Question

The average time a subscriber spends reading the local newspaper is 49 minutes. For the 10% who spend the most time reading the paper, how much time do they spend?

Assume the standard deviation is 16 minutes and that the times are normally distributed

Answer 1

`69.48` minutes.

Explanation:

basically we are asked to find the 90th percentile which can be found as such

1) look up the .90 p-value from a z table and get the z score `rightarrow 1.28`

2) use the formula to compute z score but now solve from x

`z = (x-mu)/sigma`
`zsigma+mu = x`
`1.28*16+49 = 69.48`

this means that the top 10% read more than `69.48` minutes.