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 9.6 minutes and a standard deviation of 2.3 minutes. For a randomly received emergency call, find the following probabilities. (For each answer, enter a number. Round your answers to four decimal places.)

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

Answer 1

a) 0.5447

b) 0.0228

c) 0.4325

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.

Normal distribution with a mean of 9.6 minutes and a standard deviation of 2.3 minutes.

This means that
`\mu = 9.6, \sigma = 2.3`

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. So

X = 10

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

`Z = (10 - 9.6)/(2.3)`

`Z = 0.17`

`Z = 0.17` has a p-value of 0.5675

X = 5

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

`Z = (5 - 9.6)/(2.3)`

`Z = -2`

`Z = -2` has a p-value of 0.0228

0.5675 - 0.0228 = 0.5447 probability that a randomly received emergency call is between 5 and 10 minutes.

b. less than 5 min

p-value of Z when X = 5, which from item a), is 0.0228, so 0.0228 probability that a randomly received emergency call is of less than 5 minutes.

c. more than 10 min

1 subtracted by the p-value of Z when X = 10, which, from item a), is of 0.5675.

1 - 0.5675 = 0.4325

0.4325 probability that a randomly received emergency call is of more than 10 minutes.