A catalog company administered a survey 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 3.5 minutes.What proportion of customers having to hold more than 5.0 minutes will hang up before placing an order?
A)0.2397
B)0.5714
C)0.4286
D)0.7603

Question

Quizplus · Accepted Answer

Let X be the length of time a customer waits on hold. We know that X follows an exponential distribution with mean 3.5 minutes, i.e. E(X) = 3.5.

We need to find P(X > 5 | X > 5), i.e. the probability that a customer hangs up given that they have already waited more than 5 minutes.

Using the memoryless property of the exponential distribution, we have:

P(X > 5 | X > 5) = P(X > 5)/P(X > 5) = e^(-5/3.5) / e^(-5/3.5) = e^(-5/3.5) ≈ 0.2397.

Therefore, the proportion of customers hanging up after waiting more than 5.0 minutes is approximately 0.2397, which corresponds to choice A.

A Catalog Company Administered a Survey to Determine How Long