Question

25.
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. find the proportion of response times between 14 and 29 minutes
b. calculate and interpret the 80th percentile of the distribution of response times.
c. determine whether 37 minutes is considered an outlier. show your work.

Explanation:

Step1: Define given parameters

Mean $\mu = 22$ minutes, standard deviation $\sigma = 3.9$ minutes

Step2 (Part a): Calculate z-scores

For $x_1=14$: $z_1=\frac{14-22}{3.9}\approx-2.05$
For $x_2=29$: $z_2=\frac{29-22}{3.9}\approx1.79$

Step3 (Part a): Find cumulative probabilities

$P(Z < -2.05) \approx 0.0202$, $P(Z < 1.79) \approx 0.9633$
Subtract: $0.9633 - 0.0202 = 0.9431$

Step4 (Part b): Find z for 80th percentile

The z-score for 80th percentile is $z \approx 0.84$

Step5 (Part b): Calculate response time

$x = \mu + z\sigma = 22 + (0.84)(3.9) \approx 25.28$ minutes

Step6 (Part c): Use IQR outlier rule

First, find $Q_1$ (25th percentile, $z=-0.67$) and $Q_3$ (75th percentile, $z=0.67$):
$Q_1 = 22 + (-0.67)(3.9) \approx 19.39$
$Q_3 = 22 + (0.67)(3.9) \approx 24.61$
IQR = $24.61 - 19.39 = 5.22$
Upper fence = $Q_3 + 1.5(IQR) = 24.61 + 1.5(5.22) = 32.44$

Answer:

a. 0.9431
b. The 80th percentile is approximately 25.28 minutes. This means 80% of fire response times are 25.28 minutes or less, and 20% are longer.
c. Since 37 > 32.44, 37 minutes is an outlier.