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?

0.0918

Step-by-step explanation:

We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

The mean and standard deviation of average spending of sample size 25 are

μxbar=μ=95.25

σxbar=σ/√n=27.32/√25=27.32/5=5.464.

So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

The z-score associated with average spending $102.5

Z=[Xbar-μxbar]/σxbar

Z=[102.5-95.25]/5.464

Z=7.25/5.464

Z=1.3269=1.33

We have to find P (Xbar>102.5).

P (Xbar>102.5) = P (Z>1.33)

P (Xbar>102.5) = P (0
P (Xbar>102.5) = 0.5-0.4082

P (Xbar>102.5) = 0.0918.

Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.