# Poisson Distribution

```Poisson Distribution
This is a distribution useful for modeling of the number of events occuring in an interval of space or time,
for example, the number of automobiles arriving at a toll booth during a 10-minute period or the number of
weak points in a square meter of plastic sheeting. The expected number of events that occur in one period
or region is constant and is called the rate parameter and is usually denoted by λ. The requirements for a
Poisson distribution are that (a) no two events can occur simultaneously, (b) events occur independently
in different intervals, and (c) that the expected number of events in each time interval remain constant.
The probability mass function of a Poisson random variable X is given by
p(x) =
e−λ λx
,
x!
x = 0, 1, 2, . . .
where p(x) = P (X = x), that is the probability that there are exactly x occurrences of an event in a
specified interval, given that the expected number of occurrences of the event in the interval is λ. For
example, if the events occur on average 3 times per minute, and if one is interested in the number of events
occurring in a 5-minute period, you would use a Poisson distribution with λ = 3 &times; 5 = 15 to model the
number of occurrences in a 5-minute period.
It can be easily shown that E[X] = λ and V ar[X] = λ. Thus Poisson is one theoretical discrete distribution
that could be used to model count data that display the property that the variance is the same as the
expected number.
Example: Prob 3.27 (Baron)
Messages arrive at an electronic message center at random times, with an average of 9 messages per hour.
1. What is the probability of receiving at least five messages during the next hour?
We can verify that conditions for assuming a Poisson model for X, the number of messages received
in an hour, are satisfied. The average number of messages receieved in an hour = E[X] = λ = 9.
Thus X ∼ P o(9) and
P (X ≥ 5) = 1 − P (X ≤ 4) = 1 − .055 = .945
from Poisson Table A3 (from Baron) or see link on Stat 330 homepage for tables.
2. What is the probability of receiving exactly five messages during the next hour?
P (X = 5) =
e−λ λx
e−9 &middot; 95
=
= .0607
x!
5!
3. What is the probability that more than 10 messages will be received within the next two hours?
We have to model Y , the number of messages received within the next two hours as another Poisson
r.v. Y ∼ P o(18) that is λ = 18 for a 2-hour period.
P (Y &gt; 10) = 1 − P (X ≤ 10) = 1 − .03 = .97
```