a fire department in a rural county reports that its response time to fires is approximately normally…

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.)

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.)

Answer

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)