A fire department in a rural county reports that its response time to fires is approximately Normally distributed with a mean of 22 minutes and a standard deviation of 11.9 minutes. Approximately what proportion of their response times is over 30 minutes?

May 22, 2018

P(>30 " min") = .2514; " or " ~~25%

Explanation:

Given: Normally distributed with mean $= \mu = 22$ min.; standard deviation $= \sigma = 11.9$ min. Find $x > 30$ min.

Calculate the $z -$score:

$z = \frac{30 - 22}{11.9} = .6723$

Look up the probability from a $z$-table:

$P \left(\le 30 \text{ min}\right) = .7486$

P(>30 " min") = 1 - .7476 = .2514; " or " ~~25%