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 7.2 minutes and a standard deviation of 2.1 minutes. For a randomly received emergency call, find the following probabilities. (For each answer, enter a number. Round your answers to four decimal places.)

Options are;

(a) the response time is between 5 and 10 minutes

(b) the response time is less than 5 minutes

(c) the response time is more than 10 minutes

A) 0.7613

B) 0.1469

C) 0.0918

Step-by-step explanation:

We are given;

mean; μ = 7.2 minutes

Standard deviation; σ = 2.1 minutes

To answer the question, we will use the z-score formula which is;

z = (x¯ - μ)/σ

a) probability that the response time is between 5 and 10 minutes.

Z_5 = (5 - 7.2)/2.1

Z_5 ≈ -1.05

Z_10 = (10 - 7.2)/2.1

Z_10 = 1.33

From z-distribution table, p-value of z = -1.05 is p = 0.1469

Also,for z = 1.33, p-value = 0.9082

Thus; P(5 < x¯ < 10) = 0.9082 - 0.1469 = 0.7613

B) the response time is less than 5 minutes.

Thus, x¯ = 5

Like seen above;

Z_5 = (5 - 7.2)/2.1

Z_5 ≈ -1.05

P-value from z-distribution table is;

p = 0.1469

C) the response time is more than 10 minutes.

Like seen in option A above, P(x¯ < 10) = 0.9082

Therefore, P(x¯ > 10) = 1 - 0.9082 = 0.0918

i think the answer is b

the weight of the object inside the box!