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 6.0 minutes and a standard deviation of 1.5 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 7 minutes

(b) the response time is less than 3 minutes

(c) the response time is more than 7 minutes

Answer 1

Solution

Let X = Police response time (in minutes) to an emergency call. We are given:

X ~ N(6.0, 1.5) .................................................................................................................................................................. (1)

Back-up Theory

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ², then,

Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution and hence

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .…..........................................….........…………...…(2)

Probability values for the Standard Normal Variable, Z, can be directly read off from Standard Normal Tables............. (2a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) ............................................…(2b)

Now, to work out the solution,

Part (a)

Probability that the response time is between 3 and 7 minutes

= P(3 < X < 7)

= P[{(3 - 6)/1.5} < Z < {(7 - 6)/1.5}] [vide (2) and (1)]

= P(- 2.0 < Z < 0.6667)

= P(Z < 0.6667) – P(Z < - 2.0)

= 0.7475 – 0.0227 [vide (2b)]

= 0.7248   Answer 1

Part (b)

Probability that the response time is less than 3 minutes

= P(X < 3)

= P[Z < {(3 - 6)/1.5}]   [vide (2) and (1)]

= P(Z < - 2.0)

= 0.0227 [vide (2b)]   Answer 2

Part (c)

Probability that the response time is more than 7 minutes

= P(X > 7)

= P[Z > {(7 - 6)/1.5}]   [vide (2) and (1)]

= P(Z > 0.6667)

= 0.2525 [vide (2b)]   Answer 3

DONE

[Going beyond,

Note that sum of answers 1, 2 and 3 is 1.]