Question

Police 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 8.1 minutes and a standard deviation of 2.0 minutes. For a randomly received emergency call, find the following probabilities.

a. between 5 and 10 min
b. less than 5 min
c. more than 10 min

Answer 1

a) 0.7683 = 76.83% probability that a randomly selected emergency call is between 5 and 10 minutes.

b) 0.0606 = 6.06% probability that a randomly received emergency call is of less than 5 min.

c) 0.1711 = 17.11% probability that a randomly received emergency call is of more than 10 min.

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 8.1 minutes and a standard deviation of 2.0 minutes.

This means that
`\mu = 8.1, \sigma = 2`

a. between 5 and 10 min

This is the p-value of Z when X = 10 subtracted by the p-value of Z when X = 5.

X = 10

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

`Z = (10 - 8.1)/(2)`

`Z = 0.95`

`Z = 0.95` has a p-value of 0.8289

X = 5

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

`Z = (5 - 8.1)/(2)`

`Z = -1.55`

`Z = -1.55` has a p-value of 0.0606

0.8289 - 0.0606 = 0.7683

0.7683 = 76.83% probability that a randomly selected emergency call is between 5 and 10 minutes.

b. less than 5 min

p-value of Z when X = 5, which, found from item a, is of 0.0606

0.0606 = 6.06% probability that a randomly received emergency call is of less than 5 min.

c. more than 10 min

1 subtracted by the p-value of Z when X = 10, which, from item a, is of 0.8289

1 - 0.8289 = 0.1711

0.1711 = 17.11% probability that a randomly received emergency call is of more than 10 min.