The number e question?

Customer Service

The service time distribution describes the probability P that the service time of the customer will be no more than t hours. If m is the mean number of customers serviced in an hours then P=1-e^-mt

A. Suppose a computer technical support representative can answer calls from 6 customers in 1 hour. What is the probability that a customer will be on hold less than 30 minutes?

This is what I did P=1-e^(-6)(30) Would this be correct?


b.A credit card customer service department average 34 calls per month. Determine that amount of time after which it is 50 percent likely that a customer has been served?

I am unsure of how to star part b

Any assistance would be very beneficent. (Giggle)

Quote:

---

Originally Posted by homeylova223

Customer Service

The service time distribution describes the probability P that the service time of the customer will be no more than t hours. If m is the mean number of customers serviced in an hours then P=1-e^-mt

A. Suppose a computer technical support representative can answer calls from 6 customers in 1 hour. What is the probability that a customer will be on hold less than 30 minutes?

This is what I did P=1-e^(-6)(30) Would this be correct?


b.A credit card customer service department average 34 calls per month. Determine that amount of time after which it is 50 percent likely that a customer has been served?

I am unsure of how to star part b

Any assistance would be very beneficent. (Giggle)

---

You need to be careful with your units. In part a t (time) should be input in hours. 30min= one half hour!

For number 2 set P=.50 fifty percent and solve for t.