##### Document Text Contents

Page 1

ANSWERS

Chapter 1

1.5 a 2.45 − 2.65, 2.65 − 2.85

b 7/30

c 16/30

1.9 a Approx. .68

b Approx. .95

c Approx. .815

d Approx. 0

1.13 a ȳ = 9.79; s = 4.14

b k = 1: (5.65, 13.93); k = 2: (1.51,

18.07); k = 3: (−2.63, 22.21)

1.15 a ȳ = 4.39; s = 1.87

b k = 1: (2.52, 6.26); k = 2: (0.65,

8.13); k = 3: (−1.22, 10)

1.17 For Ex. 1.2, range/4 = 7.35; s = 4.14;

for Ex. 1.3, range/4 = 3.04; s = 3.17;

for Ex. 1.4, range/4 = 2.32, s = 1.87.

1.19 ȳ − s = −19 < 0

1.21 .84

1.23 a 16%

b Approx. 95%

1.25 a 177

c ȳ = 210.8; s = 162.17

d k = 1: (48.6, 373); k = 2:

(−113.5, 535.1); k = 3: (−275.7,

697.3)

1.27 68% or 231 scores; 95% or 323 scores

1.29 .05

1.31 .025

1.33 (0.5, 10.5)

1.35 a (172 − 108)/4 = 16

b ȳ = 136.1; s = 17.1

c a = 136.1 − 2(17.1) = 101.9;

b = 136.1 + 2(17.1) = 170.3

Chapter 2

2.7 A = {two males} = {(M1, M2),

(M1,M3), (M2,M3)}

B = {at least one female} = {(M1,W1),

(M2,W1), (M3,W1), (M1,W2), (M2,W2),

(M3,W2), (W1,W2)}

B̄ = {no females} = A; A ∪ B = S;

A ∩ B = null; A ∩ B̄ = A

2.9 S = {A+, B+, AB+, O+, A−, B−,

AB−, O−}

2.11 a P(E5) = .10; P(E4) = .20

b p = .2

2.13 a E1 = very likely (VL); E2 =

somewhat likely (SL); E3 =

unlikely (U); E4 = other (O)

b No; P(VL) = .24, P(SL) = .24,

P(U) = .40, P(O) = .12

c .48

2.15 a .09

b .19

2.17 a .08

b .16

c .14

d .84

2.19 a (V1, V1), (V1, V2), (V1, V3),

(V2, V1), (V2, V2), (V2, V3),

(V3, V1), (V3, V2), (V3, V3)

b If equally likely, all have

probability of 1/9.

c P(A) = 1/3; P(B) = 5/9;

P(A ∪ B) = 7/9;

P(A ∩ B) = 1/9

2.27 a S = {CC, CR, CL, RC, RR, RL,

LC, LR, LL}

b 5/9

c 5/9

877

Page 2

878 Answers

2.29 c 1/15

2.31 a 3/5; 1/15

b 14/15; 2/5

2.33 c 11/16; 3/8; 1/4

2.35 42

2.37 a 6! = 720

b .5

2.39 a 36

b 1/6

2.41 9(10)6

2.43 504 ways

2.45 408,408

2.49 a 8385

b 18,252

c 8515 required

d Yes

2.51 a 4/19,600

b 276/19,600

c 4140/19,600

d 15180/19,600

2.53 a 60 sample points

b 36/60 = .6

2.55 a

(

90

10

)

b

(

20

4

)(

70

6

)/(

90

10

)

= .111

2.57 (4 × 12)/1326 = .0362

2.59 a .000394

b .00355

2.61 a

364n

365n

b .5005

2.63 1/56

2.65 5/162

2.67 a P(A) = .0605

b .001344

c .00029

2.71 a 1/3

b 1/5

c 5/7

d 1

e 1/7

2.73 a 3/4

b 3/4

c 2/3

2.77 a .40 b .37 c .10

d .67 e .6 f .33

g .90 h .27 i .25

2.93 .364

2.95 a .1

b .9

c .6

d 2/3

2.97 a .999

b .9009

2.101 .05

2.103 a .001

b .000125

2.105 .90

2.109 P(A) ≥ .9833

2.111 .149

2.113 (.98)3(.02)

2.115 (.75)4

2.117 a 4(.5)4 = .25

b (.5)4 = 1/16

2.119 a 1/4

b 1/3

2.121 a 1/n

b

1

n

;

1

n

c

3

7

2.125 1/12

2.127 a .857

c No; .8696

d Yes

2.129 .4

2.133 .9412

2.135 a .57

b .18

c .3158

d .90

2.137 a 2/5

b 3/20

2.139 P(Y = 0) = (.02)3;

P(Y = 1) = 3(.02)2(.98);

P(Y = 2) = 3(.02)(.98)2;

P(Y = 3) = (.98)3

2.141 P(Y = 2) = 1/15; P(Y = 3) = 2/15;

P(Y = 4) = 3/15; P(Y = 5) = 4/15;

P(Y = 6) = 5/15

2.145 18!

2.147 .0083

2.149 a .4

b .6

c .25

2.151 4[p4(1 − p) + p(1 − p)4]

2.153 .313

2.155 a .5

b .15

c .10

d .875

Page 10

886 Answers

5.163 b F(y1, y2) =

y1 y2[1 − α(1 − y1)(1 − y2)]

c f (y1, y2) =

1 − α[(1 − 2y1)(1 − 2y2)],

0 ≤ y1 ≤ 1, 0 ≤ y2 ≤ 1

d Choose two different values for α

with −1 ≤ α ≤ 1.

5.165 a (p1et1 + p2et2 + p3et3)n

b m(t , 0, 0)

c Cov(X1, X2) = −np1 p2

Chapter 6

6.1 a

1 − u

2

, −1 ≤ u ≤ 1

b

u + 1

2

, −1 ≤ u ≤ 1

c

1√

u

− 1, 0 ≤ u ≤ 1

d E(U1) = −1/3, E(U2) =

1/3, E(U3) = 1/6

e E(2Y −1) = −1/3, E(1−2Y ) =

1/3, E(Y 2) = 1/6

6.3 b fU (u) ={

(u + 4)/100, −4 ≤ u ≤ 6

1/10, 6 < u ≤ 11

c 5.5833

6.5 fU (u) =

1

16

(

u − 3

2

)−1/2

,

5 ≤ u ≥ 53

6.7 a fU (u) =

1

√

π

√

2

u−1/2e−u/2,

u ≥ 0

b U has a gamma distribution with

α = 1/2 and β = 2 (recall that

�(1/2) = √π).

6.9 a fU (u) = 2u, 0 ≤ u ≤ 1

b E(U ) = 2/3

c E(Y1 + Y2) = 2/3

6.11 a fU (u) = 4ue−2u , u ≥ 0, a gamma

density with α = 2

and β = 1/2

b E(U ) = 1, V (U ) = 1/2

6.13 fU (u) = F ′U (u) =

u

β2

e−u/β , u > 0

6.15 [−ln(1 − U )]1/2

6.17 a f (y) = αy

α−1

θα

, 0 ≤ y ≤ θ

b Y = θU 1/α

c y = 4√u. The values are 2.0785,

3.229, 1.5036, 1.5610, 2.403.

6.25 fU (u) = 4ue−2u for u ≥ 0

6.27 a fY (y) =

2

β

we−w

2/β , w ≥ 0, which

is Weibull density with m = 2.

b E(Y k/2) = �

(

k

2

+ 1

)

βk/2

6.29 a fW (w) =

1

�

(

3

2

)

(kT )3/2

w1/2e−w/kT w > 0

b E(W ) = 3

2

kT

6.31 fU (u) =

2

(1 + u)3 , u ≥ 0

6.33 fU (u) = 4(80 − 31u + 3u2),

4.5 ≤ u ≤ 5

6.35 fU (u) = − ln(u), 0 ≤ u ≤ 1

6.37 a mY1(t) = 1 − p + pet

b mW (t) = E(etW ) = [1− p + pet ]n

6.39 fU (u) = 4ue−2u , u ≥ 0

6.43 a Ȳ has a normal distribution

with mean µ and variance σ 2/n

b P(|Ȳ − µ| ≤ 1) = .7888

c The probabilities are .8664, .9544,

.9756. So, as the sample size

increases, so does the probability

that P(|Ȳ − µ| ≤ 1)

6.45 c = $190.27

6.47 P(U > 16.0128) = .025

6.51 The distribution of Y1 + (n2 − Y2) is

binomial with n1 + n2 trials and success

probability p = .2

6.53 a Binomial (nm, p) where

ni = m

b Binomial (n1 = n2 + · · · nn , p)

c Hypergeometric (r = n,

N = n1 + n2 + · · · nn)

6.55 P(Y ≥ 20) = .077

6.65 a f (u1, u2) =

1

2π

e−[u

2

1+(u2−u1)2]/2 =

1

2π

e−(2u

2

1−2u1u2+u22)/2

b E(U1) = E(Z1) = 0,

E(U2) = E(Z1 + Z2) = 0,

V (U1) = V (Z1) = 1,

V (U2) = V (Z1 + Z2) =

V (Z1) + V (Z2) = 2,

Cov(U1, U2) = E(Z 21) = 1

Page 11

Answers 887

c Not independent since

ρ

= 0.

d This is the bivariate normal

distribution with µ1 = µ2 = 0,

σ 21 = 1, σ 22 = 2, and ρ =

1√

2

6.69 a f (y1, y2) =

1

y21 y

2

2

, y1 > 1,

y2 > 1

e No

6.73 a g(2)(u) = 2u, 0 ≤ u ≤ 1

b E(U2) = 2/3, V (U2) = 1/18

6.75 (10/15)5

6.77 a

n!

( j − 1)!(k − 1 − j)!(n − k)!

y

j−1

j [yk − y j ]k−1− j [θ − yk]n−k

θ n

,

0 ≤ y j < yk ≤ θ

b

(n − k + 1) j

(n + 1)2(n + 2) θ

2

c

(n − k + j + 1)(k − j)

(n + 1)2(n + 2) θ

2

6.81 b 1 − e−9

6.83 1 − (.5)n

6.85 .5

6.87 a g(1)(y) = e−(y−4), y ≥ 4

b E(Y(1)) = 5

6.89 fR(r) = n(n − 1)rn−2(1 − r),

0 ≤ r ≤ 1

6.93 f (w) = 2

3

(

1√

w

− w

)

, 0 ≤ w ≤ 1

6.95 a fU1(u) =

1

2

0 ≤ u ≤ 1

1

2u2

u > 1

b fU2(u) = ue−u , 0 ≤ u

c Same as Ex. 6.35.

6.97 p(W = 0) = p(0) = .0512,

p(1) = .2048, p(2) = .3264,

p(3) = .2656, p(4) = .1186,

p(5) = .0294, p(6) = .0038,

p(7) = .0002

6.101 fU (u) = 1, 0 ≤ u ≤ 1 Therefore, U has

a uniform distribution on (0, 1)

6.103

1

π(1 + u21)

, ∞ < u1 < ∞

6.105

1

B(α, β)

uβ−1(1 − u)α−1, 0 < u < 1

6.107 fU (u) =

1

4

√

u

0 ≤ u < 1

1

8

√

u

1 ≤ u ≤ 9

6.109 P(U = C1 − C3) = .4156;

P(U = C2 − C3) = .5844

Chapter 7

7.9 a .7698

b For n = 25, 36, 69, and 64, the

probabilities are (respectively)

.8664, .9284, .9642, .9836.

c The probabilities increase with n.

d Yes

7.11 .8664

7.13 .9876

7.15 a E(X̄ − Ȳ ) = µ1 − µ2

b V (X̄ − Ȳ ) = σ 21 /m + σ 22 /n

c The two sample sizes should be at

least 18.

7.17 P

(∑6

i=1 Z

2

i ≤ 6

)

= .57681

7.19 P(S2 ≥ .065) = .10

7.21 a b = 2.42

b a = .656

c .95

7.27 a .17271

b .23041

d .40312

7.31 a 5.99, 4.89, 4.02, 3.65, 3.48, 3.32

c 13.2767

d 13.2767/3.32 ≈ 4

7.35 a E(F) = 1.029

b V (F) = .076

c 3 is 7.15 standard deviations above

this mean; unlikely value.

7.39 a normal, E(θ̂) = θ =

c1µ1 + c2µ2 + · · · + ckµk

V (θ̂) =

(

c21

n1

+ c

2

2

n2

+ · · · + c

2

k

nk

)

σ 2

b χ2 with n1 + n2 + · · · + nk − k df

c t with n1 + n2 + · · · + nk − k df

7.43 .9544

7.45 .0548

7.47 153

7.49 .0217

7.51 664

7.53 b Ȳ is approximately normal: .0132.

7.55 a random sample; approximately 1.

b .1271

Page 20

Answers 895

15.49 a .0256

b An usually small number of runs

(judged at α = .05) would imply a

clustering of defective items in

time; do not reject.

15.51 R = 13, do not reject

15.53 rS = .911818; yes.

15.55 a rS = −.8449887

b Reject

15.57 rS = .6768, use two-tailed test, reject

15.59 rS = 0; p–value < .005

15.61 a Randomized block design

b No

c p–value = .04076, yes

15.63 T = 73.5, do not reject, consistent with

Ex. 15.62

15.65 U = 17.5, fail to reject H0

15.67 .0159

15.69 H = 7.154, reject

15.71 Fr = 6.21, do not reject

15.73 .10

Chapter 16

16.1 a β(10, 30)

b n = 25

c β(10, 30), n = 25

d Yes

e Posterior for the β(1, 3) prior.

16.3 c Means get closer to .4, std dev

decreases.

e Looks more and more like normal

distribution.

16.7 a

Y + 1

n + 4

b

np + 1

n + 4 ;

np(1 − p)

(n + 4)2

16.9 b

α + 1

α + β + Y ;

(α + 1)(β + Y − 1)

(α + β + Y + 1)(α + β + Y )

16.11 e Ȳ

(

nβ

nβ + 1

)

+ αβ

(

1

nβ + 1

)

16.13 a (.099, .710)

b Both probabilities are .025

c P(.099 < p < .710) = .95

h Shorter for larger n.

16.15 (.06064, .32665)

16.17 (.38475, .66183)

16.19 (5.95889, 8.01066)

16.21 Posterior probabilities of null and

alternative are .9526 and .0474,

respectively, accept H0.

16.23 Posterior probabilities of null and

alternative are .1275 and .8725,

respectively, accept Ha .

16.25 Posterior probabilities of null and

alternative are .9700 and .0300,

respectively, accept H0.

ANSWERS

Chapter 1

1.5 a 2.45 − 2.65, 2.65 − 2.85

b 7/30

c 16/30

1.9 a Approx. .68

b Approx. .95

c Approx. .815

d Approx. 0

1.13 a ȳ = 9.79; s = 4.14

b k = 1: (5.65, 13.93); k = 2: (1.51,

18.07); k = 3: (−2.63, 22.21)

1.15 a ȳ = 4.39; s = 1.87

b k = 1: (2.52, 6.26); k = 2: (0.65,

8.13); k = 3: (−1.22, 10)

1.17 For Ex. 1.2, range/4 = 7.35; s = 4.14;

for Ex. 1.3, range/4 = 3.04; s = 3.17;

for Ex. 1.4, range/4 = 2.32, s = 1.87.

1.19 ȳ − s = −19 < 0

1.21 .84

1.23 a 16%

b Approx. 95%

1.25 a 177

c ȳ = 210.8; s = 162.17

d k = 1: (48.6, 373); k = 2:

(−113.5, 535.1); k = 3: (−275.7,

697.3)

1.27 68% or 231 scores; 95% or 323 scores

1.29 .05

1.31 .025

1.33 (0.5, 10.5)

1.35 a (172 − 108)/4 = 16

b ȳ = 136.1; s = 17.1

c a = 136.1 − 2(17.1) = 101.9;

b = 136.1 + 2(17.1) = 170.3

Chapter 2

2.7 A = {two males} = {(M1, M2),

(M1,M3), (M2,M3)}

B = {at least one female} = {(M1,W1),

(M2,W1), (M3,W1), (M1,W2), (M2,W2),

(M3,W2), (W1,W2)}

B̄ = {no females} = A; A ∪ B = S;

A ∩ B = null; A ∩ B̄ = A

2.9 S = {A+, B+, AB+, O+, A−, B−,

AB−, O−}

2.11 a P(E5) = .10; P(E4) = .20

b p = .2

2.13 a E1 = very likely (VL); E2 =

somewhat likely (SL); E3 =

unlikely (U); E4 = other (O)

b No; P(VL) = .24, P(SL) = .24,

P(U) = .40, P(O) = .12

c .48

2.15 a .09

b .19

2.17 a .08

b .16

c .14

d .84

2.19 a (V1, V1), (V1, V2), (V1, V3),

(V2, V1), (V2, V2), (V2, V3),

(V3, V1), (V3, V2), (V3, V3)

b If equally likely, all have

probability of 1/9.

c P(A) = 1/3; P(B) = 5/9;

P(A ∪ B) = 7/9;

P(A ∩ B) = 1/9

2.27 a S = {CC, CR, CL, RC, RR, RL,

LC, LR, LL}

b 5/9

c 5/9

877

Page 2

878 Answers

2.29 c 1/15

2.31 a 3/5; 1/15

b 14/15; 2/5

2.33 c 11/16; 3/8; 1/4

2.35 42

2.37 a 6! = 720

b .5

2.39 a 36

b 1/6

2.41 9(10)6

2.43 504 ways

2.45 408,408

2.49 a 8385

b 18,252

c 8515 required

d Yes

2.51 a 4/19,600

b 276/19,600

c 4140/19,600

d 15180/19,600

2.53 a 60 sample points

b 36/60 = .6

2.55 a

(

90

10

)

b

(

20

4

)(

70

6

)/(

90

10

)

= .111

2.57 (4 × 12)/1326 = .0362

2.59 a .000394

b .00355

2.61 a

364n

365n

b .5005

2.63 1/56

2.65 5/162

2.67 a P(A) = .0605

b .001344

c .00029

2.71 a 1/3

b 1/5

c 5/7

d 1

e 1/7

2.73 a 3/4

b 3/4

c 2/3

2.77 a .40 b .37 c .10

d .67 e .6 f .33

g .90 h .27 i .25

2.93 .364

2.95 a .1

b .9

c .6

d 2/3

2.97 a .999

b .9009

2.101 .05

2.103 a .001

b .000125

2.105 .90

2.109 P(A) ≥ .9833

2.111 .149

2.113 (.98)3(.02)

2.115 (.75)4

2.117 a 4(.5)4 = .25

b (.5)4 = 1/16

2.119 a 1/4

b 1/3

2.121 a 1/n

b

1

n

;

1

n

c

3

7

2.125 1/12

2.127 a .857

c No; .8696

d Yes

2.129 .4

2.133 .9412

2.135 a .57

b .18

c .3158

d .90

2.137 a 2/5

b 3/20

2.139 P(Y = 0) = (.02)3;

P(Y = 1) = 3(.02)2(.98);

P(Y = 2) = 3(.02)(.98)2;

P(Y = 3) = (.98)3

2.141 P(Y = 2) = 1/15; P(Y = 3) = 2/15;

P(Y = 4) = 3/15; P(Y = 5) = 4/15;

P(Y = 6) = 5/15

2.145 18!

2.147 .0083

2.149 a .4

b .6

c .25

2.151 4[p4(1 − p) + p(1 − p)4]

2.153 .313

2.155 a .5

b .15

c .10

d .875

Page 10

886 Answers

5.163 b F(y1, y2) =

y1 y2[1 − α(1 − y1)(1 − y2)]

c f (y1, y2) =

1 − α[(1 − 2y1)(1 − 2y2)],

0 ≤ y1 ≤ 1, 0 ≤ y2 ≤ 1

d Choose two different values for α

with −1 ≤ α ≤ 1.

5.165 a (p1et1 + p2et2 + p3et3)n

b m(t , 0, 0)

c Cov(X1, X2) = −np1 p2

Chapter 6

6.1 a

1 − u

2

, −1 ≤ u ≤ 1

b

u + 1

2

, −1 ≤ u ≤ 1

c

1√

u

− 1, 0 ≤ u ≤ 1

d E(U1) = −1/3, E(U2) =

1/3, E(U3) = 1/6

e E(2Y −1) = −1/3, E(1−2Y ) =

1/3, E(Y 2) = 1/6

6.3 b fU (u) ={

(u + 4)/100, −4 ≤ u ≤ 6

1/10, 6 < u ≤ 11

c 5.5833

6.5 fU (u) =

1

16

(

u − 3

2

)−1/2

,

5 ≤ u ≥ 53

6.7 a fU (u) =

1

√

π

√

2

u−1/2e−u/2,

u ≥ 0

b U has a gamma distribution with

α = 1/2 and β = 2 (recall that

�(1/2) = √π).

6.9 a fU (u) = 2u, 0 ≤ u ≤ 1

b E(U ) = 2/3

c E(Y1 + Y2) = 2/3

6.11 a fU (u) = 4ue−2u , u ≥ 0, a gamma

density with α = 2

and β = 1/2

b E(U ) = 1, V (U ) = 1/2

6.13 fU (u) = F ′U (u) =

u

β2

e−u/β , u > 0

6.15 [−ln(1 − U )]1/2

6.17 a f (y) = αy

α−1

θα

, 0 ≤ y ≤ θ

b Y = θU 1/α

c y = 4√u. The values are 2.0785,

3.229, 1.5036, 1.5610, 2.403.

6.25 fU (u) = 4ue−2u for u ≥ 0

6.27 a fY (y) =

2

β

we−w

2/β , w ≥ 0, which

is Weibull density with m = 2.

b E(Y k/2) = �

(

k

2

+ 1

)

βk/2

6.29 a fW (w) =

1

�

(

3

2

)

(kT )3/2

w1/2e−w/kT w > 0

b E(W ) = 3

2

kT

6.31 fU (u) =

2

(1 + u)3 , u ≥ 0

6.33 fU (u) = 4(80 − 31u + 3u2),

4.5 ≤ u ≤ 5

6.35 fU (u) = − ln(u), 0 ≤ u ≤ 1

6.37 a mY1(t) = 1 − p + pet

b mW (t) = E(etW ) = [1− p + pet ]n

6.39 fU (u) = 4ue−2u , u ≥ 0

6.43 a Ȳ has a normal distribution

with mean µ and variance σ 2/n

b P(|Ȳ − µ| ≤ 1) = .7888

c The probabilities are .8664, .9544,

.9756. So, as the sample size

increases, so does the probability

that P(|Ȳ − µ| ≤ 1)

6.45 c = $190.27

6.47 P(U > 16.0128) = .025

6.51 The distribution of Y1 + (n2 − Y2) is

binomial with n1 + n2 trials and success

probability p = .2

6.53 a Binomial (nm, p) where

ni = m

b Binomial (n1 = n2 + · · · nn , p)

c Hypergeometric (r = n,

N = n1 + n2 + · · · nn)

6.55 P(Y ≥ 20) = .077

6.65 a f (u1, u2) =

1

2π

e−[u

2

1+(u2−u1)2]/2 =

1

2π

e−(2u

2

1−2u1u2+u22)/2

b E(U1) = E(Z1) = 0,

E(U2) = E(Z1 + Z2) = 0,

V (U1) = V (Z1) = 1,

V (U2) = V (Z1 + Z2) =

V (Z1) + V (Z2) = 2,

Cov(U1, U2) = E(Z 21) = 1

Page 11

Answers 887

c Not independent since

ρ

= 0.

d This is the bivariate normal

distribution with µ1 = µ2 = 0,

σ 21 = 1, σ 22 = 2, and ρ =

1√

2

6.69 a f (y1, y2) =

1

y21 y

2

2

, y1 > 1,

y2 > 1

e No

6.73 a g(2)(u) = 2u, 0 ≤ u ≤ 1

b E(U2) = 2/3, V (U2) = 1/18

6.75 (10/15)5

6.77 a

n!

( j − 1)!(k − 1 − j)!(n − k)!

y

j−1

j [yk − y j ]k−1− j [θ − yk]n−k

θ n

,

0 ≤ y j < yk ≤ θ

b

(n − k + 1) j

(n + 1)2(n + 2) θ

2

c

(n − k + j + 1)(k − j)

(n + 1)2(n + 2) θ

2

6.81 b 1 − e−9

6.83 1 − (.5)n

6.85 .5

6.87 a g(1)(y) = e−(y−4), y ≥ 4

b E(Y(1)) = 5

6.89 fR(r) = n(n − 1)rn−2(1 − r),

0 ≤ r ≤ 1

6.93 f (w) = 2

3

(

1√

w

− w

)

, 0 ≤ w ≤ 1

6.95 a fU1(u) =

1

2

0 ≤ u ≤ 1

1

2u2

u > 1

b fU2(u) = ue−u , 0 ≤ u

c Same as Ex. 6.35.

6.97 p(W = 0) = p(0) = .0512,

p(1) = .2048, p(2) = .3264,

p(3) = .2656, p(4) = .1186,

p(5) = .0294, p(6) = .0038,

p(7) = .0002

6.101 fU (u) = 1, 0 ≤ u ≤ 1 Therefore, U has

a uniform distribution on (0, 1)

6.103

1

π(1 + u21)

, ∞ < u1 < ∞

6.105

1

B(α, β)

uβ−1(1 − u)α−1, 0 < u < 1

6.107 fU (u) =

1

4

√

u

0 ≤ u < 1

1

8

√

u

1 ≤ u ≤ 9

6.109 P(U = C1 − C3) = .4156;

P(U = C2 − C3) = .5844

Chapter 7

7.9 a .7698

b For n = 25, 36, 69, and 64, the

probabilities are (respectively)

.8664, .9284, .9642, .9836.

c The probabilities increase with n.

d Yes

7.11 .8664

7.13 .9876

7.15 a E(X̄ − Ȳ ) = µ1 − µ2

b V (X̄ − Ȳ ) = σ 21 /m + σ 22 /n

c The two sample sizes should be at

least 18.

7.17 P

(∑6

i=1 Z

2

i ≤ 6

)

= .57681

7.19 P(S2 ≥ .065) = .10

7.21 a b = 2.42

b a = .656

c .95

7.27 a .17271

b .23041

d .40312

7.31 a 5.99, 4.89, 4.02, 3.65, 3.48, 3.32

c 13.2767

d 13.2767/3.32 ≈ 4

7.35 a E(F) = 1.029

b V (F) = .076

c 3 is 7.15 standard deviations above

this mean; unlikely value.

7.39 a normal, E(θ̂) = θ =

c1µ1 + c2µ2 + · · · + ckµk

V (θ̂) =

(

c21

n1

+ c

2

2

n2

+ · · · + c

2

k

nk

)

σ 2

b χ2 with n1 + n2 + · · · + nk − k df

c t with n1 + n2 + · · · + nk − k df

7.43 .9544

7.45 .0548

7.47 153

7.49 .0217

7.51 664

7.53 b Ȳ is approximately normal: .0132.

7.55 a random sample; approximately 1.

b .1271

Page 20

Answers 895

15.49 a .0256

b An usually small number of runs

(judged at α = .05) would imply a

clustering of defective items in

time; do not reject.

15.51 R = 13, do not reject

15.53 rS = .911818; yes.

15.55 a rS = −.8449887

b Reject

15.57 rS = .6768, use two-tailed test, reject

15.59 rS = 0; p–value < .005

15.61 a Randomized block design

b No

c p–value = .04076, yes

15.63 T = 73.5, do not reject, consistent with

Ex. 15.62

15.65 U = 17.5, fail to reject H0

15.67 .0159

15.69 H = 7.154, reject

15.71 Fr = 6.21, do not reject

15.73 .10

Chapter 16

16.1 a β(10, 30)

b n = 25

c β(10, 30), n = 25

d Yes

e Posterior for the β(1, 3) prior.

16.3 c Means get closer to .4, std dev

decreases.

e Looks more and more like normal

distribution.

16.7 a

Y + 1

n + 4

b

np + 1

n + 4 ;

np(1 − p)

(n + 4)2

16.9 b

α + 1

α + β + Y ;

(α + 1)(β + Y − 1)

(α + β + Y + 1)(α + β + Y )

16.11 e Ȳ

(

nβ

nβ + 1

)

+ αβ

(

1

nβ + 1

)

16.13 a (.099, .710)

b Both probabilities are .025

c P(.099 < p < .710) = .95

h Shorter for larger n.

16.15 (.06064, .32665)

16.17 (.38475, .66183)

16.19 (5.95889, 8.01066)

16.21 Posterior probabilities of null and

alternative are .9526 and .0474,

respectively, accept H0.

16.23 Posterior probabilities of null and

alternative are .1275 and .8725,

respectively, accept Ha .

16.25 Posterior probabilities of null and

alternative are .9700 and .0300,

respectively, accept H0.