Question

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. The manager of a fast-food restaurant determines that the average time that her customers wait for service is 3.5 minutes. Exercise (a) Find the probability that a customer has to wait more than 5 minutes.

Answer 1

23.97% probability that a customer has to wait more than 5 minutes.

Explanation:

The exponential probability distribution, with mean m, is described by the following equation:

`f(x) = \mu e^(-\mu x)`

In which
`\mu = (1)/(m)` is the decay parameter.

The probability that x is lower or equal to a is given by:

`P(X \leq x) = \int\limits^a_0 {f(x)} \, dx`

Which has the following solution:

`P(X \leq x) = 1 - e^(-\mu x)`

The manager of a fast-food restaurant determines that the average time that her customers wait for service is 3.5 minutes.

This means that
`m = 3.5`. So

`\mu = (1)/(m) = (1)/(3.5) = 0.2857`

Find the probability that a customer has to wait more than 5 minutes.

Either the customer has to wait for 5 minutes or less, or he has to wait for more than 5 minutes. The sum of the probabilities of these events is decimal 1. So

`P(X \leq 5) + P(X > 5) = 1`

We want
`P(X > 5)`. So

`P(X > 5) = 1 - P(X \leq 5)`

In which

`P(X \leq x) = 1 - e^(-\mu x)`

`P(X \leq 5) = 1 - e^(-0.2857*5) = 0.7603`

`P(X > 5) = 1 - P(X \leq 5) = 1 - 0.7603 = 0.2397`

23.97% probability that a customer has to wait more than 5 minutes.