Chapter 7:
Sampling and Sampling Distributions
7.1
a. Probability distribution for one die:
Die outcome
1
2
3
4
5
6
Probability
1/6
1/6
1/6
1/6
1/6
1/6
b. Sampling distribution of the sample means from rolling a pair of dice:
x
Total
Sample
Prob. of x
2
11
1
1/36
3
1 2, 2 1
1.5
2/36
4
1 3, 3 1, 2 2
2
3/36
5
1 4, 4 1, 2 3, 3 2
2.5
4/36
6
1 5, 5 1, 2 4, 4 2, 3 3
3
5/36
7
1 6, 6 1, 2 5, 5 2, 3 4, 4 3
3.5
6/36
8
2 6, 6 2, 3 5, 5 3, 4 4
4
5/36
9
3 6, 6 3, 4 5, 5 4
4.5
4/36
10
4 6, 6 4, 5 5
5
3/36
11
5 6, 6 5
5.5
2/36
12
66
6
1/36
7.2
a. Binomial random variable with n = 2, p = .5
Probability Density Function
Binomial with n = 2 and p = 0.5
x P( X = x )
0
0.25
1
0.50
2
0.25
b. Binomial random variable with n = 4, p = .5
Probability Density Function
Binomial with n = 4 and p = 0.5
x P( X = x )
0
0.0625
1
0.2500
2
0.3750
3
0.2500
4
0.0625
Chapter 7: Sampling and Sampling Distributions
c. Binomial random variable with n = 10, p = .5
Probability Density Function
Binomial with n = 10 and p = 0.5
x P( X = x )
0
0.000977
1
0.009766
2
0.043945
3
0.117188
4
0.205078
5
0.246094
6
0.205078
7
0.117188
8
0.043945
9
0.009766
10
0.000977
7.3 The sampling distribution of the sample mean can be generated by listing out
all possible samples of size n, calculate each possible x , determine the
probability of each possible x and generate the sampling distribution.
Alternatively, the probabilities of each x can be generated by use of the
binomial formula.
a. When n = 5: Use the binomial formula for x = 0, x = 1, etc.:
x
X
P(X)
0
.07776
0
1
.25920
.2
2
.34560
.4
3
.23040
.6
4
.07680
.8
5
.01024
1.0
p (1  p )
.4(.6)
2
E ( px ) = np = (5)(.4) = 2.0,  p 
=
= .048,  p  .2191
n
5
b. Using the result from part a:
E ( px ) = np = (100)(.4) = 40,  p 
2
 p  .04899
p (1  p )
.4(.6)
= 
n
100
= .0024,
7.4 The response should note that there will be errors in taking a census of the
entire population as well as errors in taking a sample. Improved accuracy can
be achieved via sampling methods versus taking a complete census (see
reference to Hogan, 90). By using sample information, we can make valid
inferences about the entire population without the time and expense involved
in taking a census.
7.5
a. mean and variance of the sampling distribution for the sample mean
 x    100
 2 x   n  81 25  3.24  x   x 2  3.24
2
145
146
Statistics for Business & Economics, 6th edition
102  100
= 1.11 1 – Fz(1.11) = .1335
3.24
98  100
c. Probability that 98  x  101 z x 
= -1.11 Fz = 1 – Fz(1.11)
3.24
= .1335
101  100
= .56
Fz(.56) – [1-Fx(1.11)] = .7123 - .1335 = .5788
zx 
3.24
101.5  100
d. Probability that x  101.5 z x 
= .83 Fz = .7967
3.24
b. Probability that x  102 z x 
7.6
a. mean and variance of the sampling distribution for the sample mean
 x    100
 2 x   n  900 30  30  x   x 2  30
109  100
b. Probability that x  109 z x 
= 1.64 1 – Fz(1.64) = .0505
30
96  100
c. Probability that 96  x  110 z x 
= -.73 1 – Fz(.73) = .2327
30
110  100
= 1.83 Fz = .9664. .9664 - .2327 = .7337
zx 
30
107  100
d. Probability that x  107 z x 
= 1.28 Fz = .8997
30
2
7.7
a. mean and variance of the sampling distribution for the sample mean
x    200
 2 x   n  625 25  25  x   x 2  25
209  200
b. Probability that x  209 z x 
= 1.80 1 – Fz(1.80) = .0359
25
198  200
c. Probability that 198  x  211 z x 
= -.40 1 – Fz(.40) =
25
.3446
211  200
zx 
= 2.20 Fz(2.20) = .9861. .9861 - .3446 = .6415
25
202  200
d. Probability that x  202 z x 
= .40 Fz = .6554
25
2
7.8
a. mean and variance of the sampling distribution for the sample mean
x    400
2
 2 x   n  1600 35  45.7143  x   x 2  45.7143
Chapter 7: Sampling and Sampling Distributions
412  400
= 1.77 1 – Fz(1.77) = .0384
45.7143
407  400
c. Probability that 393  x  407 z x 
= 1.04 Fz(1.04) =
45.7143
.8508
393  400
= -1.04 1 – Fz(1.04) = .1492. .8508 - .1492 =
zx 
45.7143
.7016
389  400
d. Probability that x  389 z x 
= -1.63 1-Fz(1.63) = 1-.9484
45.7143
= .0516
b. Probability that x  412 z x 
7.9
a. E( X ) =  x = 92.
b.  x =
2
2
n
=
 3.6 
c.  x =

n
2
= 3.24
4
=
3.6
= 1.8
2
93  92
) = P(Z > .56) =
1.8
d. P(Z >
.2877
7.10
a. E( X ) =  x = 1,200
b.  x =
2
2
n
=
 400 
9
c.  x =
2
= 17,778
d. P(Z<

n
=
400
= 133.33
3
1, 050  1, 200
) = P(Z<-1.13)
133.33
=.1292
7.11 a. i) P(Z >
24  25
) = P(Z < -.5) = .3085.
2
ii) P(Z <
24  25
) = P(Z < -1)
2 4
=.1587
24  25
iii) P(Z <
) = P(Z < -2) = .0228
2 16
b. As the sample size increases, the standard error of the sampling
distribution will decrease. That is, as the sample size increases, the
sampling distribution of the sample means will clump up tighter
around the true population mean. The graph would show a tighter
distribution with less area in the tails.
7.12
110  115
) = P(Z < -2) = .9772
25 100
113  115
117  115
b. P(
<Z<
) = P(-.8 < Z < .8) = .5762
25 100
25 100
a. P(Z >
147
148
Statistics for Business & Economics, 6th edition
114  115
116  115
<Z<
) = P(-.4 < Z < .4) = .3108
25 100
25 100
d. $114,000 - $116,000
e. Even with non-normal populations, the sampling distribution of the
sample means will be normal for sufficient sample n. Since n is  30,
the sampling distribution of the sample means can assumed to be a
normal distribution.
c. P(
7.13
7.14
60
= 20
9
270  280
b. P(Z <
) = P(Z < -.5) = 1 - .6915 = .3085
20
250  280
c. P(Z >
) = P(Z > -1.5) = .9332
20
d. If the population standard deviation is smaller, then the standard error
of the sampling distribution of the means will also be smaller. Since Z
is higher, tail areas are smaller and the probabilities calculated for parts
a and b will both be smaller.
a.  x =
b.
c.
d.
e.
7.15
22
= 5.5
16
100  87
P(Z <
) = P(Z < 2.36) = .9909
5.5
80  87
P(Z >
) = P(Z > -1.27) = .8980
5.5
85  87
95  87
P(
>Z>
) = P(-.36 > Z > 1.45) = .4329
5.5
5.5
Higher, higher, lower. The graph will show that the standard error of
the sample means will decrease with an increased sample size.
a.  x =
.6
= .3
4
19.7  20
b. P(Z <
) = P(Z < -1) = .1587
.3
20.6  20
c. P(Z >
) = P(Z > 2) = .0228
.3
19.5  20
20.5  20
d. P(
<Z<
) = P(-1.67 < Z < 1.67) = .905
.3
.3
19.5  20
20.5  20
e. P(
<Z<
) = P(-1.18 < Z < 1.18) = .762
.6 2
.6 2
a.  x =
Chapter 7: Sampling and Sampling Distributions
40
=4
100
b. P(Z > 5/4) = P(Z > 1.25) = .1056
c. P(Z < -4/4) = P(Z < -1) = .1587
d. P(-3/4 > Z > 3/4) = P(-.75 > Z > .75) = .4532
7.16
a.  x =
7.17
a.  x =
7.18
a.  x =
8
= 4. P(Z > 2/4) = P(Z > .5) = .3085
4
b. P(Z < -3/4) = P(Z < -.75) = .2266
c. P(-4/4 > Z > 4/4) = P(-1 > Z > 1) = .3174
d. Lower, lower, lower
Difference
1.6
= .16, P(Z>1.645) =.05, 1.645 =
, Difference =
.16
100
±.2632
Difference
, Difference = -.2048
.16
Difference
1.44 =
, Difference = .2304
.16
b. P(Z < -1.28) = .1, -1.28 =
c. P(Z > 1.44) = .075,
1
, n = 39.075, take n = 40
3.8 n
c. larger
7.19 a. P(Z > 1.645) = .10, 1.645 =
b. larger
7.20
a. P(Z > 1.96) = .025, 1.96 =
2
, n = 67.766, take n = 68
8.4 n
b. smaller
c. larger
1
 xi  Nx 2  205  (6)(5.5)2  23.5  47
  ( X i  X )2 
N
N
6
6
12
2
7.21
a.  x
2
b.
(4.5  5.5) 2 2(4.75  5.5) 2 2(5  5.5) 2



15
15
15
2(5.25  5.5)2 1(5.5  5.5) 2 3(5.75  5.5) 2 (6  5.5) 2





15
15
15
15
2(6.25  5.5) 2 (6.75  5.5) 2 47



15
15
120
2
 N  n 47 12 6  4 47


c.  x 2 
n N 1
4 6  1 120
 x 2   ( X i   )2 Px ( x ) 
149
150
Statistics for Business & Economics, 6th edition
0
19
20
N = 40, correction factor =
39
80
N = 100, correction factor =
99
980
N = 1,000, correction factor =
999
9,980
N = 10,000, correction factor =
9,999
b. When the population size (N) equals the sample size (n), then there is no
variation away from the population mean and the standard error will be
zero. As the sample size becomes relatively small compared to the
population size, the correction factor tends towards 1 and the correction
factor becomes less significant in the calculation of the standard error
c. The correction factor tends toward a value of 1 and becomes
progressively less important as a modifying factor when the sample size
decreases relative to the population size
7.22 a. N = 20, correction factor =
7.23
7.24
300, 000 400
= 26,859,689
499
100
825, 000  800, 000
b. P(Z >
)= P(Z > .93) = .1762
26,859.689
780, 000  800, 000
c. P(Z >
)= P(Z > -.74) = .7704
26,859.689
790, 000  800, 000
820, 000  800, 000
d. P(
<Z<
)= P(-.37 < Z < .74) =
26,859.689
26,859.689
.4147
a.  x 
200
= 3.8023
249
2.5
a. P(Z >
)= P(Z > .66) = .2546
3.8023
5
b. P(Z <
)= P(Z < -1.31) = .0951
3.8023
10
10
c. P(
<Z<
)= P(-2.63 < Z < 2.63) = 1 - .9914 = .0086
3.8023
3.8023
x 
30
50
Chapter 7: Sampling and Sampling Distributions
7.25
a.
b.
c.
d.
7.26
b.
c.
d.
7.28
(.4)(.6)
= .04899
100
Probability that the sample proportion is greater than .45
.45  .4
z
= P(Z >1.02) = .1539
.04899
Probability that the sample proportion is less than .29
.29  .4
z
= P(Z< -2.25) = .0122
.04899
Probability that the sample proportion is between .35 and .51
.35  .4
.51  .4
P(
<Z<
) = P(-1.02 < Z < 2.25) = .8339
.04899
.04899
E ( pˆ ) = .4  pˆ 
a.
7.27
10
450
= .7077
150 599
31  32
P(Z >
)= P(Z > -1.41) = .9207
.7077
33  32
P(Z <
)= P(Z < 1.41) = .9207
.7077
Normal probability graph. Due to the property of symmetry, the area
in the tails of the normal probability distribution are the same.
31  32
33  32
P(Z <
) or P(Z >
) = P(Z < -1.41) or P(Z > 1.41) =
.7077
.7077
.1586
x 
(.25)(.75)
= .0306186
200
a. Probability that the sample proportion is greater than .31
.31  .25
z
= P(Z > 1.96) = .0250
.0306186
b. Probability that the sample proportion is less than .14
.14  .25
z
= P(Z < -3.59) = .0002
.0306186
c. Probability that the sample proportion is between .24 and .40
.24  .25
.4  .25
P(
<Z<
) = P(-.33 < Z < 4.90) = .6293
.0306186
.0306186
E ( pˆ ) = .25  pˆ 
(.6)(.4)
= .04899
100
a. Probability that the sample proportion is greater than .66
.66  .6
z
= P(Z > 1.22) = .1112
.04899
E ( pˆ ) = .60  pˆ 
151
152
Statistics for Business & Economics, 6th edition
b. Probability that the sample proportion is less than .48
.48  .6
z
= P(Z < -2.45) = .0071
.04899
c. Probability that the sample proportion is between .52 and .66
.52  .6
.66  .6
P( z 
<Z< z
) = P(-1.63 < Z < 1.22) = .8372
.04899
.04899
7.29
7.30
(.5)(.5)
= .01667
900
a. Probability that the sample proportion is greater than .52
.52  .5
z
= P(Z > 1.20) = .1152
.01667
b. Probability that the sample proportion is less than .46
.46  .5
z
= P(Z < -2.40) = .0082
.01667
c. Probability that the sample proportion is between .47 and .53
.47  .5
.53  .5
P( z 
<Z< z
) = P(-1.80 < Z < 1.80) = .9282
.01667
.01667
E ( pˆ ) = .50  pˆ 
a. E ( pˆ ) = .424
(.424)(.576)
b.  pˆ 2 
= .00244
100
c.  pˆ  .0494
d. P(Z >
.5  .424
)= P(Z > 1.54) = .0618
.0494
7.31 a. E ( pˆ ) = .75
(.75)(.25)
b.  pˆ 2 
= .001875
100
c.  pˆ  .0433
d. P(Z >
7.32
.8  .75
)= P(Z > 1.15) = .1251
.0433
a. E ( pˆ ) = .20
(.2)(.8)
b.  pˆ 2 
= .000889
180
c.  pˆ  .0298
d. P(Z <
.15  .2
)= P(Z < -1.68) = .0465
.0298
Chapter 7: Sampling and Sampling Distributions
7.33
7.34
(.3)(.7)
= .0324
200
.25  .3
b. P(Z <
)= P(Z < -1.54) = .0618
.0324
.33  .3
c. P(Z >
)= P(Z > .93) = .1762
.0324
.27  .3
.33  .3
d. P(
<Z<
)= P(-.93 < Z < .93) = .6476
.0324
.0324
a.  pˆ 
(.4)(.6)
 .0447
120
.35  .4
.45  .4
P(
<Z<
)= P(-1.12 < Z < 1.12) = .7372
.0447
.0447
 pˆ 
(.42)(.58)
= .0285
300
.5  .42
b. P(Z >
)= P(Z > 2.81) = .0025
.0285
.4  .42
.45  .42
c. P(
<Z<
)= P(-.7 < Z < 1.05) = .6111
.0285
.0285
d. .41 - .43
7.35
a.  pˆ 
7.36
a.  pˆ 
7.37
a.  pˆ 
(.2)(.8)
= .0351
130
.15  .2
b. P(Z >
) = P(Z > -1.42) = .9222
.0351
.18  .2
.22  .2
c. P(
<Z<
)= P(-.57 < Z < .57) = .4314
.0351
.0351
d. Higher, higher
(.3)(.7)
= .02739
280
.32  .3
b. P(Z <
)= P(Z < .73) = .7673
.02739
c. .29 - .31
153
154
Statistics for Business & Economics, 6th edition
7.38
The largest value for  p̂ is when p = .5. In this case,  pˆ 
7.39
P(Z > 1.96) = .025, .03 = 1.96
(.5)(.5)
= .05
100
(.5)(.5)
, solving for n = 1067.11. Take a
n
sample of size 1,068.
(.25)(.75)
= .0395
120
Difference
b. P(Z > 1.28), 1.28 =
, Difference = .0506
.0395
Difference
c. P(Z < -1.645), -1.645 =
, Difference = .065
.0395
Difference
d. P(Z > 1.036), 1.036 =
, Difference = .0409
.0395
7.40
a.  pˆ 
7.41
 pˆ 
.56  .5
(.5)(.5)
= .04082, P(Z >
)= P(Z > 1.47) = .0708
.04082
150
7.42
 pˆ 
.58  .5
(.5)(.5)
= .03162, P(Z >
)= P(Z > 2.53) = .0057
.03162
250
7.43
7.44
(.55)(.45) 419
= .05065
81
499
.5  .55
P(Z <
)= P(Z < -.99) = .1611
.05065
 pˆ 
211
= .3996
528
(.3996)(.6004) 408
= .03934
 pˆ 
120
527
.33  .3996
b. P(Z <
)= P(Z < -1.77) = .0384
.03934
.5  .3996
.6  .3996
c. P(
<Z<
)= P(2.55 < Z < 5.09) = .5000 - .4946 =
.03934
.03934
.0054
a. p̂ =
Chapter 7: Sampling and Sampling Distributions
239
(.5457)(.4543) 358
= .5457,  pˆ 
= .05038
438
80
437
.5  .5457
b. P(Z <
)= P(Z < -.91) = .1814
.05038
.5  .5457
.6  .5457
c. P(
<Z<
)= P(-.91 < Z < 1.08) = .6785
.05038
.05038
7.45
a. p̂ =
7.46
P(Z <
7.47
a. Find the probability that the sample mean is > 101.
101  100
Probability that x  101 z x 
= .80 1 – Fz(.80) = .2119
5
16
b. Find the probability that the sample variance is > 45
(n  1) s 2 15(45)
P(

)  P(  2(15)  27) = between .05 and .025
2

25
c. Find the probability that the sample variance is > 60
(n  1) s 2 15(60)
P(

)  P(  2(15)  36) = Less than .005
2
25
7.48
a. Probability that the sample mean is > 200.
200  198
Probability that x  200 z x 
= 1.00 1 – Fz(1.00) = .1587
10
25
b. 5% of the sample variances would be less than this value
 (n  1) s 2  2
24s 2
2
 13.85 s 2  57.702
P( s  k )  P 
 24,.95  13.85

2
100
 

c. 5% of the samples variances would be greater than this value
 (n  1) s 2  2
24s 2
2
 36.42 s 2  151.879
P( s  k )  P 
 24,.05  36.42

2
100
 

7.49
a. Probability that the sample mean is > 50
50  46
Probability that x  50 z x 
= 2.40 1 – Fz(2.40) = .0082
7.07107
18
b. 5% of the sample variances would be less than this value
 (n  1) s 2  2
17 s 2
 8.67 s 2  25.50
P( s 2  k )  P 


8.67
17,.95

2
50
 

.1  .122
(.2709)(.7291)
= P(Z < -.61) = .2709,  pˆ 
= .04969
.036
81
.5  .2709
P(Z >
) = P(Z > 4.61)  .0000
.04969
155
156
Statistics for Business & Economics, 6th edition
c. 5% of the samples variances would be greater than this value
 (n  1) s 2  2
17 s 2
2
 27.59 s 2  81.147
P( s  k )  P 
 17,.05  27.59

2
50
 

7.50 P(
(n  1) s 2

2

19(3.1)
)  P(  2(19)  33.66) = between .01 and .025 (.0201
1.75
exactly)
(n  1)s 2
11(2.5)2
7.51.1 a. P(

)  P(  2(11)  23.79) = between .975 and .99
2
2

(1.7)
(.9864 exactly)
(n  1) s 2 11(1)1
b. P(

)  P(  2(11)  3.81) = between .975 and .99 (.9751
2
(1.7)2
exactly)
7.52 a. P(
(n  1)s 2
2

15(3,000)2
)  P(  2(15)  21.6) = greater than .1 (.1187
(2,500)2
exactly)
(n  1)s 2 15(1,500)2
b. P(

)  P(  2(15)  5.4) = between .01 and .025
2
2

(2,500)
(.0118 exactly)
7.53
7.54
7.55
(n  1) s 2
19(100)
)  P(  2(19)  7.6) = about .01 ( .0097 exactly)

250
(n  1) s 2 19(500)

)  P(  2(19)  38) = between .005 and .01 (.0059
b. P(
2

250
exactly)
a. P(
2

(n  1) s 2
24(75)2
a. P(

)  P(  2(24)  13.5) = between .025 and .05
2
2

(100)
(.0428 exactly)
(n  1)s 2 24(150)2
b. P(

)  P(  2(24)  54) = less than .005 (.0004
2
(100)2
exactly)
(n  1)s 2
29(3.5)2
a. P(

)  P(  2(29)  17.54) = between .95 and .975
2
2

(4.5)
(.9531 exactly) – yes
Chapter 7: Sampling and Sampling Distributions
b. P(
(n  1) s 2
29(6)2
)  P(  2(29)  51.56) = between .99 and .995
2
2

(4.5)
(.9939 exactly) - yes

7.56
Descriptive Statistics: C20, C21, C22, C23, C24, C25, C26, C27, ...
Variable
C20
C21
C22
C23
C24
C25
C26
C27
C28
C29
C30
C31
C32
C33
C34
Mean
3.00
4.00
4.00
4.50
5.00
5.00
5.00
6.00
6.0000
6.500
7.00
6.500
7.00
7.500
5.50
Variance
2.00
8.00
8.00
12.50
18.00
2.00
2.00
8.00
0.000000000
0.500
2.00
0.500
2.00
0.500
4.50
Descriptive Statistics: Variance
Variable
Variance
x
Mean
4.72
StDev
5.26
Variance
27.62
Sum
70.80
70.8
 4.72 E ( s 2 )  15(3.91667)
 4.1964
(14)
15
7.57
a. P(  2(5)  11.07)  .05, 11.07 = 5(Difference), Difference = 2.214
(221.4%)
b. P(  2(5)  1.61)  .1, 1.61 = 5(Difference), Difference = .322 (32.2%)
7.58
a. P(  2(9)  14.68)  .10, 14.68 = 9(Difference), Difference = 1.6311
(163.11%)
b. P(  2(9)  2.7)  .025, P(  2(9)  19.02)  .025,
2.7 = 9a, a = .3, 19.02 = 9b, b = 2.1133
The probability is .95 that the sample variance is between 30% and
211.33% of the population variance
c. The interval in part b. will be smaller
7.59
a. P ( 
2
(14 )
b. P (  2 (14 )
c. P (  2 (14 )
14 s 2 x
 29.14) = .01, 29.14 =
, s x = 2.597
(1.8) 2
14 s 2 x
 5.63) = .025, 5.63 =
, s x = 1.141
(1.8) 2
 6.57) = .05, P (  2 (14 )  23.68) = .05
between 6.57 =
14 s 2 a
14 s 2b
s
,
=
1.233,
and
23.68
=
, sb = 2.341
a
(1.8) 2
(1.8) 2
157
158
Statistics for Business & Economics, 6th edition
7.60
a. P (  2 (11)  4.57) = .95, 4.57 = 11Difference, Difference = .4155
(41.55%)
b. P(  2 (11)  5.58) = .90, 5.58 = 11Difference, Difference = .5073
(50.73%)
c. P(  2 (11)  3.82) = .025, P (  2 (11)  21.92) = .025,
3.82 = 11a, a = .34727, 21.92 = 11b, b = 1.9927
The probability is .95 that the sample variance is between 34.727% and
199.27% of the population variance
7.61
P(
(n  1) s 2

2

19(2.05)
)  P(  2(19)  25.97) = more than 10% (.1310
1.5

24(12.2)
)  P(  2(24)  19.01) = less than .90 (.5438 exactly)
15.4
exactly)
7.62
P(
(n  1) s 2

2
7.63 A sample mean can be thought of as a random variable since there are a very
large number of possible samples, sample size n, that can be drawn from a
population. Each of those samples is likely to have a different sample mean.
Therefore, all possible sample means, sample size n, will have its own
probability distribution. This probability distribution is made up of all
possible sample means calculated from all possible samples of a certain size
drawn from a specific population
7.64
6!
= 15 possible samples
2!4!
b. (41, 39), (41, 35), (41, 35), (41, 33), (41, 38), (39, 35), (39, 35), (39,
33), (39, 38), (35, 35), (35, 33), (35, 38), (35, 33), (35, 38), (33, 38)
2
c. 34 PX (34)  34  4.5333
15
35
35PX (35) 
 2.3333
15
35.5
35.5 PX (35.5) 
 2.3667
15
36
36 PX (36) 
 2.4
15
2
36.5 PX (36.5)  36.5  4.8667
15
3
37 PX (37)  37  7.4
15
a. C26 =
Chapter 7: Sampling and Sampling Distributions
2
 5.0667
15
38.5
38.5 PX (38.5) 
 2.5667
15
39.5
39.5PX (39.5) 
 2.6333
15
40
40 PX (40) 
 2.6667
15
d. The mean of the sampling distribution of the sample mean is
 xPx ( x )  36.8333 which is exactly equal to the population mean:
38PX (38)  38
1
 xi  36.8333 . This is the result expected from the Central Limit
N
Theorem.
7.65 The central limit theorem states that as the sample size increases, the
sampling distribution of the sample mean tends toward the normal
probability distribution, allowing use of the normal probability distribution
for estimating population means.
450  420
) = P(Z > 1.5) = .0668
100 25
400  420
450  420
b. P(
<Z<
) = P(-1 < Z < 1.5) = .7745
100 25
100 25
x  420
c. P(Z > 1.28) = .1, 1.28 =
, x = 445.6
100 25
x  420
d. P(Z < -1.28) = .1, -1.28 =
, x = 394.4
100 25
7.66 a. P ( Z 
24s 2
, s = 123.1868
(100) 2
24s 2
2
f. P(  (24) < 13.85) = .05, 13.85 =
, s = 75.966
(100) 2
g. Smaller. A larger sample size would lead to a smaller standard error
and the graph of the normal distribution would be tighter with less area
in the tails.
e. P(  2 (24) > 36.42) = .05, 36.42 =
65  60
) = P(Z > 1) = .1587
10 4
Xi  60
b. P(Z < -1.28) = .1, -1.28 =
, Xi = 53.6
10 4
7.67 a. P(Z >
159
160
Statistics for Business & Economics, 6th edition
3s 2
, s = 14.4337
(10)2
3s 2 x
2
d. P(  (3) < .584) = .1, .584 =
, s = 4.4121
(10)2
65  60
e. P(Z >
) = P(Z > 1.0) = .1587
10 4
Use the binomial formula: P(X >2) = P(X = 3) + P(X = 4)
C34 (.8413)3 (.1587)1  C44 (.8413) 4 (.1587) 0 = .87896
c. P(  2(3) > 6.25) = .1, 6.25 =
19  14.8
) = P(Z > 2) = .0228
6.3 9
10.6  14.8
19  14.8
b. P(
<Z<
) = P(-2 < Z < 2) = .9544
6.3 9
6.3 9
X  14.8
c. P(Z < -.675) = .25, -.675 = i
, Xi = 13.3825
6.3 9
7.68 a. P(Z >
d. P(  2(8) > 13.36) = .1, 13.36 =
8s 2
, s = 8.1414
(6.3) 2
e. Smaller
1,500  1, 600
) = P(Z > -1.00) = .8413
400 16
X  1,600
b. P(Z > 1.04) = .15, 1.04 = i
, Xi = 1,704
400 16
7.69 a. P(Z >
P(  2 (15)  22.31) = .1, 22.31 =
c.
15s 2
, s = 487.825
(400)2
7.70 Let n = N, then X   x :
E[ i 1 ( X i  X ) ]  n
N

2
2
x
n
 2x N  n
n N 1
 n 2 x 
N n 2
 x
N 1
N x
(n  1)
N 1
N 1
1
1
N 2 x
2
2
(
X

X
)
]

E
[
(
X

X
)
]

Therefore, E[
 i
 i
n 1
n 1
N 1
2
x
(nN  n  N  n) 
2
120  100
) = P(Z > 2) = .0228
30 9
X  100
b. P(Z < -.843) = .20, -.843 = i
, Xi = 91.57
30 9
7.71 a. P(Z >
Chapter 7: Sampling and Sampling Distributions
c. P(  2 (8)  2.73) = .05, 2.73 =
8s 2
, s = 17.525
(30)2
.7  .8
) = P(Z < -1.94) = .0262
(.8)(.2) / 60
b. Use the binomial formula: 6(.7) = 4.2, P(X  4) = 1 – P(X > 4) = 16(.8)5(.2)-(.8)6 = .3446
30, 000  29, 000
c. P(Z >
) = P(Z > .61) = .2709
4, 000 6
30, 000  29, 000
d. (.8)P(Z >
) = (.8)P(Z > .25) = (.8)(.4013) = .3210
4, 000
7.72 a. P(Z <
15s 2
7.73 a. P (  (15)  30.58) = .01, 30.58 =
, s = 2.5701
(1.8) 2
Difference
b. P(Z > 1.04) = .15, 1.04 =
, Difference = .468
1.8 16
Difference
c. P(Z > 1.96) = .025, 1.96 =
, Difference = ±.882
1.8 16
2
7.74 P(
(n  1) s 2

2
> 20(2)) = P (  2 ( 20 )  40) = .005
.6  .5
) = P(Z > 2) = .0228
(.5)(.5) /100
.4  .5
.55  .5
b. P(
<Z<
) = P(-1 < Z < 1) = .6826
(.5)(.5) /100
(.5)(.5) /100
c. Replace 100 with 10 in parts a. and b. The answer will be larger for
part a. and smaller for part b.
7.75 a. P(Z >
7.76 10 <  x < X + 10, -10 < X -  x < 10
10
10
P(
<Z<
) = P(-1 < Z < 1) = .6826
40 16
40 16
7.77
P(Z <
29.5  30
) = P(Z < -1.54) = .0618
1.3 16
161
162
7.78
Statistics for Business & Economics, 6th edition
(.4)(.6)
= .03098
250
a.  x 
p  .4
, p = .3739
.03098
p  .4
b. P(Z < 1.28) = .9, 1.28 =
, p = .4397
.03098
Difference
c. P(Z > 1.04) = .35, 1.04 =
, Difference = ±.0322
.03098
P(Z > -.843) = .8, -.843 =
.28  .2
) = P(Z < 3.46) = .9997
(.2)(.8) / 300
.28  .4
b. P(Z <
) = P(Z < -4.24)  .0000
(.4)(.6) / 300
7.79
a. P(Z >
7.80
a. P(
(n  1)s 2
2

24(4,000)2
)  P(  2(24)  8.82) = more than .99 (.9979
2
(6,600)
exactly)
(n  1) s 2 24(8,000) 2
b. P(

)  P(  2(24)  35.62) = between .9 and .95
2
2

(6,600)
(.9354 exactly)
7.81
P(
(n  1) s 2
2
exactly)
19(2.5)2

)  P(  2(19)  29.69) = between .05 and .1 (.0559
2
(2)