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.
参考答案：A
解析：
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.