Question

The manager of a fast-food restaurant determines that the average time that her customers wait for service is 3.5 minutes.The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. But she doesn't want to give away free hamburgers to more than 1% of her customers. What number of minutes should the advertisement use

Answer 1

The advertisement should use 16 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)`

The probability of finding a value higher than x is:

`P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^(-\mu x)) = 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, \mu = (1)/(3.5) = 0.2857`

What number of minutes should the advertisement use?

The values of x for which:

`P(X > x) = 0.01`

So

`e^(-\mu x) = 0.01`

`e^(-0.2857x) = 0.01`

`\ln{e^(-0.2857x)} = ln(0.01)`

`-0.2857x = ln(0.01)`

`x = -(ln(0.01))/(0.2857)`

`x = 16.12`

Rounding to the nearest number, the advertisement should use 16 minutes.