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. For the 10% who spend the most time reading the paper,how much time do they spend?
A)11.72
B)28.52
C)86.28
D)69.48

Question

Quizplus · Accepted Answer

To find the time spent by the top 10% of readers, we need to find the z-score that corresponds to the 90th percentile of a standard normal distribution, which is approximately 1.28. Using the formula: time = mean + (z-score * standard deviation), we calculate: 49 + (1.28 * 16) = 69.48 minutes.

The Average Time a Subscriber Spends Reading the Local Newspaper