Question

olice response time to an emergency call is the difference between the time the call is first received by the dispatcher and the time a patrol car radios that it has arrived at the scene. Over a long period of time, it has been determined that the police response time has a normal distribution with a mean of 7.2 minutes and a standard deviation of 2.1 minutes. For a randomly received emergency call, find the probability that the response time is between 3 and 9 minutes. (Round your answer to four decimal places.)

Answer 1

0.7823 = 78.23% probability that the response time is between 3 and 9 minutes.

Explanation:

Normal Probability Distribution:

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

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

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value 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 7.2 minutes and a standard deviation of 2.1 minutes.

This means that
`\mu = 7.2, \sigma = 2.1`

For a randomly received emergency call, find the probability that the response time is between 3 and 9 minutes.

This is the pvalue of Z when X = 9 subtracted by the pvalue of Z when X = 3.

X = 9

`Z = (X - \mu)/(\sigma)`

`Z = (9 - 7.2)/(2.1)`

`Z = 0.86`

`Z = 0.86` has a pvalue of 0.8051

X = 3

`Z = (X - \mu)/(\sigma)`

`Z = (3 - 7.2)/(2.1)`

`Z = -2`

`Z = -2` has a pvalue of 0.0228

0.8051 - 0.0228 = 0.7823

0.7823 = 78.23% probability that the response time is between 3 and 9 minutes.