 V x a = = = σ = V(x) = 1 16. ( ) ( ) ( )2 1 110 2 2 10 ; 10 2 f x e x π = − − μ = 110, σ = 10 ( ) ( ) ( ) 120 100 100 120 0.683 P x P x f x dx μ − σ < < μ + σ = < < = ∫ ≈ 0 0 6 0.8 [Show More]

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