Question

Aircraft arrive at the Mckinnon Airport with a Poisson distribution at a rate of one aircraft arrival every 25 minutes. What is the likelihood that at least 1 aircraft arrives in a 45-minute window of time?

Answer 1

The random variable X(t) = number of successes in time , follow Poisson distribution or X(t) ~ Poisson[$\fn_cm \lambda$= at]

The pmf of X(t) is :

$\fn_cm p[x,\lambda ] =\begin{Bmatrix} \frac{e^{-\lambda .\lambda ^{x}}}{x!} , x=0,1,2,3,4....... \\ 0 , o/w \end{Bmatrix}$

Given

X = Number of aircraft that arrive in t hours = 1 aircraft

a = 1 aircraft per 25 minute = 1 *60/25 aircraft per hour = 4 aircraft per hour

t = 45 minute = 0.75 hour

$\fn_cm X\sim Poisson [\lambda = at = 4\times \frac{45}{60} hours = 0.75 ]$

Now, probability that  at least 1 aircraft arrives in a 45-minute window of time:

$\fn_cm P\left (X(t) =0.75) \geq 1 \right ) = 1 - P (0)$

$\fn_cm =1 - \frac{e^{0.75 }\times (0.75)^{0}}{0!}$

$\fn_cm =1 - e^{0.75 }$

  $\fn_cm =1 - 0.4734$

  $\fn_cm =0.5276$