Question

8. In deciding how many customer service representatives to hire and in planning their schedules, a firm that

offers smartphone replacement screens studies repair times. One such study revealed that repair times have

an approximately exponential distribution with a mean of 21 minutes.

(a) Find the probability that a randomly selected repair time will be between 13 and 25 minutes.

(b) The charge for screen repairs is $40 for each half hour (or part thereof) for labor. What is the proba-

bility that a repair job will result in a charge for labor of $75?

(c) In planning schedules, how much time should the firm allow for each repair to ensure that the change

of any one repair time’s exceeding this allowed time is only 0.05?

(d) Find the probability distribution, the mean, and the variance of the time taken to serve three waiting

customers.

(e) Find the probability that the total service time for three customers is less than 1 hour.

Answer 1

Given the repair times follows exponentila distribution .

a) The probability

b) The charge for screen repairs is $40 for each half hour or $4/3 for a minute. $75 means

The probability that a repair job will result in a charge for labor of $75 or more is

c) We need to find

d) The distribution if 3 Identical exponential distributions is Gamma distribution.

. The mean of the this Gamma distribution is . The variance is

e) The probability,

Use R to calculate the probability, it is .

The R command is

> pgamma(60,shape = 3, scale = 21)
[1] 0.5440553