_newbold_ism_06.pdf

Chapter 6: Continuous Random Variables and Probability Distributions

6.1 P(1.4 < X < 1.8) = F(1.8) – F(1.4) = (.5)(1.8) – (.5)(1.4) = 0.20 6.2 P(1.0 < X < 1.9) = F(1.9) – F(1.0) = (.5)(1.9) – (.5)(1.0) = 0.45 6.3 P(X < 1.4) = F(1.4) = (.5)(1.4) = 0.7 6.4 P(X > 1.3) = F(1.3) = (.5)(2.0) – (.5)(1.3) = 0.35 6.5 a.

Probability Density Function: f(x) 1.5 f(x)

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b. Cumulative distribution function: F(x) F(x)

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c. P(X < .25) = .25 d. P(X >.75) = 1-P(X < .75) = 1-.75 = .25 e. P(.2 < X < .8) = P(X <.8) – P(X <.2) = .8 - .2 = .6 6.6 a.

Probability density function: f(x) 0.75 f(x)

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Statistics for Business & Economics, 6 edition

b. Cumulative density function: F(x) 1.0

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c. P(x < 1) = .25 d. P(X < .5) + P(X > 3.5)=P(X < .5) + 1 – P(X < 3.5) = .25 6.7 a. P(60,000 < X< 72,000) = P(X < 72,000) – P(X < 60,000) = .6 - .5 = .1 b. P(X < 60,000) < P(X < 65,000) < P(X < 72,000); .5 < P(X < 65,000) < .6 6.8 a. P(380 < X < 460) = P(X < 460) – P(X < 380) = .6 - .4 = .2 b. P(X < 380) < (PX< 400) < P(X < 460); .4 < P(X < 400) < .6 6.9 W = a + bX. If TC = 1000 + 2X where X = number of units produced, find the mean and variance of the total cost if the mean and variance for the number of units produced are 500 and 900 respectively. μW = a + bμ x = 1000 + 2(500) = 2000. σ 2W = b 2σ 2 X = (2)2(900) = 3600. 6.10 W = a + bX. If Available Funds = 1000 - 2X where X = number of units produced, find the mean and variance of the profit if the mean and variance for the number of units produced are 50 and 90 respectively. μW = a + bμ x = 1000 - 2(50) = 900. σ 2W = b 2σ 2 X = (-2)2(90) = 360. 6.11 W = a + bX. If Available Funds = 2000 - 2X where X = number of units produced, find the mean and variance of the profit if the mean and variance for the number of units produced are 500 and 900 respectively. μW = a + bμ x = 2000 - 2(500) = 1000. σ 2W = b 2σ 2 X = (-2)2(900) = 3600.


6.12

W = a + bX. If Available Funds = 6000 - 3X where X = number of units produced, find the mean and variance of the profit if the mean and variance for the number of units produced are 1000 and 900 respectively. μW = a + bμ x = 6000 - 2(1000) = 4000. σ 2W = b 2σ 2 X = (-3)2(900) = 8100

6.13

μY = 10,000 + 1.5 μX = 10,000 + 1.5 (30,000) = $55,000 σY = |1.5| σX = 1.5 (8,000) = $12,000

6.14

μY = 20 + μX = 20 + 4 = $24 million Bid = 1.1 μY =1.1(24) = $26.4 million, σπ = $1 million

6.15

μY = 60 + .2 μX = 60 + 140 = $200 σY = |.2| σX = .2 (130) = $26

6.16

μY = 6,000 + .08 μX = 6,000 + 48,000 = $54,000 σY = |.08| σX = .08(180,000) = $14,400

6.17

a. b. c. d. e. f. g.

P(Z < 1.20) = .8849 P(Z > 1.33) = 1 – Fz(1.33) = 1 - .9082 = .0918 P(Z < -1.70) = 1 – Fz(1.70) = 1 - .9554 = .0446 P(Z > -1.00) = Fz(1) = .8413 P(1.20 < Z < 1.33) = Fz(1.33) – Fz(1.20) = .9082 - .8849 = .0233 P(-1.70 < Z < 1.20) = Fz(1.20) – [1 - Fz(1.70)] = .8849 – .0446 = .8403 P(-1.70 < Z < -1.00) = Fz(1.70) – Fz(1.00) = .9554 - .8413 = .1141

6.18

a. b. c. d.

Find Z0 such that P(Z < Z0) = .7, closest value of Z0 = .52 Find Z0 such that P(Z < Z0) = .25, closest value of Z0 = -.67 Find Z0 such that P(Z > Z0) = .2, closest value of Z0 = .84 Find Z0 such that P(Z > Z0) = .6, closest value of Z0 = -.25

6.19

X follows a normal distribution with µ = 50 and σ2 = 64 60 − 50 a. Find P(X > 60). P(Z > ) = P(Z > 1.25) 8 = .5 - .3944 = .1056 35 − 50 62 − 50 b. Find P(35 < X < 62). P(
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e. Probability is .05 that X is in the symmetric interval about the mean X − 50 between? Z = +/- .06. ±.06 = . X = 49.52 and 50.48. 8 6.20

X follows a normal distribution with µ = 80 and σ2 = 100 60 − 80 a. Find P(X > 60). P(Z > ) = P(Z > -2.00) = .5 + .4772 = .9772 10 72 − 80 82 − 80 b. Find P(72 < X < 82). P(
6.21

X follows a normal distribution with µ = .2 and σ2 = .0025 .4 − .2 a. Find P(X > .4). P(Z > ) = P(Z > 4.00) = .5 - .5 = .0000 .05 .15 − .2 .28 − .2 b. Find P(.15 < X < .28). P(
6.22

400 − 380 ) = P(Z < .4) = .6554 50 360 − 380 b. P(Z > ) = P(Z > -.4) = FZ(.4) = .6554 50 c. The graph should show the property of symmetry – the area in the tails equidistant from the mean will be equal. 300 − 380 400 − 380 d. P(

e. The area under the normal curve is equal to .8 for an infinite number of ranges – merely start at a point that is marginally higher. The shortest range will be the one that is centered on the z of zero. The z that corresponds to an area of .8 centered on the mean is a Z of ±1.28. This yields an interval of the mean plus and minus $64: [$316, $444] 6.23

6.24

6.25

6.26

1, 000 − 1, 200 ) = P(Z > -2) =FZ(2) = .9772 100 1,100 − 1, 200 1,300 − 1, 200 1.28) = .1, plug into the z-formula all of the known information Xi − 1, 200 and solve for the unknown: 1.28 = . Solve algebraically 100 for Xi = 1,328 a. P(Z >

38 − 35 ) = P(Z > .75) = 1 - FZ(.75) = .2266 4 32 − 35 b. P(Z < ) = P(Z < -.75) = 1 - FZ(.75) = .2266 4 32 − 35 38 − 35

20 − 12.2 ) = P(Z > 1.08) = 1 – Fz (1.08) = .1401 7.2 0 − 12.2 ) = P(Z < -1.69) = 1 – Fz (1.69) = .0455 b. P(Z < 7.2 5 − 12.2 15 − 12.2 c. P(

10 − 12.2 ) = P(Z < - .79) = 1 – Fz (.79) = .2148 2.8 15 − 12.2 b. P(Z > ) = P(Z > 1) = 1 – Fz (1) = .1587 2.8 12 − 12.2 15 − 12.2 c. P(
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d. The answer to a. will be larger because 10 grams is closer to the mean than is 15 grams. Thus, there would be a greater area remaining less than 10 grams than will be the area above 15 grams. 460 − 500 540 − 500 1.96) = .025, 1.96 = . Xi = 598. The 50 Xi − 500 lower value of the interval will be –1.96 = which is Xi = 50 $402 (thousand dollars). Therefore, the shortest range will be 598 – 402 = $196 (thousand dollars).

6.27

a. P(

6.28

P(Z > 1.5) = 1 - Fz(1.5) = .0668

6.29

P(Z < -1.28) = .1, –1.28 =

6.30

P(Z > .67) = .25, .67σ = 17.8 - μ P(Z > 1.03) = .15, 1.04σ = 19.2 - μ Solving for μ, σ: μ = 15.265, σ2 = (3.7838)2 = 14.317

6.31

6.32

Xi − 18.2 Xi = 16.152 1.6

820 − 700 ) = P(Z> 1) = 1 – Fz (1) = .1587 120 730 − 700 820 − 700 b. P(

For Investment A, the probability of a return higher than 10%: 10 − 10.4 P(Z > ) = P(Z > -.33) = FZ(.33) = .6293 1.2 For Investment B, the probability of a return higher than 10% 10 − 11.0 P(Z > ) = P(Z > -.25) = FZ(.25) = .5987 4 Therefore, Investment A is a better choice


6.33

6.34

6.35

6.36

6.37

5 − 4.4 ) = P(Z < 1.5) = .9332 .4 5 − 4.2 For Supplier B: P(Z < ) = P(Z < 1.33) = .9082 .6 Therefore, Supplier A has a greater probability of achieving less than 5% impurity and is hence the better choice For Supplier A: P(Z <

Xi − 150 , Xi = 98.8 40 Xi − 150 b. P(Z < .84) = .8, .84 = , Xi = 183.6 40 120 − 150 2 c. P(X ≥ 1) = 1 – P(X = 0) = 1-[P(Z< )] = 1 – [P(Z < -.75)]2 = 40 2 1 – (.2266) = .9487

a. P(Z > -1.28) = .9, -1.28 =

60 − 75 ) = P(Z < -.75) = .2266 20 90 − 75 ) = P(Z >.75) = .2266 b. P(Z > 20 c. The graph should show that 60 minutes and 90 minutes are equidistant from the mean of 75 minutes. Therefore, the areas above 90 minutes and below 60 minutes by the property of symmetry must be equal. Xi − 75 d. P(Z > 1.28) = .1, 1.28 = , Xi = 100.6 20 a. P(Z <

400 − 420 480 − 420 1.28) = .1, 1.28 = , Xi = 522.4 80 c. 400 – 439 d. 520 – 559 500 − 420 2 e. P(X ≥1) = 1 –P(X = 0 ) = 1 – [P(Z< )] = 1 – (.8413)2 = 80 .2922 a. P(

180 − 200 < Z < 0) = .5 – [1- Fz (1)] = .5 -.1587 = .3413 20 245 − 200 b. P(Z > ) = 1 – FZ(2.25) = .0122 20 c. Smaller Xi − 200 d. P(Z < -1.28) = .1, -1.28 = , Xi = 174.4 20 a. P(

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P(Z < 1.5) = .9332, 1.5 =

85 − 70

σ

, σ = 10

80 − 70 ) = P(Z > 1) = .1587 10 P(X ≥ 1) = 1 – P(X=0) = 1 – [FZ(1)]4 = 1 – (.8413)4 = .4990

P(Z >

6.39

n = 900 from a binomial probability distribution with P = .50 a. Find P(X > 500). E[X] = μ = 900(.5) = 450, σ = (900)(.5)(.5) = 15 500 − 450 P(Z > ) = P(Z > 3.33) = 1 – FZ(3.33) = .0004 15 430 − 450 b. Find P(X < 430). P(Z < ) = P(Z < -1.33) = 1 - FZ(1.33) = 15 .0918 440 − 450 480 − 450 c. P(
6.40

n = 1600 from a binomial probability distribution with P = .40 a. Find P(X > 1650). E[X] = μ = 1600(.4) = 640, σ = (1600)(.4)(.6) = 19.5959 P(Z > 1650 − 1600 ) = P(Z > 2.55) = 1 – FZ(2.55) = 19.5959 .0054 1530 − 1600 b. Find P(X < 1530). P(Z < ) = P(Z 19.5959 < -3.57) = 1 - FZ(3.57) = .0002 1550 − 1600 1650 − 1600 c. P(

e.

Probability is .20 the number of successes is X − 1600 greater than? Z = .84. .84 = . 19.5959 X = 1616.46 ≈1,616 successes

6.41

n = 900 from a binomial probability distribution with P = .10 a. Find P(X > 110). E[X] = μ = 900(.1) = 90, σ 110 − 90 = (900)(.1)(.9) = 9 P(Z > ) 9 = P(Z > 2.22) = 1 – FZ(2.22) = .0132 53 − 90 b. Find P(X < 53). P(Z < ) = P(Z < 9 4.11) = 1 - FZ(4.11) = .0000 55 − 90 120 − 90 c. P(
6.42

n = 1600 from a binomial probability distribution with P = .40 a. Find P(P > .45). E[P] = μ = P = .40, σ = P (1 − P) .4(1 − .4) = .01225 P(Z > = n 1600 .45 − .40 ) = P(Z > 4.082) = 1 – FZ(4.082) = .01225 .0000 .36 − .40 b. Find P(P < .36). P(Z < ) = P(Z < .01225 3.27) = 1 - FZ(3.27) = .0005 .44 − .40 .37 − .40 c. P(
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e. Probability is .09 the percentage of successes X − .40 is greater than? Z = 1.34. 1.34 = . P .01225 = 41.642% 6.43

6.44

6.45

n = 400 from a binomial probability distribution with P = .20 a. Find P(P > .25). E[P] = μ = P = .20, σ = P (1 − P) .2(1 − .8) = .02 P(Z > = n 400 .25 − .20 ) = P(Z > 2.50) = 1 – FZ(2.50) = 1 .02 .4938 = .0062 .16 − .20 b. Find P(P < .16). P(Z < ) = P(Z < .02 2.00) = 1 - FZ(2.00) = .0228 .17 − .20 .24 − .20 c. P( ) = P(Z > 1.67) = 1 – FZ(1.67) = .0475 12 175 − 180 b. P(Z < ) = P(Z < -.42) = 1 - FZ(.42) = .3372 12 a. E[X] = μ = 400(.1) = 40, σ = (400)(.1)(.9) = 6 35 − 40 P(Z > ) = P(Z > -.83) = FZ(.83) = .7967 6 40 − 40 50 − 40

34 − 40 48 − 40
6.46

E[X] = (100)(.6) = 60, σ = P(Z <

6.47

6.48

6.49

6.50

(100)(.6)(.4) = 4.899

50 − 60 ) = P(Z < -2.04) = 1 – FZ(2.04) = 1- .9793 = .0207 4.899

a. E[X] = (450)(.25) = 112.5, σ = (450)(.25)(.75) = 9.1856 100 − 112.5 P(Z < ) = P(Z < -1.36) = 1 - FZ(1.36) = 1 - .9131 = .0869 9.1856 120 − 112.5 150 − 112.5 b. P( .75) = 1 - FZ(.75) = 1 - .7734 = .2266 4 E[X] = 100(.2266) = 22.66, σ = (100)(.2266)(.7734) = 4.1863 25 − 22.66 P(Z > ) = P(Z > .56) = 1 - FZ(.56) = 1 - .7123 = .2877 4.1863 P(Z >

10 − 12.2 ) = P(Z < -.79) = 1 - FZ(.79) = 1 - .7852 = .2148 2.8 E[X] = 400(.2148) = 85.92, σ = (400)(.2148)(.7852) = 8.2137 100 − 85.92 P(Z > ) = P(Z > 1.71) = 1 - FZ(1.71) = 1 - .9564 = .0436 8.2137 P(Z ≤

λ = 1.0, what is the probability that an arrival occurs in the first t=2 time units? Cumulative Distribution Function Exponential with mean = 1 x P( X <= x ) 0 0.000000 1 0.632121 2 0.864665 3 0.950213 4 0.981684 5 0.993262

P(T < 2) = .864665

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6.51 λ = 8.0, what is the probability that an arrival occurs in the first t=7 time units? Cumulative Distribution Function Exponential with mean = 8 x P( X <= x ) 0 0.000000 1 0.117503 2 0.221199 3 0.312711 4 0.393469 5 0.464739 6 0.527633 7 0.583138 8 0.632121

P(T < 7) = .583138 6.52

λ = 5.0, what is the probability that an arrival occurs after t=7 time units?

Cumulative Distribution Function Exponential with mean = 5 x P( X <= x ) 0 0.000000 1 0.181269 2 0.329680 3 0.451188 4 0.550671 5 0.632121 6 0.698806 7 0.753403 8 0.798103

P(T>7) = 1-[P(T ≤ 8)] = 1 - .7981 = .2019 6.53


Cumulative Distribution Function Exponential with mean = 6 x P( X <= x ) 0 0.000000 1 0.153518 2 0.283469 3 0.393469 4 0.486583 5 0.565402 6 0.632121

P(T>5) = 1-[P(T≤6)] = 1 - .6321 = .3679 6.54


Cumulative Distribution Function Exponential with mean = 3 x P( X <= x ) 0 0.000000 1 0.283469 2 0.486583 3 0.632121

P(T<2) = .4866 6.55 a. P(X < 20) = 1 - e− (20 /10) = .8647 b. P(X > 5) = 1 – [1 - e − (5/10) ] = e − (5/10) = .6065 c. P(10 < X < 15) = (1- e− (15 /10) - (1 - e− (10 /10) ) = e −1 - e −1.5 = .1447


6.56 P(X > 18) = e − (18 /15) = .3012 6.57 P(X > 2) = e − (2)(.8) = .2019 6.58 a. P(X > 3) = 1 – [1 - e − (3/ μ ) ] = e −3λ since λ = 1 / μ b. P(X > 6) = 1 – [1 - e − (6 / μ ) ] = e − (6 / μ ) = e−6 λ c. P(X>6|X>3) = P(X > 6)/P(X > 3) = e −6 λ / e −3λ ] = e −3λ The probability of an occurrence within a specified time in the future is not related to how much time has passed since the most recent occurrence. 6.59 Find the mean and variance of the random variable: W = 5X + 4Y with correlation = .5 μW = aμ x + bμ y = 5(100) + 4(200) = 1300

σ 2W = a 2σ 2 X + b 2σ 2Y + 2abCorr ( X , Y )σ X σ Y = 52(100) + 42(400) + 2(5)(4)(.5)(10)(20) = 12,900 6.60 Find the mean and variance of the random variable: W = 5X + 4Y with correlation = -.5 μW = aμ x + bμ y = 5(100) + 4(200) = 1300

σ 2W = a 2σ 2 X + b 2σ 2Y + 2abCorr ( X , Y )σ X σ Y = 52(100) + 42(400) + 2(5)(4)(-.5)(10)(20) = 4,900 6.61 Find the mean and variance of the random variable: W = 5X – 4Y with correlation = .5. μW = aμ x − bμ y = 5(100) – 4(200) = -300

σ 2W = a 2σ 2 X + b 2σ 2Y − 2abCorr ( X , Y )σ X σ Y = 52(100) + 42(400) – 2(5)(4)(.5)(10)(20) = 4900 6.62 Find the mean and variance of the random variable: W = 5X – 4Y with correlation = .5. μW = aμ x − bμ y = 5(500) – 4(200) = 1700

σ 2W = a 2σ 2 X + b 2σ 2Y − 2abCorr ( X , Y )σ X σ Y = 52(100) + 42(400) – 2(5)(4)(.5)(10)(20) = 4900

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6.63 Find the mean and variance of the random variable: W = 5X – 4Y with correlation of -.5. μW = aμ x − bμ y = 5(100) – 4(200) = -300

σ 2W = a 2σ 2 X + b 2σ 2Y − 2abCorr ( X , Y )σ X σ Y = 52(500) + 42(400) – 2(5)(4)(-.5)(22.3607)(20) = 27,844.28 6.64

μ Z = 100,000(.1) + 100,000(.18). μ x = 10,000 + 18,000 = 28,000 σZ = 0. Note that the first investment yields a certain profit of 10% which is a zero standard deviation. σx = 100,000(.06) = 6,000

6.65 Assume that costs are independent across years μ Z = 5 μ x = 5(200) = 1,000 σZ =

6.66

2

5σ x =

5(3, 600) = 134.16

μ Z = μ1 + μ 2 + μ3 = 50,000 + 72,000 + 40,000 = 162,000 2

2

2

σZ = σ 1 + σ 2 + σ 3 = 6.67

(10, 000) 2 + (12, 000) 2 + (9, 000) 2 = 18,027.76

μ Z = μ1 + μ 2 + μ3 = 20,000 + 25,000 + 15,000 = 60,000 2

2

2

σZ = σ 1 + σ 2 + σ 3 =

(2, 000) 2 + (5, 000) 2 + (4, 000) 2 = 6,708.2

6.68 The calculation of the mean is correct, but the standard deviations of two random variables cannot be summed. To get the correct standard deviation, add the variances together and then take the square root. The standard deviation: σ = 5(16) 2 = 35.7771 6.69

μ Z = (16 μ x ) / 16 = μ x = 28 2

σZ = 16σ x / 16 =

(2.4) 2 / 16 = 2.4 / 4 = .6

6.70 a. Compute the mean and variance of the portfolio with correlation of +.5 μW = aμ x + bμ y = 50(25) + 40(40) = 2850

σ 2W = a 2σ 2 X + b 2σ 2Y + 2abCorr ( X , Y )σ X σ Y = 502(121) + 402(225) + 2(50)(40)(.5)(11)(15) = 992,500 b. Recompute with correlation of -.5 μW = aμ x + bμ y = 50(25) + 40(40) = 2850

σ 2W = a 2σ 2 X + b 2σ 2Y + 2abCorr ( X , Y )σ X σ Y = 502(121) + 402(225) + 2(50)(40)(-.5)(11)(15) = 332,500


6.71

a. Find the probability that total revenue is greater than total cost W = aX – bY = 10X –[7Y+25)] μW = aμ x − bμ y = 10(100) – [7(100) + 250] = 50

σ 2W = a 2σ 2 X + b 2σ 2Y − 2abCorr ( X , Y )σ X σ Y = 102(64) + 72(625) – 2(10)(7)(.6)(8)(25) = 20,225 σ W = 20, 225 = 142.2146 P(Z >

0 − 50 ) = P(Z > -.35) = FZ(.35) = .6368 142.2146

b. 95% acceptance interval = 50 ± 1.96 (142.2146) = 50 ± 278.7406 = 228.7406 to 328.7406 6.72 a. W = aX – bY = 10X – 10Y μW = aμ x − bμ y = 10(100) – 10(90) = 100

σ 2W = a 2σ 2 X + b 2σ 2Y − 2abCorr ( X , Y )σ X σ Y =102(100) + 102(400) – 2(10)(10)(-.4)(10)(20) =66,000 σ W = 66, 000 =256.90465 b. P(Z <

0 − 100 ) = P(Z < -.39) = 1 – FZ(.39) = 1 – .6517 = .3483 256.90465

6.73 W = aX – bY = 10X – 4Y μW = aμ x − bμ y = 10(400) – 4(400) = 2400

σ 2W = a 2σ 2 X + b 2σ 2Y − 2abCorr ( X , Y )σ X σ Y =102(900) + 42(1600) – 2(10)(4)(.5)(30)(40) = 67,600 σ W = 67, 600 =260 P(Z >

2000 − 2400 ) = P(Z > -1.54) = FZ(1.54) = .9382 260

6.74 a. W = aX – bY = 1X – 1Y μW = aμ x − bμ y = 1(100) – 1(105) = -5

σ 2W = a 2σ 2 X + b 2σ 2Y − 2abCorr ( X , Y )σ X σ Y =12(900) + 12(625) – 2(1)(1)(.7)(30)(25) = 475 σ W = 475 =21.79449 b. P(Z > 6.75

0 − (−5) ) = P(Z > .23) = 1 – FZ(.23) = 1 – .5910 = .4090 21.79449

a. P(X < 10) = (10/12) – (8/12) = 1/6 b. P(X > 12) = (20/12) – (12/12) = 8/12 = 2/3

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c. E[π] = π(20/12 – 12/12) = 2(2/3) = 1.333 d. To jointly maximize the probability of getting the contract and the profit from that contract, maximize the following function: max E[π] = (B – 10)(20/12 – B/12). Where B is the value of the bid. To determine the value for B that maximizes the expected profit, an iterative approach can be used. The value of B is 15. 6.76 a.

Probability Density Function: f(x) f(x)

0.033333

0.000000 30

35

40

45

50

55

60

65

70

X

b. Cumulative density function Cumulative density function: F(x) 1.0

F(x)

0.8

0.6

0.4

0.2

0.0 35

40

45

50

55 X

c. P(40 < X < 50) = (50/30) – (40/30) = 10/30 65 + 35 d. E[X] = = 50 2

60

65


6.77

a. The probability density function f(x):

Probability density function: f(x) 1.50 f(x)

1.00

0.5

0.00 0

.5

1

1.5

3

X

b. Fx(x) ≥ 0 for all x. The area under fx(x) = 2[½(base x height)] = 1 .52 .52 c. P(.5 < X < 1.5 ) = (.5 ) + (.5 ) = .375 + .375 = .75 2 2 6.78

a. μ Y = 2000(1.1) + 1000(1+ μ x ) = 2,200 + 1,160 = 3,360 b. σY = |1000| σx = 1000(.08) = 80

6.79

a. μ R = 1.45 μ x = 1.45(530) = 768.5 b. σR = |1.45| σx = 1.45(69) = 100.05 c. π = R – C = .5X – 100, E[π] = .5 μ x -100 = 165, σπ = |.5| σx = .5(69) = 34.5

6.80 Given that the variance of both predicted earnings and forecast error are both positive and given that the variance of actual earnings is equal to the sum of the variances of predicted earnings and forecast error, then the Variance of predicted earnings must be less than the variance of actual earnings 6.81 Cov[(X1 + X2), (X1 – X2)] = E[(X1 + X2)(X1 – X2)] – E[X1 + X2] E[X1 – X2] = E[X12 - X22]– E[(X1) + E(X2)][E(X1) – E(X2)] = E(X12) – E(X22) - [(E(X1))2 – (E(X2)2] = Var (X1) – Var (X2) Which is 0 if and only if Var (X1) = Var (X2)

139

140

th


3 − 2.6 ) = P(Z > .8) = 1 – FZ(.8) = .2119 .5 2.25 − 2.6 2.75 − 2.6 1.28) = .1, 1.28 = .5 P(Xi > 3) = .2119 (from part a) E[X] = 400(.2119) = 84.76, σx = (400)(.2119)(.7881) = 8.173 80 − 84.76 P(Z > ) = P(Z > -.58) = FZ(.58) = .7190 8.173 P(X ≥ 1) = 1 – P(X = 0) = 1 – (.7881)2 = .3789

6.82 a. P(Z > b.

c. d.

e.

6.83

6.84

65 − 60 ) = P(Z > .5) = 1 – FZ(.5) = .3085 10 50 − 60 70 − 60 b. P( 1.96) = .025, 1.96 = , Xi = 79.6 10 d. P(Z > .675) = .025, .675 = The shortest range will be the interval Xi − 60 centered on the mean. Since the P(Z > .675) = .025, .675 = . 10 Xi − 60 Xi = 66.75. The lower value of the interval will be –.675 = 10 which is Xi = 53.25. Therefore, the shortest range will be 66.75 – 53.25 = 13.5. This is by definition the InterQuartile Range (IQR). d. P(X > 65) = .3085 (from part a) Use the binomial formula: P(X = 2) = C24 (.3085) 2 (.6915) 2 = 0.2731 a. P(Z >

85 − 100 ) = P(Z < -.5) = .3085 30 70 − 100 130 − 100 b. P( 1.645) = .05, 1.645 = , Xi = 149.35 30 60 − 100 d. P(Z > ) = P(Z > -1.33) = FZ(1.33) = .9032 30 P(X ≥ 1) = 1 – P(X = 0) = 1 – (.0918)2 = .9916 a. P(Z <


e. Use the binomial formula: P(X = 2) = C24 (.9082) 2 (.0918) 2 = 0.0417 f. 90 – 109 g. 130 - 149

6.85

b. c. d.

e. f. 6.86

15 − 20 25 − 20 2.5) =1 - Fz (2.5) = .0062 P(Z > 4 P(X ≥ 1) = 1 – P(X = 0) = 1 – [FZ(2.5)]5 = .0306 Xi − 20 , Xi = 22.1 The shortest range will be P(Z > .525) = .3, .525 = 4 the interval centered on the mean. The lower value of the interval will Xi − 20 be –.525 = which is Xi = 17.9. Therefore, the shortest range 4 is 22.1 – 17.9 = 4.2. 19 – 21 21 – 23

a. P(

P(Z > 1.28) = .1, 1.28 = P(Z >

6.87

σ

, σ = 23.4375

140 − 100 ) = P(Z > 1.71) = 1 – FZ(1.71) = .0436 23.4375

P(Z > 1.28) = .1, 1.28 = P(Z <

130 − 100

25 − μ , μ = 21.8 2.5

20 − 21.8 ) = P(Z < -.72) = 1 – FZ(.72) = .2358 2.5

6.88

E[X] = 1000(.4) = 400, σx = (1000)(.4)(.6) = 15.4919 500 − 400 P(Z < ) = P(Z < 6.45) ≈ 1.0000 15.4919

6.89

E[X] = 400(.6) = 240, σx = (400)(.6)(.4) = 9.798 200 − 240 P(Z > ) = P(Z > -4.08) ≈ 1.0000 9.798

6.90

P(Z <

50 − 70 ) = P(Z < -2.39) = 1 – FZ (2.39) = .0084 70

141

142

th


6.91

a. P(X = 6) =

e −6 66 = .1606 6!

b. 20 minutes = 1/3 hours, P(X > 1/3) = e

−

6 3

= .1353 −

6

c. 5 minutes = 1/12 hour, P(X < 1/12) = 1 - e 12 = .3935 d. 30 minutes = .5 hour, P(X > .5) = e − (.5)(6) = .0498 6.92

a. E[X] = 600(.4) = 240, σx = (600)(.4)(.6) = 12 260 − 240 P(Z > ) = P(Z > 1.67) = 1 – FZ(1.67) = .0475 12 Xi − 240 b. P(Z > -.254) = .6, -.254 = , Xi = 236.95 (237 listeners) 12

6.93

a. P(

120 − 132 150 − 132
= .7745 b. P(Z > .44) = .33, .44 =

Xi − 132 , Xi = 137.28 12

120 − 132 ) = P(Z < -1) = 1 – FZ(1) = .1587 12 d. E[X] = 100(.1587) = 15.87, σx = (100)(.1587)(.8413) = 3.654 25 − 15.87 P(Z > ) = P(Z > 2.5) = 1 – FZ(2.5) = .0062 3.654

c. P(Z <

6.94

P(Z>1.28)=.1, 1.28=

3.5 − 2.4

3+ hours on task: P(Z > 400(.242) = 96.8, σx =

σ

, σ=.8594. Probability that 1 exec spends

3 − 2.4 ) = P(Z > .7) = 1 – FZ(.7) = .242. E[X] = .8594 80 − 96.8 )=P(Z>(400)(.242)(.758) = 8.566. P(Z > 8.566

1.96) = FZ(1.96)=.975 6.95 Portfolio consists of 10 shares of stock A and 8 shares of stock B. a. Find the mean and variance of the portfolio value: W = 10X + 8Y with correlation of .3. μW = aμ x + bμ y = 10(10) + 8(12) = 196

σ 2W = a 2σ 2 X + b 2σ 2Y + 2abCorr ( X , Y )σ X σ Y = 102(16) + 82(9) + 2(10)(8)(.3)(4)(3) = 2,752 b. Option 1: Stock 1 with mean of 10, variance of 25, correlation of -.2. σ 2W = a 2σ 2 X + b 2σ 2Y + 2abCorr ( X , Y )σ X σ Y = 102(25) + 82(9) + 2(10)(8)(-.2)(5)(3) = 2,596


Option 2: Stock 2 with mean of 10, variance of 9, correlation of .6. = 102(25) + 82(9) + 2(10)(8)(.6)(5)(3) = 2,340 To reduce the variance of the porfolio, select Option 2 6.96 Portfolio consists of 10 shares of stock A and 8 shares of stock B a. Find the mean and variance of the portfolio value: W = 10X + 8Y with correlation of .3. μW = aμ x + bμ y = 10(12) + 8(10) = 200

σ 2W = a 2σ 2 X + b 2σ 2Y + 2abCorr ( X , Y )σ X σ Y = 102(14) + 82(12) + 2(10)(8)(.5)(3.74166)(3.4641) = 3,204.919 b. Option 1: Stock 1 with mean of 12, variance of 25, correlation of -.2. σ 2W = a 2σ 2 X + b 2σ 2Y + 2abCorr ( X , Y )σ X σ Y = 122(25) + 82(12) + 2(10)(8)(-.2)(5)(3.4641) = 3,813.744 Option 2: Stock 2 with mean of 10, variance of 9, correlation of .6. = 102(9) + 82(12) + 2(10)(8)(.6)(3)(3.4641) = 2,665.66 To reduce the variance of the porfolio, select Option 2 6.97

μW = aμ x + bμ y = 1(800000) + 1(60000) = 140000 σ 2W = a 2σ 2 X + b 2σ 2Y + 2abCorr ( X , Y )σ X σ Y = 12(1000000) + 12(810000) + 2(1)(1)(.4)(1000)(900) = 2,530,000 σ W = 2,530, 000 = 1590.597372 Probability that the weight is between 138,000 and 141,000: 138, 000 − 140, 000 141, 000 − 140, 000 = -1.26 fz = .3962, = .63 fz = .2357 1590.597372 1590.597372 .3962 + .2357 = .6319 6.98 a. μW = aμ x + bμ y = 1(40) + 1(35) = 75

σ 2W = a 2σ 2 X + b 2σ 2Y + 2abCorr ( X , Y )σ X σ Y = 12(100) + 12(144) + 2(1)(1)(.6)(10)(12) = 388 σ W = 388 = 19.69772 Probability that all seats are filled: 100 − 75 = 1.27 Fz = .8980. 1 - .8980 = .1020 19.69772 b. Probability that between 75 and 90 seats will be filled: 90 − 75 = .76 .5 – Fz(.76) = .2764 19.69772

143

_newbold_ism_06.pdf

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