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 7.6 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)

the response time is between 3 and 9 minutes

(b)

the response time is less than 3 minutes

(c)

the response time is more than 9 minutes

Answer 1

Solution :

Let X be a random variable which represents the police response times to an emergency call.

Given that, X ~ N(7.6, 2.3²)

Mean (μ) = 7.6

SD (σ) = 2.3

a) We have to find P(3 < X < 9).

P(3 < X < 9) = P(X < 9) - P(X ≤ 3)

We know that if X ~ N(μ, σ²) then,

Using "pnorm" function of R we get,

P(Z < 0.6087) = 0.7286 and P(Z ≤ -2) = 0.0227

The probability that the response time is between 3 and 9 minutes is 0.7059.

b) We have to find P(X < 3).

In part (a) we have already found that, P(X < 3) = 0.0228

Hence, the probability that the response time is less than 3 minutes is 0.0228.

c) We have to find P(X > 9).

P(X > 9) = 1 - P(X ≤ 9)

From part (a) we have, P(X ≤ 9) = 0.7286

Hence, P(X > 9) = 1 - 0.7286

P(X > 9) = 0.2714

Hence, the probability that the response time is more than 9 minutes is 0.2714.