The Ransin Sports Company has noted that the size of individual customer orders is normally distributed with a mean of $112 and a standard deviation of $9. If a soccer team of 11 players were to make the next batch of orders, what is the probability that the mean purchase would exceed $116?
A) 11%
B) 3%
C) 33%
D) 7%

Question

Quizplus · Accepted Answer

To find the probability that the mean purchase exceeds $116 for a sample of 11 players, we use the formula for the standard error of the mean: $ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} $, where $ \sigma $ is the population standard deviation and $ n $ is the sample size. Here, $ \sigma = $9 $ and $ n = 11 $, so $ \sigma_{\bar{x}} = \frac{9}{\sqrt{11}} \approx 2.71 $. Next, we calculate the z-score for a mean purchase of $116: $ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{116 - 112}{2.71} \approx 1.48 $. Looking up the z-score of 1.48 in the standard normal distribution table or using a calculator, we find that the area to the right of this z-score (which represents the probability that the mean purchase exceeds $116) is approximately 7%.

The Ransin Sports Company Has Noted That the Size of Individual