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?

1 Answer
May 22, 2018

Answer:

#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 = (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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