Question

The caller times at a customer service center has an exponential distribution with an average of 22 seconds. Find the probability that a randomly selected call time will be less than 30 seconds? (Round to 4 decimal places.)

Answer 1

The probability that a randomly selected call time will be less than 30 seconds is 0.7443.

Explanation:

We are given that the caller times at a customer service center has an exponential distribution with an average of 22 seconds.

Let X = caller times at a customer service center

The probability distribution (pdf) of the exponential distribution is given by;

`f(x) = \lambda e^(-\lambda x) ; x > 0`

Here,
`\lambda` = exponential parameter

Now, the mean of the exponential distribution is given by;

Mean =
`(1)/(\lambda)`

So,
`22=(1)/(\lambda)` ⇒
`\lambda=(1)/(22)`

SO, X ~ Exp(
`\lambda=(1)/(22)`)

To find the given probability we will use cumulative distribution function (cdf) of the exponential distribution, i.e;

`P(X\leq x) = 1 - e^(-\lambda x)` ; x > 0

Now, the probability that a randomly selected call time will be less than 30 seconds is given by = P(X < 30 seconds)

P(X < 30) =
`1 - e^{-(1)/(22) * 30}`

= 1 - 0.2557

= 0.7443