203
Chapter 9
9.1 (a) µ =2.5003 is a parameter (related to the population of all the ball bearings in the container and x
=2.5009 is a statistic (related to the sample of 100 ball bearings). (b) ˆ =7.2% is a statistic (related to the sample of registered voters who were unemployed).
9.2 (a) ˆ = 48% is a statistic; = 52% is a parameter. (b) Both x control
= 335 and x experimental
=
289 are statistics.
9.3 (a) Since the proportion of times the toast will land butter-side down is 0.5, the result of 20 coin flips will simulate the outcomes of 20 pieces of falling toast (landing butter-side up or butter-side down). (b) Answers will vary. A histogram for one simulation is shown below (on the left). The center of the distribution is close to 0.5. (c) Answers will vary. A histogram based on pooling the work of 25 students (250 simulated values of ˆ ) is shown below (on the right).
As expected, the simulated distribution of ˆ is approximately Normal with a center at 0.5.
60
3.0
50
2.5
40
2.0
30
1.5
1.0
20
0.5
10
0.0
0.35
0.40
0.45
0.50
0.55
Simulated proportion
0.60
0.65
0
0.300
0.375
0.450
0.525
0.600
Simulated proportion
0.675
0.750
(d) Answers will vary, but the standard deviation will be close to 0.1118
. The
20 simulation above for the pooled results for 25 students produced a standard deviation of 0.1072.
(e) By combining the results from many students, he can get a more accurate estimate of the value of p since the value of ˆ approaches p as the sample size increases.
9.4 (a) A histogram is shown below. The center of the histogram is close to 0.5, but there is considerable variation in the overall shape of the histograms for different simulations with only
10 repetitions.

204 Chapter 9
(b) A histogram for 100 repetitions is shown below (on the left). The distribution is approximately Normal with a center at 0.50.
(c) The mean and the median are extremely close to one another. See the plot above (on the right). (d) The spread of the distribution will not change. To decrease the spread, the sample size should be increased.
9.5 (a) The scores will vary depending on the starting row. Note that the smallest possible mean is 61.75 (from the sample 58, 62, 62, 65) and the largest is 77.25 (from 73, 74, 80, 82). One simulation produced a sample of 73, 82, 74, and 62 with a mean of x
= 72.75
. (b) Answers will vary. A histogram for one set of 10 simulated means is shown below (left). The center of the simulated distribution is slightly higher than 69.4. (c) Answers will vary. A histogram for a set of 250 simulated means is shown below (right). The simulated distribution is approximately
Normal with a mean very close to 69.4. The average of the 205 simulated means is 69.262.
40
4
30
3
20
2
1 10
0
68 69 70 71 72
Simulated Mean
73 74 75
0
62 64 66 68 70
Simulated mean
72 74 76

Sampling Distributions 205
7
6
5
4
3
2
9.6 (a) There are 45 possible samples of size 2 that can be drawn. The frequency table below shows the values of the sample mean for samples of size 2, the number of times that mean occurs in the 45 samples, and the corresponding percent.
Mean2 Count Percent
60.0 2 4.44
61.5 1 2.22
62.0 2 4.44
63.5 2 4.44
64.0 2 4.44
65.0 1 2.22
65.5 2 4.44
66.0 1 2.22
67.0 2 4.44
67.5 2 4.44
68.0 2 4.44
68.5 1 2.22
69.0 3 6.67
69.5 2 4.44
70.0 2 4.44
71.0 2 4.44
72.0 2 4.44
72.5 2 4.44
73.0 2 4.44
73.5 2 4.44
74.0 1 2.22
76.0 1 2.22
76.5 1 2.22
77.0 2 4.44
77.5 1 2.22
78.0 1 2.22
81.0 1 2.22
A probability histrogram is shown below.
1
0
60
.0
61
.5
62
.0
63
.5
64
.0
65
.0
65
.5
66
.0
67
.0
67
.5
68
.0
68
.5
69
.0
69
.5
70
.0
71
.0
72
.0
72
.5
73
Mean of 2 observations
.0
73
.5
74
.0
76
.0
76
.5
77
.0
77
.5
78
.0
81
.0
(b) The shapes and spreads are different, but the centers are the same. The distribution for n = 2 is roughly symmetric, but it does not have the general shape of a Normal distribution. For n = 4, the distribution of the sample mean resembles the shape of a Normal distribution. Both distributions are roughly centered at 69.4. However, the spread is a little larger for the distribution corresponding to n = 2.
9.7 (a) A histogram is shown below (on the left). (b) µ 141.847
days. (c) Means will vary with samples. The mean of our first sample was 120.833. (d) Our four additional samples produced means of 183.4, 212.8, 127.3, and 119.7. It would be unlikely (though not impossible) for all

206 Chapter 9 five values to fall on the same side of µ . This is one implication of the unbiasedness of x
: Some values will be higher and some lower than µ , but not necessarily a 50/50 split. (e) The mean of the (theoretical) sampling distribution is µ 141.847
. (f) Answers will vary. A histogram for 100 sample means of size 12 is shown below (on the right). The distribution of sample mean looks more Normal than the population distribution of survival times, but it is still skewed to the right.
The center of the distribution is about 145.9 and the means ranged from a low of 88.75 to a high of 260.42. The standard deviation of the 100 means was 33.02.
35
30
25
20
15
10
5 mean
141.8
10
8
14
12
18
16
6
4
2
0
120 240 360
Survival time (days)
480 600
0
90 120 150 180
Sample mean
210 240
9.8 The table below shows the count, sample proportion, frequency, and percent for each distinct value. A probability histogram is also provided.
Count Sample_Prop Frequency Percent
9 0.045 1 1
13 0.065 3 3
14 0.070 2 2
15 0.075 5 5
16 0.080 11 11
17 0.085 12 12
18 0.090 12 12
19 0.095 9 9
20 0.100 7 7
21 0.105 5 5
22 0.110 6 6
23 0.115 7 7
12
10
8
6
4 mean = 0.0981
2
24 0.120 10 10
25 0.125 4 4
26 0.130 1 1
0
0.
04
5
0.
06
5
0.
07
0
0.
07
5
0.
08
0
0.
08
5
0.
09
0
09
5
10
0
0.
10
5
0.
11
0
0.
0.
0.
Sample Proportion
11
5
0.
12
0
0.
12
5
0.
13
0
0.
13
5
0.
14
0
0.
15
0
27 0.135 2 2
28 0.140 2 2
30 0.150 1 1
(b) The distribution is bimodal, with a very small bias. (c) The mean of the 100 observed values of the sample proportion is 0.0981. The center of the sampling distribution is about 0.0019 below where we would expect it to be, so there appears to be a very small bias. (d) The mean of the sampling distribution of ˆ is 0.10. (e) By increasing the sample size from 200 to 1000, the mean of ˆ p would be smaller.
9.9 (a) Since the smallest number of total tax returns (i.e., the smallest population) is still more than 100 times the sample size, the variability will be (approximately) the same for all states. (b)
Yes, it will change—the sample taken from Wyoming will be about the same size, but the

Sampling Distributions 207 sample in, e.g., California will be considerably larger, and therefore the variability will be smaller.
9.10 (a) Large bias, large variability. (b) Small bias, small variability. (c) Small bias, large variability. (d) Large bias, small variability.
9.11 Both ˆ = 40.2% and ˆ = 31.7% are statistics (related, respectively, to the sample of smallclass and regular-size-class black students).
9.12 The sample mean x
= 64.5
inches is a statistic and the population mean µ = 63 inches is a parameter.
9.13 (a) If we choose many samples, the average of the x
-values from these samples will be close to µ . In other words, the sampling distribution of x
is centered at the population mean µ we are trying to estimate. (b) The larger sample will give more information, and therefore more precise results. The variability in the distribution of the sample average decreases as the sample size increases.
9.14 (a) Use digits 0 and 1 (or any other 2 of the 10 digits) to represent the presence of egg masses. Reading the first 10 digits from line 116, for example, gives YNNNN NNYNN—2 square yards with egg masses, 8 without—so ˆ = 0.2. (b) The numbers and proportions of square yards with egg masses are shown below. A stemplot is also shown below (right). The mean of this approximately Normal distribution is 0.2.
Eggmasses p-hat
0 0.0
3 0.3
2 0.2
2 0.2
2 0.2
4 0.4
3 0.3
3 0.3
2 0.2
1 0.1
3 0.3
Stem-and-leaf of p-hat N = 20
1 0.1
Leaf Unit = 0.010
1 0.1
1 0.1
2 0.2
1 0 0
1 0.1
8 1 0000000
1 0.1 (5) 2 00000
4 0.4
7 3 00000
3 0.3
2 4 00
1 0.1
(c) The mean of the sampling distribution of ˆ is 0.2. (d) The mean of the sampling distribution of ˆ is 0.4 in this other field.
9.15 (a) A probability histogram for the number of dots on the upward facing side is shown below. The mean is µ = 3.5 dots and the standard deviation is σ
=
1.708 dots.

208 Chapter 9
0.10
0.08
0.06
0.04
0.18
0.16
0.14
0.12
0.02
0.00
1 2 3 4
Number of dots
5 6 and the sample averages are shown below.
(b) This is equivalent to rolling a pair of fair, six-sided dice. (c) The 36 possible SRSs of size 2
1,1; x
=1.0
2,1; 1,2; x
=1.5
3,1; 2,2; 1,3;
4,1; 2,3; 3,2; 1,4;
5,1; 2,4; 3,3; 4,2; 1,5; x x x
=2.0
=2.5
=3.0
6,1; 2,5; 3,4; 4,3; 5,2; 1,6; x
=3.5
6,2; 3,5; 4,4; 5,3; 2,6; x
=4.0
6,3; 4,5; 5,4; 3,6; x
=4.5
6,4; 5,5; 4,6
6,5; 5,6; x x
=5.0
=5.5
6,6; x
=6.0
(d) The sampling distribution of x
is shown below. The center is identical to that of the
( population distribution, the shape is symmetrical, single-peaked, and bell shaped and the spread
σ 1.2076
) is smaller than that of the population distribution.
0.10
0.08
0.06
0.04
0.18
0.16
0.14
0.12
0.02
0.00
1.0
1.5
2.0
2.5
3.0
3.5
4.0
Sample mean
4.5
5.0
5.5
6.0
9.16 Answers will vary. A histogram of the x
3
-values is shown below. While the center of the distribution remains the same, the spread in this distribution is smaller than the spread in the sampling distribution of x
for samples of size n = 2.

Sampling Distributions 209
15
10
5
30
25
20
0
1.6
2.4
3.2
Sample average
4.0
4.8
5.6
9.17 Assuming that the poll’s sample size was less than 870,000—10% of the population of
New Jersey—the variability would be practically the same for either population. (The sample size for this poll would have been considerably less than 870,000.)
9.18 (a) The digits 1 to 41 are assigned to adults who say that they have watched
Survivor:
Guatemala . The program outputs a proportion of “Yes” answers. For (b), (c), (d), and (e), answers will vary; however, as the sample size increases from 5 to 25 to 100, the variability in the sampling distributions of the sample proportion will decrease.
9.19 (a) The mean is µ = = 0.7 and the standard deviation is
σ = p
(1 − p
)
= 0.0144
. (b) The population (all U.S. adults) is clearly at least 10 n 1012 times as large as the sample (the 1012 surveyed adults). (c) The two conditions, np
= 1012×0.7
= 708.4 > 10 and n
(1 − p
) = 1012×0.3 = 303.6 > 10, are both satisfied. (d) P( ) = P(Z ≤
− 2.08) = 0.0188. This is a fairly unusual result if 70% of the population actually drinks the cereal milk. (e) To half the standard deviation of the sample proportion, multiply the sample size by 4; we would need to sample 1012×4 = 4048 adults. (f) It would probably be higher, since teenagers (and children in general) have a greater tendency to drink the cereal milk.
9.20 (a) The mean is µ = = 0.4 and the standard deviation is σ = 0.0116
. (b)
1785
The population (all adults) is considerably larger than 10 times the sample size ( n
= 1785 adults).
(c) The two conditions, np
= 1785×0.4 = 714 > 10 and n
(1 − p
) = 1785×0.6 = 1071 > 10, are both satisfied. (d)
P
(
0.37
0.43
)
=
P ⎜
⎛
⎝ 0.0116
0.0116
⎟
⎞
⎠
P
( − 2.59
2.59
)
=
0.9952 − 0.0048 = 0.9904. Over 99% of all samples of size n = 1785 will produce a sample proportion ˆ within ±0.03 of the true population proportion.
9.21 For n
= 300, the standard deviation is σ approximately equal to
P ⎜
⎛
⎝ 0.0283
=
0.0283
0.0283
and the probability is
⎟
⎞
⎠
300
P
( − 1.06
1.06
) = 0.8554 − 0.1446 =

210 Chapter 9
0.7108. For n = 1200, the standard deviation is σ = 0.0141
and the probability is approximately equal to
P ⎜
⎛
⎝ 0.0141
0.0141
⎟
⎞
⎠
P
(
1200
− 2.13
2.13
) = 0.9834 − 0.0166 =
0.9668. For n
= 4800, the standard deviation is σ = 0.0071
and the probability is
4800 approximately equal to
P ⎜
⎛
⎝ 0.0071
0.0071
⎟
⎞
⎠
P
( − 4.23
4.23
)
1 . Larger sample sizes produce sampling distributions of the sample proportion that are more concentrated about the true proportion.
9.22 (a) The distribution of the sample proportion is approximately normal with mean 0.14 and standard deviation σ = 0.0155
. (b) 20% or more Harley owners is unlikely:
500
( )
⎝
⎜ ≥
0.0155
⎟
⎞
⎠
= ≥ 3.87
) < 0.0002
. (c) There is a fairly good chance of finding at least 15% Harley owners: ( )
⎝
≥
0.0155
⎟
⎞
⎠
= ≥ 0.64
) = 1 − 0.7389 = 0.2611.
9.23 (a) The sample proportion is 86/100 = 0.86 or 86%. (b) We can use the normal approximation, but Rule of Thumb 2 is just barely satisfied: n (1 − p ) = 10. The standard deviation of the sample proportion is σ = 0.03
and the probability is
100
( ≤ )
⎝
⎜
⎛
P Z
≤
0.03
⎟
⎞
⎠
= ≤ − 1.33
)
= 0.0918. (Note: The exact probability is 0.1239.)
(c) If the claim is correct, then we can expect to observe 86% or fewer orders shipped on time in about 12.5% of the samples of this size. Getting a sample proportion at or below 0.86 is not an unlikely event.
9.24 If ˆ p is approximately Normal with mean 0.7 and standard deviation σ = 0.02804
. The probability is
267 approximately equal to
( ≥ )
⎝
⎜
⎛
P Z
≥
0.02804
⎟
⎞
⎠
= ≥ 1.78
)
= 1 − 0.9625 = 0.0375.
(Software gives 0.0373). Alternatively, as ˆ ≥ 0.75 is equivalent to 201 or more dieters in the sample, we can compute this probability using the binomial distribution. The exact probability is
( ≥ = − ( ≤ 200
)
= 1 − 0.9671= 0.0329.
9.25 (a) The mean is µ = = 0.15 and the standard deviation is σ = 0.0091
. (b)
1540
The population (all adults) is considerably larger than 10 times the sample size ( n
= 1540). (c)
The two conditions, np = 1540×0.15 = 231 ≥ 10, n (1 − p ) = 1540×0.85 = 1309 ≥ 10, are both

Sampling Distributions 211 satisfied. (d)
P
(
0.13
0.17
)
=
P ⎜
⎛
⎝ 0.0091
0.0091
⎟
⎞
⎠
P
( − 2.20
2.20
) = 0.9861
0.0139 = 0.9722. (Software gives 0.972054.) (e) To reduce the standard deviation in (a) by a third, we need a sample nine times as large; n
= 13,860.
−
9.26 For n = 200, the standard deviation is σ = approximately equal to
P ⎜
⎛
⎝ 0.0252
0.0252
0.0252
and the probability is
200
⎟
⎞
⎠
P
( − 0.79
0.79
) = 0.7852 − 0.2148 =
0.5704. For n
= 800, the standard deviation is σ = approximately equal to
P ⎜
⎛
⎝ 0.0126
0.0126
⎟
⎞
⎠
800
P
( − 1.59
0.0126
and the probability is
1.59
)
= 0.9441 − 0.0559 =
0.8882. For n
= 3200, the standard deviation is σ = 0.0063
and the probability is
3200 approximately equal to
P ⎜
⎛
⎝ 0.0063
0.0063
⎟
⎞
⎠
P
( − 3.17
3.17
) = 0.9992 − 0.0008 =
0.9984. Larger sample sizes produce sampling distributions of the sample proportion that are more concentrated about the true proportion.
9.27 (a) The sample proportion is ˆ = 62/100 = 0.62. (b) The mean is µ = = 0.67 and the standard deviation is σ =
100
0.0470
. ( )
⎝
⎜ ≤ could arise by chance even if the true campus proportion is the same.
0.047
⎟
⎞
⎠
≤ − 1.06
)
= 0.1446. (c) Getting a sample proportion at or below 0.62 is not an unlikely event. The sample results are lower than the national percentage, but the sample was so small that such a difference
9.28 (a) The mean is µ = = 0.52 and the standard deviation is σ = 0.0223
(b)
500
The population (all residential telephone customers in Los Angeles) is considerably larger than
10 times the sample size ( n = 500). The two conditions, np = 500×0.52 = 260 > 10, n (1 − p ) =
500×0.48 = 240 > 10, are both satisfied. ( )
⎝
⎜ ≥
0.0223
⎟
⎞
⎠
= ≥ − 0.90
) = 1 −
0.1841 = 0.8159.
9.29 (a) The mean is µ = 0.75 and the standard deviation is σ = = 0.0433.
100
( )
⎝
⎜ ≤
0.0433
⎟
⎞
⎠
= ≤ − 1.15
) = 0.1251. (b) The mean is µ = 0.75 and the standard deviation is σ =
250
0.0274
. ( )
⎝
⎜
≤
0.0274
⎟
⎞
⎠
= ≤ − 1.82
) =
0.0344. (c) To reduce the standard deviation for a 100-item test by a fourth, we need a sample

212 Chapter 9 sixteen times as large; n = 1600. (d) Yes, the answer is the same for Laura. Taking a sample sixteen times as large will cut the standard deviation by a fourth, for all values of p
.
9.30 (a) One of the two conditions in Rule of Thumb 2 is not satisfied; np = 15×0.3 = 4.5 < 10.
(b) Rule of Thumb 1 is not satisfied. The population size (316) is not at least 10 times as large as the sample size ( n
= 50). (c)
( ≤ 3
)
= binomcdf(15, 0.3, 3) 0.2969.
9.31 (a) The mean is
(
(
>
>
5
5
)
)
⎝
⎜
⎛
P Z
⎝
⎜
⎛
P Z
>
>
µ x
26
=
µ
= -
3.5% and the standard deviation is σ x
=
σ n
=
11.6276
⎟
⎞
⎠
=
⎟
⎞
⎠
= >
> 0.33
0.73
)
) = 1 – 0.6293 = 0.3707. (c)
= 1 – 0.7673 = 0.2327. (d)
26
5
11.6276% . (b)
( < 0
) <
⎝ 11.6276
portfolios lost money.
9.32 (a) > 21
)
⎝
⎜
⎛
P Z
>
⎟
⎞
⎠
=
5.9
< 0.30
⎟
⎞
⎠
=
)
= 0.6179. Approximately 62% of all five-stock
> 0.41
)
= 1 – 0.6591 = 0.3409. (Software gives
0.3421.) (b) The mean is µ x
= 18.6
and the standard deviation is σ x
=
5.9
0.8344
. These
50 results do not depend on the distribution of the individual scores. (c)
> 21
)
⎝
>
0.8344
⎟
⎞
⎠
= > 2.88
) = 1 – 0.9980 = 0.002.
9.33 (a) The standard deviation is σ x
=
10
3
5.7735
milligrams. (b) Solve
10
= 3; n = 11.1, so n n
= 12. There is less variability in the average of several measurements than there is in a single measurement. Also, the average of several measurements is more likely to be close to the true mean than a single measurement.
9.34 (a) If x is the mean number of strikes per square kilometer, then µ x
= 6 strikes / km 2 and
σ x
=
2.4
10
0.7589
strikes
/ km 2 . (b) We cannot calculate the probability because we do not know the distribution of the number of lightning strikes. If we were told the population is Normal, then we would be able to compute the probability.
9.35 The sample mean x has approximately a
N
(
1.6,1.2
200 0.0849
) distribution. The probability is approximately equal to ( > 2
)
⎝
>
0.0849
⎟
⎞
⎠
= > 4 .71
) , which is essentially
0.

Sampling Distributions 213
9.36 The central limit theorem says that over 40 years, the mean return x
is approximately
Normal with mean µ x
=13.2% and standard deviation σ x
=
17.5
40
2.7670% . Therefore,
> 15%
)
⎝
>
2.767
⎟
⎞
⎠
= > 0.65
) = 1 – 0.7422 = 0.2578 and
< 10%
)
⎝
⎜ <
2.767
⎟
⎞
⎠
= < − 1.16
) = 0.1230.
9.37 (a) No this probability cannot be calculated, because we do not know the distribution of the weights. (b) If
W
is total weight and x
=
W
20 , the central limit theorem says that x
is
35 approximately Normal with mean 190 lb and standard deviation σ = x
7.8262
lb. Thus,
20
> 4000
) = ( > 200
)
⎝
⎜
⎛
P Z
>
7.8262
⎟
⎞
⎠
= > 1.28
)
= 1 – 0.8997 = 0.1003. There is about a 10% chance that the total weight exceeds the limit of 4000 lb.
9.38 (a) The mean is µ = x
40.125
mm and the standard deviation is σ = x
0.002
= 0.001
mm. These
4 results do not depend on the distribution of the individual axel diameters. (b) No, the probability cannot be calculated because we do not know the distribution of the population, and n
= 4 is too small for the central limit theorem to provide a reasonable approximation.
9.39 (a) Let
X
denote Sheila’s glucose measurement.
> 140
) =
⎝
>
10
⎟
⎞
⎠
= > 1.5
) = 1 – 0.9332 = 0.0668. (b) If x
is the mean of four measurements (assumed to be independent), then x
has a
N
(
125,10 4
)
=
N
(
125mg/dl,5mg/dl
) distribution and > 140
) =
⎝
⎜
⎛
P Z
>
5
⎟
⎞
⎠
= > 3.0
) =1 – 0.9987 =
0.0013. (c) No, Sheila’s glucose levels follow a Normal distribution, so there is no need to use the central limit theorem.
9.40 The mean of four measurements has a
N
(
125mg/dl,5mg/dl
) distribution, and
(
> 1.645
)
= 0.05
if Z is N (0, 1), so L = 125 + 1 .
645×5 = 133 .
225 mg / dl.
9.41 (a) ) Let
X
denote the amount of cola in a bottle.
< 295
) =
⎝
⎜ <
3
⎟
⎞
⎠
= < − 1
) = 0.1587. (b) If x
is the mean contents of six bottles
(assumed to be independent), then x
has a
N
(
298,3 6
)
=
N
(
298ml,1.2247ml
) distribution and < 295
) =
⎝
⎜
⎛
P Z
<
1.2247
⎟
⎞
⎠
< − 2.45
)
=0.0071.

214 Chapter 9
9.42 (a) The sample mean lifetime has a
N
(
55000, 4500 8
)
N
(
55000 miles, 1591 miles
) distribution. (b) < 51,800
) =
⎝
⎜
⎛
P Z
<
1591
⎟
⎞
⎠
< − 2.01
) = 0.0222.
9.43 (a) The approximate distribution of the mean number of accidents x
is
N
(
2.2,1.4
52
)
N
(
2.2, 0.1941
)
. (b)
( < 2
) =
⎝
⎜
⎛
P Z
<
0.1941
⎟
⎞
⎠
< − 1 .03
)
X
denotes the number of accidents at an intersection per year, then
= 0.1515. (c) If
< 100
) ( < 100 / 52
) =
⎝
⎜ <
0.1941
⎟
⎞
⎠
< − 1.43
) = 0.0764.
9.44 The mean of 22 measurements has a
N
(
13.6,3.1 22
) distribution, and
( < − 1.645
) = 0.05
if Z is N (0, 1), so L = 13.6 − 1 .
645×0.6609 = 12.5128 points.
9.45 The mean loss from fire, by definition, is the long-term average of many observations of the random variable X = fire loss. The behavior of X is much less predictable if only a small number of observations are made. If only 12 policies were sold, then the company would have no protection against the large expense that would be incurred if at least one of the 12 policyholders happened to lose his or her home. If thousands of policies were sold, then the average fire loss for these policies would be far more likely to be close to µ , and the company’s profit would not be endangered by the few large fire-loss payments that it would have to make.
9.46 The approximate distribution of the average loss x
is
N
(
$250,$300 10, 000
)
=
N
(
$250,$3
)
and > $260
) =
⎝
>
3
⎟
⎞
⎠
> 3.33
)
0.9996 = 0.0004.
CASE CLOSED!
(1) The central limit theorem suggests that the mean of 30 lifetimes will be approximately
= 1 –
0.8
Normal with mean µ x
= 17 hours and the standard deviation of σ x
= 0.1461
hours. (2)
30
< 16.7
)
⎝
⎜
⎛
P Z
<
0.1461
⎟
⎞
⎠
= < − 2.05
)
= 0.0202. (3) There is only about a 2% chance of observing a sample average as small or smaller than the one observed
( x
≤ 16.7
) if the process is working properly. This small probability provides evidence that the process is not working properly. (4) Let ˆ denote the sample proportion of unsuitable batteries among a random sample of n
= 480 batteries. The mean is µ = = 0.1. The population (all batteries produced) is clearly at least 10 times as large as the sample ( n
= 480), so Rule of Thumb 1 suggests that the standard deviation σ = 0.0137
is appropriate. The two conditions, np
= 480×0.1 =
480
48 > 10 and n
(1 − p
) = 480×0.9 = 432 > 10, are both satisfied, so Rule of Thumb 2 suggests that

Sampling Distributions 215 the distribution of ˆ is approximately Normal. (5)
⎝
⎜
⎛
P p
≤
40
480
⎟
⎠
⎞ ⎛
P Z
≤
0.0137
⎟
⎞
⎠
= ≤ − 1.22
)
= 0.1112. (6) There is about an 11% chance of observing a sample proportion this small or smaller, if the true proportion of unsuitable batteries produced is 0.1. A good plant manager would require a smaller probability and evaluate the overall process before sending this shipment.
9.47 (a) p
= 0.68 or 68% is a parameter; ˆ = 0.73 or 73% is a statistic. (b) The mean is µ = p
=
0.68 and the standard deviation is σ = 0.0381
. (c)
150
( > )
⎝
⎜
⎛
P Z
>
0.0381
⎟
⎞
⎠
= > 1.31
)
= 1 – 0.9049 = 0.0951. There is about a 10% chance of getting a sample proportion of 0.73 or greater, if the population proportion is 0.68.
Thus, the random digit device appears to be working fine.
9.48 (a), (b) Answers will vary. In one simulation, we obtained a total of 8 simulated values of less than or equal to 0.65, a percentage of 16%. This estimated probability suggests that the customs agents have given us reasonable information about the chance of getting a green light.
(c) Let ˆ denote the sample proportion of passengers who get the green light among a random sample of n
= 100 travelers going through security at Guadalajara airport. The mean is µ = 0.7.
The population (all travelers going through security at Guadalajara airport) is clearly at least 10 times as large as the sample ( n = 100), so Rule of Thumb 1 suggests that the standard deviation
σ = 0.0458
is appropriate. The two conditions, np = 100×0.7 = 70 > 10 and n (1 −
100 p
) = 100×0.3 = 30 > 10, are both satisfied, so Rule of Thumb 2 suggests that the distribution of
ˆ is approximately Normal. (d)
( )
⎝
≤
0.0458
⎟
⎞
⎠
= ≤ − 1.09
)
= = 0.1379. This probability (or 14% chance) is reasonably close to the 16% obtained in our simulation. (e) For n
= 1000, ˆ is again approximately normal, with mean µ = 0.7 and standard deviation
σ = 0.0145
. (The conditions for Rule of Thumb 2 are clearly satisfied, but some
1000 students may wonder whether or not the standard deviation is reasonable in this situation because the may worry about having less than 10,000 travelers pass through security at this airport.)
( )
⎝
≤
0.0145
⎟
⎞
⎠
= ≤ − 3.45
) = 0.0003. The sample proportion ˆ p is less variable for larger sample sizes, so the probability of seeing a value of less than or equal to 0.65 decreases.
9.49 (a) Let
X
denote the WAIS score for a randomly selected individual.
≥ 105
) =
⎝
≥
15
⎟
⎞
⎠
≥ 0.33
) = 1 – 0.6293 = 0.3707. (Software gives 0.3695.)

216 Chapter 9
(b) The mean is µ x
=100 and the standard deviation is σ x
=
15
60
1.9365
. (c)
≥ 105
) =
⎝
≥
1.9365
⎟
⎞
⎠
≥ 2.582
) = 1 – 0.9951 = 0.0049 (d) The answer to (a) could be quite different; the answer to (b) would be the same (it does not depend on normality at all).
The answer we gave for (c) would still be fairly reliable because of the central limit theorem.
9.50 (a) Let ˆ denote the sample proportion of women who think they do not get enough time for themselves in a random sample of n
= 1025 adult women The mean is µ = 0.47. The population (all adult women) is clearly at least 10 times as large as the sample ( n
= 1025), so
Rule of Thumb 1 suggests that the standard deviation σ = 0.0156
is appropriate.
1025
The two conditions, np
= 1025×0.47 = 481.75 > 10 and n
(1 − p
) = 1025×0.53 = 543.25 > 10, are both satisfied, so Rule of Thumb 2 suggests that the distribution of ˆ is approximately Normal.
(b) The middle 95% of all sample results will fall within two standard deviations (2×0.0156 =
0.0312) of 0.47 or in the interval (0.4388, 0.5012). (c)
( )
⎝
⎜ ≤
0.0156
⎟
⎞
⎠
= ≤ − 1.28
) = 0.1003. (Software gives 0.0998.)
9.51 (a) The mean is µ x
= 0.5 and the standard deviation is σ x
=
0.7
50
0.0990
. (b) Because this distribution is only approximately normal, it would be quite reasonable to use the 68–95–99.7 rule to give a rough estimate: 0.6 is about one standard deviation above the mean, so the probability should be about 0.16 (half of the 32% that falls outside ±1 standard deviation).
Alternatively, ≥ 0.6
)
⎝
≥
0.0990
⎟
⎞
⎠
= ≥ 1.01
)
= 1 – 0.8438 = 0.1562.
9.52 (a) Let
X
denote the number of high school dropouts who will receive a flyer. The mean is
µ = =
X np
. (b) The standard deviation is
σ =
X np (1 − p ) = × × and
≥ 5000
)
⎝
⎜
⎛
P Z
≥ − 0.7876
)
0.7845
9.53 (a) Let ˆ denote the sample proportion of Internet users who have posted a photo online in a random sample of
≥ n
63.4815
⎟
⎞
⎠
=
= 1555 Internet users. The mean is µ
.
= 0.2. The population (all Internet users) is clearly at least 10 times as large as the sample ( n
= 1555), so Rule of Thumb 1 suggests that the standard deviation σ = 0.0101
is appropriate. The two conditions, np
=
1555
1555×0.2 = 311 > 10 and n
(1 − p
) = 1555×0.8 = 1244 > 10, are both satisfied, so Rule of Thumb
2 suggests that the distribution of ˆ is approximately Normal. (b) Let
X
= the number of in the sample who have posted photos online.
X
has a binomial distribution with n
= 1555 and p
= 0.2.

Sampling Distributions 217
( ≤ 300
)
=
⎝
⎜ ≤
300
1555
⎞
⎟ P Z
⎠ ⎝
⎜ ≤
0.0101
⎟
⎞
⎠
= ≤ − 0.7
) = 0.2420. (The exact probability is 0.25395.) Note:
Actually, X has a hypergeometric distribution, but the size of the population
(all Internet users) is so much larger than the sample that the binomial distribution is an extremely good approximation.
9.54 (a) Let
X
denote the level of nitrogen oxides (NOX) for a randomly selected car.
> 0.3
) =
⎝
≥
0.05
⎟
⎞
⎠
= ≥ 2.0
)
= 1 – 0.9772 = 0.0228 (or 0.025, using the 68–95–99.7 rule). (b) The mean NOX level for these 25 cars is µ x
= 0.2 g/mi, the standard deviation is
σ x
=
0.05
25
= 0.0100
g/mi, and ≥ 0.3
) =
⎝
≥
0.01
⎟
⎞
⎠
= ≥ 10
) , which is basically 0.
9.55 The mean NOX level for 25 cars has a N (0.2, 0.01) distribution, and P ( Z > 2 .
33) = 0 .
01 if
Z is
N
(0, 1), so
L
= 0.2 + 2.33×0.01 = 0.2233 g/mi.
9.56 (a) No—a count only takes on whole-number values, so it cannot be normally distributed.
(b) The approximate distribution is Normal with mean µ x
=1.5 people and standard deviation
σ x
=
0.75
0.0283
. (c)
700
0.8962 = 0.1038.
( > 1075) > 1.5357
) =
⎝
⎜
⎛
P Z
>
0.0283
⎟
⎞
⎠
= > 1.26
)
= 1 –
9.57 (a) The mean is µ = 0
.
5, and the standard deviation is σ = 0.0041
. (b)
14941
P
(
0.49
0.51
)
P ⎜
⎛
⎝ 0.0041
0.0041
⎟
⎞
⎠
=
P
( − 2.44
2.44
)
= 0.9927 – 0.0073 =
0.9854.
9.58 (a) If samples of size n
= 219 were obtained over and over again and the sample mean was computed for each sample, the center of the distribution of these means would be at µ grams.
In other words, the expected value of the sample mean is µ grams. (b) No, the population distribution of birth weights among ELBW babies is likely to be skewed to the left. Most ELBW babies will be above a certain value, but some babies will even be below this value. (c) Yes, the central limit theorem suggests that the mean birth weight of 219 babies will be approximately
Normal.