Chapter 7 Introduction to Sampling Distributions
Section 7.1
1. Answers vary. Students should identify the individuals (subjects) and variable involved. Answers may
include: A population is a set of measurements or counts either existing or conceptual. For example, the
population of all ages of all people in Colorado; the population of weights of all students in your school;
the population count of all antelope in Wyoming.
2. See Section 1.2. Answer may include:
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 and
(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 µ, the population
mean; σ, the population standard deviation; σ2, the population variance; p, the population proportion;
ρ (rho) the population correlation coefficient for those who have already studied linear regression from
Chapter 10.
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; r, the sample correlation
coefficient for those who have already studied linear regression from Chapter 10.
5. A statistical inference is a conclusion about the value of a population parameter based on information
about the corresponding sample statistic and probability. 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.
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421
422
Instructor’s Resource Guide Understandable Statistics, 8th Edition
Section 7.2
Note: Answers may vary slightly depending on the number of digits carried in the standard deviation.
1. (a) µ x = µ = 15
σx =
σ
n
=
14
49
= 2.0
Because n = 49 ≥ 30, by the central limit theorem, we can assume that the distribution of x is
approximately normal.
z=
x −µ
σx
=
x − 15
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) µ x = µ = 15
σx =
σ
n
=
14
64
= 1.75
Because n = 64 ≥ 30, by the central limit theorem, we can assume that the distribution of x is
approximately normal.
z=
x −µ
σx
=
x − 15
1.75
15 − 15
=0
1.75
17 − 15
x = 17 converts to z =
= 1.14
1.75
x = 15 converts to z =
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 µ x is narrower in part (b).
2. (a) µ x = µ = 100
σ
48
σx =
=
= 5.33
n
81
Because n = 81 ≥ 30, by the central limit theorem, we can assume that the distribution of x is
approximately normal.
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Part IV: Complete Solutions, Chapter 7
z=
x −µ
σx
=
423
x − 100
5.33
92 − 100
= −1.50
5.33
100 − 100
x = 100 converts to z =
=0
5.33
x = 92 converts to z =
P ( 92 ≤ x ≤ 100 ) = P ( −1.50 ≤ z ≤ 0 )
= P ( z ≤ 0 ) − P ( z ≤ −1.50 )
= 0.5000 − 0.0668
= 0.4332
(b) µ x = µ = 100
σ
48
=
= 4.36
σx =
n
121
Because n = 121 ≥ 30, by the central limit theorem, we can assume that the distribution of x is
approximately normal.
z=
x −µ
σx
=
x − 100
4.36
92 − 100
= −1.83
4.36
100 − 100
x = 100 converts to z =
=0
4.36
x = 92 converts to z =
P ( 92 ≤ x ≤ 100 ) = P ( −1.83 ≤ z ≤ 0 )
= P ( z ≤ 0 ) − P ( z ≤ −1.83)
= 0.5000 − 0.0336
= 0.4664
(c) The probability of part (b) is greater than that of part (a). The standard deviation of part (b) is smaller
because of the larger sample size. Therefore, the distribution about µ x is narrower in part (b).
3. (a) No, we cannot say anything about the distribution of sample means because the sample size is only 9
and so it is too small to apply the central limit theorem.
(b) Yes, now we can say that the x distribution will also be normal with
µ x = µ = 25 and σ x =
z=
x −µ
σx
=
σ
n
=
3.5
= 1.17.
9
x − 25
1.17
26 − 25 ⎞
⎛ 23 − 25
P ( 23 ≤ x ≤ 26 ) = P ⎜
≤z≤
⎟
1.17 ⎠
⎝ 1.17
= P ( −1.71 ≤ z ≤ 0.86 )
= P ( z ≤ 0.86 ) − P ( z ≤ −1.71)
= 0.8051 − 0.0436
= 0.7615
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Instructor’s Resource Guide Understandable Statistics, 8th Edition
4. (a) No, we cannot say anything about the distribution of sample means because the sample size is only 16
and so it is too small to apply the central limit theorem.
(b) Yes, now we can say that the x distribution will also be normal with
µ x = µ = 72 and σ x =
z=
x −µ
σx
=
σ
n
=
8
= 2.
16
x − 72
2
73 − 72 ⎞
⎛ 68 − 72
P ( 68 ≤ x ≤ 73) = P ⎜
≤z≤
⎟
2
2 ⎠
⎝
= ( −2 ≤ z ≤ 0.5 )
= P ( z ≤ 0.5 ) − P ( z ≤ −2 )
= 0.6915 − 0.0228
= 0.6687
5. (a) µ = 75, σ = 0.8
74.5 − 75 ⎞
⎛
P ( x < 74.5 ) = P ⎜ z <
⎟
0.8 ⎠
⎝
= P ( z < −0.63)
= 0.2643
(b) µ x = 75, σ x =
σ
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 cannot 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. As we see in part (b), the probability of this happening is very low if the
loader is correctly adjusted.
6. (a) µ = 68, σ = 3
69 − 68 ⎞
⎛ 67 − 68
≤z≤
P ( 67 ≤ x ≤ 69 ) = P ⎜
⎟
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) µ x = 68, σ x =
σ
n
=
425
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.
7. (a) µ = 85, σ = 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 µ x = 85; σ x =
σ
n
=
25
2
= 17.68.
40 − 85 ⎞
⎛
P ( x < 40 ) = P ⎜ z <
⎟
17.68 ⎠
⎝
= P ( z < −2.55 )
= 0.0054
(c) µ x = 85, σ x =
σ
n
=
25
= 14.43
3
40 − 85 ⎞
⎛
P ( x < 40 ) = P ⎜ z <
⎟
14.43 ⎠
⎝
= P ( z < −3.12 )
= 0.0009
(d) µ x = 85, σ x =
σ
n
=
25
= 11.2
5
40 − 85 ⎞
⎛
P ( x < 40 ) = P ⎜ z <
⎟
11.2 ⎠
⎝
= P ( z < −4.02 )
< 0.0002
(e) Yes; If the average value based on five tests were less than 40, the patient is almost certain to have
excess insulin.
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Instructor’s Resource Guide Understandable Statistics, 8th Edition
8. µ = 7500, σ = 1750
3500 − 7500 ⎞
⎛
(a) P ( x < 3500 ) = P ⎜ z <
⎟
1750
⎝
⎠
= P ( z < −2.29 )
= 0.0110
(b) The probability distribution of x is approximately normal with
µ x = 7500; σ x =
σ
n
=
1750
2
= 1237.44.
3500 − 7500 ⎞
⎛
P ( x < 3500 ) = P ⎜ z <
⎟
1237.44 ⎠
⎝
= P ( z < −3.23)
= 0.0006
(c) µ x = 7500, σ x =
σ
=
n
1750
= 1010.36
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 3500 purely by chance; the person probably has leukopenia.
9. (a) µ = 63.0, σ = 7.1
54 − 63.0 ⎞
⎛
P ( x < 54 ) = P ⎜ z <
⎟
7.1 ⎠
⎝
= P ( z < −1.27 )
= 0.1020
(b) The expected number undernourished is 2200(0.1020) = 224.4, or about 224.
(c) µ x = 63.0, σ x =
σ
n
=
7.1
= 1.004
50
60 − 63.0 ⎞
⎛
P ( x < 60 ) = P ⎜ z <
⎟
1.004 ⎠
⎝
= P ( z < −2.99 )
= 0.0014
(d) µ x = 63.0, σ 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.
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Part IV: Complete Solutions, Chapter 7
427
10. (a) From the Central Limit Theorem, we expect the x distribution to be approximately normal with the
σ
2
=
= 0.3651.
mean µ x = µ = 16 and standard deviation σ x =
30
n
(b) µ x = 16, σ x = 0.3651
17 − 16 ⎞
⎛ 16 − 16
P (16 ≤ x ≤ 17 ) = P ⎜
≤z≤
⎟
0.3651 ⎠
⎝ 0.3651
= P ( 0 ≤ z ≤ 2.74 )
= P ( z ≤ 2.74 ) − P ( z ≤ 0 )
= 0.9969 − 0.5000
= 0.4969
(c) µ x = 16, σ x = 0.3651
15 − 16 ⎞
⎛
P ( x < 15 ) = P ⎜ z <
⎟
0.3651 ⎠
⎝
= P ( z < −2.74 )
= 0.0031
11. (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 stocks or bonds, x has a
distribution that is approximately normal (Central Limit Theorem).
(b) µ x = 1.6%, σ x =
σ
n
=
0.9%
= 0.367%
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
Note: It does not matter whether you solve the problem using percents or their decimal equivalents as
long as you are consistent.
(c) Note: 2 years = 24 months; x is monthly percentage return.
σ
0.9%
µ x = 1.6%, σ x =
=
= 0.1837%
24
n
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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Instructor’s Resource Guide Understandable Statistics, 8th Edition
(e) µ x = 1.6%, σ x = 0.1837%
1% − 1.6% ⎞
⎛
P ( x < 1% ) = P ⎜ z <
⎟
0.1837% ⎠
⎝
= P ( z < −3.27 )
= 0.0005
This is very unlikely if µ = 1.6%. One would suspect that µ has slipped below 1.6%.
12. (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 stocks, x has a distribution that
is approximately normal (Central Limit Theorem).
(b) µ x = 1.4%, σ x =
σ
n
=
0.8%
= 0.2667%
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
Note: It does not matter whether you solve the problem using percents or their decimal equivalents as
long as you are consistent.
(c) µ x = 1.4%, σ x =
σ
n
=
0.8%
= 0.1886%
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) µ x = 1.4%, σ 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 µ = 1.4%. One would suspect that the European stock market may be heating
up, i.e., µ is greater than 1.4%.
13. (a) Since x itself represents a sample mean from a large n ≈ 80 (random) sample of bonds, x is
approximately normally distributed according to the Central Limit Theorem.
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Part IV: Complete Solutions, Chapter 7
(b) µ x = 10.8%, σ x =
σ
n
=
429
4.9%
5
= 2.19%
6% − 10.8% ⎞
⎛
P ( x < 6% ) = P ⎜ z <
⎟
2.19% ⎠
⎝
= P ( z < −2.19 )
= 0.0143
Yes. Since this probability is so small, it is very unlikely that x would be less than 6% if µ = 10.8%.
The junk bond market appears to be weaker, i.e., µ is less than 10.8%.
(c) µ x = 10.8%, σ x = 2.19%
16% − 10.8% ⎞
⎛
P ( x > 16% ) = P ⎜ z >
⎟
2.19% ⎠
⎝
= P ( z > 2.37 )
= 1 − P ( z ≤ 2.37 )
= 1 − 0.9911
= 0.0089
Yes. Since this probability is so small, it is very unlikely that x would be greater than 16% if µ =
10.8%. The junk bond market may be heating up, i.e., µ is greater than 10.8%.
14. (a) µ x = 6.4, σ x =
σ
n
=
1.5
= 0.2372
40
7 − 6.4 ⎞
⎛ 6 − 6.4
P (6 ≤ x ≤ 7) = P ⎜
≤z≤
⎟
0.2372
0.2372
⎝
⎠
= P ( −1.69 ≤ z ≤ 2.53)
= P ( z ≤ 2.53) − P ( z ≤ −1.69 )
= 0.9943 − 0.0455
= 0.9488
(b) µ x = 6.4, σ x =
σ
n
=
1.5
= 0.1677
80
7 − 6.4 ⎞
⎛ 6 − 6.4
P (6 ≤ x ≤ 7) = P ⎜
≤z≤
⎟
0.1677
0.1677
⎝
⎠
= P ( −2.39 ≤ z ≤ 3.58 )
= P ( z ≤ 3.58 ) − P ( z ≤ −2.39 )
≈ 1 − 0.0084
= 0.9916
(c) Yes. Since this is such a large probability, the chances of x not being in this time interval is extremely
unlikely. A second security guard should drop in for a look.
15. (a) The sample size should be 30 or more.
(b) No. If the distribution of x is normal, the distribution of x is also normal, regardless of the sample
size.
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Instructor’s Resource Guide Understandable Statistics, 8th Edition
16. (a) By the Central Limit Theorem, the sampling distribution of x is approximately normal with
σ
$7
mean µ x = µ = $20 and standard error σ x =
=
= $0.70. It is not necessary to make any
100
n
assumption about the x distribution because n is large.
(b) µ x = $20, σ 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) µ x = $20, σ = $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, σ n , will decrease as n increases. The standard deviation of x , a.k.a. the standard error
of x , measures the spread of the x values; the smaller σ n is, the less variability there is in
the x values. The less variability there is in the values of x , the more reliable x is as an estimate or
predictor of µ. For large n, approximately 95% of the possible values of x are within 2σ n of µ.
The amount x a typical customer spends on impulse buys also estimates µ (recall µ x = µ x = µ ), but
approximately 95% of individual impulse buys x are within 2σ of µ (using either the Empirical Rule
for somewhat mound-shaped data, or assuming x has a distribution that is approximately normal). 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.
17. (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).
(b) If we divide both sides of w < 90 by 30, we get
so
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 µ x = µ = 2.7 and standard
deviation σ x =
σ
n
=
0.6
30
= 0.1095.
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Part IV: Complete Solutions, Chapter 7
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, i.e., P(w < 90) = 0.9969.
18. Let w = x1 + x2 + … + x36.
σ
2.5
(a) w < 320 is equivalent to w < 320 or x < 8.889. µ x = µ = 8.5, σ x =
=
= 0.4167
36 36
36
n
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. µ x = 8.5, σ 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
19. Let w = x1 + x2 + … + x45.
σ
84
(a) w < 9500 is equivalent to w < 9500 or x < 211.111. µ x = 240, σ x =
=
= 12.522
45
45
45
n
P ( w < 9500 ) = P ( x < 211.111)
211.111 − 240 ⎞
⎛
= P⎜ z <
⎟
12.522
⎝
⎠
= P ( z < −2.31)
= 0.0104
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431
432
Instructor’s Resource Guide Understandable Statistics, 8th Edition
12, 000
(b) w < 12,000 is equivalent to w >
or x > 266.667. µ x = 240, σ x = 12.522
45
45
P ( w > 12, 000 ) = P ( x > 266.667 )
266.667 − 240 ⎞
⎛
= P⎜z >
⎟
12.522
⎝
⎠
= P ( z > 2.13)
= 1 − P ( z ≤ 2.13)
= 1 − 0.9834
= 0.0166
(c) P ( 9500 < w < 12, 000 ) = P ( 211.111 < x < 266.667 )
= P ( −2.31 < z < 2.13)
= P ( z < 2.13) − P ( z < −2.31)
= 0.9834 − 0.0104
= 0.9730
20. (a) Let w = x1 + x2 + … + x9. µ x = µ = 6.3, σ x =
σ
n
=
1.2
9
= 0.4
⎛ w 60 ⎞
P ( w < 60 ) = P ⎜ < ⎟
9 ⎠
⎝9
= P ( x < 6.667 )
6.667 − 6.3 ⎞
⎛
= P⎜ z <
⎟
0.4
⎝
⎠
= P ( z < 0.92 )
= 0.8212
⎛ w 65 ⎞
P ( w > 65 ) = P ⎜ > ⎟
9 ⎠
⎝9
= P ( x > 7.222 )
7.222 − 6.3 ⎞
⎛
= P⎜z >
⎟
0.4
⎝
⎠
= P ( z > 2.31)
= 1 − P ( z ≤ 2.31)
= 1 − 0.9896
= 0.0104
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Part IV: Complete Solutions, Chapter 7
433
(b) Let w = x1 + x2 + … + x50. µ x = µ = 6.3, σ x =
σ
n
=
1.2
50
= 0.170
⎛ w 342 ⎞
P ( w < 342 ) = P ⎜ <
⎟
⎝ 50 50 ⎠
P ( x < 6.84 )
6.84 − 6.3 ⎞
⎛
= P⎜ z <
⎟
0.170 ⎠
⎝
= P ( z < 3.18 )
= 0.9993
No. By the Central Limit Theorem the sample size is large enough so the sampling distribution of x is
approximately normal.
21. (a) Let w = x1 + x2 +
+ x5 .
σ
3.3
=
= 1.476
n
5
⎛ w 90 ⎞
P ( w > 90) = P ⎜ > ⎟
5 ⎠
⎝5
= P ( x > 18)
µ x = µ = 17, σ x =
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. (a) Answers vary.
(b) The random variable p̂ can be approximated by a normal random variable when both np and nq
exceed 5.
µ pˆ = p, σ pˆ =
pq
n
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Instructor’s Resource Guide Understandable Statistics, 8th Edition
(c) np = 33 ( 0.21) = 6.93, nq = 33 ( 0.79 ) = 26.07
Yes, p̂ can be approximated by a normal random variable since both np and nq exceed 5.
0.21( 0.79 )
≈ 0.071
33
µ pˆ = p = 0.21, σ pˆ =
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
(d) No; np = 25(0.15) = 3.75 which does not exceed 5.
(e) np = 48 ( 0.15 ) = 7.2, nq = 48 ( 0.85 ) = 40.8
Yes, p̂ can be approximated by a normal random variable since both np and nq exceed 5.
µ pˆ = p = 0.15, σ pˆ =
continuity correction =
0.15 ( 0.85 )
≈ 0.052
48
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
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Part IV: Complete Solutions, Chapter 7
435
2. (a) n = 50, p = 0.36
np = 50 ( 0.36 ) = 18, nq = 50 ( 0.64 ) = 32
Approximate p̂ by a normal random variable since both np and nq exceed 5.
µ pˆ = p = 0.36, σ pˆ =
continuity correction =
0.36 ( 0.64 )
≈ 0.068
50
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 since both np and nq exceed 5.
µ pˆ = p = 0.25, σ pˆ =
continuity correction =
0.25 ( 0.75 )
≈ 0.070
38
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 since np < 5.
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3. n = 30, p = 0.60
np = 30 ( 0.60 ) = 18, nq = 30 ( 0.40 ) = 12
Approximate p̂ by a normal random variable since both np and nq exceed 5.
µ pˆ = p = 0.6, σ pˆ =
0.6 ( 0.4 )
≈ 0.089
30
continuity correction =
0.5 0.5
=
= 0.017
n
30
(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.
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Part IV: Complete Solutions, Chapter 7
437
4. (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 since both np and nq exceed 5.
µ pˆ = p = 0.73, σ pˆ =
0.73 ( 0.27 )
≈ 0.072
38
continuity correction =
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 since both np and nq exceed 5.
µ pˆ = p = 0.86, σ pˆ =
0.86 ( 0.14 )
≈ 0.052
45
continuity correction =
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.
5. n = 55, p = 0.11
np = 55 ( 0.11) = 6.05, nq = 55 ( 0.89 ) = 48.95
Approximate p̂ by a normal random variable since both np and nq exceed 5.
µ pˆ = p = 0.11, σ pˆ =
0.11( 0.89 )
≈ 0.042
55
continuity correction =
0.5 0.5
=
= 0.009
n
55
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438
Instructor’s Resource Guide Understandable Statistics, 8th Edition
(a) P ( pˆ ≤ 0.15 ) ≈ P ( x ≤ 0.15 + 0.009 )
= P ( x ≤ 0.159 )
0.159 − 0.11 ⎞
⎛
= P⎜z ≤
⎟
0.042 ⎠
⎝
= P ( z ≤ 1.17 )
= 0.8790
(b) P ( 0.10 ≤ pˆ ≤ 0.15 ) ≈ P ( 0.10 − 0.009 ≤ x ≤ 0.15 + 0.009 )
= P ( 0.091 ≤ x ≤ 0.159 )
0.159 − 0.11 ⎞
⎛ 0.091 − 0.11
= P⎜
≤z≤
⎟
0.042
0.042 ⎠
⎝
= P ( −0.45 ≤ z ≤ 1.17 )
= P ( z ≤ 1.17 ) − P ( z ≤ −0.45 )
= 0.8790 − 0.3264
= 0.5526
(c) Yes, both np and nq exceed 5.
6. n = 28, p = 0.31
np = 28 ( 0.31) = 8.68, nq = 28 ( 0.69 ) = 19.32
Approximate p̂ by a normal random variable since both np and nq exceed 5.
µ pˆ = p = 0.31, σ pˆ =
0.31( 0.69 )
≈ 0.087
28
continuity correction =
0.5 0.5
=
= 0.018
n
28
(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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Part IV: Complete Solutions, Chapter 7
439
7. (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 since both np and nq exceed 5.
µ pˆ = p = 0.06, σ pˆ =
0.06 ( 0.94 )
≈ 0.024
100
continuity correction =
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, since this probability is so small, it should rarely occur. The machine might need an adjustment.
8. (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 since both np and nq exceed 5.
µ pˆ = p = 0.565, σ pˆ =
0.565 ( 0.435 )
≈ 0.070
50
continuity correction =
0.5 0.5
=
= 0.01
50
n
(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
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Instructor’s Resource Guide Understandable Statistics, 8th Edition
(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%.
9.
total number of successes from all 12 quarters
total number of families from all 12 quarters
11 + 14 + … + 19
=
12 ( 92 )
p=
206
1104
= 0.1866
=
q = 1 − p = 1 − 0.1866 = 0.8134
µ pˆ = p ≈ p = 0.1866
σ 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
441
P Chart for r
3.0SL=0.3084
0.3
Proportion
2.0SL=0.2678
0.2
–
P=0.1866
0.1
–3.0SL=0.1054
–3.0SL=0.06474
5
Sample Number
0
10
There are no out-of-control signals.
10.
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
µ pˆ = p ≈ p = 0.08061
σ 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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Instructor’s Resource Guide Understandable Statistics, 8th Edition
P Chart for r
3.0SL=0.1585
0.15
Proportion
2.0SL=0.1325
0.10
–
P=0.08061
0.05
–3.0SL=0.02869
0.00
–3.0SL=0.002738
0
5
10
Sample Number
15
There are no out-of-control signals. It appears that the production process is in reasonable control.
11.
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
µ pˆ = p ≈ p = 0.7751
σ 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
443
P Chart for r
1
3.0SL=0.9197
Proportion
0.9
2.0SL=0.8715
0.8
–
P=0.7751
0.7
–3.0SL=0.6787
0.6
–3.0SL=0.6305
1
0
5
10
Sample Number
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 µ x of the x distribution equals the mean µ of the x distribution, regardless of the sample
size.
(c) The standard deviation σ x of the sampling distribution equals
σ
n
, where σ 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
σ
50
and
σ
100
respectively.
2. All the x distributions will be normal with mean µ x = µ = 15. The standard deviations will be:
n = 4: σ x =
n = 16: σ x =
n = 100: σ x =
σ
n
σ
n
σ
n
=
=
=
3
4
3
16
3
=
3
2
=
3
4
100
=
3
10
3. (a) µ = 35, σ = 7
40 − 35 ⎞
⎛
P ( x ≥ 40 ) = P ⎜ z ≥
⎟
7 ⎠
⎝
= P ( z ≥ 0.71)
= 0.2389
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Instructor’s Resource Guide Understandable Statistics, 8th Edition
(b) µ x = µ = 35, σ x =
σ
n
=
7
7
=
9 3
40 − 35 ⎞
⎛
P ( x ≥ 40 ) = P ⎜ z ≥
⎟
7
⎜
⎟
3
⎝
⎠
= P ( z ≥ 2.14 )
= 0.0162
4. (a) µ = 38, σ = 5
35 − 38 ⎞
⎛
P ( x ≤ 35 ) = P ⎜ z ≤
⎟
5 ⎠
⎝
= P ( z ≤ −0.6 )
= 0.2743
(b) µ x = µ = 38, σ x =
σ
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. µ x = µ = 100, σ x =
σ = 15 = 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. µ x = µ = 15, σ x =
σ = 2 = 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
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Part IV: Complete Solutions, Chapter 7
7. µ x = µ = 750, σ x =
445
σ = 20 = 2.5
n
64
750 − 750 ⎞
⎛
(a) P ( x ≥ 750 ) = P ⎜ z ≥
⎟
2.5 ⎠
⎝
= P ( z ≥ 0)
= 0.5000
755 − 750 ⎞
⎛ 745 − 750
≤z≤
(b) P ( 745 ≤ x ≤ 755 ) = P ⎜
⎟
2.5
2.5 ⎠
⎝
= P ( −2 ≤ z ≤ 2 )
= P ( z ≤ 2 ) − P ( z ≤ −2 )
= 0.9772 − 0.0228
= 0.9544
8. (a) Miami: µ = 76, σ = 1.9
77 − 76 ⎞
⎛
P ( x < 77 ) = P ⎜ z <
⎟
1.9 ⎠
⎝
= P ( z < 0.53)
= 0.7019
Fairbanks: µ = 0, σ = 5.3
3−0⎞
⎛
P ( x < 3) = P ⎜ z <
⎟
5.3 ⎠
⎝
= P ( z < 0.57 )
= 0.7157
(b) Since x has a normal distribution, the sampling distribution of x is also normal regardless of the
sample size.
σ = 1.9 = 0.718
Miami: µ x = µ = 76, σ x =
n
7
77 − 76 ⎞
⎛
P ( x < 77 ) = P ⎜ z <
⎟
0.718 ⎠
⎝
= P ( z < 1.39 )
= 0.9177
Fairbanks: µ x = µ = 0, σ x =
σ = 5.3 = 2.003
n
7
3−0 ⎞
⎛
P ( x < 3) = P ⎜ z <
⎟
2.003 ⎠
⎝
= P ( z < 1.50 )
= 0.9332
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Instructor’s Resource Guide Understandable Statistics, 8th Edition
(c) We cannot say anything about the probability distribution of x , because the sample size is not 30 or
greater. Consider using all 31 days.
σ = 1.9 = 0.341
Miami: µ x = µ = 76, σ x =
n
31
77 − 76 ⎞
⎛
P ( x < 77 ) = P ⎜ z <
⎟
0.341 ⎠
⎝
= P ( z < 2.93)
= 0.9983
Fairbanks: µ x = µ = 0, σ x =
σ = 5.3 = 0.952
n
31
3−0 ⎞
⎛
P ( x < 3) = P ⎜ z <
⎟
0.952 ⎠
⎝
= P ( z < 3.15 )
= 0.9992
9. (a) n = 50, p = 0.22
np = 50 ( 0.22 ) = 11, nq = 50 ( 0.78 ) = 39
Approximate p̂ by a normal random variable since both np and nq exceed 5.
µ pˆ = p = 0.22, σ pˆ =
continuity correction =
0.22 ( 0.78 )
50
≈ 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
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Part IV: Complete Solutions, Chapter 7
447
(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 since both np and nq exceed 5.
µ pˆ = p = 0.27, σ pˆ =
continuity correction =
( 0.27 )( 0.73)
38
≈ 0.0720
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 since np < 5.
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