In a recent survey of college professors,it was found that the average amount of money spent on entertainment each week was normally distributed with a mean of $95.25 and a standard deviation of $27.32.What is the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.50?
A)0.0918
B)0.1064
C)0.3936
D)0.4082

Question

Quizplus · Accepted Answer

To find the probability that the average spending of a sample of 25 professors exceeds $102.50, we use the formula for the standard error of the mean: SE = σ / √n, where σ is the standard deviation and n is the sample size. Here, SE = 27.32 / √25 = 27.32 / 5 = 5.464. We then find the Z-score for $102.50 using the formula Z = (X - μ) / SE, where X is the value we're comparing to the mean, and μ is the mean. So, Z = (102.50 - 95.25) / 5.464 = 7.25 / 5.464 ≈ 1.327. Looking up 1.327 on the Z-table gives us the area to the left of Z, so we subtract this value from 1 to find the area to the right, which represents the probability of exceeding $102.50. The Z-table value for 1.327 is approximately 0.9082, so the probability is 1 - 0.9082 = 0.0918.

In a Recent Survey of College Professors,it Was Found That