Part IV: Complete Solutions, Chapter 7
420
Chapter 7: Introduction to Sampling Distributions
Section 7.1
1.
Answers vary. Students should identify the individuals (subjects) and variable involved. For example, the
population of all ages of all people in Colorado, the population of weights of all students in your school, or
the population count of all antelope in Wyoming.
2.
Answers vary. A simple random sample of n measurements from a population is a subset of the population
selected in a manner such that
(a) Every sample of size n from the population has an equal chance of being selected.
(b) Every member of the population has an equal chance of being included in the sample.
3.
A population parameter is a numerical descriptive measure of a population, such as , the population
mean, , the population standard deviation, 2, the population variance, p, the population proportion, the
population maximum and minimum, etc.
4.
A sample statistic is a numerical descriptive measure of a sample, such as x , the sample mean, s, the
sample standard deviation, s2, the sample variance, pˆ, the sample proportion, the sample maximum and
minimum, etc.
5.
A statistical inference refers to conclusions about the value of a population parameter based on
information from the corresponding sample statistic and the associated probability distributions. We will
do both estimation and testing.
6.
A sampling distribution is a probability distribution for a sample statistic.
7.
They help us visualize the sampling distribution by using tables and graphs that approximately represent
the sampling distribution.
8.
Relative frequencies can be thought of as a measure or estimate of the likelihood of a certain statistic
falling within the class bounds.
9.
We studied the sampling distribution of mean trout lengths based on samples of size 5. Other such
sampling distributions abound. Notice that the sample size remains the same for each sample in a sampling
distribution.
Section 7.2
Note: Answers may vary slightly depending on the number of digits carried in the standard deviation.
1.
The standard error is simply the standard deviation for a sampling distribution.
2.
The standard error
3.
x is an unbiased estimator for μ, and p̂ is an unbiased estimator for p.
4.
As the sample size increases, the variability decreases.
5.
(a) The required sample size is n ≥ 30.
(b) No. If the original distribution is normal, then x is distributed normally for any sample size.
6.
(a) No, because we require a sample size of n ≥ 30 if the original distribution is not normal.
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Part IV: Complete Solutions, Chapter 7
421
(b) Yes, the x distribution is normal with mean  x  72 and  x 
8
16
2.
P(68  x  73)  P(2  z  0.50)  0.6687
z
7.
68  72
 2
2
z
73  72
 0.50
2
The distribution with n = 225 will have a smaller standard error. Since  x 

, dividing by the square
n
root of 225 will result in a small standard error regardless of the value of σ.
8.
(a) We require n = 36 because
(b) We require n = 144 because
9.
12
12
2.
6
12

 1.
144 12
36
12

(a)  x    15
x 


14
 2.0
n
49
Because n = 49  30, by the central limit theorem, we can assume that the distribution of x is
approximately normal.
x   x  15
z

x
2.0
15  15
0
2.0
17  15
x  17 converts to z 
1
2.0
x  15 converts to z 
P 15  x  17   P  0  z  1
 P  z  1  P  z  0 
 0.8413  0.5000
 0.3413
(b)  x    15
x 


14

x  15
1.75
 1.75
n
64
Because n = 64  30, by the central limit theorem, we can assume that the distribution of x is
approximately normal.
z
x 
x
15  15
0
1.75
17  15
x  17 converts to z 
 1.14
1.75
x  15 converts to z 
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Part IV: Complete Solutions, Chapter 7
422
P 15  x  17   P  0  z  1.14 
 P  z  1.14   P  z  0 
 0.8729  0.5000
 0.3729
(c) The standard deviation of part (b) is smaller because of the larger sample size. Therefore, the
distribution about  x is narrower.
10. (a) For both distributions, the mean will be  x  5 .
(b) The distribution with n = 81 because the standard deviation will be smaller. This distribution will be
less spread out around its mean.
(c) The distribution with n = 81 because the standard deviation will be smaller. This distribution will be
less spread out around its mean.
11. (a)   75,   0.8
74.5  75 

P  x  74.5  P  z 

0.8 

 P  z  0.63
 0.2643
(b)  x  75,  x 

n

0.8
 0.179
20
74.5  75 

P  x  74.5   P  z 

0.179 

 P  z  2.79 
 0.0026
(c) No. If the weight of only one car were less than 74.5 tons, we could not conclude that the loader is out
of adjustment. If the mean weight for a sample of 20 cars were less than 74.5 tons, we would suspect
that the loader is malfunctioning because the probability of this event occurring is 0.26% if indeed the
distribution is correct.
12. (a)   68,   3
69  68 
 67  68
P  67  x  69   P 
z

3 
 3
 P  0.33  z  0.33
 P  z  0.33  P  z  0.33
 0.6293  0.3707
 0.2586
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Part IV: Complete Solutions, Chapter 7
(b)  x  68,  x 

n

423
3
1
9
69  68 
 67  68
P  67  x  69   P 
z

1
1 

 P  1  z  1
 P  z  1  P  z  1
 0.8413  0.1587
 0.6826
(c) The probability in part (b) is much higher because the standard deviation is smaller for
the x distribution.
13. (a)   85,   25
40  85 

P  x  40   P  z 

25 

 P  z  1.8 
 0.0359
(b) The probability distribution of x is approximately normal with  x  85;  x 

n

25
2
 17.68.
40  85 

P  x  40   P  z 

17.68 

 P  z  2.55
 0.0054

25

 14.43
(c)  x  85,  x 
n
3
40  85 

P  x  40   P  z 

14.43 

 P  z  3.12 
 0.0009

25

 11.2
(d)  x  85,  x 
n
5
40  85 

P  x  40   P  z 

11.2 

 P  z  4.02 
 0.0002
(e) Yes. The more tests a patient completes, the stronger is the evidence for excess insulin. If the average
value based on five tests were less than 40, the patient is almost certain to have excess insulin.
14.   7500,   1750
3500  7500 

(a) P  x  3500   P  z 

1750


 P  z  2.29 
 0.0110
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Part IV: Complete Solutions, Chapter 7
424
(b) The probability distribution of x is approximately normal with
 x  7500;  x 

n

1750
2
 1237.44.
3500  7500 

P  x  3500   P  z 

1237.44 

 P  z  3.23
 0.0006
 1750

 1010.36
(c)  x  7500,  x 
n
3
3500  7500 

P  x  3500   P  z 

1010.36 

 P  z  3.96 
 0.0002
(d) The probabilities decreased as n increased. It would be an extremely rare event for a person to have
two or three tests below 3,500 purely by chance. The person probably has leukopenia.
15. (a)   63.0,   7.1
54  63.0 

P  x  54   P  z 

7.1 

 P  z  1.27 
 0.1020
(b) The expected number undernourished is 2,200 × 0.1020 = 224.4, or about 224.

7.1

 1.004
(c)  x  63.0,  x 
n
50
60  63.0 

P  x  60   P  z 

1.004 

 P  z  2.99 
 0.0014
(d)  x  63.0,  x  1.004
64.2  63.0 

P  x  64.2   P  z 

1.004 

 P  z  1.20 
 0.8849
Since the sample average is above the mean, it is quite unlikely that the doe population is
undernourished.
16. (a) By the central limit theorem, the sampling distribution of x is approximately normal with

$7
mean  x    $20 and standard error  x 

 $0.70. It is not necessary to make any
n
100
assumption about the x distribution because n is large.
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Part IV: Complete Solutions, Chapter 7
425
(b)  x  $20,  x  $0.70
$22  $20 
 $18  $20
P  $18  x  $22   P 
z

$0.70 
 $0.70
 P  2.86  z  2.86 
 P  z  2.86   P  z  2.86 
 0.9979  0.0021
 0.9958
(c)  x  $20,   $7
$22  $20 
 $18  $20
P  $18  x  $22   P 
z

$7
$7


 P  0.29  z  0.29 
 0.6141  0.3859
 0.2282
(d) We expect the probability in part (b) to be much higher than the probability in part (c) because the
standard deviation is smaller for the x distribution than it is for the x distribution. By the central limit
theorem, the sampling distribution of x will be approximately normal as n increases, and its standard
deviation  n will decrease as n increases. For a fixed interval, such as $18 to $22, centered at the
mean, $20 in this case, the proportion of the possible x values within the interval will be greater than
the proportion of the possible x values within the same interval. A sample of 100 customers contains
much more information about purchasing tendencies than a single customer, so averages are much
more predictable than a single observation.
17. (a) The random variable x is itself an average based on the number of stocks or bonds in the fund. Since x
itself represents a sample mean return based on a large (random) sample of size n = 250 of stocks or
bonds, x has a distribution that is approximately normal (central limit theorem).

0.9%

 0.367%
(b)  x  1.6%,  x 
n
6
2%  1.6% 
 1%  1.6%
P 1%  x  2%   P 
z

0.367% 
 0.367%
 P  1.63  z  1.09 
 P  z  1.09   P  z  1.63
 0.8621  0.0516
 0.8105
(c) Note: 2 years = 24 months; x is monthly percentage return.

0.9%
 x  1.6%,  x 

 0.1837%
n
24
2%  1.6% 
 1%  1.6%
P 1%  x  2%   P 
z

0.1837% 
 0.1837%
 P  3.27  z  2.18 
 P  z  2.18   P  z  3.27 
 0.9854  0.0005
 0.9849
(d) Yes. The probability increases as the standard deviation decreases. The standard deviation decreases as
the sample size increases.
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Part IV: Complete Solutions, Chapter 7
426
(e)  x  1.6%,  x  0.1837%
1%  1.6% 

P  x  1%   P  z 

0.1837% 

 P  z  3.27 
 0.0005
This is very unlikely if  = 1.6%. One would suspect that  has slipped below 1.6%.
18. (a) The random variable x is itself an average based on the number of stocks in the fund. Since x itself
represents a sample mean return based on a large (random) sample of size n = 100 of stocks, x has a
distribution that is approximately normal (central limit theorem).

0.8%

 0.2667%
(b)  x  1.4%,  x 
n
9
2%  1.4% 
 1%  1.4%
P 1%  x  2%   P 
z

0.2667% 
 0.2667%
 P  1.50  z  2.25 
 P  z  2.25   P  z  1.50 
 0.9878  0.0668
 0.9210

0.8%

 0.1886%
(c)  x  1.4%,  x 
n
18
2%  1.4% 
 1%  1.4%
P 1%  x  2%   P 
z

0.1886%
0.1886% 

 P  2.12  z  3.18 
 P  z  3.18   P  z  2.12 
 0.9993  0.0170
 0.9823
(d) Yes. The probability increases as the standard deviation decreases. The standard deviation decreases as
the sample size increases.
(e)  x  1.4%,  x  0.1886%
2%  1.4% 

P  x  2%   P  z 

0.1886% 

 P  z  3.18 
 1  P  z  3.18 
 1  0.9993
 0.0007
This is very unlikely if  = 1.4%. One would suspect that the European stock market may be heating
up; i.e.,  is greater than 1.4%.
19. (a) The total checkout time for 30 customers is the sum of the checkout times for each individual
customer. Thus w = x1 + x2 + … + x30, and the probability that the total checkout time for the next 30
customers is less than 90 is P(w < 90).
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Part IV: Complete Solutions, Chapter 7
(b) If we divide both sides of w < 90 by 30, we obtain
427
w
< 3. However, w is the sum of 30 waiting times,
30
w
is x . Therefore, P  w  90  P  x  3 .
30
(c) The probability distribution of x is approximately normal with mean  x    2.7 and standard
so
deviation  x 

n

0.6
30
 0.1095.
3  2.7 

(d) P  x  3  P  z 

0.1095


 P  z  2.74 
 0.9969
The probability that the total checkout time for the next 30 customers is less than 90 minutes is
0.9969.
20. Let w = x1 + x2 +  + x36.

2.5
(a) w < 320 is equivalent to w  320 or x  8.889.  x    8.5,  x 

 0.4167.
36 36
n
36
P  w  320   P  x  8.889 
8.889  8.5 

 P z 

0.4167 

 P  z  0.93
 0.8238
(b) w > 275 is equivalent to w  275 or x  7.639.  x  8.5,  x  0.4167.
36 36
P  w  275   P  x  7.639 
7.639  8.5 

 Pz 

0.4167 

 P  z  2.07 
 1  P  z  2.07 
 1  0.0192
 0.9808
(c) P  275  w  320   P  7.639  x  8.889 
 P  2.07  z  0.93
 P  z  0.93  P  z  2.07 
 0.8238  0.0192
 0.8046
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Part IV: Complete Solutions, Chapter 7
428
21. (a) Let w  x1  x2 
 x5 .
 x    17,  x 


3.3
n
5
 w 90 
P ( w  90)  P   
5 
5
 P ( x  18)
 1.476
18  17 

 P z 

1.476 

 P ( z  0.68)
 1  0.7517
 0.2483
 w 80 
(b) P ( w  80)  P   
5 5 
 P ( x  16)
16  17 

 P z 

1.476 

 P ( z  0.68)
 0.2483
(c) P(80  w  90)  P(16  x  18)
 P(0.68  z  0.68)
 P( z  0.68)  P( z  0.68)
 0.7517  0.2483
 0.5034
Section 7.3
1.
We must check that np > 5 and nq > 5.
2.
 p̂ 
3.
Yes, it is unbiased. The mean of the distribution for p̂ is p.
4.
(a) 0.5/25 = 0.02
(b) 0.5/100 = 0.005
(c) As n increases, the continuity correction decreases.
5.
pq
and  p̂  p
n
(a) np  33  0.21  6.93, nq  33  0.79   26.07
Yes, p̂ can be approximated by a normal random variable because both np and nq exceed 5.
 pˆ  p  0.21,  pˆ 
0.21 0.79 
 0.071
33
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Part IV: Complete Solutions, Chapter 7
Continuity correction 
0.5 0.5

 0.015
n
33
P  0.15  pˆ  0.25   P  0.15  0.015  x  0.25  0.015 
 P  0.135  x  0.265 
0.265  0.21 
 0.135  0.21
 P
z

0.071
0.071 

 P  1.06  z  0.77 
 P  z  0.77   P  z  1.06 
 0.7794  0.1446
 0.6348
(b) No, because np = 25 × 0.15 = 3.75 does not exceed 5.
(c) np = 48  0.15 = 7.2, nq = 48  0.85 = 40.8
Yes, p̂ can be approximated by a normal random variable because both np and nq exceed 5.
 pˆ  p  0.15,  pˆ 
0.15  0.85 
 0.052
48
Continuity correction 
0.5 0.5

 0.010
n
45
P  pˆ  0.22   P  x  0.22  0.010 
 P  x  0.21
0.21  0.15 

 P z 

0.052 

 P  z  1.15 
 1  P  z  1.15 
 1  0.8749
 0.1251
6.
(a) n  50, p  0.36
np  50  0.36   18, nq  50  0.64   32
Approximate p̂ by a normal random variable because both np and nq exceed 5.
 pˆ  p  0.36,  pˆ 
0.36  0.64 
 0.068
50
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429
Part IV: Complete Solutions, Chapter 7
430
Continuity correction 
0.5 0.5

 0.01
n
50
P  0.30  pˆ  0.45   P  0.30  0.01  x  0.45  0.01
 P  0.29  x  0.46 
0.46  0.36 
 0.29  0.36
 P
z

0.068
0.068 

 P  1.03  z  1.47 
 P  z  1.47   P  z  1.03
 0.9292  0.1515
 0.7777
(b) n  38, p  0.25
np  38  0.25  9.5, nq  38  0.75  28.5
Approximate p̂ by a normal random variable because both np and nq exceed 5.
 pˆ  p  0.25,  pˆ 
0.25  0.75 
 0.070
38
Continuity correction 
0.5 0.5

 0.013
n
38
P  pˆ  0.35   P  x  0.35  0.013
 P  x  0.337 
0.337  0.25 

 P z 

0.070 

 P  z  1.24 
 1  P  z  1.24 
 1  0.8925
 0.1075
(c) n  41, p  0.09
np  41 0.09   3.69
We cannot approximate p̂ by a normal random variable because np < 5.
7.
n  30, p  0.60
np  30  0.60   18, nq  30  0.40   12
Approximate p̂ by a normal random variable because both np and nq exceed 5.
 pˆ  p  0.6,  pˆ 
0.6  0.4 
Continuity correction 
30
 0.089
0.5 0.5

 0.017
n
30
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Part IV: Complete Solutions, Chapter 7
431
(a) P  pˆ  0.5   P  x  0.5  0.017 
 P  x  0.483
0.483  0.6 

 Pz 

0.089 

 P  z  1.31
 0.9049
(b) P  pˆ  0.667   P  x  0.667  0.017 
 P  x  0.65 
0.65  0.6 

 Pz 

0.089 

 P  z  0.56 
 0.2877
(c) P  pˆ  0.333  P  x  0.333  0.017 
 P  x  0.35 
0.35  0.6 

 Pz 

0.089 

 P  z  2.81
 0.0025
(d) Yes, both np and nq exceed 5.
8.
(a) n  38, p  0.73
np  38  0.73  27.74, nq  38  0.27   10.26
Approximate p̂ by a normal random variable because both np and nq exceed 5.
 pˆ  p  0.73,  pˆ 
0.73  0.27 
Continuity correction 
38
 0.072
0.5 0.5

 0.013
n
38
P  pˆ  0.667   P  x  0.667  0.013
 P  x  0.654 
0.654  0.73 

 Pz 

0.072 

 P  z  1.06 
 0.8554
(b) n  45, p  0.86
np  45  0.86   38.7, nq  45  0.14   6.3
Approximate p̂ by a normal random variable because both np and
nq exceed 5.
Copyright © Houghton Mifflin Company. All rights reserved.
Part IV: Complete Solutions, Chapter 7
432
 pˆ  p  0.86,  pˆ 
0.86  0.14 
Continuity correction 
45
 0.052
0.5 0.5

 0.011
n
45
P  pˆ  0.667   P  x  0.667  0.011
 P  x  0.656 
0.656  0.86 

 Pz 

0.052 

 P  z  3.92 
1
(c) Yes, both np and nq exceed 5 for men and for women.
9.
(a) n  100, p  0.06
np  100  0.06   6, nq  100  0.94   94
p̂ can be approximated by a normal random variable because both np and nq exceed 5.
 pˆ  p  0.06,  pˆ 
0.06  0.94 
Continuity correction 
100
 0.024
0.5
 0.005
100
(b) P  pˆ  0.07   P  x  0.07  0.005 
 P  x  0.065 
0.065  0.06 

 Pz 

0.024 

 P  z  0.21
 0.4168
(c) P  pˆ  0.11  P  x  0.11  0.005 
 P  x  0.105 
0.105  0.06 

 Pz 

0.024 

 P  z  1.88 
 0.0301
Yes; because this probability is so small, it should rarely occur. The machine might need an
adjustment.
10. (a) n  50, p  0.565
np  50  0.565  28.25, nq  50  0.435  21.75
p̂ can be approximated by a normal random variable because both np and nq exceed 5.
 pˆ  p  0.565,  pˆ 
Continuity correction 
0.565  0.435 
50
 0.070
0.5 0.5

 0.01
n
50
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Part IV: Complete Solutions, Chapter 7
433
(b) P  pˆ  0.53  P  x  0.53  0.01
 P  x  0.54 
0.54  0.565 

 Pz 

0.070 

 P  z  0.36 
 0.3594
(c) P  pˆ  0.41  P  x  0.41  0.01
 P  x  0.42 
0.42  0.565 

 Pz 

0.070 

 P  z  2.07 
 0.0192
(d) Meredith has the more serious case because the probability of having such a low reading in a healthy
person is less than 2%.
11.
total number of successes from all 12 quarters
total number of families from all 12 quarters
11  14   19

12  92 
p
206
1,104
 0.1866

q  1  p  1  0.1866  0.8134
 pˆ  p  p  0.1866
 pˆ 
pq

n
0.1866  0.8134 
pq

 0.0406
n
92
Check: np  92  0.1866  17.2, nq  92  0.8134   74.8
Since both np and nq exceed 5, the normal approximation should be reasonably good.
Center line  p  0.1866
Control limits at p  2
pq
n
 0.1866  2  0.0406
 0.1866  0.0812
or 0.1054 and 0.2678
Control limits at p  3
pq
n
 0.1866  3  0.0406
 0.1866  0.1218
or 0.0648 and 0.3084
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Part IV: Complete Solutions, Chapter 7
434
P Chart of Victims
+3SL=0.3084
0.30
+2SL=0.2678
Proportion
0.25
0.20
_
P=0.1866
0.15
-2SL=0.1054
0.10
-3SL=0.0647
0.05
1
2
3
4
5
6
7
Quarter
8
9
10
11
12
There are no out-of-control signals.
12.
total number of defective cans
total number of cans
8  11   10

110 15 
p
133
1650
 0.08061

q  1  p  1  0.08061  0.91939
 pˆ  p  p  0.08061
 pˆ 
pq

n
pq

n
 0.08061 0.91939   0.02596
110
Check: np  110  0.08061  8.9, nq  110 0.91939   101.1
Since both np and nq exceed 5, the normal approximation should be reasonably good.
Center line  p  0.08061
Control limits at p  2
pq
n
 0.08061  2  0.02596
 0.08061  0.05192, or 0.02869 and 0.1325.
Control limits at p  3
pq
n
 0.08061  3  0.02596
 0.08061  0.07788, or 0.00273 and 0.1585.
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Part IV: Complete Solutions, Chapter 7
435
P Chart of Defective Cans
0.16
+3SL=0.1585
0.14
+2SL=0.1325
Proportion
0.12
0.10
_
P=0.0806
0.08
0.06
0.04
-2SL=0.0287
0.02
-3SL=0.0027
0.00
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15
Test Sheet
There are no out-of-control signals. It appears that the production process is in reasonable control.
13.
total number who got jobs
total number of people
60  53   58

75 15 
p
872
1125
 0.7751

q  1  p  1  0.7751  0.2249
 pˆ  p  p  0.7751
 pˆ 
pq

n
pq

n
 0.7751 0.2249   0.0482
75
Check: np  75  0.7751  58.1, nq  75  0.2249  16.9
Since both np and nq exceed 5, the normal approximation should be reasonably good.
Center line  p  0.7751
Control limits at p  2
pq
n
 0.7751  2  0.0482
 0.7751  0.0964, or 0.6787 to 0.8715.
Control limits at p  3
pq
n
 0.7751  3  0.0482
 0.7751  0.1446, or 0.6305 to 0.9197.
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Part IV: Complete Solutions, Chapter 7
436
P Chart of Jobs
0.95
1
+3SL=0.9197
0.90
+2SL=0.8715
Proportion
0.85
0.80
_
P=0.7751
0.75
0.70
-2SL=0.6787
0.65
-3SL=0.6305
0.60
1
1
2
3
4
5
6
7
8
9
Day
10 11 12 13 14 15
Out-of-control signal III occurs on days 4 and 5; out-of-control signal I occurs on day 11 on the low side
and day 14 on the high side. Out-of-control signals on the low side are of most concern for the homeless
seeking work. The foundation should look to see what happened on that day. The foundation might take a
look at the out-of-control periods on the high side to see if there is a possibility of cultivating more jobs.
Chapter 7 Review
1.
(a) The x distribution approaches a normal distribution.
(b) The mean  x of the x distribution equals the mean  of the x distribution regardless of the sample size.
(c) The standard deviation  x of the sampling distribution equals

n
, where  is the standard
deviation of the x distribution and n is the sample size.
(d) They will both be approximately normal with the same mean, but the standard deviations will
be

and
50
2.
100
, respectively.
All the x distributions will be normal with mean  x    15. The standard deviations will be
n  4:  x 
n  16:  x =
n  100:  x =
3.


n

n

n
3

4
3


3
2

3
4
16
3

100

3
10
(a)   35,   7
40  35 

P  x  40   P  z 

7 

 P  z  0.71
 0.2389
(b)  x    35,  x 

n

7
7

9 3
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Part IV: Complete Solutions, Chapter 7
40  35 

P  x  40   P  z 

7


3


 P  z  2.14 
 0.0162
4.
(a)   38,   5
35  38 

P  x  35  P  z 

5 

 P  z  0.6 
 0.2743
(b)  x    38,  x 

n

5
 1.58
10
35  38 

P  x  35   P  z 

1.58 

 P  z  1.90 
 0.0287
(c) The probability in part (b) is much smaller because the standard deviation is smaller for
the x distribution.
5.
15
 x    100,  x   
 1.5
n
100
P 100  2  x  100  2   P  98  x  102 
102  100 
 98  100
 P
z

1.5
1.5 

 P  1.33  z  1.33
 P  z  1.33  P  z  1.33
 0.9082  0.0918
 0.8164
6.
2
 x    15,  x   
 0.333
n
36
P 15  0.5  x  15  0.5   P 14.5  x  15.5 
15.5  15 
 14.5  15
 P
z

0.333 
 0.333
 P  1.5  z  1.5 
 P  z  1.5   P  z  1.5 
 0.9332  0.0668
 0.8664
7.
(a) n  50, p  0.22
np  50  0.22   11, nq  50  0.78  39
Approximate p̂ by a normal random variable because both np and nq exceed 5.
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437
Part IV: Complete Solutions, Chapter 7
438
 pˆ  p  0.22,  pˆ 
0.22  0.78
50
Continuity correction 
 0.0586
0.5 0.5

 0.01
n
50
P  0.20  pˆ  0.25   P  0.20  0.01  x  0.25  0.01
 P  0.19  x  0.26 
0.26  0.22 
 0.19  0.22
 P
z

0.0586 
 0.0586
 P  0.51  z  0.68 
 P  z  0.68   P  z  0.51
 0.7517  0.3050
 0.4467
(b) n  38, p  0.27
np  38  0.27   10.26, nq  38  0.73  27.74
Approximate p̂ by a normal random variable because both np and nq exceed 5.
 pˆ  p  0.27,  pˆ 
 0.27  0.73  0.0720
38
Continuity correction 
0.5 0.5

 0.013
n
38
P  pˆ  0.35   P  x  0.35  0.013
 P  x  0.337 
0.337  0.27 

 P z 

0.0720 

 P  z  0.93
 0.1762
(c) n  51, p  0.05
np  51 0.05  2.55
No, we cannot approximate p̂ by a normal random variable because np < 5.
8.
n  28, p  0.31
np  28  0.31  8.68, nq  28  0.69   19.32
Approximate p̂ by a normal random variable because both np and nq exceed 5.
 pˆ  p  0.31,  pˆ 
0.31 0.69 
Continuity correction 
28
 0.087
0.5 0.5

 0.018
n
28
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Part IV: Complete Solutions, Chapter 7
(a) P  pˆ  0.25   P  x  0.25  0.018 
 P  x  0.232 
0.232  0.31 

 Pz 

0.087 

 P  z  0.90 
 0.8159
(b) P  0.25  pˆ  0.50   P  0.25  0.018  x  0.50  0.018 
 P  0.232  x  0.518 
0.518  0.31 
 0.232  0.31
 P
z

0.087 
 0.087
 P  0.90  z  2.39 
 P  z  2.39   P  z  0.90 
 0.9916  0.1841
 0.8075
(c) Yes, both np and nq exceed 5.
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