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 3.9 minutes.

a) Make an accurate sketch of the distribution of response time.

b) Use the empirical rule to find the proportion of response times between 14.2 and 29.8 minutes.

c) Calculate and interpret the 80th percentile of the distribution of response times.
d)The national mean response time is 18.5 minutes. What proportion of all response times for this fire department are below the national mean response time?

c) The 80th percentile of response times is of 25.3 minutes, which means that 80% of the response times are less than 25.3 minutes and 100 - 80 = 20% are more.

d) 0.1841 = 18.41% of all response times for this fire department are below the national mean response time.

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Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

It measures how many standard deviations the measure is from the mean.
Each z-score has a p-value associated, which is the probability that the value of the measure is smaller than X, that is, the percentile of X.
Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 22 means that [tex]\mu = 22[/tex]
Standard deviation of 3.9 means that [tex]\sigma = 3.9[/tex]

Item b:

The Empirical Rule states that for a normal variable, 68% of the measures are within 1 standard deviation, 95% are within 2 and 99.7% are within 3.
14.2 = 22 - 2(3.9)
29.8 = 22 + 2(3.9)
Within 2 standard deviations of the mean, thus, 0.95 = 95%.

Item c:

The 80th percentile is the value of X when Z has a p-value of 0.8, so X when Z = 0.84.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.84 = \frac{X - 22}{3.9}[/tex]

[tex]X - 22 = 0.84(3.9)[/tex]

[tex]X = 25.3[/tex]

The 80th percentile of response times is of 25.3 minutes, which means that 80% of the response times are less than 25.3 minutes and 100 - 80 = 20% are more.

Item d:

This proportion is the p-value of Z when X = 18.5, so:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{18.5 - 22}{3.9}[/tex]

[tex]Z = -0.9[/tex]

[tex]Z = -0.9[/tex] has a p-value of 0.1841.

0.1841 = 18.41% of all response times for this fire department are below the national mean response time.

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