Question

The manager of a fast-food restaurant determines that the average time that her customers wait for service is 2.5 minutes. This process can be described using an exponential probability density function with a mean of 2.5.

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

a) 0.2018

b)0.329

c)x=5

Explanation:

We are given that This process can be described using an exponential probability density function with a mean of 2.5.

So,
`\lambda = 2.5`

`f(x)=(1)/(\lambda)e^{-(1)/(\lambda) x}`

Where x>0 ,
`\lambda` > 0

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

`P(X>4)=(1)/(\lambda) \int\limits^(\infty) _(4) {e^{(-1)/(\lambda) x} \, dx`

`P(X>4)=1-(1-e^{(-4)/(\lambda)})\\P(X>4)=e^{(-4)/(2.5)}\\P(X>4)=0.2018`

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

`P(0<x<1)=1-e^{(-1)/(2.5)}=0.329`

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?

`P(X>x) \leq 0.01\\e^{(-x)/(2.5)} \leq 0.01\\(-x)/(2.5) \leq log (0.01)\\x=-2.5 * log (0.01)\\x=5`

Answer 2

Explanation:

a)P( a customer has to wait more than 4 minutes )=P(X>4)=e-4/2.5 =0.202

b)probability that a customer is served within the first minute =P(X<1)=1-e-1/2.5 =0.3297

c) here 99th percentile corresponding value =-2.5*ln(1-0.99)=11.51 minutes ~ 12minutes