Question

The manager of a fast-food restaurant determines that the average time that her customers wait for service is 2.5 minutes. (a) Find the probability that a customer has to wait more than 4 minutes. (Round your answer to three decimal places.) (b) Find the probability that a customer is served within the first minute. (Round your answer to three decimal places.) (c) 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:

a) 0.202 = 20.2% probability that a customer has to wait more than 4 minutes.

b) 0.33 = 33% probability that a customer is served within the first minute.

c) 11.5 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:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

In this question:

$m = 2.5, \mu = \frac{1}{2.5} = 0.4$

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

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

$P(X > 4) = e^{-0.4*4} = 0.202$

0.202 = 20.2% probability that a customer has to wait more than 4 minutes.

(b) Find the probability that a customer is served within the first minute.

$P(X \leq x) = 1 - e^{-\mu x}$

$P(X \leq 1) = 1 - e^{-0.4*1} = 0.33$

0.33 = 33% probability that a customer is served within the first minute.

(c) 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

We have to find x for which:

$P(X > x) = 0.01$

So

$P(X > x) = e^{-0.4x}$

Then

$e^{-0.4x} = 0.01$

$\ln{e^{-0.4x}} = \ln{0.01}$

$-0.4x = \ln{0.01}$

$x = -\frac{\ln{0.01}}{0.4}$

$x = 11.5$

So 11.5 minutes.