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 proportion of callers is put on hold longer than 2.8 minutes?
A) 0.50
B) 0.367879
C) 0.60810
D) 0.632121

Question

Quizplus · Accepted Answer

We know that the length of time customers are willing to wait on hold before ordering a product follows an exponential distribution with a mean of 2.8 minutes.

The probability density function (PDF) of an exponential distribution is given by:

f(x) = λe^(-λx)

where λ is the rate parameter. The mean of an exponential distribution is equal to 1/λ.

In this case, we know that the mean is 2.8 minutes, so:

2.8 = 1/λ

λ = 1/2.8

λ = 0.357

Now we need to find the probability that a caller is put on hold for longer than 2.8 minutes.

P(X > 2.8) = ∫2.8∞ λe^(-λx) dx

P(X > 2.8) = e^(-λ(2.8))

P(X > 2.8) = e^(-0.357(2.8))

P(X > 2.8) = e^(-0.9996)

P(X > 2.8) = 0.367879

Therefore, the proportion of callers put on hold longer than 2.8 minutes is 0.367879 or approximately 0.37.

The answer is B.

A Catalog Company That Receives the Majority of Its Orders