The average time a subscriber spends reading the local newspaper is 49 minutes.Assume the standard deviation is 16 minutes and that the times are normally distributed.What is the probability a subscriber will spend at least 1 hour reading the paper?
A).9987
B).7549
C).2451
D).0013

Question

Quizplus · Accepted Answer

To find the probability that a subscriber spends at least 1 hour (60 minutes) reading the paper, we first calculate the z-score: z = (60 - 49) / 16 = 0.6875. Using the standard normal distribution table, the probability of a z-score less than 0.6875 is approximately 0.7549. Therefore, the probability of spending at least 60 minutes is 1 - 0.7549 = 0.2451.

The Average Time a Subscriber Spends Reading the Local Newspaper