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Vital Statistics: Probability and Statistics for Economics and Business Answers to Selected Exercises William H. Sandholm and Brett A. Saraniti

Chapter 2 Answers 2.1.1. a. Ex ante. b. Ex ante. c. Ex post. d. Ex ante. 2.2.1. a. {8}. b. {0}. c. {0, 1, 2, 3, 4, 5, 6, 7}. d. {0, 1, 2, 3}.

2.2.3. a. No airline raises its fare. b. Both airlines raise their fares. c. {RR, LL, UU, RL, LR, LU, UL}. d. {RR, RU, LL, UR, UU}.

2.2.5. a. {r, o, b, i, v}. b. ∅. c. {r, o, y, g}. d. {b}. e. {r, y, g, b, i, v}. f. {r, y, g}. g. ∅. h. {r}. i. {b, i, v}. j. ∅. k. A and B are disjoint. 2.2.7. (A ∪ B)C = AC ∩ BC is the unshaded region in the figure below.

(A ∩ B)C = AC ∪ BC is the unshaded region in the figure below.

2.3.1. a. Consistent. b. (A1) and (A2) are both violated. c. (A2) is violated. 2.3.3. a.

1 . 100

b.

1 . 10

2.3.5. .10. 2.4.1. a. 16 . b. 61 . c. 41 . d. 0. 2.4.3.

relax regulations raise minimum wage .20 a. don’t raise minimum wage .30 total .50 b. 31 . c. 45 .

2.4.5. a. 23 . b. 43 . 2.4.7. .0784. 2.5.1. a. .0004. b. .3643. 2.5.3. b., c. 2.5.5. .1785. 2.5.7. 1

don’t relax regulations .40 .10 .50

total .60 .40

a. Since A = (A ∩ B) ∪ (A ∩ BC ) additivity implies that P(A) = P(A ∩ B) + P(A ∩ BC ). Then by the independence of A and B and the complement rule, P(A ∩ BC ) = P(A) − P(A ∩ B) = P(A) − P(A)P(B) = P(A)(1 − P(B)) = P(A)P(BC ). b. Since part (a) did not place any restrictions on A or B, it tells us that for any pair of independent events, each member of the pair is independent of the complement of the other member. 2.5.9. a. Yes. b. No. c. Pairwise independence means that learning that one of the events occurred provides no information about whether any other single event occurred, but does not rule out the possibility of providing information about whether combinations of other events occurred. 2.6.1. a. Certainly false. b. Possibly true and possibly false. c. Possibly true and possibly false. d. Certainly true. e. Possibly true and possibly false. 2.C.1. Let I = “the stock’s value increased” and D = “the stock’s value decreased”. a. S = {III, DII, IDI, DDI, IID, DID, IDD, DDD}, where the first, second, and third terms of each element represent the outcomes of XO, Yeti, and Zoolander, respectively. b. A = {III, DII, IDI, IID}. c. B = {III, IDI, IID, IDD}. d. The event that XO and at least one other stock increases in value. 2.C.3. a. False. b. False. c. False. d. False. e. False. 2.C.5. a. P(A) =

7. 29

P(B) =

11 . 29

P(C) =

11 . 29

b. P(C|B ∪ C) = 12 . c. P(B|A ∪ B) =

11 . 18

1 . c. 9.7656. d. It is not strong evidence. Since there are many funds, 2.C.7. a. 321 . b. 1024 we would expect some of the funds to have long streaks of beating the market even if the performances of different funds and of each fund over different years were completely random.

2.C.9. c. 2.C.11. Let D = “owns a U.S. car” and F = “owns a foreign car”. D DC total F .38 .23 .61 a. FC .37 .02 .39 total .75 .25 b. c. d. e. f.

.3863. .9816. .5115. .6306. No.

2.C.13. a.

10 . 37

b.

12 . 29

2

Chapter 3 Answers 3.1.1. a. Discrete. b. Continuous. c. Discrete. d. Continuous. 3.1.3. a. b. c. d. e.

{sn, so, cn}. {sn, so}. {cn, co}. P({cn, co}) = .2. P({sn, cn}) = .75. P({sn}) = .6. P({co}) = .05. P({cn}) = .15. The distribution of X is calculated using the probabilities in part (d): x 30 50 180

P(X = x) .80 .15 .05

3.1.5. a. S = {ac, ad, bc, bd}. P(ac) = .006. P(bc) = .036. P(ad) = .094. P(bd) = .864. b. The definition of X is: state realization of X ac X(ac) = $5000 ad X(ad) = $5000 bc X(bc) = $5000 bd X(bd) = $0 The distribution of X is calculated using part (a) and the definition of X: x 0 $5000

P(X = x) .864 .136

3.2.1. E(X) = 1.15. Var(X) = 1.3275. SD(X) = 1.1522. 3.2.3. a. 2.6. b. 1.14. 3.2.5. E(X) = 2.9. Var(X) = 1.6095. SD(X) = 1.2687. 3.3.1. a. E(S) = 65,600. Var(S) = 41,400,000. b. E(V) = 15.1. Var(V) = 1.15. 3.3.3. a. Ci = 5,000 + .45Mi . b. E(Ci ) = !13,100. Var(Ci ) = 1,012,500. ! 5 5 P P Ci = 5,062,500. Ci = 65,500. Var c. E i=1

i=1

3.3.5. E(Y) = 530. 3.4.1. Yes.

3

$500 3.4.3. X $1,000 $2,000

$500 .09 .165 .045 .30

Y $1,000 .165 .3025 .0825 .55

$2,000 .045 .0825 .0225 .15

.30 .55 .15

3.4.5. a. .2597. b. At least 139 trucks. 3.4.7. E(T) = 23,100. 3.4.9. a. 9.7. b. 2.2113. 3.C.1. Let y = “yes” and n = “no”. a. S = {yyy, nyy, yny, nny, yyn, nyn, ynn, nnn}, where the first, second, and third terms of each element represent the applicant’s answer to the first, second, and third questions, respectively. s P(s) yyy .027 nyy .063 yny .063 nny .147

s P(s) yyn .063 nyn .147 ynn .147 nnn .343

b. The random variable as a function of the state space is: state yyy nyy yny nny

realization of X 3 2 2 1

state yyn nyn ynn nnn

realization of X 2 1 1 0

The distribution of X is: x 0 1 2 3

P(X = x) .343 .147 + .147 + .147 = .441 .063 + .063 + .063 = .189 .027

c. E(X) = .9. Var(X) = .63. SD(X) = .7937. d. .027. 3.C.3.

a.

H

D 45 50 55 60 75 .04 .08 .06 .02 .20 80 .10 .20 .15 .05 .50 85 .06 .12 .09 .03 .30 .20 .40 .30 .10

b. .55. 4

3.C.5. a.

X1

0 1

X2 0 1 .64 .16 .80 .16 .04 .20 .80 .20

b. Let Ai = {Xi = 1}. Then the event that the first four customers sampled have loyalty cards is A1 ∩ A2 ∩ A3 ∩ A4 . P(A1 ∩ A2 ∩ A3 ∩ A4 ) = .4096.

3.C.7. E(Y) = 80,000. SD(Y) = 30,000.

3.C.9. a. E(N3 ) = 1.6639. b. E(N4 ) = 2.1344. c. E(N5 ) = 2.7045. 3.C.11. a. E(Y) = 2.23. Var(Y) = .9411. SD(Y) = .9701. b. E(Z) = 26.76. Var(Z) = 135.8184. SD(Z) = 11.6541.

5

Chapter 4 Answers 4.1.1. The answers are in millions of dollars. Y 1 3 4 4 0 0 .30 .30 a. X 6 0 .50 0 .50 10 .20 0 0 .20 .20 .50 .30 b. E(X) = 6.2. Var(X) = 4.36. c. E(Y) = 2.9. Var(Y) = 1.09. 4.1.3. a.

x 0 1 2 3

P(X = x) .10 .30 .40 .20

b.

x 0 1 2 3

P(X = x) .10 .30 .40 .20

c.

y 0 1 2 3

P(Y = y) .10 .30 .40 .20

y 0 1 2 3

P(Y = y) .10 .30 .40 .20

P (X = x|Y = 3) .15 .15 .20 .50   P Y = y|X = 0 .20 .20 .30 .30

4.1.5. The answers are in millions of dollars. state s probability P(s) realization X(s) d .15 8 a. r .65 11 i .20 15

realization Y(s) .50 .15 -.30

b. E(X) = 11.35. Var(X) = 4.4275. SD(X) = 2.1042. c. E(Y) = .1125. Var(Y) = .0575. SD(Y) = .2397. 4.1.7. a. b. c. d.

E(X) = .7. Var(X) = .61. SD(X) = .781. E(X = x|Y = 0) = .6667. Var(X = x|Y = 0) = .5556. SD(X = x|Y = 0) = .7454. E(X = x|Y = 2) = 1.0. Var(X = x|Y = 2) = .6667. SD(X = x|Y = 2) = .8165. The differences are due to the fact that knowing the realization of Y gives us information about the probabilities of the realizations of X.

4.1.9. 6

a.

x 0 1

P(X = x) .15 .85

b.

y 0 1

P(Y = y) .25 .75

c.

y 0 1

P(Y = y) .25 .75

y 0 1

P(Y = y) .25 .75   P Y = y|X = 0 .6667 .3333   P Y = y|X = 1 .1765 .8235

4.2.1. a. Var(X) = 4.36. Var(Y) = 1.09. b. Cov(X, Y) = −2.18. Corr(X, Y) = −1.

4.2.3. Corr(M, S) = √

Cov(M,S)

Var(M)

√

Var(S)

=

√ 0√ .5 .5

= 0.

4.2.5. Let {±x1 , ±x2 , . . . , ±xn } be the set of realizations of the random variable X. a. Using the = xi ) = P(X = −xi ) allows us to calculate: P fact that P(XP Pn n E(X) = x xP(X = x) = i=1 0 = 0. ((xi × P(X = xi ) − xi × P(X = xi )) = i=1 Since X has a distribution that is symmetric about 0, it follows that X3 is symmetric about 0. So it follows from the previous result that E(X3 ) = 0. b. Cov(X, Y) = E(XY) − E(X)E(Y) = E(X3 ) − E(X)E(X2 ) = 0 − 0 × E(X2 ) = 0. 4.3.1.

a.

z 7 11 15

P(Z = z) .15 .05 0

z 27 31 35

z P(Z = z) 127 0 131 0 135 .2

P(Z = z) .05 .5 .05

b. E(Z) = 47.2. Var(X) = 2,008.76. 4.3.3. a. E(aX + bY) = aE(X) + bE(Y). b. Var(aX + bY) = a2 Var(X) + b2 Var(Y) + 2ab Cov(X, Y). 4.3.5. a. 482.25. b. 9,282.195. 4.4.1. a. Yes. b. Yes. 4.4.3. σA,B < .004. 4.4.5. a. E(.9IBM + .1GE) = .2453. SD(.9IBM + .1GE) = .3101. b. (.2611, .7389). c. The minimum variance portfolio is (.5197, .4803). E(.5197IBM + .4803GE) = .3072. SD(.5197IBM + .4803GE) = .2757. d. No because it is dominated by the portfolio (.5197, .4803). 4.C.1.

a.

X

Y -40 0 80 -10 .40 0 0 .40 -5 .05 .05 0 .10 20 0 .20 .30 .50 .45 .25 .30

7

b. E(X) = 5.5. E(Y) = 6.0. Var(X) = 212.25. Var(Y) = 2,604.0. SD(X) = 14.5688. SD(Y) = 51.0294. c. Cov(X, Y) = 617.0. Corr(X, Y) = .8299. x P(X = x) P (X = x|Y = 0) -10 .40 0 d. -5 .10 .20 20 .50 .80 4.C.3.

a.

40% A 20% 0%

B 20% 10% 0% .08 .02 0 .10 .07 .10 .03 .20 .05 .48 .17 .70 .20 .60 .20

b. E(A) = 8.0. E(B) = 10.0. Var(A)=176.0. Var(B) = 40.0. SD(A) = 13.2665. SD(B) = 6.3246. c. No. a P(A = a) P (A = a|B = 20%) 40% .10 .40 d. 20% .20 .35 0% .70 .25 Learning that the returns of Biloxi are good will indicate that the returns on Acme are likely to be good. b P(B = b) P (B = b|A = 40%) 20% .20 .80 e. 10% .60 .20 0% .20 0 E(B|A = 40%) = 18. f. Cov(A, B) = 40.0. Corr(A, B) = .4767. 4.C.5. a. 500V − 15. b. 1000W . c. 500V + 1000W − 15. d. .01. e. .0016. f. .0032. g. .01. h. 1. i. 45. j. 2500. k. 50. l. 90. m. 1600. n. 40. o. 135. p. 1600. q. 7300. r. 85.44. s. .8. t. The correlations are identical because the 500V − 15 and 1000W are the dollar returns of Vox and Wyck and they have the same relationship with each other as the percentage returns V and W. 4.C.7. Yes. 4.C.9. a. No. b. 0. c. Parts (a) and (b) demonstrate that 0 covariance does not imply independence.

8

Chapter 5 Answers 5.1.1. a. Let F ={A, 2, 3, 4, 5}.   1 if s ∈ F; IF (s) =   0 if s < F.

b. Let B ={2, 4, 6, 8, 10}.   1 if s ∈ B; IB (s ) =   0 if s < B. c. Let C ={A}.   1 if s ∈ C; IC (s) =   0 if s < C.

d. Let D ={A, J, Q, K}.    1 if s ∈ D; ID (s ) =    0 if s < D.

5.1.3. E(IA ) = .70. Var(IA ) = .21. 10 5.2.1. a. {Xi }i=1 ∼ BTP(10, .92) where Xi is a random variable that equals 1 if the ith copy of software is fake and is 0 otherwise. b. .4344. c. 1.0000.

5.2.3. a. Let Xi be a random variable that equals 1 if ith car passes the safety check and is 4 ∼ BTP(4, .96) is the Bernoulli 0 otherwise. Then the sequence of random variables {Xi }i=1 trials process. b. .8493. 5.3.1. 240. 5.3.3. a. 105 . b. 1022. 5.3.5. 20,475. 5.3.7. a. 390,700,800. b. 15,504. 5.4.1. a. .1478. b. .0343. c. .9599. d. E(X) = 9.2. Var(X) = .736. 5.4.3. a. X ∼ binomial(16, .20). b. E(X) = 3.2. Var(X) = 2.56. c. .2001.

5.4.5. a. X ∼ binomial(6, .7). b. E(X) = 4.2. Var(X) = 1.26. c. .4202. d. The answer would change because it would no longer be reasonable to assume that each friend’s actions are independent of the others. 5.4.7. Answers will vary. 5.5.1. a. The probability will be larger because there is an additional date on which someone’s birthday could occur. b. .997264. c. .991806. P500 5.C.1. P500(.60). E(Xi ) = .60. Var(Xi ) = .24. b. i=1 Xi ∼ binomial(500, .60). P500 a. Xi ∼ Bernoulli E( i=1 Xi ) = 300. Var( i=1 Xi ) = 120. 9

P20 5.C.3. a. E(Xi ) = .40. Var(Xi ) = .24. b. S20 = i=1 Xi ∼ binomial(20, .40). c. .1659. 1 ¯ ¯ d. E(S20) = 8. Var(S20 ) = 4.8. e. X20 = 20 S20. E( X20) = .40. Var(X¯ 20) = .012. 5.C.5. a. E(X) = 160. Var(X) = 32. b. .0704. c. .0017. 5.C.7. a. 173.18. b. 263.36. c. 734.48. d. 1010.82. 5.C.9. a.

b.

T1 s 0 1 2

0 1

T2 0 1 .81 .09 .09 .01 .9 .1

.9 .1

P(S = s) .81 .18 .01

c. Yes. d. T1 and T2 are independent Bernoulli(.1). S is the sum of these two random variables, so it is binomially distributed. 5.C.11. 2,300. 5.C.13. a. 2.8. b. .0253. c. .1875. d. .4278.

10

Chapter 6 Answers 6.2.1. The table is: x 0 1 2 3 4 5

P(X = x) .1160 .3124 .3364 .1811 .0488 .0053

P(X ≤ x) .1160 .4284 .7648 .9460 .9947 1

6.2.3. a. .49. b. .49. c. 1. d. .5. e. 0. 6.2.5. .066. 6.3.1. a. .15. b. .40. c. 10. d. 33.3333. 6.3.3. .05. 6.3.5. a. .2. b. .35. c. uniform(.0284, .0852). d. uniform(.5, 1.5). 6.3.7. a. X ∼ uniform(0, 30). E(X) = 15. Var(X) = 75. b. Y ∼ uniform(0, .5). E(Y) = .25. Var(X) = .0208. 6.3.9. x =

c(h−l) 100

+ l.

6.4.1. N(162.5, 900). 6.4.3. a. −1.2816. b. 1.6449. c. 2.3263.

6.4.5. a. .6827. b. .9545. c. .9973. 6.4.7. a. .8416. b. 1.2816.

6.4.9. a. N(63, 256). b. .5744. c. Trip distances cannot be exactly normally distributed because distances cannot be negative. 6.4.11. a. N(90000, 756250000). b. .3581. c. .0018. 6.5.1. a. .2734. b. .4147. c. .4147. d. .0143. e. .3594. 6.5.3. a. .0344. b. .5793. c. .1562. d. .3084. 6.5.5. a. .6826. b. .9544. c. .9974. 6.5.7. 117 minutes. 6.6.1. a. A + B ∼ N(950, 3600). b. .2023. 6.6.3. a. N(250, 490). b. .3257. 6.6.5. a. .0127. b. .0067. 6.6.7. a. .1574. b. .0036. 6.C.1. a. X ∼ uniform(0, 30). b. A = {X ∈ [0, 6]}. P(A) = 51 . c. B = {X ∈ [15, 18]}. P(B) = 11

1 . 10

6.C.3. a.

1. 10

b. 52.

6.C.5. a. .0228. b. 21.7275 thousand gallons. 6.C.7. a. .2. b. .2. 6.C.9. The 10th percentile is 15.5749 minutes and the 90th percentile is 40.4251 minutes. 6.C.11. a. uniform(500, 937.5). b.

11 . 35

6.C.13. a. P(X1 < $10,000) = .1587. P(X2 < $10,000) = .0008. P(X3 < $10,000) = .3618. b. P(X1 > $18,000) = .0014. P(X2 > $18,000) = .7365. P(X3 > $18,000) = .0067. c. No. This is because the probabilities of seeing events at the tails of the distribution depends on the variance.

12

Chapter 7 Answers 7.1.1. a., c., d. 7.1.3. a. A and B are independent and identically distributed. b. A and C are independent and are not identically distributed. c. A and D are not independent and are identically distributed. P64 1 Xi . E(S64) = 256,000. SD(S64 ) = 12,000. b. X¯ 64 = 64 7.2.1. a. S64 = i=1 S64. E(X¯ 64) = 4,000. ¯ 64) = 187.5. SD( X 7.2.3. a. E(S25) = 8,000. Var(S25) = 490,000. SD(S25 ) = 700. b. E(X¯ 25) = 320. Var(X¯ 25) = 784. ¯ 25) = 28. SD( X 7.2.5. a. E(S100) = 4,800. Var(S100 ) = 40,000. SD(S100 ) = 200. The expected value, variance, and standard deviation are larger than in Exercise 7.2.4. This is because there are more employees in the sample, so the total amount of time wasted is very likely to be larger, ¯ 100) = 2. The expected and to exhibit more dispersion. b. E(X¯100) = 48. Var(X¯100) = 4. SD( X value is identical to the expected value in Exercise 7.2.4. The sample mean is measured per employee, so its expected value is independent of the number of employees. The variance and standard deviation are smaller because the as the size of the sample increases, values of the sample mean that are further from the mean become less likely. 7.2.7. a. E(S600) = 138. Var(S600) = 106.26. SD(S600 ) = 10.3082. b. E(B¯600) = .23. ¯ 600) = .0003. SD(B¯ 600) = .0172. Var( B 7.3.1. a. Var(X¯ 10) = .9. b. Var(X¯ 100) = .09. c. Var(X¯ 1,000) = .009. d. Var(X¯ 10,000) = .0009. e. Var(X¯ 100,000) = .00009. ¯ 50 < (292.00, 298.00)) = .1573. This calculation 7.3.3. a. P(X¯ 50 < (292.00, 298.00)) ≤ .5. b. P( X relies on the fact that the sum of independent normally distributed random variables is a normally distributed random variable. ¯ 85 ≈ N(19, 1.1765). b. P(X¯85 > 20) ≈ .1753. 7.4.1. a. X 7.4.3. a. S60 ≈ N(162, 114). b. P(S85 ∈ [159.5, 160.5]) ≈ .0367. c. P(S85 ∈ [169.5, 170.5]) ≈ .0282. d. P(S85 ≥ 160.5) ≈ .5559. 7.4.5. a. S64 ≈ N(256000, 144000000). b. P(S64 > 240,000) ≈ .9088. c.X¯64 ≈ N(4000, 35156.25). d. P(X¯ 64 < 3,750) ≈ .0912. 7.4.7. a. X¯ 500 ≈ N(48, .8). b. P( X¯ 500 > 49) ≈ .1318. c. X¯ 2500 ≈ N(48, .16). d. P(X¯ 2500 > 49) ≈ .0062. 7.5.1. a. .8336. b. .0049. 7.5.3. P(X¯ 100 > 1.60) ≈ .0228. This event is not likely to occur. 7.5.5. a. .1954. b. .8936. 7.C.1. a. E(S50,000) = 800,000. Var(S50,000 ) = 2,000. b. N(800000, 2000). c. E(X¯ 50) = 16. ¯ 50) = .0008. d. N(16, .0008). Var( X

13

P 7.C.3. a. S100 = 100 i=1 Xi is a random variable that represents the total number of returns that require an audit. S100 ≈ N(40, 24). b. .4099. c. .0160. ¯ n ) = .10. Var(R¯n ) = .25/n. b. .1587. c. By the law of large numbers, we know 7.C.5. a. E( R that the portfolio returns converge in probability to the mean return E(Ri ) = .10 of an asset as the number of assets approaches infinity. d. This is a true statement. 7.C.7. Let SEn and SnW be random variables that represent the total sales over n days of the restaurants on the east side and west side, respectively. a. SnE≈ N(n × 890, n × 1902 ) and 2 SW n ≈ N(n × 905, n × 225 ), where n is the number of days in a month. b. .3901. c. .1652.

7.C.9. a. .7000. b. .4127. c. This is not strong evidence because it is reasonable to observe at least 90 acceptances if the true probability is .08. In particular, the probability of seeing at least 90 acceptances is approximately .1341. Pt 7.C.11. a. Pt = Pt−1 + Xt = Pt−2 + Xt−1 + Xt = · · · = P0 + i=1 Xi . b. E(P40 − P0 ) = 0. ¯ 100) = 1 . Var(P40 − P0 ) = 40. c. E(X¯ 100) = 0. Var( X 100

14

Chapter 8 Answers 8.1.1. a. 1,080,000. b. 1,082,999.51. c. 1,083,277.57. d. 1,083,287.07.  P  P 400 400 Xi = 240. SD i=1 Xi = 8.1.3. a. E(Xi ) = .6. SD(Xi ) = .7746. b. .5488. c. .0231. d. E i=1 15.4919. 8.1.5. a. .0984. b. .1771. c. .6973. 8.1.7. a. .0994. b. .0993. 8.1.9. a. P(N = 0) = .1353. P(N = 1) = .2707. P(N = 2) = .2707 b. E(N) = 2. Var(N) = 2.0000. c. P(N = 0) ≈ .1353. P(N = 1) ≈ .2707. P(N = 2) ≈ .2707. d. E(N) ≈ 2. Var(N) ≈ 2.

8.2.1. a. E(T) = 50. Var(T) = 2500. SD(T) = 50. b.

1 . minutes

8.2.3. a. .0084. b. .0555. c. .1842. 8.2.5. a. .4493. b. .2231. c. .8647. d. .1410. 8.2.7. a. exponential(.0000004). b. E(C) = $2,500,000. SD(C) = $2,500,000. c. .3297. d. .3012. 8.2.9. a. 2.2222. b. .1054. c. .4444. d. By memorylessness, the answers to (a) and (b) are the same if understood as referring to the additional time. (Alternatively, to have answers in terms of the time since the passenger ahead of you was served, add 7 to the answers to (a) and (b).) The answer to (c) is unchanged, also because of memorylessness. 8.3.1. a., b., d. 8.3.3. a. .0009. b. .2704. c. .0220. 8.3.5. a. .1353. b. .0498. c. .1954. d. .5470. 8.C.1. a. .1653. b. .2678. c. .0364. 8.C.3. a. 4.8. b. .0082. c. .9523. 8.C.5. a. .1175. b. .2760. c. .3127. 8.C.7. a. .7163. b. .2857. c. .0823. d. This would not affect the answers to parts (a)–(c). This follows from the fact that the exponential distribution is memoryless. 8.C.9. a. .4724. b. .2784. c. .7769.

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Chapter 9 Answers 9.Q.1. If sequence B occurs it must be that sequence A occurred so sequence B is less likely as it imposes an additional restriction. Thus, choosing sequence B is not optimal. 9.Q.3. a. The interval of IQ values should be centered at a number above 75, since this person’s “random error” was probably negative. b. The interval of IQ values should be centered at 100. 9.Q.5. a. Summing choices C and F results in a 75% chance of losing $760 and a 25% chance of gaining $240. Summing choices D and E results in a 25% chance of gaining $250 and a 75% chance of losing $750. b. Choosing choices C and F implies that the person prefers choice A over choice B. However, the person chose B, which is inconsistent with selecting choices C and F. c. Ma...