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?

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Answer 1 · 2018-05-22T15:33:08

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

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

Calculate the $z-$ score:

$z = (30 -22)/11.9 = .6723$

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

$P(<= 30 " min") = .7486$

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

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