# Solved: Suppose the waiting time to get food after placing an order at a fast-food restaurant is exponentially distributed with a mean of 60 seconds

#### ByDr. Raju Chaudhari

Feb 24, 2021

Suppose the waiting time to get food after placing an order at a fast-food restaurant is exponentially distributed with a mean of 60 seconds. If a randomly selected customer orders food at the restaurant, what is the probability that the customer will wait at least:

a. 90 seconds?
b. Two minutes?

## Solution

The waiting time to get food after placing an order at a fast-food restaurant is exponentially distributed with a mean of 60 seconds = 1 min.

The pdf of waiting time $X$ is

$$f(x)=e^{-x}, x>0$$

and the distribution function of $X$ is $F(x)=1-e^{-x}$.

(a) The probability that the customer will wait at least 90 seconds is

 \begin{aligned} P(X\geq 90\text{ second}) &=P(X\geq 1.5 \text{min})\\ &= P(X\geq 1.5)\\ &=1-P(X< 1.5)\\ &=1- F(1.5)\\ &= 1- \bigg[1- e^{-1.5}\bigg]\\ &=e^{-1.5}\\ &= 0.2231302. \end{aligned}

(b) The probability that the customer will wait at least 2 minutes is

 \begin{aligned} P(X\geq 2)&=1-P(X< 2)\\ &=1- F(2)\\ &= 1- \bigg[1- e^{-2}\bigg]\\ &=e^{-2}\\ &= 0.1353353. \end{aligned}