A catalogue 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) 0.60810
B) 0.50
C) 0.367879
D) 0.632121

Question

Quizplus · Accepted Answer

We are told that the length of time customers are willing to wait on hold is best described by an exponential distribution with a mean of 2.8 minutes. The probability density function of an exponential distribution with mean 𝜇 is:

f(x) = (1/𝜇) * e^(-x/𝜇)

We want to find the proportion of callers who are put on hold longer than 2.8 minutes, which is the same as finding:

P(X > 2.8) = ∫2.8∞ f(x)dx

Plugging in the mean 𝜇 = 2.8, we get:

P(X > 2.8) = ∫2.8∞ (1/2.8) * e^(-x/2.8) dx

Using a software or a calculator, we can evaluate this integral to get:

P(X > 2.8) = 0.367879

Therefore, the answer is choice C.

A Catalogue Company That Receives the Majority of Its Orders