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 percent 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: X = μ + zσ, where μ = 49, σ = 16, and z = 1.28, we calculate X = 49 + (1.28)(16) = 69.48 minutes.

The Average Time a Subscriber Spends Reading the Local Newspaper