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

Answer :

a) Pr(5<x<10) = 0.6832

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<a) = 0.5 - Φ(z)

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<X<10)

Pr(5<X<10) = 1 - Pr(X>10) - Pr(X<5)

= 1 - 0.2912 - 0.0256

= 1 - 0.3168

= 0.6832

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