Question

In: Statistics and Probability

The Customer Service Center for a large airline has determined that the amount of time spent...

The Customer Service Center for a large airline has determined that the amount of time spent with a customer about a complaint is normally distributed, with a mean of 9.3 minutes and a standard deviation of 2.9 minutes.

(a) What is the probability that, for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be less than 10 minutes? (Round your answers to four decimal places.)


(b) What is the probability that, for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be between 8 and 15 minutes? (Round your answers to four decimal places.)


(c) Only 20% of the complaints will take longer than what amount of time to resolve? (Round your answers to four decimal places.)
minutes

Solutions

Expert Solution

X : amount of time spent with a customer about a complaint

X follows  normally distribution, with a mean of 9.3 minutes and a standard deviation of 2.9 minutes.

(a)

Probability that, for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be less than 10 minutes = P(X<10)

Z-score for 10 = (10-9.3)/2.9 = (0.7)/2.9 = 0.24

From standard normal tables,

P(Z<0.24) = 0.5948

P(X<10) = P(Z<0.24) = 0.5948

Probability that, for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be less than 10 minutes = 0.5948

(b) Probability that, for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be between 8 and 15 minutes = P(8<X<15)

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

Z-score for 15 = (15-9.3)/2.9 = 1.97 ; Z-score for 8 = (8-9.3)/2.9 = -0.45

From standard normal tables,

P(Z < 1.97) =0.9756 ; P(Z<-0.45) = 0.3264

P(X<15) = P(Z < 1.97) =0.9756

P(X<8) = P(Z<-0.45) = 0.3264

P(8<X<15) = P(X<15) - P(X<8) = 0.9756-0.3264 = 0.6492

Probability that, for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be between 8 and 15 minutes = 0.6492

(c) Only 20% of the complaints will take longer than what amount of time to resolve

Let X20 be the time to resolve such that 20% of the complaints will take longer than X20

i.e

P(X>X20) = 20/100 =0.20

P(X>X20) = 1-P(X X20) = 0.20

P(X X20) = 1-0.20 =0.80

Let Z20 be the z-score for X20

Z20 = (X20 - Mean)/Standard deviation = (X20 - 9.3)/2.9

X20 = 9.3 + 2.9Z20

P(Z Z20) = 0.80

From standard normal tables,

P(Z 0.84 ) = 0.7995

Z20 = 0.84

X20 = 9.3 + 2.9Z20 =9.3 + 2.9 x 0.84 = 9.3+2.436=11.736

X20 = 11.736

Only 20% of the complaints will take longer than 11.736 minutes  to resolve


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