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 10.3 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

Given,

= 10.3 , = 2.8

We convert this to standard normal as

P( X < x) = P( Z < x - / )

a)

P( X < 10) = P( Z < 10 - 10.3 / 2.8)

= P( Z < -0.1071)

= 1 - P( Z < 0.1071)

= 1 - 0.5427 (From Z table)

= 0.4573

b)

P( X > 5) = P( Z > 5 - 10.3 / 2.8)

= P( Z > -1.8929)

= P( Z < 1.8929)

= 0.9708 (From Z table)

c)

P( 8 < X < 15) = P( X < 15) - P( X < 8)

= P( Z < 15 - 10.3 / 2.8) - P( Z < 8 - 10.3 / 2.8)

= P( Z < 1.6786) - P( Z < -0.8214)

= P( Z < 1.6786) - (1 - P( Z < 0.8214) )

= 0.9534 - ( 1 - 0.7943 ) (From Z table)

= 0.7477