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 6.9 minutes. assume that this claim is true. one percent of response times take at least how many minutes? minutes (round to 2 decimal places.)

Explanation:

Step1: Find the z - score

We want to find the value \(x\) such that \(P(X\geq x)=0.01\), which is equivalent to \(P(X < x)=0.99\). Looking up in the standard normal distribution table (z - table), the z - score \(z\) corresponding to a cumulative probability of \(0.99\) is approximately \(z = 2.33\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 22\) (mean), \(\sigma=6.9\) (standard deviation), and \(z = 2.33\). We need to solve for \(x\). Rearranging the formula gives \(x=\mu+z\sigma\).
Substitute \(\mu = 22\), \(z = 2.33\), and \(\sigma = 6.9\) into the formula: \(x=22+2.33\times6.9\).
First, calculate \(2.33\times6.9 = 2.33\times(7 - 0.1)=2.33\times7-2.33\times0.1=16.31 - 0.233 = 16.077\).
Then, \(x=22 + 16.077=38.077\).

Answer:

\(38.08\)