07 ch ken black solution

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. Determine when to use sampling instead of a census.
2. Distinguish between random and nonrandom sampling.
3. Decide when and how to use various sampling techniques.
4. Be aware of the different types of error that can occur in a study.
5. Understand the impact of the central limit theorem on statistical analysis.
6. Use the sampling distributions of x and pˆ .
CHAPTER TEACHING STRATEGY
Virtually every analysis discussed in this text deals with sample data. It is
important, therefore, that students are exposed to the ways and means that samples are
gathered. The first portion of chapter 7 deals with sampling. Reasons for sampling
versus taking a census are given. Most of these reasons are tied to the fact that taking a
census costs more than sampling if the same measurements are being gathered. Students
are then exposed to the idea of random versus nonrandom sampling. Random sampling
appeals to their concepts of fairness and equal opportunity. This text emphasizes that
nonrandom samples are nonprobability samples and cannot be used in inferential analysis
because levels of confidence and/or probability cannot be assigned. It should be
emphasized throughout the discussion of sampling techniques that as future business
managers (most students will end up as some sort of supervisor/manager) students should

be aware of where and how data are gathered for studies. This will help to assure that
they will not make poor decisions based on inaccurate and poorly gathered data.
The central limit theorem opens up opportunities to analyze data with a host of
techniques using the normal curve. Section 7.2 is presented by showing a population
(randomly generated and presented in histogram form) that is uniformly distributed and
one that is exponentially distributed. Histograms of the means for various random
samples of varying sizes are presented. Note that the distributions of means “pile up” in
the middle and begin to approximate the normal curve shape as sample size increases.
Note also by observing the values on the bottom axis that the dispersion of means gets
smaller and smaller as sample size increases thus underscoring the formula for the
standard error of the mean (σ/ n). As the student sees the central limit theorem unfold,
he/she begins to see that if the sample size is large enough sample means can be
analyzed using the normal curve regardless of the shape of the population.
Chapter 7 presents formulas derived from the central limit theorem for both
sample means and sample proportions. Taking the time to introduce these techniques in
this chapter can expedite the presentation of material in chapters 8 and 9.
CHAPTER OUTLINE
7.1 Sampling
Reasons for Sampling
Reasons for Taking a Census
Frame
Random Versus Nonrandom Sampling
Random Sampling Techniques
Simple Random Sampling
Stratified Random Sampling
Systematic Sampling
Cluster or Area Sampling
Nonrandom Sampling
Convenience Sampling
Judgment Sampling
Quota Sampling
Snowball Sampling
Sampling Error
Nonsampling Errors
7.2 Sampling Distribution of x
Sampling from a Finite Population
7.3 Sampling Distribution of pˆ

KEY TERMS
Central Limit Theorem Quota Sampling
Cluster (or Area) Sampling Random Sampling
Convenience Sampling Sample Proportion
Disproportionate Stratified Random Sampling Sampling Error
Finite Correction Factor Simple Random Sampling
Frame Snowball Sampling
Judgment Sampling Standard Error of the Mean
Nonrandom Sampling Standard Error of the Proportion
Nonrandom Sampling Techniques Stratified Random Sampling
Nonsampling Errors Systematic Sampling
Proportionate Stratified Random Sampling Two-Stage Sampling
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.
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 = 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.
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
attorney's 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, σ = 10, n = 64
a) Prob( x > 52):
z =
64
10
5052 −
=
−
n
x
σ
µ
= 1.6
from Table A.5 Prob. = .4452
Prob( x > 52) = .5000 - .4452 = .0548
b) Prob( x< 51):
z =
64
10
5051−
=
−
n
x
σ
µ
= 0.80
from Table A.5 prob. = .2881
Prob( x < 51) = .5000 + .2881 = .7881
c) Prob( x < 47):
z =
64
10
5047 −
=
−
n
x
σ
µ
= -2.40

Prob( x < 47) = .5000 - .4918 = .0082
d) Prob(48.5 < x < 52.4):
z =
64
10
505.48 −
=
−
n
x
σ
µ
= -1.20
z =
64
10
504.52 −
=
−
n
x
σ
µ
= 1.92
Prob(48.5 < x < 52.4) = .3849 + .4726 = .8575
e) Prob(50.6 < x < 51.3):
z =
64
10
506.50 −
=
−
n
x
σ
µ
= 0.48
from Table A.5, prob. = .1844
z =
64
10
503.51 −
=
−
n
x
σ
µ
Prob(50.6 < x < 51.3) = .3508 - .1844 = .1644
7.14 µ = 23.45 σ = 3.8
a) n = 10, Prob( x > 22):

z =
10
8.3
45.2322 −
=
−
n
x
σ
µ
= -1.21
Prob( x > 22) = .3869 + .5000 = .8869
b) n = 4, Prob( x > 26):
z =
4
8.3
45.2326 −
=
−
n
x
σ
µ
= 1.34
Prob( 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 σ :
z =
n
x
σ
µ−
1.08 =
36
278280
σ
−
1.08
6
σ
= 2
1.08σ = 12

σ =
08.1
12
= 11.11
7.16 n = 81 σ = 12 Prob( x > 300) = .18
.5000 - .1800 = .3200
from Table A.5, z.3200 = 0.92
Solving for µ:
z =
n
x
σ
µ−
0.92 =
81
12
300 µ−
0.92
9
12
= 300 - µ
1.2267 = 300 - µ
µ = 300 - 1.2267 = 298.77
7.17 a) N = 1,000 n = 60 µ = 75 σ = 6
Prob( x < 76.5):
z =
11000
601000
60
6
755.76
1 −
−
−
=
−
−
−
N
nN
n
x
σ
µ
= 2.00

Prob( x < 76.5) = .4772 + .5000 = .9772
b) N = 90 n = 36 µ = 108 σ = 3.46
Prob(107 < x < 107.7):
z =
190
3690
36
46.3
108107
1 −
−
−
=
−
−
−
N
nN
n
x
σ
µ
= -2.23
z =
190
3690
36
46.3
1087.107
1 −
−
−
=
−
−
−
N
nN
n
x
σ
µ
= -0.67
Prob(107 < x < 107.7) = .4871 - .2486 = .2385
c) N = 250 n = 100 µ = 35.6 σ = 4.89
Prob( x > 36):
z =
1250
100250
100
89.4
6.3536
1 −
−
−
=
−
−
−
N
nN
n
x
σ
µ
= 1.05
Prob( x > 36) = .5000 - .3531 = .1469
d) N = 5000 n = 60 µ = 125 σ = 13.4
Prob( x < 123):

z =
15000
605000
60
4.13
125123
1 −
−
−
=
−
−
−
N
nN
n
x
σ
µ
= -1.16
Prob( x < 123) = .5000 - .3770 = .1230
7.18 µ = 99.9 σ = 30 n = 38
a) Prob( x < 90):
z =
38
30
9.9990
1
−
=
−
−
−
N
nN
n
x
σ
µ
= -2. 03
from table A.5, area = .4788
Prob( x < 90) = .5000 - .4788 = .0212
b) Prob(98 < x < 105):
z =
38
30
9.99105
1
−
=
−
−
−
N
nN
n
x
σ
µ
= 1.05
z =
38
30
9.9998
1
−
=
−
−
−
N
nN
n
x
σ
µ
= -0.39
Prob(98 < x < 105) = .3531 + .1517 = .5048
c) Prob( x < 112):

z =
38
30
9.99112
1
−
=
−
−
−
N
nN
n
x
σ
µ
= 2.49
Prob( x < 112) = .5000 - .4936 = .0064
d) Prob(93 < x < 96):
z =
38
30
9.9993
1
−
=
−
−
−
N
nN
n
x
σ
µ
= -1.42
z =
38
30
9.9996
1
−
=
−
−
−
N
nN
n
x
σ
µ
= -0.80
Prob(93 < x < 96) = .4222 - .2881 = .1341
7.19 N = 1500 n = 100 µ = 177,000 σ = 8,500
Prob( X > $185,000):
z =
11500
1001500
100
500,8
000,177000,185
1 −
−
−
=
−
−
−
N
nN
n
X
σ
µ
= 9.74
Prob( X > $185,000) = .5000 - .5000 = .0000

7.20 µ = $65.12 σ = $21.45 n = 45
Prob( x > 0x ) = .2300
Prob. x lies between 0x and µ = .5000 - .2300 = .2700
from Table A.5, z.2700 = 0.74
Solving for 0x :
z =
n
x
σ
µ−0
0.74 =
45
45.21
12.650 −x
2.366 = 0x - 65.12
0x = 65.12 + 2.366 = 67.486
7.21 µ = 50.4 σ = 11.8 n = 42
a) Prob( x > 52):
z =
42
8.11
4.5052 −
=
−
n
x
σ
µ
= 0.88
from Table A.5, the area for z = 0.88 is .3106
Prob( x > 52) = .5000 - .3106 = .1894
b) Prob( x < 47.5):

z =
42
8.11
4.505.47 −
=
−
n
x
σ
µ
= -1.59
from Table A.5, the area for z = -1.59 is .4441
Prob( x < 47.5) = .5000 - .4441 = .0559
c) Prob( x < 40):
z =
42
8.11
4.5040 −
=
−
n
x
σ
µ
= -5.71
Prob( 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
z =
n
x
σ
µ−
-0.55 =
42
4.5049
σ
−
σ = 16.4964
7.22 P = .25
a) n = 110 Prob( pˆ < .21):

z =
110
)75)(.25(.
25.21.ˆ −
=
⋅
−
n
QP
Pp
= -0.97
Prob( pˆ < .21) = .5000 - .3340 = .1660
b) n = 33 Prob( pˆ > .24):
z =
33
)75)(.25(.
25.24.ˆ −
=
⋅
−
n
QP
Pp
= -0.13
Prob( pˆ > .24) = .5000 + .0517 = .5517
c) n = 59 Prob(.24 < pˆ < .27):
z =
59
)75)(.25(.
25.24.ˆ −
=
⋅
−
n
QP
Pp
= -0.18
z =
59
)75)(.25(.
25.27.ˆ −
=
⋅
−
n
QP
Pp
= 0.35
Prob(.24 < pˆ < .27) = .0714 + .1368 = .2082
d) n = 80 Prob( pˆ > .30):
z =
80
)75)(.25(.
25.30.ˆ −
=
⋅
−
n
QP
Pp
= 1.03

Prob( pˆ > .30) = .5000 - .3485 = .1515
e) n = 800 Prob( pˆ > .30):
z =
800
)75)(.25(.
25.30.ˆ −
=
⋅
−
n
QP
Pp
= 3.27
Prob( pˆ > .30) = .5000 - .4995 = .0005
7.23 P = .58 n = 660
a) Prob( pˆ > .60):
z =
660
)42)(.58(.
58.60.ˆ −
=
⋅
−
n
QP
Pp
= 1.04
Prob( pˆ > .60) = .5000 - .3508 = .1492
b) Prob(.55 < pˆ < .65):
z =
660
)42)(.58(.
58.65.ˆ −
=
⋅
−
n
QP
Pp
= 3.64
z =
660
)42)(.58(.
58.55.ˆ −
=
⋅
−
n
QP
Pp
= 1.56
Prob(.55 < pˆ < .65) = .4998 + .4406 = .9404

c) Prob( pˆ > .57):
z =
660
)42)(.58(.
58.57.ˆ −
=
⋅
−
n
QP
Pp
= 0.52
d) Prob(.53 < pˆ < .56):
z =
660
)42)(.58(.
58.56.ˆ −
=
⋅
−
n
QP
Pp
= 1.04
z =
660
)42)(.58(.
58.53.ˆ −
=
⋅
−
n
QP
Pp
= 2.60
Prob(.53 < pˆ < .56) = .4953 - .3508 = .1445
e) Prob( pˆ < .48):
z =
660
)42)(.58(.
58.48.ˆ −
=
⋅
−
n
QP
Pp
= 5.21
Prob( pˆ < .48) = .5000 - .5000 = .0000
7.24 P = .40 Prob.( pˆ > .35) = .8000
Prob(.35 < pˆ < .40) = .8000 - .5000 = .3000
from Table A.5, z.3000 = -0.84
Solving for n:

z =
n
QP
Pp
⋅
−ˆ
-0.84 =
n
)60)(.40(.
40.35. −
-0.84 =
n
24.
05.−
n=
−
−
05.
24.84.0
8.23 = n
n = 67.73 ≈≈≈≈ 68
7.25 P = .28 n = 140 Prob( pˆ < 0
ˆp ) = .3000
Prob( pˆ < 0
ˆp < .28) = .5000 - .3000 = .2000
from Table A.5, z.2000 = -0.52
Solving for 0
ˆp :
z =
n
QP
Pp
⋅
−0
ˆ
-0.52 =
140
)72)(.28(.
28.ˆ0 −p
-.02 = 0
ˆp - .28
0
ˆp = .28 - .02 = .26
7.26 Prob(x > 150): n = 600 P = .21 x = 150

pˆ =
600
150
= .25
z =
600
)79)(.21(.
21.25.ˆ −
=
⋅
−
n
QP
Pp
= 2.41
Prob(x > 150) = .5000 - .4920 = .0080
7.27 P = .48 n = 200
a) Prob(x < 90):
pˆ =
200
90
= .45
z =
200
)52)(.48(.
48.45.ˆ −
=
⋅
−
n
QP
Pp
= -0.85
Prob(x < 90) = .5000 - .3023 = .1977
b) Prob(x > 100):
pˆ =
200
100
= .50
z =
200
)52)(.48(.
48.50.ˆ −
=
⋅
−
n
QP
Pp
= 0.57
Prob(x > 100) = .5000 - .2157 = .2843
c) Prob(x > 80):

pˆ =
200
80
= .40
z =
200
)52)(.48(.
48.40.ˆ −
=
⋅
−
n
QP
Pp
= -2.26
Prob(x > 80) = .5000 + .4881 = .9881
7.28 P = .19 n = 950
a) Prob( pˆ > .25):
z =
950
)89)(.19(.
19.25.ˆ −
=
⋅
−
n
QP
Pp
= 4.71
from Table A.5, area = .5000
Prob( pˆ > .25) = .5000 - .5000 = .0000
b) Prob(.15 < pˆ < .20):
z =
950
)81)(.19(.
19.15.ˆ −
=
⋅
−
n
QP
Pp
= -3.14
z =
950
)89)(.19(.
19.20.ˆ −
=
⋅
−
n
QP
Pp
= 0.79
from Table A.5, area for z = -3.14 is .4992
from Table A.5, area for z = 0.79 is .2852
Prob(.15 < pˆ < .20) = .4992 + .2852 = .7844
c) Prob(133 < x < 171):
1
ˆp =
950
133
= .14 2
ˆp =
950
171
= .18

Prob(.14 < pˆ < .18):
z =
950
)81)(.19(.
19.14.ˆ −
=
⋅
−
n
QP
Pp
= -3.93
z =
950
)81)(.19(.
19.18.ˆ −
=
⋅
−
n
QP
Pp
= -0.79
the area for z = -0.79 is .2852
P(133 < x < 171) = .49997 - .2852 = .21477
7.29 µ = 76, σ = 14
a) n = 35, Prob( x > 79):
z =
35
14
7679 −
=
−
n
x
σ
µ
= 1.27
Prob( x > 79) = .5000 - .3980 = .1020
b) n = 140, Prob(74 < x < 77):
z =
140
14
7674 −
=
−
n
x
σ
µ
= -1.69
z =
140
14
7677 −
=
−
n
x
σ
µ
= 0.85

P(74 < x < 77) = .4545 + .3023 = .7568
c) n = 219, Prob( x < 76.5):
z =
219
14
765.76 −
=
−
n
x
σ
µ
= 0.53
Prob( x < 76.5) = .5000 - .2019 = .2981
7.30 P = .46
a) n = 60
Prob(.41 < pˆ < .53):
z =
60
)54)(.46(.
46.53.ˆ −
=
⋅
−
n
QP
Pp
= 1.09
z =
60
)54)(.46(.
46.41.ˆ −
=
⋅
−
n
QP
Pp
= 0.78
Prob(.41 < pˆ < .53) = .3621 + .2823 = .6444
b) n = 458 Prob( pˆ < .40):
z =
458
)54)(.46(.
46.40.ˆ −
=
⋅
−
n
QP
Pp

Prob( pˆ < .40) = .5000 - .4951 = .0049
c) n = 1350 Prob( pˆ > .49):
z =
1350
)54)(.46(.
46.49.ˆ −
=
⋅
−
n
QP
Pp
= 2.21
Prob( pˆ > .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
pˆ =
600
298
=
n
x
= .497
Prob( pˆ < .497):
z =
600
)45)(.55(.
55.497.ˆ −
=
⋅
−
n
QP
Pp
= -2.61
from Table A.5, Prob. = .4955
Prob( pˆ < .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.
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 σ = $ 650 n = 30 N = 120
Prob( x < 17,500):
z =
1120
30120
30
650
755,17500,17
−
−
−
= -2.47
Prob( 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 σ2
= $525
Prob( x > $110):
z =
32
525
125110 −
=
−
n
x
σ
µ
= -3.70
from Table A.5, Prob.= .5000
Prob( x > $110) = .5000 + .5000 = 1.0000
Prob( x > $135):
z =
32
525
125135 −
=
−
n
x
σ
µ
= 2.47

Prob( x > $135) = .5000 - .4932 = .0068
Prob($120 < x < $130):
z =
32
525
125120 −
=
−
n
x
σ
µ
= -1.23
z =
32
525
125130 −
=
−
n
x
σ
µ
= 1.23
Prob($120 < x < $130) = .3907 + .3907 = .7814
7.37 n = 1100
a) x > 810, P = .73
pˆ =
1100
810
=
n
x
z =
1100
)27)(.73(.
73.7364.ˆ −
=
⋅
−
n
QP
Pp
= 0.48
Prob(x > 810) = .5000 - .1844 = .3156
b) x < 1030, P = .96,
pˆ =
1100
1030
=
n
x
= .9364

z =
1100
)04)(.96(.
96.9364.ˆ −
=
⋅
−
n
QP
Pp
= -3.99
Prob(x < 1030) = .5000 - .49997 = .00003
c) P = .85
Prob(.82 < pˆ < .84):
z =
1100
)15)(.85(.
85.82.ˆ −
=
⋅
−
n
QP
Pp
= -2.79
z =
1100
)15)(.85(.
85.84.ˆ −
=
⋅
−
n
QP
Pp
= -0.93
Prob(.82 < pˆ < .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. Randomly sample from each of the selected
area distribution centers and retail outlets.
7.40 N=12,080 n=300
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) Prob(x > 339):
pˆ =
565
339
=
n
x
= .60
z =
565
)46)(.54(.
54.60.ˆ −
=
⋅
−
n
QP
Pp
= 2.86
Prob(x > 339) = .5000 - .4979 = .0021

b) Prob(x > 288):
pˆ =
565
288
=
n
x
= .5097
z =
565
)46)(.54(.
54.5097.ˆ −
=
⋅
−
n
QP
Pp
= -1.45
Prob(x > 288) = .5000 + .4265 = .9265
c) Prob( pˆ < .50):
z =
565
)46)(.54(.
54.50.ˆ −
=
⋅
−
n
QP
Pp
= -1.91
Prob( pˆ < .50) = .5000 - .4719 = .0281
7.42 µ = $550 n = 50 σ = $100
Prob( x < $530):
z =
50
100
550530 −
=
−
n
x
σ
µ
= -1.41
from Table A.5, Prob.=.4207
Prob(x < $530) = .5000 - .4207 = .0793
7.43 µ = 56.8 n = 51 σ = 12.3
a) Prob( x > 60):

z =
51
3.12
8.5660 −
=
−
n
x
σ
µ
= 1.86
from Table A.5, Prob. = .4686
Prob( x > 60) = .5000 - .4686 = .0314
b) Prob( x > 58):
z =
51
3.12
8.5658 −
=
−
n
x
σ
µ
= 0.70
Prob( x > 58) = .5000 - .2580 = .2420
c) Prob(56 < x < 57):
z =
51
3.12
8.5656 −
=
−
n
x
σ
µ
= -0.46
z =
51
3.12
8.5657 −
=
−
n
x
σ
µ
= 0.12
Prob(56 < x < 57) = .1772 + .0478 = .2250
d) Prob( x < 55):
z =
51
3.12
8.5655 −
=
−
n
x
σ
µ
= -1.05

Prob( x < 55) = .5000 - .3531 = .1469
e) Prob( x < 50):
z =
51
3.12
8.5650 −
=
−
n
x
σ
µ
= -3.95
Prob( x < 50) = .5000 - .5000 = .0000
7.45 P = .73 n = 300
a) Prob(210 < x < 234):
1
ˆp =
300
210
=
n
x
= .70 2
ˆp =
300
234
=
n
x
= .78
z =
300
)27)(.73(.
73.70.ˆ −
=
−
n
PQ
Pp
= -1.17
z =
300
)27)(.73(.
73.78.ˆ −
=
−
n
PQ
Pp
= 1.95
the area for z = 1.95 is .4744
Prob(210 < x < 234) = .3790 + .4744 = .8534
b) Prob( pˆ > .78):
z =
300
)27)(.73(.
73.78.ˆ −
=
−
n
PQ
Pp
= 1.95

Prob( pˆ > .78) = .5000 - .4744 = .0256
c) P = .73 n = 800 Prob( pˆ > .78):
z =
800
)27)(.73(.
73.78.ˆ −
=
−
n
PQ
Pp
= 3.19
Prob( pˆ > .78) = .5000 - .4993 = .0007
7.46 n = 140 Prob(x > 35):
pˆ =
140
35
= .25 P = .22
z =
140
)78)(.22(.
22.25.ˆ −
=
−
n
PQ
Pp
= 0.86
Prob(x > 35) = .5000 - .3051 = .1949
Prob(x < 21):
pˆ =
140
21
= .15
z =
140
)78)(.22(.
22.15.ˆ −
=
−
n
PQ
Pp
= 2.00
Prob(x < 21) = .5000 - .4772 = .0228
n = 300 P = .20
Prob(.18 < pˆ < .25):

z =
300
)80)(.20(.
20.18.ˆ −
=
−
n
PQ
Pp
= -0.87
z =
300
)80)(.20(.
20.25.ˆ −
=
−
n
PQ
Pp
= 2.17
Prob(.18 < pˆ < .25) = .3078 + .4850 = .7928
7.47 By taking a sample, there is potential for more detailed information to be
obtained. 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
Prob( pˆ > .80):
z =
150
)25)(.75(.
75.80.ˆ −
=
−
n
PQ
Pp
= 1.41
Prob( pˆ > .80) = .5000 - .4207 = .0793
7.49 Switzerland: n = 40 µ = $ 21.24 σ = $ 3
Prob(21 < x < 22):

z =
40
3
24.2121−
=
−
n
x
σ
µ
= -0.51
z =
40
3
24.2122 −
=
−
n
x
σ
µ
= 1.60
the area for z = 1.60 is .4452
Prob(21 < x < 22) = .1950 + .4452 = .6402
Japan: n = 35 µ = $ 22.00 σ = $3
Prob( x > 23):
z =
35
3
2223 −
=
−
n
x
σ
µ
= 2.11
P( x > 23) = .5000 - .4826 = .0174
U.S.: n = 50 µ = $ 19.86 σ = $ 3
Prob( X < 18.90):
z =
50
3
86.1990.18 −
=
−
n
x
σ
µ
= -2.02
Prob( X < 18.90) = .5000 - .4783 = .0217
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 σ = $47
P( x > $273):
z =
65
47
281273 −
=
−
n
x
σ
µ
= -1.37
from Table A.5 the area for z = -1.37 is .4147
Prob.( x > $273) = .5000 + .4147 = .9147

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