# answers to exercises - mathematical statistics with applications (7th edition).pdf

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Chapter 1

1.5 a 2.45 2.65, 2.65 2.85b 7/30c 16/30

1.9 a Approx. .68b Approx. .95c Approx. .815d Approx. 0

1.13 a y = 9.79; s = 4.14b k = 1: (5.65, 13.93); k = 2: (1.51,

18.07); k = 3: (2.63, 22.21)1.15 a y = 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 y s = 19 < 0

1.21 .841.23 a 16%

b Approx. 95%1.25 a 177

c y = 210.8; s = 162.17d 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 scores1.29 .051.31 .0251.33 (0.5, 10.5)1.35 a (172 108)/4 = 16

b y = 136.1; s = 17.1c 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)}B = {no females} = A; A B = S;A B = null; A B = A

2.9 S = {A+, B+, AB+, O+, A, B,AB, O}

2.11 a P(E5) = .10; P(E4) = .20b 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 .09b .19

2.17 a .08b .16c .14d .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 haveprobability 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 5/9c 5/9

877

2.29 c 1/152.31 a 3/5; 1/15

b 14/15; 2/52.33 c 11/16; 3/8; 1/42.35 422.37 a 6! = 720

b .52.39 a 36

b 1/62.41 9(10)62.43 504 ways2.45 408,4082.49 a 8385

b 18,252c 8515 requiredd Yes

2.51 a 4/19,600b 276/19,600c 4140/19,600d 15180/19,600

2.53 a 60 sample pointsb 36/60 = .6

2.55 a(

9010

)b

(204

)(706

)/(9010

)= .111

2.57 (4 12)/1326 = .03622.59 a .000394

b .003552.61 a

364n

365nb .5005

2.63 1/562.65 5/1622.67 a P(A) = .0605

b .001344c .00029

2.71 a 1/3b 1/5c 5/7d 1e 1/7

2.73 a 3/4b 3/4c 2/3

2.77 a .40 b .37 c .10d .67 e .6 f .33g .90 h .27 i .25

2.93 .3642.95 a .1

b .9

c .6d 2/3

2.97 a .999b .9009

2.101 .052.103 a .001

b .0001252.105 .902.109 P(A) .98332.111 .1492.113 (.98)3(.02)2.115 (.75)42.117 a 4(.5)4 = .25

b (.5)4 = 1/162.119 a 1/4

b 1/32.121 a 1/n

b1n

;1n

c37

2.125 1/122.127 a .857

c No; .8696d Yes

2.129 .42.133 .94122.135 a .57

b .18c .3158d .90

2.137 a 2/5b 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 .00832.149 a .4

b .6c .25

2.151 4[p4(1 p) + p(1 p)4]2.153 .3132.155 a .5

b .15c .10d .875

2.157 .0212.161 P(R 3) = 12/662.163 P(A) = 0.9801

P(B) = .96392.165 .9162.167 P(Y = 1) = 35/70 = .5;

P(Y = 2) = 20/70 = 2/7;P(Y = 3) = 10/70;P(Y = 4) = 4/70; P(Y = 5) = 1/70

2.169 a (4!)3 = 13,824

b 3456/13,824 = .252.173 .252.177 a .364

b .636c (49/50)n .60, so n is at most 25

2.179 a 20(

12

)6= .3125

b 27(

12

)10

Chapter 3

3.1 P(Y = 0) = .2, P(Y = 1) = .7,P(Y = 2) = .1

3.3 p(2) = 16

, p(3) = 26

, p(4) = 12

3.5 p(0) = 26

, p(1) = 36

, p(3) = 16

3.7 p(0) = 3!27

= 627

, p(2) = 327

,

p(1) = 1 627

327

= 1827

3.9 a P(Y = 3) = .000125,P(Y = 2) = .007125,P(Y = 1) = .135375,P(Y = 0) = .857375

c P(Y > 1) = .007253.11 P(X = 0) = 8

27, P(X = 1) = 12

27,

P(X = 2) = 627

, P(X = 3) = 127

,

P(Y = 0) = 27443375

,

P(Y = 1) = 5883375

,

P(Y = 2) = 143375

,

P(Y = 3) = 13375

, Z = X + Y ,

P(Z = 0) = 27125

, P(Z = 1) = 54125

,

P(Z = 2) = 36125

, P(Z = 3) = 8125

3.13 E(Y ) = 14

, E(Y 2) = 74

, V (Y ) = 2716

,

cost = 14

3.15 a P(Y = 0) = .1106,P(Y = 1) = .3594,

P(Y = 2) = .3894,P(Y = 3) = .1406

c P(Y = 1) = .3594d = E(Y ) = 1.56, 2 = .7488,

= 0.8653e (.1706, 3.2906),

P(0 Y 3) = 13.17 = E(Y ) = .889,

2 = V (Y ) = E(Y 2)[E(Y )]2 = .321, = 0.567, ( 2 , + 2) = (.245, 2.023),P(0 Y 2) = 1

3.19 C = \$853.21 13,800.3883.23 \$.313.25 Firm I : E (prot) = \$60,000

E(total prot) = \$120,0003.27 \$5103.35 .4; .39993.39 a .1536;

b .97283.41 .0003.43 a .1681

b .52823.45 P(alarm functions) = 0.9923.49 a .151

b .3023.51 a .51775

b .49143.53 a .0156

b .4219c 25%

3.57 \$185,0003.59 \$8403.61 a .672

b .672c 8

3.67 .072033.69 Y is geometric with p = .593.73 a .009

b .013.75 a .081

b .813.81 2

3.831n

(n 1

n

)53.87 E

(1Y

)= p ln(p)

1 p3.91 \$150; 45003.93 a .04374

b .991443.95 .13.97 a .128

b .049c = 15, 2 = 60

3.99 p(x) = y!(r 1)!(y r + 1)! p

r q y+1r ,

y = r 1, r , r + 1, . . .3.101 a

511

br

y0

3.1031

423.105 b .7143

c = 1.875, = .7087

3.107 hypergeometric with N = 6, n = 2,and r = 4.

3.109 a .0238b .9762c .9762

3.111 a p(0) = 1430

, p(1) = 1430

,

p(2) = 230

b p(0) = 530

, p(1) = 1530

,

p(2) = 930

, p(3) = 130

3.113 P(Y 1) = .1873.115 p(0) = 1

5, p(1) = 3

5, p(2) = 1

53.117 a P(Y = 0) = .553

b E(T ) = 9.5, V (T ) = 28.755, = 5.362

3.119 .0163.121 a .090

b .143c .857d .241

3.123 .18393.125 E(S) = 7, V (S) = 700; no3.127 .62883.129 23 seconds3.131 .55783.133 .17453.135 .95243.137 .15123.139 403.141 \$13003.149 Binomial, n = 3 and p = .63.151 Binomial, n = 10 and p = .7,

P(Y 5) = .15033.153 a Binomial, n = 5 and p = .1

b Geometric, p = 12

c Poisson, = 23.155 a E(Y ) = 7

3b V (Y ) = 5

9c p(1) = 1

6, p(2) = 2

6, p(3) = 3

63.167 a .64

b C = 103.169 d p(1) = 1/(2k2),

p(0) = 1 (1/k2), p(1) = 1(2k2)3.171 (85, 115)3.173 a p(0) = 1

8, p(1) = 3

8, p(2) = 3

8,

p(3) = 18

c E(Y ) = 1.5, V (Y ) = .75, = .866

3.175 a 38.4b 5.11

3.177 (61.03, 98.97)3.179 No, P(Y 350) 1

(2.98)2= .1126.

3.181p = Fraction defective P(acceptance)

a 0 1b .10 .5905c .30 .1681d .50 .0312e 1.0 0

3.185 a .2277b Not unlikely

3.187 a .023b 1.2c \$1.25

3.189 1 (.99999)10,0003.191 V (Y ) = .43.193 .476

3.195 a .982b P(W 1) = 1 e12

3.197 a .9997b n = 2

3.199 a .300b .037

3.201 (18.35, 181.65)3.203 a E[Y (t)] = k(e2t et )

b 3.2974, 2.1393.205 .007223.207 a p(2) = .084, P(Y 2) = .125

b P(Y > 10) = .0143.209 .08373.211 33.213 a .1192

b .1173.215 a n[1 + k(1 .95k)]

b g(k) is minimized at k = 5 andg(5) = .4262.

c .5738N

Chapter 4

4.7 a P(2 Y < 5) = 0.591,P(2 < Y < 5) = .289, sonot equal

b P(2 Y 5) = 0.618,P(2 < Y 5) = 0.316, sonot equal

c Y is not a continuous randomvariable, so the earlier resultsdo not hold.

4.9 a Y is a discrete random variableb These values are 2, 2.5, 4, 5.5, 6,

and 7.

c p(2) = 18

, p(2.5) = 116

,

p(4) = 516

, p(5.5) = 18

,

p(6) = 116

, p(7) = 516

d .5 = 44.11 a c = 1

2b F(y) = y

2

4, 0 y 2

d .75e .75

4.13 a F(y) =

0 y < 0y2

20 y 1

y 12

1 < y 1.51 y > 1.5

b .125c .575

4.15 a For b 0, f (y) 0; also,

f (y) = 1

b F(y) = 1 by

, for y b;0 elsewhere.

cb

(b + c)d

(b + c)(b + d)

4.17 a c = 32

b F(y) = y3

2+ y

2

2, for 0 y 1

d F(1) = 0, F(0) = 0, F(1) = 1e

316

f104123

4.19 a f (y) =

0 y 0.125 0 < y < 2.125y 2 y < 40 y 4

b7

16

c1316

d79

4.21 E(Y ) = .708, V (Y ) = .04874.25 E(Y ) = 31/12, V (Y ) = 1.1604.27 \$4.65, .0124.29 E(Y ) = 60, V (Y ) = 1

34.31 E(Y ) = 44.33 a E(Y ) = 5.5, V (Y ) = .15

b Using Tchebysheffs theorem,the interval is (5, 6.275).

c Yes; P(Y ) = .57814.37 E(Y ) = 04.39 .5; .254.45 a P(Y < 22) = 2

5= .4

b P(Y > 24) = 15

= .2

4.47 a P(Y > 2) = 34

b c0 + c1[

43

+ 9]

4.4934

4.5113

4.53 a18

b18

c14

4.55 a27

b = .015, V (Y ) = .000414.57 E

(6

D3)

= .0000065 ,V(

6D3)

= .000352524.59 a z0 = 0

b z0 = 1.10c z0 = 1.645d z0 = 2.576

4.63 a P(Z > 1) = .1587b The same answer is obtained.

4.65 \$425.604.67 = 3.000 in.4.69 .26604.71 a .9544

b .82974.73 a .406

b 960.5 mm4.75 = 7.3014.77 a 0.758

b 22.24.87 a .05 = .70369.

b .05 = .351854.89 a = .8

b P(Y 1.7) = .88064.91 a .1353

b 460.52 cfs4.93 a .5057

b 19364.97 .36794.99 a .7358

4.101 a E(Y ) = 1.92b P(Y > 3) = .21036d P(2 Y 3) = .12943

4.103 E(A) = 200 , V (A) = 200,00024.105 a E(Y ) = 3.2, V (Y ) = 6.4

b P(Y > 4) = .289554.107 a (0, 9.657), because Y must

be positive.b P(Y < 9.657) = .95338

4.109 E(L) = 276, V(L) = 47,6644.111 d

( + 1

2

)/() if > 0

e1

( 1) if > 1,( 12 )

()

if >12

,

12( 1)( 2)

if > 24.123 a k = 60

b .95 = 0.846844.125 E(Y ) = 3

5, V (Y ) = 1