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 waiting time was found to be a random variable best approximated by an exponential distribution with a mean length of waiting time equal to 2.8 minutes (i.e. the mean number of calls answered in a minute is 1/2.8). What proportion of callers is put on hold longer than 2.8 minutes?
A) 0.3679
B) 0.50
C) 0.60810
D) 0.6321

Question

Quizplus · Accepted Answer

Since the waiting time follows an exponential distribution with mean 2.8 minutes, we can use the cumulative distribution function of the exponential distribution to find the probability that a customer will wait longer than 2.8 minutes on hold:

P(X > 2.8) = 1 - P(X ≤ 2.8) = 1 - F(2.8)

where F(2.8) is the cumulative distribution function of the exponential distribution with mean 2.8 evaluated at 2.8.

Using the formula for the cumulative distribution function of the exponential distribution with mean λ:

F(x) = 1 - e^(-λx)

we can find that:

F(2.8) = 1 - e^(-2.8/2.8) = 1 - e^(-1) = 1 - 0.3679 = 0.6321

Therefore, the proportion of callers put on hold longer than 2.8 minutes is:

P(X > 2.8) = 1 - F(2.8) = 1 - 0.6321 = 0.3679

Hence, the answer is A.

A Catalog Company That Receives the Majority of Its Orders