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

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

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators