Chapter 8
8.1a. P(X > 45) 
(60  45 )  2 (75  60 )  2

= .0800
50  15
50  15
b. P(10 < X < 40) 
c. P(X < 25) 
(15  [30 ])  6 (0  [15 ])  10 (15  0)  17 (25  15 )  7



50 15
50  15
50  15
50  15
d. P(35 < X < 65) 
8.2a. P(X > 45) 
= .7533
(45  35 )  6 (60  45)  2 (65  60 )  2


= .1333
50  15
50  15
50  15
(60  45 )  3 (75  60 )  3

= .1200
50  15
50  15
b. P(10 < X < 40) 
c. P(X < 25) 
(15  10 )  16 (30  15 )  8 (40  30 )  8


= .4800
50  15
50  15
50  15
(15  10 )  17 (30  15 )  7 (40  30 )  6


50  15
50  15
50  15
= .3333
(30  [45 ])  5 (15  [30 ])  5 (0  [15 ])  2 (15  0)  16 (25  15 )  8




= .6667
50  15
50  15
50  15
50  15
50  15
d. P(35 < X < 65) 
(45  35 )  8 (60  45 )  3 (65  60 )  3


= .1867
50  15
50  15
50  15
8.3a. P(55 < X < 80) 
(60  55 )  16 (70  60 )  5 (80  70 )  24


60  10
60  10
60  10
= .6167
b. P(X > 65) 
(70  65 )  5 (80  70 )  24 (90  80 )  7 (100  90 )  1



= .5750
60  10
60  10
60  10
60  10
c. P(X < 85) 
(50  40 )  7 (60  50 )  16 (70  60 )  5 (80  70 )  24 (85  80 )  7




60  10
60  10
60  10
60  10
60  10
d. P(75 < X < 85) 
(80  75 )  24 (85  80 )  7

= .2583
60  10
60  10
199
= .9250
8.4 a.
b. P(X > 25) = 0
c. P(10 < X < 15) = (15  10 )
1
20
d. P(5.0 < X < 5.1) = (5.1  5)
8.5a. f(x) =
1
20
1
1
=
(60  20 ) 40
b. P(35 < X < 45) = (45–35)
= .25
= .005
20 < x < 60
1
40
= .25
200
c.
8.6 f(x) =
1
1

(60  30 ) 30
a. P(X > 55) = (60  55 )
30 < x < 60
1
30
= .1667
b. P(30 < X < 40) = (40  30 )
1
30
= .3333
c. P(X = 37.23) = 0
8.7
1
 (60  30 )  7.5 ; The first quartile = 30 + 7.5 = 37.5 minutes
4
8.8 .10  (60  30)  3 ; The top decile = 60–3 = 57 minutes
8.9 f(x) =
1
1

(175  110 ) 65
a. P(X > 150) = (175  150 )
110 < x < 175
1
65
= .3846
b. P(120 < X < 160) = (160  120 )
1
65
= .6154
8.10 .20(175–110) = 13. Bottom 20% lie below (110 + 13) = 123
201
For Exercises 8.11 to 8.14 we calculate probabilities by determining the area in a triangle. That
is,
Area in a triangle = (.5)(height)(base)
8.11a.
b. P(0 < X < 2) = (.5)(2–0)(1) = 1.0
c. P(X > 1) = (.5)(2 – 1)(.5) = .25
d. P(X < .5) = 1 – P(X > .5) = 1 – (.5)(.75)(2–.5) = 1 – .5625 = .4375
e. P(X = 1.5) = 0
8.12 a
202
b. P(2 < X < 4) = P(X < 4) – P(X < 2) = (.5)(3/8)(4–1) – (.5)(1/8)(2–1) = .5625 – .0625 = .5
c. P(X < 3) = (.5)(2/8)(3–1) = .25
8.13a.
b. P(1 < X < 3) = P(X < 3) – P(X < 1) =
1 3
1 1

 (3  0)  
 (1  0) = .18 – .02 = .16
2 25
2 25
c. P(4 < X < 8) = P(4 < X < 5) + P(5 < X < 8)
P(4 < X < 5)= P(X < 5) – P(X <4) =
1 5
1 4

 (5  0)  
 (4  0) = .5
2 25
2 25
P(5 < X < 8) = P(X > 5) – P(X > 8) =
1 5
1 2

 (10  5)  
 (10  8) = .5
2 25
2 25
P(4 < X < 8) = .18 + .42 = .60
d. P(X < 7) = 1 – P(X > 7)
P(X > 7) =
– .32 = .18
1 3

 (10  7) = .18
2 25
P(X < 7) = 1 – .18 = .82
e. P(X > 3) = 1 – P(X < 3)
203
– .08 = .42
P(X < 3) =
1 3

 (3  0) = .18
2 25
P(X > 3) = 1 – .18 = .82
8.14 a. f(x) = .10 – .005x 0  x  20
b. P(X > 10) = (.5)(.05)(20–10) = .25
c. P(6 < X < 12) = P(X > 6) – PX > 12) = (.5)(.07)(20–6) – (.5)(.04)(20–12) = .49 – .16 = .33
8.15 P( Z < 1.50) = .9332
8.16 P(Z < 1.51) = .9345
8.17 P(Z < 1.55) = .9394
8.18 P(Z < −1.59) = .0559
8.19 P(Z < −1.60) = .0548
8.20 P(Z < − 2.30) = .0107
8.21 P(–1.40 < Z < 0.60) = P( Z < 0.60) − P(Z < −1.40) = .7257− .0808 = .6449
8.22 P(Z > –1.44) = 1 – P(Z < −1.44) = 1 − .0749 = .9251
8.23 P(Z < 2.03) = .9788
8.24 P(Z > 1.67) = 1 – P(Z < 1.67) = 1 – .9525 = .0475
8.25 P(Z < 2.84) = ..9977
8.26 P(1.14 < Z < 2.43) = P(Z < 2.43) – P(Z < 1.14) = .9925 – .8729 = .1196
204
8.27 P(–0.91 < Z < –0.33) = P(Z < −.33) – P(Z < −.91) = .3707 – .1814 = .1893
8.28 P(Z > 3.09) = .5 – P(0 < Z < 3.09) = .5 – .4990 = .0010
8.29 P(Z > 0) = 1 – P(Z < 0) = 1 − .5 = .5
8.30 P(Z > 4.0) = 0
8.31 P(Z < z .02 ) = 1 – .02 = .9800; z .02 = 2.05
8.32 P(Z < z .045 ) = 1 – .045 = .9550; z .045 = 1.70
8.33 P(Z < z .20 ) = 1 – .20 = .8000; z .20 = .84
X   145  100 

8.34 P(X > 145) = P
 = P(Z > 2.25) = 1 – P(Z < 2.25) = 1 – .9878 = .0122


20

8.35 P(Z < z.15 ) = 1 – .15 = .8500; z .15 = 1.04; z.15 
x  250
x
; 1.04 
; x = 291.6
40

800  1,000 X   1,100  1,000 


8.36 P(800 < X < 1100) = P
 = P(–.8 < Z < .4)

250

250

= P(Z < .4) − P(Z < −.8) = .6554 − .2119 = .4435
8.37 P(Z < z .08 ) = .0800; z .08  1.41;  z .08 
x 
x  50
;  1.41 

8
5  6.3 X   10  6.3 


 = P(–.59

2.2 
 2.2

8.38 a P(5 < X < 10) = P
; x = 38.72
< Z > 1.68)
= P(Z < 1.68) − P(Z < −.59) = .9535 − .2776 = .6759
X   7  6. 3 

b P(X > 7) = P
 = P(Z > .32) = 1 – P(Z < .32) = 1 – .6255 = .3745


2.2 
205
X   4  6.3 

c P(X < 4) = P
 = P(Z < –1.05) = .1469


2.2 
8.39 P(Z < z .10 ) = 1 – .10 = .9000; z .10 = 1.28; z .10 
x 

; 1.28 
x  6 .3
;
2 .2
x = 9.116
Calls last at least 9.116 minutes.
X   5,000  5,100 

8.40 P(X > 5,000) = P
 = P(Z > –.5) = 1 −P(Z < −.5) = 1 − .3085 = .6915


200

8.41 P(Z < z .02 ) = .02; z .02  2.05;  z .02 
 X 

8.42 a P(X > 12,000) = P
 
x 
x  5100
;  2.05 

200
12 ,000  10 ,000 
 = P(Z > .83)
2,400

; x = 4690;
= 1 – P(Z < .83) = 1 – .7967 =
.2033
 X 

b P(X < 9,000) = P
 
9,000  10 ,000 
 = P(Z <
2,400

8.43 P(Z < z .001 ) = .9990; z .001 = 3.08; z.001 
–.42) = .3372
x
x  10,000
; 3.08 
; x = 17,392

2,400
X   70  65 

8.44 a P(X > 70) = P
 = P(Z > 1.25) = 1 – P(Z < 1.25) = 1 – .8944 = .1056


4

X   60  65 

b P(X < 60) = P
 = P(Z < –1.25) = .1056


4

55  65 X   70  65 


 = P(–2.50

4 
 4

c P(55 < X < 70) = P
< Z < 1.25)
= P(Z < 1.25) − P(Z < −2.50) = .8944− .0062 = .8882
 X 

8.45 a P(X < 70,000) = P
 
70 ,000  82 ,000 
 = P(Z <
6,400

206
–1.88) = .0301
 X 

b P(X > 100,000) = P
 
100 ,000  82 ,000 
 = P(Z > 2.81)
6,400

= 1 – P(Z < 2.81) = 1 – .9975 =
.0025
8.46 Top 5%: P(Z < z .05 ) = 1 – .05 = .9500; z .05 = 1.645; z .05 
Bottom 5%: P(Z < z .05 ) = .0500; z .05  1.645;  z .05 
x 

x  32
1 .5
; 1.645 
x 
x  32
;  1.645 

1.5
; x = 34.4675
;
x = 29.5325
X   36  32 

8.47 a P(X > 36) = P
 = P(Z > 2.67) = 1 – P(Z < 2.67) = 1 – .9962 = .0038


1.5

X   34  32 

b P(X < 34) = P
 = P(Z < 1.33) = .9082


1.5

30  32 X   33  32 


c P(30 < X < 33) = P
 = P(–1.33 < Z < .67)
 1.5

1.5 
= P(Z < .67) − P(Z < −1.33) = .7486 − .0918 = .6568
X   8  7.2 

8.48 P(X > 8) = P
 = P (Z > 1.20) = 1 – P(Z < 1.20) = 1 – .8849 = .1151


.667 
8.49 P(Z < z .25 ) = .7500; z .25 = .67; z .25 
x 

; .67 
x  7.2
.667
; x = 7.65 hours
X   10  7.5 

8.50 a P(X > 10) = P
 = P Z > 1.19) = 1 – P(Z < 1.19) = 1 – .8830 = .1170


2.1

7  7. 5 X   9  7. 5 


 = P(–.24

2.1 
 2. 1

b P(7 < X < 9) = P
< Z < .71)
= P(Z < .71) − P(0 < Z < −.24) = .7611 − .4052 = .3559
X   3  7.5 

= P
2.1 
 

c P(X < 3) = P
Z < –2.14) = .0162
d P(Z < – z .05 ) = .0500; z .05 = –1.645;  z .05 
x 
x  7 .5
;  1.645 
;x

2 .1
207
= 4.05 hours
X   12 ,000  11,500 

8.51 a P(X > 12,000) = P
 = P Z > .63) = 1– P(Z < .63) = 1 – .7357 = .2643



800
X   10 ,000  11,500 

b P(X < 10,000) = P
 = P(Z < –1.88) = .0301



800
8.52 P(Z < – z .01 ) = .0100; z .05 = –2.33;  z .01 
x 

;  2.33 
x  11,500
800
; x = 9,636
24  26 X   28  26 


8.53 a P(24 < X < 28) = P
 = P(–.80 < Z < .80) = P(Z < .80) – P(Z <


2. 5
2.5 
−.80)
= .7881 − .2119 = .5762
X   28  26 


2.5 
 
= P(Z > .80) = 1 – P(Z < .80) = 1 – .7881 = .2119
X   24  26 


2.5 
 
= P(Z < –.80) =.2119
b P(X > 28) = P
c P(X < 24) = P
X   30  27 

8.54 a P(X > 30) = P
 = P(Z > .43) = 1 – P(Z < .43) = 1 – .6664 = .3336


7

X   40  27 

b P(X > 40) = P
 = P(Z > 1.86) = 1 – P(Z < 1.86) = 1 – .9686 = .0314


7

X   15  27 

c P(X < 15) = P
 = P(Z < –1.71) = .0436


7

d P(Z < z .20 ) = 1 – .20 = .8000; z .20 = .84; z .20 
x 

; .84 
x  27
7
; x = 32.88
X   4  7.5 

8.55 a P(X < 4) = P
 = P(Z < –2.92) = .0018


1.2 
7  7.5 X   10  7.5 


b P(7 < X < 10) = P
 = P(–.42 < Z < 2.08) = P(Z < 2.08) − P(Z < −.42)
 1.2

1.2

= .9812 − .3372 = .6440
X   10  16 .40 

8.56 a P(X < 10) = P
 = P(Z < –2.33) = .0099


2.75

208
b P(Z < – z .10 ) = .1000; – z .10 = –1.28;  z .10 
x 
x  16 .40
;  1.28 

2.75
8.57 A: P(Z < z .10 ) = 1– .10 = .9000; z .10 = 1.28; z .10 
B: P(Z < z .40 ) = 1 –.40 = .6000; z .40 = .25; z .40 
C: P(Z < – z .20 ) = .2000; z .20  .84;  z .20 
x 

x 

; .25 
; 1.28 
x  70
10
x 
x  70
;  .84 

10
D: P(Z < – z .05 ) = .0500; z .05  1.645;  z .05 
x 

x  70
10
; x = 82.8
; x = 72.5
; x = 61.6;
x 
x  70
;  1.645 

10
8.58 P(Z < z .02 ) = 1 – .02 = .9800; z .02 = 2.05; z .02 
; x = 12.88
; 2.05 
; x = 53.55
x  100
16
; x = 132.80
(rounded to 133)
 X 
70 ,000  61,823 

17 ,301

= P(Z > .47) = 1 – P(Z < .47) = 1 – .6808 = .3192
 X 
45,000  41,825 

13,444

= P(Z < .24) = .5948

8.59 P(X > 70,000) = P
 

8.60 P(X < 45,000) = P
 
8.61 P(Z < z .01 ) = .0100; z .01  2.33;  z .01 
 X 

8.62 P(x > 150,000) = P
 
x 

;  2.33 
x  75
8
150 ,000  99 ,700 
 = P(Z > 1.68)
30 ,000

; x = 56.36
= 1 – P(Z < 1.68) = 1 – .9535 =
.0465
8.63 P(Z < z .06 ) = 1 – .06 = .9400; z .06 = 1.55; z .06 
(rounded to 247)
209
ROP  

; 1.55 
ROP  200
30
; ROP = 246.5
8.64 P(Z < z.20 ) = 1 – .20 = .8000; z .20 = .84; z .20 
x 

8.65 P(Z < z .30 ) = 1 – .30 = .7000; z .30 = .52; z .30 
; .84 
x  150
25
; x = 171
x 

; .52 
x  850
90
; x = 896.8
x 

; .25 
x  850
90
; .x = 872.5
(rounded to 897)
8.66 P( Z < z .40 ) = 1– .40 = .6000; z .40 = .25; z .40 
(rounded to 873)
8.67 From Exercise 7.57:  = 65,  2 = 21, and  = 4.58
X   60  65 

P(X > 60) = P
 = P(Z > –1.09) = 1 − P(Z < −1.09) = 1 −.1379 = .8621


4.58 
X   150  145 

8.68 P(X < 150) = P
 = P(Z < .90) = .8159



5.57
X   25  14 

8.69 a. P(X > 25) = P
 = P(Z > .61) = 1 – P(Z < .61) = 1 – .7291 = .2709



18
X   0  14 

b. P(X < 0) = P
 = P(Z < –.78) = .2177


18 
X   0  10 .60 

8.70 a. P(X < 0) = P
 = P(Z < –.73) = .2327


14 .56 
X   20  10 .60 

b. P(X > 20) = P
 = P(Z > .65) = 1 − P(Z < .65) = 1 − .7422 = .2578


14 .56

210
8.71
8.72
8.73 a P(X  1)  e ..5(1)  e ..5 = .6065
211
a P(X  .4)  e ..5(.4)  e ..2 = .8187
c P(X  .5)  1  e ..5(..5)  1  e ..25 = 1 – .7788 = .2212
d P(X  2)  1  e .5(2)  1  e 1 = 1 – .3679 = .6321
8.74 a P(X  2)  e ..3(2)  e .6 = .5488
b P(X  4)  1  e ..3(4)  1  e 1.2 = 1 – .3012 = .6988
c P(1  X  2)  e ..3(1)  e ..3(2)  e ..3  e .6 = .7408 – .5488 = .1920
d P(X = 3) = 0
8.75  = 6 kilograms/hour = .1 kilogram/minute
P(X  15)  e .1(15)  e 1.5 = .2231
8.76   1 /   25 hours;  = .04 breakdowns/hour
P(X  50)  e .04(50)  e 2 = .1353
8.77  = 10 trucks/hour = .167 truck/minute
P(X  15)  e .167(15)  e 2.5 = .0821
8.78   1 /  = 5 minutes;  = .2 customer/minute
P(X  10)  1  e .2(10)  1  e 2 = 1– .1353 = .8647
8.79   1 /  = 2.7 minutes;  = .37 service/minute
P(X  3)  1  e .37(3)  1  e 1.11 = 1– .3296 = .6704
8.80   1 /  = 7.5 minutes;  = .133 service/minute
P(X  5)  1  e .133(5)  1  e .665 = 1– .5143 = .4857
8.81   1 /  = 125 seconds;  = .008 transactions/second = .48 transactions/minute
P(X  3)  e .48(3)  e 1.44 = .2369
212
8.82   1 /  = 6 minutes;  = .167 customers/minute
P(X  10)  e .167(10)  e 1.67 = .1889
8.83
a 1.341
b 1.319
c 1.988
d 1.653
8.84
a 2.724
b 1.282
c 2.132
d 2.528
8.85
a 1.3406
b 1.3195
c 1.9890
d 1.6527
8.86
a 1.6556
b 2.6810
c 1.9600
d 1.6602
8.87
a .0189
b .0341
c .0927
d .0324
8.88
a .1744
b .0231
c .0251
d .0267
8.89
a 9.24
b 136
c 9.39
d 37.5
8.90
a 17.3
b 50.9
c 2.71
d 53.5
8.91
a 73.3441
b 102.946
c 16.3382
d 24.7690
8.92
a 33.5705
b 866.911
c 24.3976
d 261.058
8.93
a .2688
b 1.0
c .9903
d 1.0
8.94
a .4881
b .9158
8.95
a 4.35
b 8.89
c 3.29
d 2.50
8.96
a 2.84
b 1.93
c 3.60
d 3.37
c .9988
213
d .9077
8.97
a 1.4857
b 1.7633
c 1.8200
8.98
a 1.5204
b 1.5943
c 2.8397
8.99
a .0510
b .1634
c .0222
d .2133
8.100 a .1050
b .1576
c .0001
d .0044
d 1.1587
d 1.1670
Chapter 9
9.1a. 1/6
b. 1/6
9.2 a P( X  1) =P(1,1)= 1/36
b P( X  6) = P(6,6) = 1/36
9.3a P( X = 1) = (1/6) 5 = .0001286
b P( X = 6) = (1/6) 5 = .0001286
9.4 The variance of X is smaller than the variance of X.
9.5 The sampling distribution of the mean is normal with a mean of 40 and a standard deviation
of 12/ 100 = 1.2.
9.6 No, because the sample mean is approximately normally distributed.
 X 
9.7 a P( X  1050 ) = P
/ n
 X 
b P( X  960 )  P
/ n


1050  1000 
 = P(Z > 1.00) = 1 – P(Z < 1.00) = 1 – .8413 = .1587

200 / 16 
960  1000 
 = P(Z < –.80) = .2119

200 / 16 
214
 X 
c P( X  1100 )  P
/ n

1100  1000 
 = P(Z > 2.00) = 1 – P(Z < 2.00) = 1 – .9772 = .0228

200 / 16 
 X 
9.8 a P( X  1050 ) = P
/ n
 X 
b P( X  960 )  P
/ n
 X 
c P( X  1100 )  P
/ n


1100  1000 
 = P(Z > 2.50) = 1 – P(Z < 2.50) = 1 – .9938 = .0062

200 / 25 
 X 
/ n
 X 
/ n
 X 
c P( X  1100 )  P
/ n

1050  1000 
 = P(Z > 1.25) = 1 – P(Z < 1.25) = 1 – .8944 = .1056

200 / 25 
960  1000 
 = P(Z < –1.00) = .1587

200 / 25 
9.9 a P( X  1050 ) = P
b P( X  960 )  P


1050  1000 
 = P(Z > 2.50) = 1 – P(Z < 2.50) = 1 – .9938 = .0062

200 / 100 
960  1000 
 = P(Z < –2.00) = .0228

200 / 100 

1100  1000 
 = P(Z > 5.00) = 0

200 / 100 
 49  50 X   52  50 
 = P(–.40 < Z < .80)
9.10 a P(49  X  52 )  P



/ n
 5/ 4
5/ 4 
= P(Z < .80) − P(Z < −.40) = .7881 −.3446 = .4435
 49  50
b P(49  X  52 )  P
 5 / 16

X 

/ n
52  50 
 = P(–.80 < Z < 1.60)

5 / 16 
= P(Z < 1.60) − P(Z < −.80) = .9452 −.2119 = .7333
 49  50
c P(49  X  52 )  P
 5 / 25

X 

/ n
52  50 
 = P(–1.00 < Z < 2.00)

5 / 25 
= P(Z < 2.00) − P(Z < −1.00) = .9772 −.1587 = .8185
 49  50
9.11 a P(49  X  52 )  P
 10 / 4

X 
/ n

52  50 
 = P(–.20 < Z < .40)

10 / 4 
= P(Z < .40) − P(Z < −.20) = .6554 −.4207= .2347
 49  50
b P(49  X  52 )  P
 10 / 16

X 
/ n

52  50 
 = P(–.40 < Z < .80)

10 / 16 
215
= P(Z < .80) − P(Z < −.40) = .7881 −.3446= .4435
 49  50
c P(49  X  52 )  P

 10 / 25
X 
/ n

52  50 
 = P(–.50 < Z < 1.00)

10 / 25 
= P(Z < 1.00) − P(Z < −.50) = .8413 −.3085= .5328
 49  50
X 
52  50 
 = P(–.10 < Z < .20)
9.12 a P(49  X  52 )  P



 20 / 4  / n 20 / 4 
= P(Z < .20) − P(Z < −.10) = .5793 −.4602= .1191
 49  50
b P(49  X  52 )  P

 20 / 16
X 
/ n

52  50 
 = P(–.20 < Z < .40)

20 / 16 
= P(Z < .40) − P(Z < −.20) = .6554 −.4207= .2347
 49  50
c P(49  X  52 )  P
 20 / 25

X 
/ n

52  50 
 = P(–.25 < Z < .50)

20 / 25 
= P(Z < .50) − P(Z < −.25) = .6915 −.4013= .2902
Nn
N 1
9.13 a
=
1,000  100
= .9492
1,000  1
b
Nn
N 1
=
3,000  100
= .9834
3,000  1
c
Nn
N 1
=
5,000  100
= .9900
5,000  1
d. The finite population correction factor is approximately 1.
9.14 a  x =
b x =
c x =

n

n

n
Nn
=
N 1
500
1,000
10 ,000  1,000
10 ,000  1
= 15.00
Nn
500
=
N 1
500
10,000  500
= 21.80
10,000  1
Nn
500
=
N 1
100
10,000  100
= 49.75
10,000  1
X   66  64 

9.15 a P(X > 66) = P
 = P(Z > 1.00) = 1 – P(Z < 1.00) = 1 – .8413 = .1587


2

216
 X 

66  64 
 = P(Z > 2.00) = 1 – P(Z < 2.00) = 1 – .9772 = .0228

2/ 4 

66  64 
 = P(Z > 10.00) = 0

2 / 100 
b P( X  66 )  P
/ n
 X 
c P( X  66 )  P
/ n
9.16 We can answer part (c) and possibly part (b) depending on how nonnormal the population
is.
X   120  117 

9.17 a P(X > 120) = P
 = P(Z > 0.58) = 1 – P(Z < .58) = 1 – .7190 = .2810


 X 
b P( X  120 )  P
/ n


5.2
120  117 
 = P(Z > 1.15) = 1 – P(Z < 1.15) = 1 – .8749 = .1251

5.2 / 4 
c [P(X >120)] 4 =[.2810] 4 = .00623
X   60  52 

9.18 a P(X > 60) = P
 = P(Z > 1.33) = 1 – P(Z < 1.33) = 1 – .9082 = .0918


6

 X   60  52 
 = P(Z > 2.31) = 1 – P(Z < 2.31) = 1 – .9896 = .0104
b P( X  60 )  P


/ n
6/ 3 
c [P(X >60)] 3 =[.0918] 3 = .00077
X   12  10 

9.19 a P(X > 12) = P
 = P(Z > .67) = 1 – P(Z < .67) = 1 – .7486 = .2514


3

 X 
b P( X  275 / 25)  P( X  11)  P
/ n

11  10 
 = P(Z > 1.67) = 1 – P(Z < 1.67) = 1 – .9525 =

3 / 25 
0475
X   75  78 

9.20 a P(X < 75) = P
 = P(Z < –.50) = .3085


6

 X   75  78 
 = P(Z < –3.54) = 1 – P(Z < 3.54) = 1 – 1 = 0

b P( X  75)  P

/ n
6 / 50 
217
X  76

9.21 a P(X > 7) = P
 = P(Z > .67) = 1 – P(Z < .67) = 1 – .7486 = .2514


 X 
b P( X  7)  P
/ n

1 .5 
76 
 = P(Z > 1.49) = 1 – P(Z < 1.49) = 1 – .9319 = .0681

1.5 / 5 
c [P(X >7)] 5 =[.2514] 5 = .00100
 X   5.97  6.05 
 = P(Z < –2.67) =.0038
9.22 a P( X  5.97 )  P


/ n
.18 / 36 
b It appears to be false.
 X 
9.23 P( X  10,000 / 16 )  P( X  625 )  P
/ n

625  600 
 = P(Z > .50) = 1 – P(Z < .50)

200 / 16 
= 1 – .6915 = .3085
9.24 The professor needs to know the mean and standard deviation of the population of the
weights of elevator users and that the distribution is not extremely nonnormal.
 X   71 .25  75 
 = P(Z > –1.50)
9.25 P( X  1,140 / 16 )  P( X  71 .25)  P


/ n
10 / 16 
= 1 − P(Z < −1.50) = 1 − 0668 = .9332
 X 
5  4.8 
 = P(Z > 1.19)
9.26 P(Total time > 300) = P( X  300 / 60 )  P( X  5)  P


  / n 1.3 / 60 
= 1 – P(Z < 1.19) = 1 – .8830 = .1170
9.27 No because the central limit theorem says that the sample mean is approximately normally
distributed.
 X 
9.28 P(Total number of cups > 240) = P( X  240 / 125 )  P( X  1.92 )  P
/ n
= P(Z > –1.49) = 1 − P(Z < −1.49) = 1 − .0681 = .9319
218

1.92  2.0 


.6 / 125 
 X 
9.29 P(Total number of faxes > 1500) = P( X  1500 / 5)  P( X  300 )  P
/ n

300  275 


75 / 5 
= P(Z > .75) = 1 – P(Z < .75) = 1 – .7734 = .2266

9.30a P( P̂ > .60) = P
P̂  p
 p(1  p) / n


P̂  p
b. P( P̂ > .60) = P
 p(1  p) / n


 = P(Z >
(.5)(1  .5) / 300 
.60  .5

3.46) = 0

 = P(Z > 1.74)
(.55 )(1  .55 ) / 300 
.60  .55

= 1 – P(Z < 1.74)
= 1 – .9591 = .0409

P̂  p
c. P( P̂ > .60) = P
 p(1  p) / n



 = P(Z > 0)
(.6)(1  .6) / 300 
.60  .6

P̂  p
9.31a P( P̂ < .22) = P

 p(1  p) / n


b. P( P̂ < .22) = P
P̂  p

P̂  p
 p(1  p) / n

c. P( P̂ < .22) = P
 p(1  p) / n


9.32 P( P̂ < .75) = P

 = P(Z <
(.25 )(1  .25 ) / 800 


 = P(Z <
(.25 )(1  .25 ) / 1000 
 p(1  p) / n


9.33 P( P̂ > .35)= P

 = P(Z <
(.25 )(1  .25 ) / 500 
.22  .25

P̂  p
P̂  p
 p(1  p) / n

.22  .25

–2.19) = .0143

 = P(Z <
(.80 )(1  .80 ) / 100 
.75  .80

 = P(Z >
(.40 )(1  .40 ) / 60 
.35  .40
–1.55) = .0606
–1.96) = .0250
.22  .25

= 1 – P(Z < 0) = 1 − .5 = .5
–1.25) = .1056
–.79) = .1 − P(Z < −.79)
1 − .2148= .7852

9.34 P( P̂ < .49) = P
P̂  p
 p(1  p) / n



P̂  p

 p(1  p) / n

9.35 P( P̂ > .04)= P

 = P(Z <
(.55 )(1  .55 ) / 500 
.49  .55
–2.70) = .0035

 = P(Z > 4.04)
(.02 )(1  .02 ) / 800 
.04  .02
219
= 1 – P(Z < 4.04) = 1 – 1= 0;
The defective rate appears to be larger than 2%.

P̂  p
9.36 a P( P̂ < .50) = P
 p(1  p) / n



 = P(Z <
(.53)(1  .53) / 400 
.50  .53
–1.20) =.1151; the claim may be
true

P̂  p
b P( P̂ < .50) = P
 p(1  p) / n


 = P(Z <
(.53)(1  .53) / 1,000 
.50  .53

–1.90) = .0287; the claim appears to be
false

P̂  p
9.37 P( P̂ > .10) = P

 = P(Z >
(.14 )(1  .14 ) / 100 
.10  .14

 p(1  p) / n

–1.15) = 1 – P(Z < – 1.15)
= 1 – .1251 = .8749

P̂  p

 p(1  p) / n

9.38 P( P̂ > .05)= P

 = P(Z > 2.34)
(.03)(1  .03) / 400 
.05  .03
= 1 – P(Z < 2.34) = 1 – .9904
= .0096; the commercial appears to be dishonest

P̂  p
9.39 P( P̂ > .32) = P

 = P(Z > 1.38)
(.30 )(1  .30 ) / 1,000 
.32  .30

 p(1  p) / n

= 1 – P(Z < 1.38)
= 1 – .9162 = .0838

9.40 a P( P̂ < .45) = P
P̂  p
 p(1  p) / n



 = P(Z <
(.50 )(1  .50 ) / 600 
.45  .50
–2.45) = .0071
b The claim appears to be false.

9.41 P( P̂ < .75) = P
P̂  p

P̂  p
 p(1  p) / n

9.42 P( P̂ < .70) = P
 p(1  p) / n



 = P(Z <
(.80 )(1  .80 ) / 350 
–2.34) = .0096


 = P(Z <
(.75 )(1  .75 ) / 460 
–2.48) = .0066
.75  .80
.70  .75
220

9.43 P( P̂ > .28) = P
P̂  p
 p(1  p) / n



 = P(Z >
(.25 )(1  .25 ) / 1200 
.28  .25
2.40) = 1 – P(Z < 2.40)
= 1 – .9918 = .0082
9.44 The claim appears to be false.
9.45




 ( X  X )  (   ) 25  (280  270 ) 
1
2
1
2
=
P( X1  X 2  25)  P


12  22
25 2 30 2 





10
10 
n1 n 2

P(Z > 1.21) = 1 – P(Z < 1.21)
= 1 – .8869 = .1131
9.46




 ( X  X )  (   ) 25  (280  270 ) 
2
1
2
=
P( X1  X 2  25)  P 1

2
2

1  2
25 2 30 2 





50
50 
n1 n 2

P(Z > 2.72) = 1 – P(Z < 2.72)
= 1 – .9967 = .0033
9.47




 ( X  X )  (   ) 25  (280  270 ) 
2
1
2
=
P( X1  X 2  25)  P 1

2
2

1  2
25 2 30 2 





100 100 
n1 n 2

P(Z > 3.84) = 1 – P(Z < 3.84)
= 1 – 1= 0
9.48




 ( X  X )  (   ) 0  (40  38) 
2
1
2
=
P( X1  X 2  0)  P 1

2
2
2
2 

1  2
6
8





25
25 
n
n
1
2

P(Z > –1.00) = 1 – P(Z < –1.00)
1 – .1587 = .8413
9.49




 ( X  X )  (   ) 0  (40  38) 
2
1
2
=
P( X1  X 2  0)  P 1

2
2

1  2
12 2 16 2 






25
25
n
n
1
2


221
P(Z > –.50) = 1 – P(Z < –.50)
= 1 – .3085 = .6915
9.50




 ( X  X )  (   ) 0  (140  138 ) 
2
1
2
=
P( X1  X 2  0)  P 1

2
2

1  2
62 82 





25 25 
n1 n 2

P(Z > –1.00) = 1 – P(Z < –1.00)
= 1 – .1587 = .8413
9.51




 ( X  X )  (   ) 0  (75  65) 
2
1
2
=
P( X1  X 2  0)  P 1

2
2
2
2 

1  2
20
21





5
5 
n
n
1
2

P(Z > –.77) = 1 – P(Z < −.77)
= 1 – .2206 = .7794
9.52




 ( X  X )  (   ) 0  (73  77 ) 
2
1
2
=
P( X1  X 2  0)  P 1

2
2

1  2
12 2 10 2 





4
4 
n1 n 2

P(Z > .51) = 1 – P(Z < .51)
= 1 – .6950 = .3050




 ( X  X )  (   ) 0  (18  15) 
1
2
1
2
=
9.53 P(X1  X 2  0)  P


12  22
32 32 





10 10 
n1 n 2

P(Z > –2.24) = 1– P(Z < –2.24)
= 1 – .0125 = .9875
9.54




 ( X  X )  (   ) 0  (10  15) 
2
1
2
=
P( X1  X 2  0)  P 1

2
2

1  2
32 32 





25 25 
n1 n 2

222
P(Z < 5.89) = 1
Chapter 10
10.1 A point estimator is a single value; an interval estimator is a range of values.
10.2 An unbiased estimator of a parameter is an estimator whose expected value equals the
parameter.
10.3
10.4
10.5 An unbiased estimator is consistent if the difference between the estimator and the
parameter grows smaller as the sample size grows.
10.6
223
10.7 If there are two unbiased estimators of a parameter, the one whose variance is smaller is
relatively efficient.
10.8
10.9 a. x  z  / 2  / n = 100  1.645(25/ 50 ) = 100  5.82; LCL = 94.18, UCL = 105.82
b. x  z  / 2  / n = 100  1.96(25/ 50 ) = 100  6.93; LCL = 93.07, UCL = 106.93
c. x  z  / 2  / n = 100  2.575(25/ 50 ) = 100  9.11; LCL = 90.89, UCL = 109.11
d. The interval widens.
10.10 a. x  z  / 2  / n = 200  1.96(50/ 25 ) = 200  19.60; LCL = 180.40, UCL = 219.60
b. x  z  / 2  / n = 200  1.96(25/ 25 ) = 200  9.80; LCL = 190.20, UCL = 209.80
c. x  z  / 2  / n = 200  1.96(10/ 25 ) = 200  3.92; LCL = 196.08, UCL = 203.92
d. The interval narrows.
10.11 a. x  z  / 2  / n = 80  1.96(5/ 25 ) = 80  1.96; LCL = 78.04, UCL = 81.96
b. x  z  / 2  / n = 80  1.96(5/ 100 ) = 80  .98; LCL = 79.02, UCL = 80.98
224
c. x  z  / 2  / n = 80  1.96(5/ 400 ) = 80  .49; LCL = 79.51, UCL = 80.49
d. The interval narrows.
10.12 a. x  z  / 2  / n = 500  2.33(12/ 50 ) = 500  3.95; LCL = 496.05, UCL = 503.95
b. x  z  / 2  / n = 500  1.96(12/ 50 ) = 500  3.33; LCL = 496.67, UCL = 503.33
c. x  z  / 2  / n = 500  1.645(12/ 50 ) = 500  2.79; LCL = 497.21, UCL = 502.79
d. The interval narrows.
10.13 a. x  z  / 2  / n = 500  2.575(15/ 25 ) = 500  7.73; LCL = 492.27, UCL = 507.73
b. x  z  / 2  / n = 500  2.575(30/ 25 ) = 500  15.45; LCL = 484.55, UCL = 515.45
c. x  z  / 2  / n = 500  2.575(60/ 25 ) = 500  30.91; LCL = 469.09, UCL = 530.91
d. The interval widens.
10.14 a. x  z  / 2  / n = 10  1.645(5/ 100 ) = 10  .82; LCL = 9.18, UCL = 10.82
b. x  z  / 2  / n = 10  1.645(5/ 25 ) = 10  1.64; LCL = 8.36, UCL = 11.64
c. x  z  / 2  / n = 10  1.645(5/ 10 ) = 10  2.60; LCL = 7.40, UCL = 12.60
d. The interval widens.
10.15 a. x  z  / 2  / n = 100  1.96(20/ 25 ) = 100  7.84; LCL = 92.16, UCL = 107.84
b. x  z  / 2  / n = 200  1.96(20/ 25 ) = 200  7.84; LCL = 192.16, UCL = 207.84
c. x  z  / 2  / n = 500  1.96(20/ 25 ) = 500  7.84; LCL = 492.16, UCL = 507.84
d. The width of the interval is unchanged.
10.16 a. x  z  / 2  / n = 400  2.575(5/ 100 ) = 400  1.29; LCL = 398.71, UCL = 401.29
b. x  z  / 2  / n = 200  2.575(5/ 100 ) = 200  1.29; LCL = 198.71, UCL = 201.29
c. x  z  / 2  / n = 100  2.575(5/ 100 ) = 100  1.29; LCL = 98.71, UCL = 101.29
225
d. The width of the interval is unchanged.
10.17 Yes, because the expected value of the sample median is equal to the population mean.
10.18 The variance decreases as the sample size increases, which means that the difference between the estimator
and the parameter grows smaller as the sample size grows larger.
10.19 Because the variance of the sample mean is less than the variance of the sample median,
the sample mean is relatively more efficient than the sample median.
10.20 a sample median  z  / 2
1.2533 
= 500  1.645
1.2533 (12 )
n
= 500  3.50
50
b. The 90% confidence interval estimate of the population mean using the sample mean is 500
 2.79.
The 90% confidence interval of the population mean using the sample median is wider than that
using the sample mean because the variance of the sample median is larger. The median is
calculated by placing all the observations in order. Thus, the median loses the potential
information contained in the actual values in the sample. This results in a wider interval estimate.
10.21 x  z  / 2  / n = 6.89  1.645(2/ 9 ) = 6.89  1.10; LCL = 5.79, UCL = 7.99
10.22 x  z  / 2  / n = 43.75  1.96(10/ 8 ) = 43.75  6.93; LCL = 36.82, UCL = 50.68
We estimate that the mean age of men who frequent bars lies between 36.82 and 50.68. This type
of estimate is correct 95% of the time.
10.23 x  z  / 2  / n = 22.83  1.96(12/ 12 ) = 22.83  6.79; LCL = 16.04, UCL = 29.62
10.24 x  z  / 2  / n = 9.85  1.645(8/ 20 ) = 9.85  2.94; LCL = 6.91, UCL = 12.79
10.25 x  z  / 2  / n = 68.6  1.96(15/ 15 ) = 68.6  7.59; LCL = 61.01, UCL = 76.19
226
We estimate that the mean number of cars sold annually by all used car salespersons lies between
61.01 and 76.19. This type of estimate is correct 95% of the time.
10.26 x  z  / 2  / n = 16.9  2.575(5/ 10 ) = 16.9  4.07; LCL = 12.83, UCL = 20.97
10.27 x  z  / 2  / n = 147.33  1.96(40/ 15 ) = 147.33  20.24; LCL = 127.09, UCL = 167.57
10.28 x  z  / 2  / n = 13.15  1.645(6/ 13 ) = 13.15  2.74; LCL = 10.41, UCL = 15.89
10.29 x  z  / 2  / n = 75.625  2.575(15/ 16 ) = 75.625  9.656; LCL = 65.969, UCL = 85.281
10.30 x  z  / 2  / n = 252.38  1.96(30/ 400 ) = 252.38  2.94; LCL = 249.44, UCL = 255.32
10.31 x  z  / 2  / n = 1,810.16  1.96(400/ 64 ) = 1,810.16  98.00; LCL = 1,712.16,
UCL = 1,908.16
10.32 x  z  / 2  / n = 12.10  1.645(2.1/ 200 ) = 12.10  .24; LCL = 11.86, UCL = 12.34. We
estimate that the mean rate of return on all real estate investments lies between 11.86% and
12.34%. This type of estimate is correct 90% of the time.
10.33 x  z  / 2  / n = 10.21  2.575(2.2/ 100 ) = 10.21  .57; LCL = 9.64, UCL = 10.78
10.34 x  z  / 2  / n = .510  2.575(.1/ 250 ) = .510  .016; LCL = .494, UCL = .526. We
estimate that the mean growth rate of this type of grass lies between .494 and .526 inch . This
type of estimate is correct 99% of the time.
10.35 x  z  / 2  / n = 26.81  1.96(1.3/ 50 ) = 26.81  .36; LCL = 26.45, UCL = 27.17. We
estimate that the mean time to assemble a cell phone lies between 26.45 and 27.17 minutes. This
type of estimate is correct 95% of the time.
227
10.36 x  z  / 2  / n = 19.28  1.645(6/ 250 ) = 19.28  .62; LCL = 18.66, UCL = 19.90. We
estimate that the mean leisure time per week of Japanese middle managers lies between 18.66
and 19.90 hours. This type of estimate is correct 90% of the time.
10.37 x  z  / 2  / n = 15.00  2.575(2.3/ 100 ) = 15.00  .59; LCL = 14.41, UCL = 15.59. We
estimate that the mean pulse-recovery time lies between 14.41 and 15.59 minutes. This type of
estimate is correct 99% of the time.
10.38 x  z  / 2  / n = 585,063  1.645(30,000/ 80 ) = 585,063  5,518; LCL = 579,545,
UCL = 590,581. We estimate that the mean annual income of all company presidents lies
between $579,545 and $590,581. This type of estimate is correct 90% of the time.
10.39 x  z  / 2  / n = 14.98  1.645(3/ 250 ) = 14.98  .31; LCL = 14.67, UCL = 15.29
10.40 x  z  / 2  / n = 27.19  1.96(8/ 100 ) = 27.19  1.57; LCL = 25.62, UCL = 28.76
z

2
10.41 a. n    / 2  = 

 B 

2
z 
n    / 2 
 B 
2

2
z
b. n    / 2  = 

 B 
c.
z
2
1.645  50 
 = 68
10

2
1.645 100 
 = 271
10

2
1.96  50 
= 
 = 97

d. n    / 2  = 

 B 
10

2
1.645  50 
 = 17
20

10.42 a The sample size increases.
b The sample size increases.
c The sample size decreases.
228
z

2
2
2.575  250 
 = 166
50

10.43 a. n    / 2  = 

 B 
z

2
2.575  50 
 =7
50

z

2
1.96  250 
 = 97
50

z

2
2.575  250 
 = 4,145
10

2
b. n    / 2  = 

 B 
c. n    / 2  = 

 B 
d. n    / 2  = 

 B 
2
2
10.44 a The sample size decreases.
b the sample size decreases.
c The sample size increases.
z

2
10.45 a. n    / 2  = 

 B 
2
1.645 10 
 = 271
1

b. 150  1
10.46 a. x  z  / 2
b. x  z  / 2


 150  1.645
n
5
 150  .5
271
 150  1.645
n
20
 150  2
271
10.47 a. The width of the confidence interval estimate is equal to what was specified.
b. The width of the confidence interval estimate is smaller than what was specified.
c. The width of the confidence interval estimate is larger than what was specified.
z

2
10.48 a. n    / 2  = 

 W 
2
1.96  200 
 = 1,537
10

b. 500  10
10.49 a. x  z  / 2

n
 500  1.96
100
 500  5
1537
229
b. x  z  / 2

 500  1.96
n
400
 500  20
1537
10.50 a The width of the confidence interval estimate is equal to what was specified.
b The width of the confidence interval estimate is smaller than what was specified.
c The width of the confidence interval estimate is larger than what was specified.
z

2
1.645  10 
 = 68
2

z

2
2.575  360 
 = 2,149
20

z

2
1.96  12 
 = 139
2

z

2
1.645  20 
 = 1,083
1

z

2
1.96  25 
 = 97
5

z 
n    / 2 
 B 
2
10.51 n    / 2  = 

 B 
10.52 n    / 2  = 

 B 
10.53 n    / 2  = 

 B 
10.54 n    / 2  = 

 B 
10.55 n    / 2  = 

 B 
10.56
2
2
2
2
2
2
1.96  15 
= 
 = 217
2


Chapter 11
230
11.1
H 0 : The drug is not safe and effective
H 1 : The drug is safe and effective
11.2
H 0 : I will complete the Ph.D.
H 1 : I will not be able to complete the Ph.D.
11.3
H 0 : The batter will hit one deep
H 1 : The batter will not hit one deep
11.4
H 0 : Risky investment is more successful
H 1 : Risky investment is not more successful
11.5
H 1 : The plane is on fire
H 1 : The plane is not on fire
11.6 The defendant in both cases was O. J. Simpson. The verdicts were logical because in the
criminal trial the amount of evidence to convict is greater than the amount of evidence required
in a civil trial. The two juries concluded that there was enough (preponderance of) evidence in
the civil trial, but not enough evidence (beyond a reasonable doubt) in the criminal trial.
All p-values and probabilities of Type II errors were calculated manually using Table 3 in
Appendix B.
11.7 Rejection region: z < z.005  2.575 or z > z.005 = 2.575
z
x 
/ n

980  1000
 1.00
200 / 100
p-value = 2P(Z < –1.00) = 2(.1587) = .3174
There is not enough evidence to infer that   1000.
231
11.8 Rejection region: z > z.03 = 1.88
z
x 
/ n

51  50
 .60
5/ 9
p-value = P(Z > .60) = 1 – .7257 = .2743
There is not enough evidence to infer that  > 50.
11.9 Rejection region: z < z.10  1.28
z
x 
/ n

14 .3  15
2 / 25
 1.75
p-value = P(Z < –1.75) = .0401
There is enough evidence to infer that  < 15.
11.10 Rejection region: z < z .025  1.96 or z > z.025 = 1.96
z
x 
/ n

100  100
10 / 100
0
232
p-value = 2P(Z > 0) = 2(.5) = 1.00
There is not enough evidence to infer that   100.
11.11 Rejection region: z > z.01 = 2.33
z
x 
/ n

80  70
20 / 100
 5.00
p-value = p(z > 5.00) = 0
There is enough evidence to infer that  > 70.
11.12 Rejection region: z < z.05  1.645
z
x 
/ n

48  50
15 / 100
 1.33
p-value = P(Z < –1.33) = .0918
There is not enough evidence to infer that  < 50.
233
11.13a. z 
x 
/ n

52  50
 1.20
5/ 9
p-value = P(Z > 1.20) = 1 – .8849 = .1151
b. z 
x 

/ n
52  50
 2.00
5 / 25
p-value = P(Z > 2.00) = .5 – .4772 = .0228.
c. z 
x 
52  50

/ n
 4.00
5 / 100
p-value = P(Z > 4.00) = 0.
d. The value of the test statistic increases and the p-value decreases.
11.14a. z 
x 
/ n

190  200
 .60
50 / 9
p-value = P(Z < –.60) = .5 – .2257 = .2743
x 
b. z 

/ n
190  200
 1.00
30 / 9
p-value = P(Z < –1.00) = .1587
c z
x 
/ n

190  200
 3.00
10 / 9
p-value = P(Z < –3.00) = .0013
d. The value of the test statistic decreases and the p-value decreases.
11.15 a. z 
x 
/ n

21  20
5 / 25
 1.00
p-value = 2P(Z > 1.00) = 2(1 – .8413) = .3174
b. z 
x 
/ n

22  20
 2.00
5 / 25
p-value = 2P(Z > 2.00) = 2(1 – .9772) = .0456
c. z 
x 
/ n

23  20
 3.00
5 / 25
p-value = 2P(Z > 3.00) = 2(1 – .9987) = .0026
d. The value of the test statistic increases and the p-value decreases.
234
11.16 a. z 
x 
/ n
99  100

8 / 100
 1.25
p-value = 2P(Z < –1.25) = 2(.1056) = .2112
b. z 
x 
/ n

99  100
 .88
8 / 50
p-value = 2P(Z < –.88) = 2(.1894) = .3788
c. z 
x 
/ n

99  100
 .56
8 / 20
p-value = 2P(Z < –.56) = 2(.2877) = .5754
d. The value of the test statistic increases and the p-value increases.
11.17 a. z 
x 
/ n

990  1000
 4.00
25 / 100
p-value = P(Z < –4.00) = 0
b. z 
x 
/ n

990  1000
 2.00
50 / 100
p-value = P(Z < –2.00) = .0228
c. z 
x 
/ n

990  1000
 1.00
100 / 100
p-value = P(Z < –1.00) = .1587
d. d. The value of the test statistic increases and the p-value increases.
11.18 a. z 
x 
/ n
72  60

 3.00
20 / 25
p-value = P(Z > 3.00) = 1 – .9987 = .0013
b. z 
x 
/ n

68  60
 2.00
20 / 25
p-value = P(Z > 2.00) = 1 – .9772 = .0228
c. z 
x 
/ n

64  60
 1.00
20 / 25
p-value = P(Z > 1.00) = 1 – .8413 = .1587
d. The value of the test statistic decreases and the p-value increases.
235
x 
11.19 a z 
/ n

178  170
65 / 200
 1.74
p-valu.e = P(Z > 1.74) = 1 – .9591 = .0409
b. z 
x 

/ n
178  170
 1.23
65 / 100
p-value = P(Z > 1.23) = 1 – .8907 = .1093
c. The value of the test statistic increases and the p-value decreases.
11.20 a z 
x 
/ n

178  170
35 / 400
 4.57
p-value = P(Z > 4.57) = 0.
b z
x 
/ n

178  170
100 / 400
 1.60
p-value = P(Z > 1.60) = 1 – .9452 = .0548
The value of the test statistic decreases and the p-value increases.
11.21 See Table 11.1 in the book.
11.22 a z 
x 
/ n

21.63  22
 .62
6 / 100
p-value = P(Z < –.62) = .2676
bz 
x 
/ n

21 .63  22
6 / 500
 1.38
p-value = P(Z < –1.38) = .0838
The value of the test statistic decreases and the p-value decreases.
11.23 a z 
x 
/ n

21 .63  22
3 / 220
 1.83
p-value = P(Z < –1.83) = .0336
bz 
x 
/ n

21.63  22
12 / 220
 .46
p-value = P(Z < –.46) = .3228
The value of the test statistic increases and the p-value increases.
236
z
11.24 x
x  22
p-value
6 / 220
22.0
21.8
21.6
21.4
21.2
21.0
20.8
20.6
20.4
11.25 a z 
0
–.49
–.99
–1.48
–1.98
–2.47
–2.97
–3.46
–3.96
x 
/ n

17 .55  17 .09
.5
.3121
.1611
.0694
.0239
.0068
.0015
0
0
 .84
3.87 / 50
p-value = 2P(Z > .84) = 2(1 – .7995) = 2(.2005) = .4010
bz 
x 
/ n

17.55  17.09
3.87 / 400
 2.38
p-value = 2P(Z > 2.38) = 2(1 – .9913) = 2(.0087) = .0174
The value of the test statistic increases and the p-value decreases.
11.26 a z 
x 
/ n

17 .55  17 .09
 2.30
2 / 100
p-value = 2P(Z > 2.30) = 2(1 – .9893) = 2(.0107) = .0214
bz 
x 
/ n

17 .55  17 .09
 .46
10 / 100
p-value = 2P(Z > .46) = 2(1 – .6772) = 2(.3228) = .6456
The value of the test statistic decreases and the p-value increases.
11.27a x
z
x  17 .09
p-value
3.87 / 100
15.0
15.5
16.0
16.5
17.0
17.5
18.0
18.5
–5.40
–4.11
–2.82
–1.52
–.23
1.06
2.35
3.64
0
0
.0048
.1286
.8180
.2892
.0188
0
237
19.0
4.94
0
11.28 H 0 :  = 5
H1 :  > 5
z
x 
/ n

65
1.5 / 10
 2.11
p-value = P(Z > 2.11) = 1 – .9826 = .0174
There is enough evidence to infer that the mean is greater than 5 cases.
11.29 H 0 :  = 50
H1 :  > 50
z
x 
/ n

59 .17  50
10 / 18
 3.89
p-value = P(Z > 3.89) = 0
There is enough evidence to infer that the mean is greater than 50 minutes.
11.30
H 0 :  = 12
H1 :  < 12
z
x 
/ n

11.00  12
3 / 15
 1.29
p-value = P(Z < –1.29) = .0985
There is enough evidence to infer that the average number of golf balls lost is less than 12.
11.31 H 0 :  = 36
H1 :  < 36
z
x 
/ n

34 .25  36
8 / 12
 .76
p-value = P(Z < –.76) = .2236
There is not enough evidence to infer that the average student spent less time than recommended.
11.32 H 0 :  = 6
H1 :  > 6
238
z
x 
/ n

6.60  6
 .95
2 / 10
p-value = P(Z > .95) = 1 – .8289 = .1711
There is not enough evidence to infer that the mean time spent putting on the 18 th green is greater than 6 minutes.
11.33 H 0 :  = .50
H1 :   .50
z
x 
/ n

.493  .50
.05 / 10
 .44
p-value = 2P(Z < –.44) = 2(.3300) = .6600
There is not enough evidence to infer that the mean diameter is not .50 inch.
11.34 H 0 :  = 25
H1 :  > 25
z
x 
/ n

30 .22  25
12 / 18
 1.85
p-value = P(Z > 1.85) = 1 – .9678 =.0322
There is not enough evidence to conclude that the manager is correct.
11.35 H 0 :  = 5,000
H1 :  > 5,000
z
x 
/ n

5,065  5,000
 1.62
400 / 100
p-value = P(Z > 1.62) = 1 – .9474 =.0526
There is not enough evidence to conclude that the claim is true.
11.36 H 0 :  = 30,000
H1 :  < 30,000
z
x 
/ n

29 ,120  30 ,000
 2.06
8,000 / 350
p-value = (P(Z < –2.06) = .0197
There is enough evidence to infer that the president is correct
239
11.37 H 0 :  = 560
H1 : 
z
x 
/ n

> 560
569 .0  560
 .80
50 / 20
p-value = P(Z > .80) = 1 – .7881 = .2119
There is not enough evidence to conclude that the dean’s claim is true.
11.38a H 0 :  = 17.85
H1 : 
z
x 
/ n

> 17.85
19 .13  17 .85
 1.65
3.87 / 25
p-value = P(Z > 1.65) = 1 – .9505 = .0495
There is enough evidence to infer that the campaign was successful.
b We must assume that the population standard deviation is unchanged.
11.39 H 0 :  = 0
H1 : 
z
x 
/ n

<0
1.20  0
 1.41
6 / 50
p-value = P(Z < –1.70) = .0793
There is not enough evidence to conclude that the safety equipment is effective.
11.40 H 0 :  = 55
H1 :  > 55
z
x 
/ n

55 .80  55
 2.26
5 / 200
p-value = P(Z > 2.26) = 1 – .9881 = .0119
There is not enough evidence to support the officer’s belief.
11.41 H 0 :  = 4
240
H1 :  > 4
z
x 
/ n

5.04  4
1.5 / 50
 4.90
p-value = P(Z > 4.90) = 0
There is enough evidence to infer that the expert is correct.
11.42 H 0 :  = 20
H1 :  < 20
z
x 
/ n

19 .39  20
 1.22
3 / 36
p-value = P(Z < –1.22) = .1112
There is not enough evidence to infer that the manager is correct.
11.43 H 0 :  = 100
H1 :  > 100
z
x 
/ n

105 .7  100
 2.25
16 / 40
p-value = P(Z > 2.25) = 1 – .9878 = .0122
There is not enough evidence to infer that the site is acceptable.
11.44 H 0 :  = 4
H1 :   4
z
x 
/ n

4.84  4
 3.33
2 / 63
p-value = 2P(Z > 3.33) = 0
There is enough evidence to infer that the average Alpine skier does not ski 4 times per year.
11.45 H 0 :  = 5
H1 :  > 5
z
x 
/ n

5.64  5
2 / 25
 1.60
241
p-value = P(Z > 1.60) = 1 – .9452 = .0548
There is enough evidence to infer that the golf professional’s claim is true.
11.46 H 0 :  = 32
H1 :  < 32
z
x 
/ n

29 .92  32
 2.73
8 / 110
p-value = P(Z < –2.73) = 1– .9968 = .0032
There is enough evidence to infer that there has been a decrease in the mean time away from
desks. A type I error occurs when we conclude that the plan decreases the mean time away from
desks when it actually does not. This error is quite expensive. Consequently we demand a low pvalue. The p-value is small enough to infer that there has been a decrease.
11.47 H 0 :  = 230
H1 :  > 230
z
x 
/ n

231 .56  230
10 / 100
 1.56
p-value = P(Z > 1.56) = 1 – .9406 = .0594
There is not enough evidence to infer that Nike is correct.
11.48 Rejection region:
x  200
x 
> z / 2 or
/ n
x  200
x 
/ n
< z  / 2
> z .025  1.96 or
< –1.96
10 / 100
10 / 100
x > 201.96 or x < 198.04
 = P(198.04 < x < 201.96 given  = 203)
 198 .04  203
= P
 10 / 100

x 
/ n
11.49 Rejection region:
x  1000

201 .96  203 
 = P( –4.96 < z < –1.04) = .1492 – 0 = .1492

10 / 100 
x 
/ n
> z
> z.01  2.33
50 / 25
x > 1023.3
242
 x   1023 .3  1050

 = P( x < 1023.3 given  = 1050) = P
50 / 25
 / n
11.50 Rejection region:
x  50
x 
/ n

 = P(z < –2.67) = .0038

< z 
< z.05  1.645
10 / 40
x < 47.40
 x 
47 .40  48 
 = P(z > –.38) = 1 − .3520 = .6480

 = P( x > 47.40 given  = 48) = P

10 / 40 
/ n
11.51
Exercise 11.48
Exercise 11.49
243
Exercise 11.50
11.52 a. Rejection region:
x  100
x 
/ n
> z
> z .10  1.28
20 / 100
x > 102.56
 x   102 .56  102 
 = P(z < .28) = .6103

 = P( x < 102.56 given  = 102) = P

20 / 100 
/ n
x 
b. Rejection region:
> z
/ n
x  100
> z .02  2.55
20 / 100
x > 104.11
 x   104 .11  102 
 = P(z < 1.06) = .8554

 = P( x < 104.11 given  = 102) = P

20 / 100 
/ n
c.  increases.
11.53 a. Rejection region:
x  40
x 
/ n
< z 
< z .05  1.645
5 / 25
x < 38.36
 x   38 .36  37 
 = P(z > 1.36) = 1 – .9131 = .0869

 = P( x > 38.36 given  = 37) = P

5 / 25 
/ n
x 
b. Rejection region:
< z 
/ n
244
x  40
< z .15  1.04
5 / 25
x < 38.96
 x   38 .96  37 
 = P(z > 1.96) = 1 – .9750 = .0250

 = P( x > 38.96 given  = 37) = P

5 / 25 
/ n
c.  decreases.
11.54
Exercise 11.52 a
Exercise 11.52 b
245
Exercise 11.53 a
Exercise 11.53 b
11.55 a. Rejection region:
x  200
x 
/ n
< z 
< z .10  1.28
30 / 25
x < 192.31
246
 x   192 .31  196 
 = P(z > –.62) = 1 − .2676 = .7324

 = P( x > 192.31 given  = 196) = P

30 / 25 
/ n
x 
b. Rejection region:
< z 
/ n
x  200
< z .10  1.28
30 / 100
x < 196.16
 x   196 .16  196 
 = P(z > .05) = 1 – .5199 = .4801

 = P( x > 196.16 given  = 196) = P

30 / 100 
/ n
c.  decreases.
11.56 a. Rejection region:
x  300
x 
/ n
> z
> z .05  1.645
50 / 81
x > 309.14
 x   309 .14  310 
 = P(z < –.15) = .4404

 = P( x < 309.14 given  = 310) = P

50 / 81 
/ n
x 
b. Rejection region:
> z
/ n
x  300
> z .05  1.645
50 / 36
x > 313.71
 x   313 .71  310 
 = P(z < .45) = .6736

 = P( x < 313.71 given  = 310) = P

50 / 36 
/ n
c.  increases.
11.57
Exercise 11.55 a
247
Exercise 11.55 b
Exercise 11.56 a
248
Exercise 11.56 b
249
11.58
250
11.59
11.60 H 0 :  = 170
H1 :  < 170
A Type I error occurs when we conclude that the new system is not cost effective when it
actually is. A Type II error occurs when we conclude that the new system is cost effective when
it actually is not.
The test statistic is the same. However, the p-value equals 1 minus the p-value calculated
Example 11.1. That is,
p-value = 1 – .0069 = .9931
We conclude that there is no evidence to infer that the mean is less than 170. That is, there is no
evidence to infer that the new system will not be cost effective.
11.61 Rejection region:
x0
6 / 50
x 
/ n
<  z
< z .10  1.28
x < –1.09
251
 x 
 1.09  (2) 
 = P(z > 1.07) = 1 – .8577 = .1423

 = P( x > –1.09 given  = –2) = P

6 / 50 
/ n
 can be decreased by increasing  and/or increasing the sample size.
11.62 Rejection region:
x  22
x 
/ n
< z 
< z .10  1.28
6 / 220
x < 21.48
 x 
21 .48  21 
 = P(z > 1.19) =1 – .8830 = .1170

 = P( x > 21.48 given  = 21) = P

6 / 220 
/ n
The company can decide whether the sample size and significance level are appropriate.
11.63 Rejection region:
x  100
x 
/ n
> z
> z .01  2.33
16 / 40
x > 105.89
 x   105 .89  104 
 = P(z < .75) = .7734

 = P( x < 105.89 given  = 104) = P

16 / 40 
/ n
11.64 Rejection region:
x  32
8 / 110
x 
/ n
<  z
< z .05  1.645
x < 30.75
 x   30 .75  30 ) 
 = P(z > .98) = 1 – .8365 = .1635

 = P( x > 30.75 given  = 30) = P

8 / 110 
/ n
 can be decreased by increasing  and/or increasing the sample size.
11.65 i Rejection region:
x  10
3 / 100
x 
/ n
<  z
< z.01  2.33
252
x < 9.30
 x 
9.30  9 
 = P(z > 1) = 1 – .8413 = .1587

 = P( x > 9.30 given  = 9) = P

  / n 3 / 100 
ii Rejection region:
x  10
3 / 75
x 
/ n
<  z
< z.05  1.645
x < 9.43
 x 
9.43  9 
 = P(z > 1.24) = 1 – .8925 = .1075

 = P( x > 9.43 given  = 9) = P

  / n 3 / 75 
iii Rejection region:
x  10
3 / 50
x 
/ n
<  z
< z.10  1.28
x < 9.46
 x   9.46  9 
 = P(z > 1.08) = 1 – .8599 = .1401

 = P( x > 9.46 given  = 9) = P

  / n 3 / 50 
Plan ii has the lowest probability of a type II error.
11.66 A Type I error occurs when we conclude that the site is feasible when it is not. The
consequence of this decision is to conduct further testing. A Type II error occurs when we do not
conclude that a site is feasible when it actually is. We will do no further testing on this site, and
as a result we will not build on a good site. If there are few other possible sits, this could be an
expensive mistake.
11.67 H 0 :   20
H1 :   25
Rejection region:
x  20
x 
/ n
> z
> z .01  2.33
8 / 25
x > 23.72
 x 
23 .72  25 
 = P(z < –.80) = .2119

 = P( x < 23.72 given  = 25) = P

8 / 25 
/ n
253
The process can be improved by increasing the sample size.
254