Question

The Customer Service Center in a large New York department store has determined that the amount of time spent with a customer about a complaint is normally distributed, with a mean of 10.1 minutes and a standard deviation of 2.8 minutes. What is the probability that for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be as follows. (Round your answers to four decimal places.)(a) less than 10 minutes (b) longer than 5 minutes (c) between 8 and 15 minutes

Answer 1

0.6395 ; 0.9657 ; 0.7333

Explanation:

Given that:

Mean (m) = 10.1

Standard deviation (s) = 2.8

Obtain the following probabilities :

a) less than 10 minutes

P(x < 10)

Z = (x - m) / s

Z = (10 - 10.1) / 2.8

Z = 0.1 / 2.8

Z = 0.0357142

P(Z < 0.357) = 0.6395 ( Z probability calculator)

(b) longer than 5 minutes

P(x > 5)

Z = (x - m) / s

Z = (5 - 10.1) / 2.8

Z = −1.821428

P(Z < - 1.821) = 0.9657 ( Z probability calculator)

(c) between 8 and 15 minutes

P(8 < x < 15)

P(x < 8)

Z = (x - m) / s

Z = (8 - 10.1) / 2.8

Z = − 0.75

P(Z < - 0.75) = 0.22663 ( Z probability calculator)

P(x < 15)

Z = (x - m) / s

Z = (15 - 10.1) / 2.8

Z = 1.75

P(Z < 1.75) = 0.95994 ( Z probability calculator)

P(Z < 1.75) - P(Z < - 0.75)

0.95994 - 0.22663 = 0.7333