Question

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?

Answer 1

Answer-

In 25.15% cases their response times is over 30 minutes.

Solution-

Hint- Calculating the z-score (how much standard deviations away from the mean) and then getting the corresponding probability value from normal distribution table we can get solution.

Applying the formula for Z-score,

$Z=\frac{X-\mu}{\sigma}$

Here,

X = score = 30,

μ = mean = 22,

σ = standard deviation = 11.9

Putting the values,

$\Rightarrow Z=\frac{30-22}{11.9}=0.67$

$\therefore P(x>30)=1-P(Z\leq 0.67)=1-0.7485=0.2515=25.15\%$

(Subtracted from 1 as the values of area to the left of the Z score is given in table, but we need the area right to Z score as response times more than 30 minutes is asked)