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) p̂ =7.2% is a statistic (related to the sample of registered voters who were unemployed). 9.2 (a) p̂ = 48% is a statistic; = 52% is a parameter. (b) Both p controlx = 335 and experimentalx = 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 p̂ ) is shown below (on the right). As expected, the simulated distribution of p̂ is approximately Normal with a center at 0.5.

Simulated proportion

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0.650.600.550.500.450.400.35

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0.7500.6750.6000.5250.4500.3750.300

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(d) Answers will vary, but the standard deviation will be close to 0.5 0.5 0.111820× . The

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 p̂ 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.

203

(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 72.75x = . (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.

Simulated Mean

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7574737271706968

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7674727068666462

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204 Chapter 9

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.

Mean of 2 observations

Perc

ent

81.0

78.0

77.5

77.0

76.5

76.0

74.0

73.5

73.0

72.5

72.0

71.0

70.0

69.5

69.0

68.5

68.0

67.5

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7

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(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) 141.847µ 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

Sampling Distributions 205

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 thanµ , but not necessarily a 50/50 split. (e) The mean of the (theoretical) sampling distribution is 141.847µ . (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.

Survival time (days)

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600480360240120

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24021018015012090

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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 24 0.120 10 10 25 0.125 4 4 26 0.130 1 1 27 0.135 2 2 28 0.140 2 2 30 0.150 1 1

Sample Proportion

Perc

ent

0.150

0.140

0.135

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0.115

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0.105

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0.095

0.090

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mean = 0.0981

(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 p̂ is 0.10. (e) By increasing the sample size from 200 to 1000, the mean of p̂ would stay the same, 0.10, but the spread in the distribution of p̂ 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

206 Chapter 9

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 p̂ = 40.2% and p̂ = 31.7% are statistics (related, respectively, to the sample of small-class and regular-size-class black students). 9.12 The sample mean 64.5x = inches is a statistic and the population mean 63µ = 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 p̂ = 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 1 0.1 1 0.1 1 0.1 2 0.2 1 0.1 1 0.1 4 0.4 3 0.3 1 0.1

Stem-and-leaf of p-hat N = 20 Leaf Unit = 0.010 1 0 0 8 1 0000000 (5) 2 00000 7 3 00000 2 4 00

(c) The mean of the sampling distribution of p̂ is 0.2. (d) The mean of the sampling distribution of p̂ 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.

Sampling Distributions 207

Number of dots

Prob

abili

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654321

0.18

0.16

0.14

0.12

0.10

0.08

0.06

0.04

0.02

0.00

(b) This is equivalent to rolling a pair of fair, six-sided dice. (c) The 36 possible SRSs of size 2 and the sample averages are shown below.

1,1; x =1.0 2,1; 1,2; x =1.5 3,1; 2,2; 1,3; x =2.0 4,1; 2,3; 3,2; 1,4; x =2.5 5,1; 2,4; 3,3; 4,2; 1,5; x =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 x =5.0 6,5; 5,6; x =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 ( 1.2076σ ) is smaller than that of the population distribution.

Sample mean

Prob

abili

ty

6.05.55.04.54.03.53.02.52.01.51.0

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0.14

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0.08

0.06

0.04

0.02

0.00

9.16 Answers will vary. A histogram of the 3x -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.

208 Chapter 9

Sample average

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5.64.84.03.22.41.6

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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 p̂ pµ = = 0.7 and the standard deviation is

p̂σ = (1 )p pn− = 0.7 0.3 0.0144

1012× . (b) The population (all U.S. adults) is clearly at least 10

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.

ˆ 0.67p ≤

9.20 (a) The mean is p̂ pµ = = 0.4 and the standard deviation is p̂σ = 0.4 0.6 0.01161785× . (b)

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) = ( )ˆ0.37 0.43P p≤ ≤ ( )0.37 0.4 0.43 0.4 2.59 2.590.0116 0.0116

P Z P Z− −⎛ ⎞≤ ≤ − ≤ ≤⎜ ⎟⎝ ⎠

=

0.9952 − 0.0048 = 0.9904. Over 99% of all samples of size n = 1785 will produce a sample proportion p̂ within ±0.03 of the true population proportion.

9.21 For n = 300, the standard deviation is ˆ0.4 0.6 0.0283

300pσ ×= and the probability is

approximately equal to (0.37 0.4 0.43 0.4 1.06 1.060.0283 0.0283

P Z P Z− −⎛ ⎞≤ ≤ − ≤ ≤⎜ ⎟⎝ ⎠

) = 0.8554 − 0.1446 =

Sampling Distributions 209

0.7108. For n = 1200, the standard deviation is ˆ0.4 0.6 0.0141

1200pσ ×= and the probability is

approximately equal to (0.37 0.4 0.43 0.4 2.13 2.130.0141 0.0141

P Z P Z− −⎛ ⎞≤ ≤ − ≤ ≤⎜ ⎟⎝ ⎠

)= 0.9834 − 0.0166 =

0.9668. For n = 4800, the standard deviation is ˆ0.4 0.6 0.0071

4800pσ ×= and the probability is

approximately equal to (0.37 0.4 0.43 0.4 4.23 4.23 10.0071 0.0071

P Z P Z− −⎛ ⎞≤ ≤ − ≤ ≤⎜ ⎟⎝ ⎠

) . 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 ˆ0.14 0.86 0.0155

500pσ ×= . (b) 20% or more Harley owners is unlikely:

( ) ( )0.2 0.14ˆ 0.2 3.87 0.00020.0155

P p P Z P Z−⎛ ⎞≥ ≥ = ≥ <⎜ ⎟⎝ ⎠

. (c) There is a fairly good chance of finding

at least 15% Harley owners: ( ) (0.15 0.14ˆ 0.2 0.640.0155

P p P Z P Z−⎛ ⎞≥ ≥ = ≥⎜ ⎟⎝ ⎠

) = 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 ˆ0.9 0.1 0.03

100pσ ×= and the probability is

( ) (0.86 0.9ˆ 0.86 1.330.03

P p P Z P Z−⎛ ⎞≤ ≤ = ≤ −⎜ ⎟⎝ ⎠

)= 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 p̂ is the sample proportion who have been on a diet, then p̂ is approximately Normal

with mean 0.7 and standard deviation ˆ0.7 0.3 0.02804

267pσ ×= . The probability is

approximately equal to ( ) (0.75 0.7ˆ 0.75 1.780.02804

P p P Z P Z−⎛ ⎞≥ ≥ = ≥⎜ ⎟⎝ ⎠

)= 1 − 0.9625 = 0.0375.

(Software gives 0.0373). Alternatively, as p̂ ≥ 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

= 1 − 0.9671= 0.0329. (( 201) 1 200P X P X≥ = − ≤ )

9.25 (a) The mean is p̂ pµ = = 0.15 and the standard deviation is ˆ0.15 0.85 0.0091

1540pσ ×= . (b)

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

210 Chapter 9

satisfied. (d) = ( )ˆ0.13 0.17P p≤ ≤ ( )0.13 0.15 0.17 0.15 2.20 2.200.0091 0.0091

P Z P Z− −⎛ ⎞≤ ≤ − ≤ ≤⎜ ⎟⎝ ⎠

= 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 ˆ0.15 0.85 0.0252

200pσ ×= and the probability is

approximately equal to (0.13 0.15 0.17 0.15 0.79 0.790.0252 0.0252

P Z P Z− −⎛ ⎞≤ ≤ − ≤ ≤⎜ ⎟⎝ ⎠

)= 0.7852 − 0.2148 =

0.5704. For n = 800, the standard deviation is ˆ0.15 0.85 0.0126

800pσ ×= and the probability is

approximately equal to (0.13 0.15 0.17 0.15 1.59 1.590.0126 0.0126

P Z P Z− −⎛ ⎞≤ ≤ − ≤ ≤⎜ ⎟⎝ ⎠

)= 0.9441 − 0.0559 =

0.8882. For n = 3200, the standard deviation is ˆ0.15 0.85 0.0063

3200pσ ×= and the probability is

approximately equal to (0.13 0.15 0.17 0.15 3.17 3.170.0063 0.0063

P Z P Z− −⎛ ⎞≤ ≤ − ≤ ≤⎜ ⎟⎝ ⎠

) = 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 p̂ = 62/100 = 0.62. (b) The mean is p̂ pµ = = 0.67 and the

standard deviation is p̂σ = 0.67 0.33 0.0470100× . ( ) ( )0.62 0.67ˆ 0.62 1.06

0.047P p P Z P Z−⎛ ⎞≤ ≤ ≤ −⎜ ⎟

⎝ ⎠

= 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 could arise by chance even if the true campus proportion is the same.

9.28 (a) The mean is p̂ pµ = = 0.52 and the standard deviation is ˆ0.52 0.48 0.0223

500pσ ×= (b)

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. ( ) ( )0.5 0.52ˆ 0.5 0.900.0223

P p P Z P Z−⎛ ⎞≥ ≥ = ≥ −⎜ ⎟⎝ ⎠

= 1 −

0.1841 = 0.8159.

9.29 (a) The mean is p̂µ = 0.75 and the standard deviation is p̂σ = 0.75 0.25100× = 0.0433.

( ) (0.7 0.75ˆ 0.7 1.150.0433

P p P Z P Z−⎛ ⎞≤ ≤ = ≤ −⎜ ⎟⎝ ⎠

)= 0.1251. (b) The mean is p̂µ = 0.75 and the

standard deviation is ˆ0.75 0.25 0.0274

250pσ ×= . ( ) ( )0.7 0.75ˆ 0.7 1.82

0.0274P p P Z P Z−⎛ ⎞≤ ≤ = ≤ −⎜ ⎟

⎝ ⎠ =

0.0344. (c) To reduce the standard deviation for a 100-item test by a fourth, we need a sample

Sampling Distributions 211

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) ( 3P X )≤ = binomcdf(15, 0.3, 3) 0.2969.

9.31 (a) The mean is xµ = µ = -3.5% and the standard deviation is 26 11.6276%5x n

σσ = = . (b)

( ) (5 ( 3.5)526

P X P Z P Z− −⎛ ⎞> > = >⎜ ⎟⎝ ⎠

)0.33 = 1 – 0.6293 = 0.3707. (c)

( ) (5 ( 3.5)511.6276

P x P Z P Z− −⎛ ⎞> > = >⎜ ⎟⎝ ⎠

)0.73 = 1 – 0.7673 = 0.2327. (d)

( ) (0 ( 3.5)011.6276

P x P Z P Z− −⎛ ⎞< < = <⎜ ⎟⎝ ⎠

)0.30 = 0.6179. Approximately 62% of all five-stock

portfolios lost money.

9.32 (a) ( ) (21 18.621 0.415.9

P X P Z P Z−⎛ ⎞> > = >⎜ ⎟⎝ ⎠

)= 1 – 0.6591 = 0.3409. (Software gives

0.3421.) (b) The mean is 18.6xµ = and the standard deviation is 5.9 0.834450xσ = . These

results do not depend on the distribution of the individual scores. (c)

( ) (21 18.621 2.880.8344

P x P Z P Z−⎛ ⎞> > = >⎜ ⎟⎝ ⎠

)= 1 – 0.9980 = 0.002.

9.33 (a) The standard deviation is 10 5.77353xσ = milligrams. (b) Solve 10 3;

n= n = 11.1, so 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 xµ = 6 strikes/km2 and

2.4 0.758910xσ = strikes/km2. (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 ( )1.6,1.2 200 0.0849N distribution. The

probability is approximately equal to ( ) (2 1.62 40.0849

P x P Z P Z−⎛ ⎞> > = >⎜ ⎟⎝ ⎠

).71 , which is essentially

0.

212 Chapter 9

9.36 The central limit theorem says that over 40 years, the mean return x is approximately

Normal with mean xµ =13.2% and standard deviation 17.5 2.7670%40xσ = . Therefore,

( ) (15 13.215% 0.652.767

P x P Z P Z−⎛ ⎞> > = >⎜ ⎟⎝ ⎠

)= 1 – 0.7422 = 0.2578 and

( ) (10 13.210% 1.162.767

P x P Z P Z−⎛ ⎞< < = < −⎜ ⎟⎝ ⎠

)= 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 20x W= , the central limit theorem says that x is

approximately Normal with mean 190 lb and standard deviation 35 7.826220xσ = lb. Thus,

( ) ( ) (200 1904000 200 1.287.8262

P W P x P Z P Z−⎛ ⎞> = > > = >⎜ ⎟⎝ ⎠

)= 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 40.125xµ = mm and the standard deviation is 0.002 0.0014xσ = = mm. These

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 125140 1.510

P X P Z P Z−⎛ ⎞> = > = >⎜ ⎟⎝ ⎠

) = 1 – 0.9332 = 0.0668. (b) If x is the mean of four

measurements (assumed to be independent), then x has a ( )125,10 4N =

distribution and( )125mg/dl,5mg/dlN ( ) (140 125140 3.05

P x P Z P Z−⎛ ⎞> = > = >⎜ ⎟⎝ ⎠

) =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 ( )125mg/dl,5mg/dlN distribution, and

( )1.645 0.05P Z > = 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 298295 13

P X P Z P Z−⎛ ⎞< = < = < −⎜ ⎟⎝ ⎠

) = 0.1587. (b) If x is the mean contents of six bottles

(assumed to be independent), then x has a ( )298,3 6N = ( )298ml,1.2247mlN distribution

and ( ) (295 298295 2.451.2247

P x P Z P Z−⎛ ⎞< = < < −⎜ ⎟⎝ ⎠

)=0.0071.

Sampling Distributions 213

9.42 (a) The sample mean lifetime has a ( ) ( )55000, 4500 8 55000 miles,1591 milesN N

distribution. (b) ( ) (51,800 55,00051,800 2.011591

P x P Z P Z−⎛ ⎞< = < < −⎜ ⎟⎝ ⎠

)= 0.0222.

9.43 (a) The approximate distribution of the mean number of accidents x is

( ) ( )2.2,1.4 52 2.2,0.1941N N . (b) ( ) (2 2.22 10.1941

P x P Z P Z−⎛ ⎞< = < < −⎜ ⎟⎝ ⎠

).03 = 0.1515. (c) If

X denotes the number of accidents at an intersection per year, then

( ) ( ) (1.9231 2.2100 100 / 52 1.430.1941

P X P x P Z P Z−⎛ ⎞< < = < < −⎜ ⎟⎝ ⎠

)= 0.0764.

9.44 The mean of 22 measurements has a ( )13.6,3.1 22N distribution, and

( )1.645 0.05P Z < − = 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 toµ , 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

( ) ( )$250,$300 10,000 $250,$3N N= and ( ) ( )260 250$260 3.333

P x P Z P Z−⎛ ⎞> = > >⎜ ⎟⎝ ⎠

= 1 –

0.9996 = 0.0004. CASE CLOSED! (1) The central limit theorem suggests that the mean of 30 lifetimes will be approximately

Normal with mean 17xµ = hours and the standard deviation of 0.8 0.146130xσ = hours. (2)

( ) (16.7 1716.7 2.050.1461

P x P Z P Z−⎛ ⎞< < = < −⎜ ⎟⎝ ⎠

) = 0.0202. (3) There is only about a 2% chance of

observing a sample average as small or smaller than the one observed ( )16.7x ≤ if the process is working properly. This small probability provides evidence that the process is not working properly. (4) Let p̂ denote the sample proportion of unsuitable batteries among a random sample of n = 480 batteries. The mean is p̂ pµ = = 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 p̂σ = 0.1 0.9 0.0137480× is appropriate. The two conditions, np = 480×0.1 =

48 > 10 and n(1 − p) = 480×0.9 = 432 > 10, are both satisfied, so Rule of Thumb 2 suggests that

214 Chapter 9

the distribution of p̂ is approximately Normal. (5)

(40 0.0833 0.1ˆ 1.22480 0.0137

P p P Z P Z−⎛ ⎞ ⎛ ⎞≤ ≤ = ≤ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

)= 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; p̂ = 0.73 or 73% is a statistic. (b) The mean is p̂µ = p =

0.68 and the standard deviation is ˆ0.68 0.32 0.0381

150pσ ×= . (c)

( ) (0.73 0.68ˆ 0.73 1.310.0381

P p P Z P Z−⎛ ⎞> > = >⎜ ⎟⎝ ⎠

)= 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 p̂ 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 p̂µ = 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

p̂σ = 0.7 0.3 0.0458100× is appropriate. The two conditions, np = 100×0.7 = 70 > 10 and n(1 −

p) = 100×0.3 = 30 > 10, are both satisfied, so Rule of Thumb 2 suggests that the distribution of

p̂ is approximately Normal. (d) ( ) (0.65 0.7ˆ 0.65 1.090.0458

P p P Z P Z−⎛ ⎞≤ ≤ = ≤ −⎜ ⎟⎝ ⎠

) = = 0.1379. This

probability (or 14% chance) is reasonably close to the 16% obtained in our simulation. (e) For n = 1000, p̂ is again approximately normal, with mean p̂µ = 0.7 and standard deviation

p̂σ = 0.7 0.3 0.01451000× . (The conditions for Rule of Thumb 2 are clearly satisfied, but some

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.)

( ) (0.65 0.7ˆ 0.65 3.450.0145

P p P Z P Z−⎛ ⎞≤ ≤ = ≤ −⎜ ⎟⎝ ⎠

)= 0.0003. The sample proportion p̂ 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 100105 0.3315

P X P Z P Z−⎛ ⎞≥ = ≥ ≥⎜ ⎟⎝ ⎠

) = 1 – 0.6293 = 0.3707. (Software gives 0.3695.)

Sampling Distributions 215

(b) The mean is xµ =100 and the standard deviation is 15 1.936560xσ = . (c)

( ) (105 100105 2.5821.9365

P x P Z P Z−⎛ ⎞≥ = ≥ ≥⎜ ⎟⎝ ⎠

)= 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 p̂ 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 p̂µ = 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 ˆ0.47 0.53 0.0156

1025pσ ×= is appropriate.

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 p̂ 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)

( ) (0.45 0.47ˆ 0.45 1.280.0156

P p P Z P Z−⎛ ⎞< ≤ = ≤ −⎜ ⎟⎝ ⎠

) = 0.1003. (Software gives 0.0998.)

9.51 (a) The mean is xµ = 0.5 and the standard deviation is 0.7 0.099050xσ = . (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 0.50.6 1.010.0990

P x P Z P Z−⎛ ⎞≥ ≥ = ≥⎜ ⎟⎝ ⎠

)= 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

25,000 0.202 5050X npµ = = × = . (b) The standard deviation is (1 ) 25,000 0.202 0.798 63.4815X np pσ = − = × × and

( ) ( )5000 50505000 0.7876 0.784563.4815

P X P Z P Z−⎛ ⎞≥ ≥ = ≥ −⎜ ⎟⎝ ⎠

.

9.53 (a) Let p̂ denote the sample proportion of Internet users who have posted a photo online in a random sample of n = 1555 Internet users. The mean is p̂µ = 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 ˆ0.2 0.8 0.0101

1555pσ ×= is appropriate. The two conditions, np =

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 p̂ 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.

216 Chapter 9

( )300P X ≤ = (300 0.1929 0.2ˆ 0.71555 0.0101

P p P Z P Z−⎛ ⎞ ⎛ ⎞≤ ≤ = ≤⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

)− = 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 0.20.3 2.00.05

P X P Z P Z−⎛ ⎞> = ≥ = ≥⎜ ⎟⎝ ⎠

) = 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 xµ = 0.2 g/mi, the standard deviation is 0.05 0.0100

25xσ = = g/mi, and ( ) (0.3 0.20.3 100.01

P x P Z P Z−⎛ ⎞≥ = ≥ = ≥⎜ ⎟⎝ ⎠

) , 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 xµ =1.5 people and standard deviation

0.75 0.0283700xσ = . (c) ( ) ( )1.5375 1.5( 1075) 1.5357 1.26

0.0283P X P x P Z P Z−⎛ ⎞> > = > = >⎜ ⎟

⎝ ⎠= 1 –

0.8962 = 0.1038.

9.57 (a) The mean is p̂µ = 0.5, and the standard deviation is ˆ0.5 0.5 0.004114941pσ ×

= . (b)

( ) (0.49 0.5 0.51 0.5ˆ0.49 0.51 2.44 2.440.0041 0.0041

P p P Z P Z− −⎛ ⎞< < < < = − < <⎜ ⎟⎝ ⎠

)= 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.

Sampling Distributions 217