付録

35. Yes, because you are sampling from a population composed of two groups (boys and girls), have a group of interest (boys), and are sampling without replacement (hence, the probabilities change with each pick, and you are not performing Bernoulli trials).

36. The group of interest is the cards that are spades, the size of the group of interest is 13, and the sample size is five.

37. A Poisson distribution models the number of events occurring in a fixed interval of time or space, when the events are independent and the average rate of the events is known.

38. X ~ P(4)

39. The domain of X = {0, 1, 2, 3, …..) i.e., any integer from 0 upwards.

41. The discrete variables are the number of books purchased, and the number of books sold after the end of the semester. The continuous variables are the amount of money spent for the books, and the amount of money received when they were sold.

42. Because for a continuous random variable, P(x = c) = 0, where c is any single value. Instead, we calculate P(c < x < d), i.e., the probability that the value of x is between the values c and d.

43. Because P(x = c) = 0 for any continuous random variable.

44. P(x > 5) = 1 – 0.35 = 0.65, because the total probability of a continuous probability function is always 1.

45. This is a uniform probability distribution. You would draw it as a rectangle with the vertical sides at 0 and 20, and the horizontal sides at $\frac{1}{10}$ and 0.

48. X ~ U(0, 15)

49. $f(x)=\frac{1}{b-a}$ for $(a\le x\le b)\text{so}f(x)=\frac{1}{30}$ for $(0\le x\le 30)$

52. X has an exponential distribution with decay parameter m and mean and standard deviation $\frac{1}{m}$. In this distribution, there will be a relatively large numbers of small values, with values becoming less common as they become larger.

54. f(x) = 0.2e^–0.2x where x ≥ 0.

55. The random variable X has a normal distribution with a mean of 100 and a standard deviation of 15.

56. X ~ N(0,1)

57. $z=\frac{x-\mu }{\sigma }$ so $z=\frac{112-109}{4.5}=0.67$

58. $z=\frac{x-\mu }{\sigma }$ so $z=\frac{100-109}{4.5}=-2.00$

59. $z=\frac{105-109}{4.5}=\text{−0.89}$
This girl is shorter than average for her age, by 0.89 standard deviations.

60. 109 + (1.5)(4.5) = 115.75 cm

61. We expect about 68 percent of the heights of girls of age five years and zero months to be between 104.5 cm and 113.5 cm.

62. We expect 99.7 percent of the heights in this distribution to be between 95.5 cm and 122.5 cm, because that range represents the values three standard deviations above and below the mean.

63. Yes, because both np and nq are greater than five.
np = (500)(0.20) = 100 and nq = 500(0.80) = 400

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.

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.

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.