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```									                                      Chapter 7: Sampling and Sampling Distributions   1

Chapter 7
Sampling and Sampling Distributions
LEARNING OBJECTIVES

The two main objectives for Chapter 7 are to give you an appreciation for
the proper application of sampling techniques and an understanding of the
sampling distributions of two statistics, thereby enabling you to:

1.     Contrast sampling to census and differentiate among different methods
of sampling, which include simple, stratified, systematic, and cluster
random sampling; and convenience, judgment, quota, and snowball
nonrandom sampling, by assessing the advantages associated with each
2.     Describe the distribution of a sample’s mean using the central limit
theorem, correcting for a finite population if necessary
3.     Describe the distribution of a sample’s proportion using the z formula
for sample proportions

SOLUTIONS TO PROBLEMS IN CHAPTER 7

7.1   a)    i.   A union membership list for the company.
ii.   A list of all employees of the company.

b)    i.   White pages of the telephone directory for Utica, New York.
ii.   Utility company list of all customers.

c)    i.   Airline company list of phone and mail purchasers of tickets from the airline during the past
six months.
ii.   A list of frequent flyer club members for the airline.

d)    i.   List of boat manufacturer's employees.
ii.   List of members of a boat owners association.

e)    i.   Cable company telephone directory.
ii.   Membership list of cable management association.
Chapter 7: Sampling and Sampling Distributions 2
7.4   a)    Size of motel (rooms), age of motel, geographic location.

b)   Gender, age, education, social class, ethnicity.

c)   Size of operation (number of bottled drinks per month), number of employees, number of different
types of drinks bottled at that location, geographic location.

d)   Size of operation (sq.ft.), geographic location, age of facility, type of process used.

7.5   a)   Under 21 years of age, 21 to 39 years of age, 40 to 55 years of age, over 55 years of age.

b)   Under \$1,000,000 sales per year, \$1,000,000 to \$4,999,999 sales per year, \$5,000,000 to
\$19,999,999 sales per year, \$20,000,000 to \$49,000,000 per year, \$50,000,000 to \$99,999,999 per
year, over \$100,000,000 per year.

c)   Less than 2,000 sq. ft., 2,000 to 4,999 sq. ft.,
5,000 to 9,999 sq. ft., over 10,000 sq. ft.

d)   East, southeast, midwest, south, southwest, west, northwest.

e)   Government worker, teacher, lawyer, physician, engineer, business person, police officer, fire
fighter, computer worker.

f) Manufacturing, finance, communications, health care, retailing, chemical, transportation.

7.6   n = N/k = 100,000/200 = 500

7.7   N = nk = 75(11) = 825

7.8   k = N/n = 3,500/175 = 20

Start between 0 and 20. The human resource department probably has a list of
company employees which can be used for the frame. Also, there might be a
company phone directory available.
Chapter 7: Sampling and Sampling Distributions 3
7.9    a)      i.         Counties
ii.        Metropolitan areas

b)      i.         States (beside which the oil wells lie)
ii.        Companies that own the wells

c)      i.         States
ii.        Counties

7.10   Go to the district attorney's office and observe the apparent activity of various
attorneys at work. Select some who are very busy and some who seem to be
less active. Select some men and some women. Select some who appear to
be older and some who are younger. Select attorneys with different ethnic
backgrounds.

7.11   Go to a conference where some of the Fortune 500 executives attend.
Approach those executives who appear to be friendly and approachable.

7.12   Suppose 40% of the sample is to be people who presently own a personal computer and 60% with
people who do not. Go to a computer show at the city's conference center and start interviewing people.
Suppose you get enough people who own personal computers but not enough interviews with those who
do not. Go to a mall and start interviewing people. Screen out personal computer owners. Interview
non personal computer owners until you meet the 60% quota.

7.13        µ = 50,           = 10,        n = 64

a) P( x > 52):

x         52  50
z =                         = 1.6
             10
n             64

from Table A.5, Prob. = .4452

P( x > 52) = .5000 - .4452 = .0548

b) P( x < 51):

x         51 50
z =                        = 0.80
            10
n          64

from Table A.5 prob. = .2881

P( x < 51) = .5000 + .2881 = .7881
Chapter 7: Sampling and Sampling Distributions 4

c) P( x < 47):

x          47  50
z =                        = -2.40
             10
n           64

from Table A.5 prob. = .4918

P( x < 47) = .5000 - .4918 =                .0082

d) P(48.5 < x < 52.4):

x           48.5  50
z =                          = -1.20
              10
n            64

from Table A.5 prob. = .3849

x           52.4  50
z =                          = 1.92
               10
n            64

from Table A.5 prob. = .4726

P(48.5 < x < 52.4) = .3849 + .4726 = .8575

e) P(50.6 < x < 51.3):

x          50.6  50
z =                          = 0.48
              10
n            64

from Table A.5, prob. = .1844

x          51.3  50
z =                           1.04
               10
n               64

from Table A.5, prob. = .3508

P(50.6 < x < 51.3) = .3508 - .1844 = .1644
Chapter 7: Sampling and Sampling Distributions 5

7.14   µ = 23.45           = 3.8

a) n = 10, P( x > 22):

x        22  23.45
z =                            = -1.21
             3.8
n            10

from Table A.5, prob. = .3869

P( x > 22) = .3869 + .5000 =           .8869

b) n = 4, P( x > 26):

x        26  23.45
z =                            = 1.34
             3.8
n             4

from Table A.5, prob. = .4099

P( x > 26) = .5000 - .4099 =              .0901

7.15      n = 36               µ = 278      P( x < 280) = .86

.3600 of the area lies between x = 280 and µ = 278. This probability is
associated with z = 1.08 from Table A.5. Solving for  :

x
z =

n

280  278
1.08 =

36


1.08       = 2
6

12
=          = 11.11
1.08

7.16     n = 81            = 12         P( x > 300) = .18
Chapter 7: Sampling and Sampling Distributions 6

.5000 - .1800 = .3200 and from Table A.5, z.3200 = 0.92

Solving for µ:

x
z =

n

300  
0.92 =
12
81

12
0.92      = 300 - 
9

1.2267 = 300 - 

µ = 300 - 1.2267 = 298.77
7.17   a) N = 1,000 n = 60      µ = 75   =6

P( x < 76.5):

x                 76.5  75
z =                                              = 2.00
      N n          6  1000  60
n     N 1          60 1000  1

from Table A.5, prob. = .4772

P( x < 76.5) = .4772 + .5000 = .9772

b) N = 90         n = 36            µ = 108          = 3.46

P(107 < x < 107.7):

x             107  108
z =                                            = -2.23
      N n         3.46 90  36
n     N 1           36 90  1

from Table A.5, prob. = .4871

x             107.7  108
z =                                           = -0.67
      N n         3.46 90  36
n     N 1           36 90  1
Chapter 7: Sampling and Sampling Distributions 7
from Table A.5, prob. = .2486

P(107 < x < 107.7) = .4871 - .2486 =                      .2385

c) N = 250              n = 100            µ = 35.6         = 4.89

P( x > 36):

x                    36  35.6
z =                                                     = 1.05
        N n            4.89     250  100
n       N 1            100       250  1

from Table A.5, prob. = .3531

P( x > 36) = .5000 - .3531 = .1469

d) N = 5000              n = 60            µ = 125         = 13.4

P( x < 123):

x                123  125
z =                                                    = -1.16
           N n        13.4 5000  60
n       N 1         60 5000  1

from Table A.5, prob. = .3770

P( x < 123) = .5000 - .3770 = .1230

7.18   µ = 99.9            = 30           n = 38

a) P( x < 90):

x          90  99.9
z =                            = -2. 03
               30
n              38

from table A.5, area = .4788

P( x < 90) = .5000 - .4788 = .0212

b) P(98 < x < 105):

x          105  99.9
z =                            = 1.05
               30
n               38
Chapter 7: Sampling and Sampling Distributions 8

from table A.5, area = .3531

x         98  99.9
z =                        = -0.39
             30
n           38

from table A.5, area = .1517

P(98 < x < 105) = .3531 + .1517 = .5048

c) P( x < 112):

x         112  99.9
z =                         = 2.49
             30
n           38

from table A.5, area = .4936

P( x < 112) = .5000 + .4936 = .9936

d) P(93 < x < 96):

x         93  99.9
z =                         = -1.42
              30
n            38

from table A.5, area = .4222

x         96  99.9
z =                         = -0.80
              30
n            38

from table A.5, area = .2881

P(93 < x < 96) = .4222 - .2881 = .1341

7.19     N = 1500            n = 100     µ = 177,000           = 8,500

P( X > \$185,000):
Chapter 7: Sampling and Sampling Distributions 9
X              185,000  177,000
z =                                                   = 9.74
        N n        8,500 1500  100
n       N 1          100 1500  1

from Table A.5, prob. = .5000

P( X > \$185,000) = .5000 - .5000 = .0000
7.20     µ = \$65.12     = \$21.45      n = 45   P( x > x 0 ) = .2300

Prob. x lies between x 0 and µ = .5000 - .2300 = .2700
from Table A.5, z.2700 = 0.74

Solving for x 0 :

x0  
z =

n

x 0  65.12
0.74 =
21.45
45

2.366 = x 0 - 65.12              and        x 0 = 65.12 + 2.366 = 67.486

7.21   µ = 50.4             = 11.8            n = 42

a) P( x > 52):

x          52  50.4
z =                           = 0.88
             11.8
n           42

from Table A.5, the area for z = 0.88 is .3106

P( x > 52) = .5000 - .3106 = .1894

b) P( x < 47.5):

x          47.5  50.4
z =                             = -1.59
             11.8
n            42

from Table A.5, the area for z = -1.59 is .4441
Chapter 7: Sampling and Sampling Distributions 10
P( x < 47.5) = .5000 - .4441 = .0559

c) P( x < 40):

x          40  50.4
z =                                 = -5.71
             11.8
n              42

from Table A.5, the area for z = -5.71 is .5000

P( x < 40) = .5000 - .5000 = .0000

d) 71% of the values are greater than 49. Therefore, 21% are between the
sample mean of 49 and the population mean, µ = 50.4.

The z value associated with the 21% of the area is -0.55

z.21 = -0.55

x
z =

n

49  50.4
-0.55 =

42

 = 16.4964

7.22      p = .25

a) n = 110                    ˆ
P( p < .21):

p p
ˆ               .21  .25
z =                                       = -0.97
pq          (. 25 )(. 75 )
n               110

from Table A.5, prob. = .3340

ˆ
P( p < .21) = .5000 - .3340 = .1660

b) n = 33                     ˆ
P( p > .24):

p p
ˆ               .24  .25
z =                                       = -0.13
pq         (. 25 )(. 75 )
n               33
Chapter 7: Sampling and Sampling Distributions 11

from Table A.5, prob. = .0517

ˆ
P( p > .24) = .5000 + .0517 = .5517

c) n = 59                   ˆ
P(.24 < p < .27):

p p
ˆ           .24  .25
z =                                  = -0.18
pq        (. 25 )(. 75 )
n              59

from Table A.5, prob. = .0714

pP
ˆ           .27  .25
z =                                 = 0.35
pq        (. 25 )(. 75 )
n              59

from Table A.5, prob. = .1368

ˆ
P(.24 < p < .27) = .0714 + .1368 = .2082

d) n = 80             ˆ
P( p > .30):

p p
ˆ           .30  .25
z =                                 = 1.03
pq        (. 25 )(. 75 )
n              80

from Table A.5, prob. = .3485

ˆ
P( p > .30) = .5000 - .3485 = .1515

e) n = 800            ˆ
P( p > .30):

p p
ˆ           .30  .25
z =                                  = 3.27
pq        (. 25 )(. 75 )
n             800

from Table A.5, prob. = .4995

ˆ
P( p > .30) = .5000 - .4995 = .0005

7.23   p = .58       n = 660

ˆ
a) P( p > .60):
Chapter 7: Sampling and Sampling Distributions 12
p p
ˆ            .60  .58
z =                                 = 1.04
pq         (. 58 )(. 42 )
n              660

from table A.5, area = .3508

ˆ
P( p > .60) = .5000 - .3508 = .1492

ˆ
b) P(.55 < p < .65):

p p
ˆ            .65  .58
z =                                  = 3.64
pq         (. 58 )(. 42 )
n              660

from table A.5, area = .4998

p p
ˆ            .55  .58
z =                                  = 1.56
pq         (. 58 )(. 42 )
n              660

from table A.5, area = .4406

ˆ
P(.55 < p < .65) = .4998 + .4406 = .9404

ˆ
c) P( p > .57):

p p
ˆ            .57  .58
z =                                 = -0.52
pq        (. 58 )(. 42 )
n             660

from table A.5, area = .1985

ˆ
P( p > .57) = .1985 + .5000 = .6985

ˆ
d) P(.53 < p < .56):

p p
ˆ            .56  .58                                      p p
ˆ              .53  .58
z =                                 = -1.04              z =                                    = -2.60
pq        (. 58 )(. 42 )                                   pq          (. 58 )(. 42 )
n             660                                           n               660

from table A.5, area for z = -1.04 is .3508
from table A.5, area for z = -2.60 is .4953

ˆ
P(.53 < p < .56) = .4953 - .3508 = .1445
Chapter 7: Sampling and Sampling Distributions 13
ˆ
e) P( p < .48):

p p
ˆ            .48  .58
z =                                 = -5.21
pq        (. 58 )(. 42 )
n             660

from table A.5, area = .5000

ˆ
P( p < .48) = .5000 - .5000 = .0000

7.24   p = .40       ˆ
P( p > .35) = .8000

ˆ
P(.35 < p < .40) = .8000 - .5000 = .3000

from Table A.5, z.3000 = -0.84

Solving for n:
p p
ˆ
z =
pq
n

.35  .40               .05
-0.84 =                         =
(. 40 )(. 60 )          .24
n                   n

 0.84 .24
 n
 .05

8.23 =        n

n = 67.73  68

7.25   p = .28          n = 140              ˆ   ˆ
P( p < p0 ) = .3000

ˆ   ˆ
P( p < p0 < .28) = .5000 - .3000 = .2000

from Table A.5, z.2000 = -0.52

ˆ
Solving for p0 :
p0  p
ˆ
z =
pq
n
Chapter 7: Sampling and Sampling Distributions 14
p0  .28
ˆ
-0.52 =
(. 28 )(. 72 )
140

ˆ
-.02 = p0 - .28

ˆ
p0 = .28 - .02 = .26
7.26     P(x > 150): n = 600 p = .21 x = 150

150
ˆ
p =       = .25
600

p p
ˆ           .25  .21
z =                                     = 2.41
pq         (. 21)(. 79 )
n              600

from table A.5, area = .4920

P(x > 150) = .5000 - .4920 = .0080

7.27     p = .48 n = 200

a) P(x < 90):

90
ˆ
p =       = .45
200

p p
ˆ            .45  .48
z =                                 = -0.85
pq         (. 48 )(. 52 )
n              200

from Table A.5, the area for z = -0.85 is .3023

P(x < 90) = .5000 - .3023 = .1977

b) P(x > 100):

100
ˆ
p =       = .50
200

p p
ˆ            .50  .48
z =                                 = 0.57
pq         (. 48 )(. 52 )
n              200

from Table A.5, the area for z = 0.57 is .2157
Chapter 7: Sampling and Sampling Distributions 15

P(x > 100) = .5000 - .2157 = .2843

c) P(x > 80):

80
ˆ
p =          = .40
200

p p
ˆ            .40  .48
z =                                   = -2.26
pq       (. 48 )(. 52 )
n            200

from Table A.5, the area for z = -2.26 is .4881

P(x > 80) = .5000 + .4881 = .9881

7.28     p = .19         n = 950

ˆ
a) P( p > .25):

p p
ˆ            .25  .19
z =                                   = 4.71
pq       (. 19 )(. 89 )
n            950

from Table A.5, area = .5000

ˆ
P( p > .25) = .5000 - .5000 = .0000

ˆ
b) P(.15 < p < .20):

p p
ˆ            .15  .19
z =                                   = -3.14
pq       (. 19 )(. 81)
n            950

p p
ˆ              .20  .19
z =                                   = 0.79
pq         (. 19 )(. 89 )
n              950

from Table A.5, area for z = -3.14 is .4992

from Table A.5, area for z = 0.79 is .2852

ˆ
P(.15 < p < .20) = .4992 + .2852 = .7844

c) P(133 < x < 171):
Chapter 7: Sampling and Sampling Distributions 16

133                                      171
ˆ
p1 =        = .14                       ˆ
p2 =         = .18
950                                      950

ˆ
P(.14 < p < .18):

p p
ˆ             .14  .19                                     p p
ˆ              .18  .19
z =                                   = -3.93             z =                                     = -0.79
pq           (. 19 )(. 81)                                  pq           (. 19 )(. 81)
n                950                                         n                950

from Table A.5, the area for z = -3.93 is .49997
the area for z = -0.79 is .2852

P(133 < x < 171) = .49997 - .2852 = .21477

7.29   µ = 76,  = 14

a) n = 35,           P( x > 79):

x          79  76
z =                        = 1.27
             14
n            35

from table A.5, area = .3980

P( x > 79) = .5000 - .3980 = .1020

b) n = 140, P(74 < x < 77):

x          74  76                                      x         77  76
z =                        = -1.69                      z =                        = 0.85
             14                                                      14
n           140                                          n            140

from table A.5, area for z = -1.69 is .4545
from table A.5, area for 0.85 is .3023

P(74 < x < 77) = .4545 + .3023 = .7568
Chapter 7: Sampling and Sampling Distributions 17
c) n = 219,           P( x < 76.5):

x           76.5  76
z =                          = 0.53
            14
n            219

from table A.5, area = .2019

P( x < 76.5) = .5000 + .2019 = .7019

7.30 p = .46

a) n = 60

ˆ
P(.41 < p < .53):

p p
ˆ               .53  .46
z =                                   = 1.09
pq         (. 46 )(. 54 )
n               60

from table A.5, area = .3621

p p
ˆ               .41  .46
z =                                   = -0.78
pq         (. 46 )(. 54 )
n               60

from table A.5, area = .2823

ˆ
P(.41 < p < .53) = .3621 + .2823 = .6444

b) n = 458            ˆ
P( p < .40):

p p
               .40 .46
z =                             = -2.58
p q          (.46)(.54)
n               458

from table A.5, area = .4951


P( p < .40) = .5000 - .4951 = .0049
Chapter 7: Sampling and Sampling Distributions 18
c) n = 1350              ˆ
P( p > .49):

p p
ˆ               .49  .46
z =                                       = 2.21
pq             (. 46 )(. 54 )
n                  1350

from table A.5, area = .4864

ˆ
P( p > .49) = .5000 - .4864 = .0136

7.31                        Under 18                   250(.22) =       55
18 – 25                    250(.18) =       45
26 – 50                    250(.36) =       90
51 – 65                    250(.10) =       25
over 65                    250(.14) =       35
n =       250

7.32    p = .55         n = 600              x = 298

x 298
ˆ
p =         = .497
n 600

ˆ
P( p < .497):

p p
ˆ           .497  .55
z =                                        = -2.61
pq           (. 55 )(. 45 )
n                600

from Table A.5, Prob. = .4955

ˆ
P( p < .497) = .5000 - .4955 = .0045

No, the probability of obtaining these sample results by chance from a population that supports the
candidate with 55% of the vote is extremely low (.0045). This is such an unlikely chance sample result
that it would cause the researcher to probably reject her claim of 55% of the vote.
Chapter 7: Sampling and Sampling Distributions 19
7.33 a) Roster of production employees secured from the human
resources department of the company.

b) Alpha/Beta store records kept at the headquarters of
their California division or merged files of store
records from regional offices across the state.

c) Membership list of Maine lobster catchers association.

7.34   µ = \$ 17,755          = \$ 650    n = 30        N = 120

P( x < 17,500):

17,500  17,755
z =                          = -2.47
650 120  30
30 120  1

from Table A.5, the area for z = -2.47 is .4932

P( x < 17,500) = .5000 - .4932 = .0068

7.35   Number the employees from 0001 to 1250. Randomly sample from the random number table until 60
different usable numbers are obtained. You cannot use numbers from 1251 to 9999.

7.36   µ = \$125            n = 32         x = \$110              2 = \$525

P( x > \$110):

x         110  125
z =                          = -3.70
            525
n           32

from Table A.5, Prob.= .5000

P( x > \$110) = .5000 + .5000 =           1.0000
Chapter 7: Sampling and Sampling Distributions 20
P( x > \$135):

x          135  125
z =                              = 2.47
                525
n            32

from Table A.5, Prob.= .4932

P( x > \$135) = .5000 - .4932 = .0068

P(\$120 < x < \$130):

x          120  125
z =                              = -1.23
                525
n            32

x          130  125
z =                              = 1.23
                525
n            32

from Table A.5, Prob.= .3907

P(\$120 < x < \$130) = .3907 + .3907 = .7814

7.37 n = 1100

a) x > 810,       p = .73

x 810
ˆ
p =    
n 1100

p p
ˆ                .7364  .73
z =                                    = 0.48
pq          (. 73)(. 27 )
n               1100

from table A.5, area = .1844

P(x > 810) = .5000 - .1844 = .3156

b) x < 1030,          p = .96,

x 1030
ˆ
p =         = .9364
n 1100
Chapter 7: Sampling and Sampling Distributions 21

p p
ˆ             .9364  .96
z =                                  = -3.99
pq        (. 96 )(. 04 )
n             1100

from table A.5, area = .49997

P(x < 1030) = .5000 - .49997 = .00003

c) p = .85

ˆ
P(.82 < p < .84):

p p
ˆ              .82  .85
z =                                  = -2.79
pq           (. 85 )(. 15 )
n                1100

from table A.5, area = .4974

p p
ˆ              .84  .85
z =                                  = -0.93
pq           (. 85 )(. 15 )
n                1100

from table A.5, area = .3238
ˆ
P(.82 < p < .84) = .4974 - .3238 = .1736

7.38   1)    The managers from some of the companies you are interested in
studying do not belong to the American Managers Association.
2)    The membership list of the American Managers Association is not up-to-date.
3)    You are not interested in studying managers from some of the companies belonging to the
American Management Association.
4)    The wrong questions are asked.
5)    The manager incorrectly interprets a question.
6)    The assistant accidentally marks the wrong answer.
7)    The wrong statistical test is used to analyze the data.
8)    An error is made in statistical calculations.
9)    The statistical results are misinterpreted.

7.39   Divide the factories into geographic regions and select a few factories to represent those regional areas
of the country. Take a random sample of employees from each selected factory. Do the same for
distribution centers and retail outlets. Divide the United States into regions of areas. Select a few areas.
Take a random sample from each of the selected area distribution centers and retail outlets.

7.40   N = 12,080          n = 300
Chapter 7: Sampling and Sampling Distributions 22
k = N/n = 12,080/300 = 40.27

Select every 40th outlet to assure n > 300 outlets.

Use a table of random numbers to select a value between 0 and 40 as a starting point.

7.41     p = .54 n = 565

a) P(x > 339):

x 339
ˆ
p =        = .60
n 565

p p
ˆ             .60  .54
z =                                  = 2.86
pq         (. 54 )(. 46 )
n              565

from Table A.5, the area for z = 2.86 is .4979

P(x > 339) = .5000 - .4979 = .0021

b) P(x > 288):

x 288
ˆ
p =           = .5097
n 565
p p
ˆ       .5097  .54
z =                            = -1.45
pq     (. 54 )(. 46 )
n          565

from Table A.5, the area for z = -1.45 is .4265

P(x > 288) = .5000 + .4265 = .9265
ˆ
c) P( p < .50):

p p
ˆ            .50  .54
z =                                  = -1.91
pq         (. 54 )(. 46 )
n              565

from Table A.5, the area for z = -1.91 is .4719

ˆ
P( p < .50) = .5000 - .4719 = .0281

7.42     µ = \$550           n = 50              = \$100
Chapter 7: Sampling and Sampling Distributions 23
P( x < \$530):

x          530  550
z =                         = -1.41
             100
n             50

from Table A.5, Prob.=.4207

P(x < \$530) = .5000 - .4207 = .0793

7.43   µ = 56.8            n = 51        = 12.3

a) P( x > 60):

x          60  56.8
z =                         = 1.86
             12.3
n             51

from Table A.5, Prob. = .4686

P( x > 60) = .5000 - .4686 = .0314

b) P( x > 58):

x          58  56.8
z =                         = 0.70
             12.3
n             51

from Table A.5, Prob.= .2580

P( x > 58) = .5000 - .2580 = .2420

c) P(56 < x < 57):

x          56  56.8                                  x          57  56.8
z =                         = -0.46                   z =                          = 0.12
             12.3                                                   12.3
n           51                                         n            51

from Table A.5, Prob. for z = -0.46 is .1772

from Table A.5, Prob. for z = 0.12 is .0478

P(56 < x < 57) = .1772 + .0478 = .2250
Chapter 7: Sampling and Sampling Distributions 24
d) P( x < 55):

x        55  56.8
z =                         = -1.05
            12.3
n             51

from Table A.5, Prob.= .3531

P( x < 55) = .5000 - .3531 = .1469

e) P( x < 50):

x        50  56.8
z =                         = -3.95
            12.3
n              51

from Table A.5, Prob.= .5000

P( x < 50) = .5000 - .5000 = .0000
7.45 p = .73 n = 300

a) P(210 < x < 234):

x 210                                            x 234
ˆ
p1 =        = .70                              ˆ
p2 =         = .78
n 300                                            n 300

p p
ˆ              .70  .73
z =                                   = -1.17
pq           (. 73)(. 27 )
n                300

p p
ˆ              .78  .73
z =                                   = 1.95
pq           (. 73)(. 27 )
n                300

from Table A.5, the area for z = -1.17 is .3790
the area for z = 1.95 is .4744

P(210 < x < 234) = .3790 + .4744 = .8534

ˆ
b) P( p > .78):

p p
ˆ              .78  .73
z =                                   = 1.95
pq           (. 73)(. 27 )
n                300

from Table A.5, the area for z = 1.95 is .4744
Chapter 7: Sampling and Sampling Distributions 25
ˆ
P( p > .78) = .5000 - .4744 = .0256

c) p = .73             n = 800               ˆ
P( p > .78):

p p
ˆ              .78  .73
z =                                   = 3.19
pq        (. 73)(. 27 )
n             800

from Table A.5, the area for z = 3.19 is .4993

ˆ
P( p > .78) = .5000 - .4993 = .0007

7.46     n = 140          P(x > 35):

35
ˆ
p =       = .25               p = .22
140

p p
ˆ              .25  .22
z =                                   = 0.86
pq        (. 22 )(. 78 )
n             140

from Table A.5, the area for z = 0.86 is .3051

P(x > 35) = .5000 - .3051 = .1949

P(x < 21):

21
ˆ
p =       = .15
140

p p
ˆ              .15  .22
z =                                   = -2.00
pq        (. 22 )(. 78 )
n             140

from Table A.5, the area for z = 2.00 is .4772

P(x < 21) = .5000 - .4772 = .0228

n = 300               p = .20

ˆ
P(.18 < p < .25):

p p
ˆ              .18  .20
z =                                   = -0.87
pq        (. 20 )(. 80 )
n             300

from Table A.5, the area for z = -0.87 is .3078
Chapter 7: Sampling and Sampling Distributions 26

p p
ˆ              .25  .20
z =                                   = 2.17
pq         (. 20 )(. 80 )
n              300

from Table A.5, the area for z = 2.17 is .4850

ˆ
P(.18 < p < .25) = .3078 + .4850 = .7928

7.47   By taking a sample, there is potential for obtaining more detailed information.
More time can be spent with each employee. Probing questions can
be asked. There is more time for trust to be built between employee and
interviewer resulting in the potential for more honest, open answers.

With a census, data is usually more general and easier to analyze because it is in a more standard format.
Decision-makers are sometimes more comfortable with a census because everyone is included and there
is no sampling error. A census appears to be a better political device because the CEO can claim that
everyone in the company has had input.

7.48   p = .75         n = 150            x = 120

ˆ
P( p > .80):

p p
ˆ              .80  .75
z =                                   = 1.41
pq         (. 75 )(. 25 )
n              150

from Table A.5, the area for z = 1.41 is .4207

ˆ
P( p > .80) = .5000 - .4207 = .0793

7.49 a) Switzerland: n = 40               µ = \$ 30.67      =\$3

P(30 < x < 31):

x           31  30.67
z                            0.70
              3
n              40

x           30  30.67
z                            1.41
              3
n              40
from Table A.5, the area for z = 0.70 is .2580
the area for z = -1.41 is .4207
Chapter 7: Sampling and Sampling Distributions 27

P(30 < x < 31) = .2580 + .4207 = .6787

b) Japan: n = 35                  µ = \$ 20.20          = \$3

P( x > 21):

x            21  20.20
z                             1.58
                 3
n             35
from Table A.5, the area for z = 1.58 is .4429

P( x > 21) = .5000 - .4429 = .0571

c) U.S.: n = 50               µ = \$ 23.82        =\$3

P( x < 22.75):

x           22.75  23.82
z                                   2.52
                  3
n              50
from Table A.5, the area for z = -2.52 is .4941

P( x < 22.75) = .5000 - .4941 = .0059

7.50     a)        Age, Ethnicity, Religion, Geographic Region, Occupation, Urban-Suburban-Rural, Party
Affiliation, Gender
b)        Age, Ethnicity, Gender, Geographic Region, Economic Class
c)        Age, Ethnicity, Gender, Economic Class, Education
d)        Age, Ethnicity, Gender, Economic Class, Geographic Location

7.51     µ = \$281            n = 65         = \$47

P( x > \$273):

x               273  281
z =                              = -1.37
               47
n              65

from Table A.5 the area for z = -1.37 is .4147

P( x > \$273) = .5000 + .4147 = .9147

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