0.2 Practice tests (1-4) and final exams  (Page 11/36)

64 . $\mu =np=(500)(0.20)=100$
$\sigma =\sqrt{npq}=\sqrt{500(0.20)(0.80)}=8.94$

65 . Fifty percent, because in a normal distribution, half the values lie above the mean.

66 . The results of our sample were two standard deviations below the mean, suggesting it is unlikely that 20 percent of the lotto tickets are winners, as claimed by the distributor, and that the true percent of winners is lower. Applying the Empirical Rule, If that claim were true, we would expect to see a result this far below the mean only about 2.5 percent of the time.

7.1: the central limit theorem for sample means (averages)

67 . The central limit theorem states that if samples of sufficient size drawn from a population, the distribution of sample means will be normal, even if the distribution of the population is not normal.

68 . The sample size of 30 is sufficiently large in this example to apply the central limit theorem. This theorem ] states that for samples of sufficient size drawn from a population, the sampling distribution of the sample mean will approach normality, regardless of the distribution of the population from which the samples were drawn.

69 . You would not expect each sample to have a mean of 50, because of sampling variability. However, you would expect the sampling distribution of the sample means to cluster around 50, with an approximately normal distribution, so that values close to 50 are more common than values further removed from 50.

70 . $\overline{X}\sim N(25,0.2)$ because $\overline{X}\sim N\left({\mu }_x,\frac{{\sigma }_x}{\sqrt{n}}\right)$

71 . The standard deviation of the sampling distribution of the sample means can be calculated using the formula $\left(\frac{{\sigma }_x}{\sqrt{n}}\right)$ , which in this case is $\left(\frac{16}{\sqrt{50}}\right)$ . The correct value for the standard deviation of the sampling distribution of the sample means is therefore 2.26.

72 . The standard error of the mean is another name for the standard deviation of the sampling distribution of the sample mean. Given samples of size n drawn from a population with standard deviation σ _x , the standard error of the mean is $\left(\frac{{\sigma }_x}{\sqrt{n}}\right)$ .

73 . X ~ N (75, 0.45)

74 . Your friend forgot to divide the standard deviation by the square root of n .

75 . $z=\frac{\overline{x}-{\mu }_x}{{\sigma }_x}=\frac{76-75}{4.5}=2.2$

76 . $z=\frac{\overline{x}-{\mu }_x}{{\sigma }_x}=\frac{74.7-75}{4.5}=\text{−0.67}$

77 . 75 + (1.5)(0.45) = 75.675

78 . The standard error of the mean will be larger, because you will be dividing by a smaller number. The standard error of the mean for samples of size n = 50 is:
$\left(\frac{{\sigma }_x}{\sqrt{n}}\right)=\frac{4.5}{\sqrt{50}}=0.64$

79 . You would expect this range to include values up to one standard deviation above or below the mean of the sample means. In this case:
$70+\frac{9}{\sqrt{60}}=71.16$ and $70-\frac{9}{\sqrt{60}}=68.84$ so you would expect 68 percent of the sample means to be between 68.84 and 71.16.

80 . $70+\frac{9}{\sqrt{100}}=70.9$ and $70-\frac{9}{\sqrt{100}}=69.1$ so you would expect 68 percent of the sample means to be between 69.1 and 70.9. Note that this is a narrower interval due to the increased sample size.

7.2: the central limit theorem for sums

81 . For a random variable X , the random variable ΣX will tend to become normally distributed as the size n of the samples used to compute the sum increases.

82 . Both rules state that the distribution of a quantity (the mean or the sum) calculated on samples drawn from a population will tend to have a normal distribution, as the sample size increases, regardless of the distribution of population from which the samples are drawn.