A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 2.8 minutes. What is the probability that a randomly selected caller is placed on hold fewer than 7 minutes?
A) 0.0009119
B) 0.082085
C) 0.917915
D) 0.9990881

Question

Quizplus · Accepted Answer

Since the length of time a customer is on hold is best approximated by an exponential distribution with a mean of 2.8 minutes, we know that the probability density function for this distribution can be expressed as lambda * e^(-lambda * x), where lambda = 1/2.8.
To find the probability that a randomly selected caller is placed on hold fewer than 7 minutes, we need to find P(X < 7), where X is the length of time a customer is on hold. 
P(X < 7) = integral from 0 to 7 of lambda * e^(-lambda * x) dx 
= integral from 0 to 7 of (1/2.8) * e^(-x/2.8) dx 
Using integration by substitution with u = x/2.8, du = 1/2.8 * dx, we can simplify the integral as follows:
P(X < 7) = integral from 0 to 7 of (1/2.8) * e^(-x/2.8) dx 
= integral from 0 to 7/2.8 of e^(-u) du 
= (-e^(-u)) evaluated from 0 to 7/2.8 
= (-e^(-7/2.8)) - (-e^0) 
= 0.917915
Therefore, the answer is C, 0.917915.

A Catalog Company That Receives the Majority of Its Orders