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

Answer:

The advertisement should use 16 minutes.

Step-by-step explanation:

Exponential distribution:

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

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ 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 $m = 3.5, \mu = \frac{1}{3.5} = 0.2857$

What number of minutes should the advertisement use?

The values of x for which:

$P(X > x) = 0.01$

So

$e^{-\mu x} = 0.01$

$e^{-0.2857x} = 0.01$

$\ln{e^{-0.2857x}} = \ln{0.01}$

$-0.2857x = \ln{0.01}$

$x = -\frac{\ln{0.01}}{0.2857}$

$x = 16.12$

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