答案和详解如下： 1、The average annual rainfall amount in Yucutat, Alaska, is normally distributed with a mean of 150 inches and a standard deviation of 20 inches. The 90% confidence interval for the annual rainfall in Yucutat is closest to: A) 110 to 190 inches. B) 137 to 163 inches. C) 144 to 156 inches. D) 117 to 183 inches. The correct answer was D) The 90% confidence interval is µ ± 1.65 standard deviations. 150 - 1.65(20) = 117 and 150 + 1.65(20) = 183. 2、Standardizing a normally distributed random variable requires the: A) mean, variance and skewness. B) mean and the standard deviation. C) natural logarithm of X. D) variance and kurtosis. The correct answer was B) All that is necessary is to know the mean and the variance. Subtracting the mean from the random variable and dividing the difference by the standard deviation standardizes the variable. 3、Which of the following represents the mean, standard deviation, and variance of a standard normal distribution? A) 1, 1, 1. B) 0, 1, 1. C) 0, 2, 4. D) 1, 2, 4. The correct answer was B) By definition, for the standard normal distribution, the mean, standard deviation, and variance are 0, 1, 1. 4、The owner of a bowling alley determined that the average weight for a bowling ball is 12 pounds with a standard deviation of 1.5 pounds. A ball denoted “heavy” should be one of the top 2 percent based on weight. Assuming the weights of bowling balls are normally distributed, at what weight (in pounds) should the “heavy” designation be used? A) 15.08 pounds. B) 14.00 pounds. C) 14.22 pounds. D) 17.36 pounds. The correct answer was A) The first step is to determine the z-score that corresponds to the top 2 percent. Since we are only concerned with the top 2 percent, we only consider the right hand of the normal distribution. Looking on the cumulative table for 0.9800 (or close to it) we find a z-score of 2.05. To answer the question, we need to use the normal distribution given: 98 percentile = sample mean + (z-score)(standard deviation) = 12 + 2.05(1.5) = 15.08. 5、Monthly sales of hot water heaters are approximately normally distributed with a mean of 21 and a standard deviation of 5. What is the probabilility of selling 12 hot water heaters or less next month? A) 3.59%. B) 1.80%. C) 96.41%. D) 98.2%. The correct answer was A) Z = (12 – 21) / 5 = -1.8 From the cumulative Z-table, the probability of being more than 1.8 standard deviations below the mean, probability x < -1.8, is 3.59%. 6、The standard normal distribution is most completely described as a: A) symmetrical distribution with a mean equal to its median. B) normal distribution with a mean of zero and a standard deviation of one. C) distribution that exhibits zero skewness and no excess kurtosis. D) distribution with 95% of its values within two standard deviations of the mean. The correct answer was B) The standard normal distribution is defined as a normal distribution that has a mean of zero and a standard deviation of one. The other choices apply to any normal distribution | 
发表于 2008-4-10 14:22
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