Question

A radio station claims that the amount of advertising each hour has a mean of 15 minutes and a standard deviation of 2.9 minutes. You listen to the radio station for 1 hour and observe that the amount of advertising time is 16 minutes. Calculate the ​z-score for this amount of advertising time. Round to the nearest hundredth.

Answer 1

0.1296

Explanation:

Use the normpdf function available on basic calculators:

normpdf(16,15,2.9) = 0.1296

This is the z-score representing 16 minutes. Note that 16 is just one point above 15, so this small positive z-score makes sense.