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 8.9 minutes and a standard deviation of 2.0 minutes. For a randomly received emergency call, find the following probabilities. (Round your answers to four decimal places.)

a) between 5 and 10 min

b) less than 5 min

c) more than 10 min

Question

Mathematics

Peyton Proctor

20 October, 15:53

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 8.9 minutes and a standard deviation of 2.0 minutes. For a randomly received emergency call, find the following probabilities. (Round your answers to four decimal places.)

a) between 5 and 10 min

b) less than 5 min

c) more than 10 min

+2

Answers (1)

Know the Answer?

Kadyn · Answer 1

a) Pr (5
b) Pr (x<5) = 0.0256

c) Pr (x>10) = 0.2912

Step-by-step explanation:

Police response time to an emergency call is the difference between the time the call is first received and the time a patrol car radios it has arrived at the scene

Mean (u) = 8.9mins

Standard deviation (α) = 2.0mins

Let X be the random variable which is a measure of the time to get a response from the police.

We first solve for b and c

b) The response time less than 5 mins = Pr (X<5)

For normal distribution

Z = (X - u) / α

For X = 5

Z = (5 - 8.9) / 2

Z = - 3.9/2

Z = - 1.95

From the normal distribution table, Z = - 1.95 is 0.4744

Φ (z) = 0.4744

When Z is negative

Pr (X
Pr (X<5) = 0.5 - 0.4744

= 0.0256

c) The response time more than 10mins = Pr (X>10)

For normal distribution

Z = (X - u) / α

For X = 10

Z = (10 - 8.9) / 2

Z = 1.1/2

Z = 0.55

From the normal distribution table, Z = 0.55 is 0.2088

Φ (z) = 0.2088

When Z is positive

Pr (X>a) = 0.5 - Φ (z)

Pr (X>5) = 0.5 - 0.2088

= 0.2912

c) The response time between 5 and 10mins = Pr (5
Pr (5
= 1 - 0.2912 - 0.0256

= 1 - 0.3168

= 0.6832