Question

The manager of a fast-food restaurant determines that the average time that her customers wait for service is 1.5 minutes. (a) Find the probability that a customer has to wait more than 4 minutes. (Round your answer to three decimal

Answer 1

0.069 = 6.9% probability that a customer has to wait more than 4 minutes.

Explanation:

Exponential distribution:

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)`

In this problem, we have that:

`m = 1.5`

So

`\mu = (1)/(1.5) = 0.6667`

`P(X \leq x) = 1 - e^(-0.667x)`

Find the probability that a customer has to wait more than 4 minutes.

Either the customer has to wait 4 minutes or less, or he has to wait more than 4 minutes. The sum of the probabilities of these events is decimal 1. So

`P(X \leq 4) + P(X > 4) = 1`

We want P(X > 4). So

`P(X > 4) = 1 - P(X \leq 4) = 1 - (1 - e^(-0.667*4)) = 0.069`

0.069 = 6.9% probability that a customer has to wait more than 4 minutes.