Answer to Question #288516 in Statistics and Probability for sima

Question

A cell phone company is interested to know more about the behaviour of customers who exceed their free minutes included on their basic cell phone contracts. They evaluate such customers and find that the excess time follows an exponential distribution with a mean of 15 minutes. For a random sample of 50 customers who exceed their free minutes’ allowance, what is the probability that their average excess time is longer than 18 minutes?

Answer 1

Consider a random sample of 50 customers who exceed their free minutes’ allowance.

Let $X =$ the excess time used by one individual cell phone customer who exceeds his contracted time allowance: $X\sim Exp(\dfrac{1}{15}).$

The mean and variance of the exponential distribution are

$$\mu=15, \sigma^2=15^2$$

Let $\bar{X}=$ the mean excess time used by a sample of $n = 50$ customers who exceed their contracted time allowance.

By the central limit theorem for sample means $\bar{X}\sim N(\mu,\sigma^2/n)$

$$P(\bar{X}>18)=1-P(\bar{X}\leq 18)$$

$$=1-P(Z\leq\dfrac{18-15}{15/\sqrt{50}})\approx1-P(Z\leq1.4142)$$

$$\approx0.07865$$

The probability that their average excess time is longer than 18 minutes is

$0.07865.$