# 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

Aug 19, 2016

$69.48$ 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

$z = \frac{x - \mu}{\sigma}$
$z \sigma + \mu = x$
$1.28 \cdot 16 + 49 = 69.48$

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