The time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour.

What is the probability that more than three aircraft arrive within an hour?

#### Solution

If the inter-arrival time is exponential with mean $\mu= 1$ hour, the arrivals must form a Poisson process with rate $\lambda = 1$.

Let $X$ denote the number of arrivals in 1 hour.

Therefore $X\sim P(1)$.

The pmf of Poisson distribution with $\lambda =1$ is

` $$ \begin{aligned} P(X=x) &= \frac{e^{-1}(1)^x}{x!},\; x=0,1,2,\cdots \end{aligned} $$ `

The probability that more than three aircraft arrive within an hour is

` $$ \begin{aligned} P(X\geq 3) &= 1- P(X\leq 2)\\ &= P(X=0) + P(X=1) +P(X=2)\\ &= \frac{e^{-1}1^{0}}{0!}+\frac{e^{-1}1^{1}}{1!}+\frac{e^{-1}1^{2}}{2!}\\ &= 0.3679+0.3679+0.1839\\ &= 0.9197 \end{aligned} $$ `

#### Further Reading

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