Question

Arrivals at a parking lot are assumed to follow the Poisson distribution. The average arrival rate is 2.8 per minute. What is the probability that during a given minute no cars will arrive

Answer 1

`P(x = 0) = 0.061`

Explanation:

Given

`Average = 2.8`

Required

Determine the probability that no car will arrive.

Since it is a Poisson distribution, the required probability is:

`P(x) = (e^(-m)m^x)/(x!)`

Where

`m = average = 2.8`

In this case;

`x = 0`

So:

`P(x) = (e^(-m)m^x)/(x!)`

`P(x = 0) = (e^(-2.8)2.8^0)/(0!)`

`P(x = 0) = (e^(-2.8)*1)/(1)`

`P(x = 0) = e^(-2.8)`

`P(x = 0) = 0.06081006262`

`P(x = 0) = 0.061` --- Approximated