Poisson Distribution
The Poisson Distribution is a discrete
distribution which takes on the values X = 0,
1, 2, 3, ... .
It is often used as a model for the number of
events (such as the number of telephone
calls at a business or the number of
accidents at an intersection) in a specific time
period. It is also useful in ecological studies,
e.g., to model the number of prairie dogs
found in a square mile of prairie.
Read Page 236 & 237 for examples
Discrete Variable Distributions
BINOMIAL DISTRIBUTION:
Used to calculate the probability of a certain number of
successes in a given number of trials
POISSON DISTRIBUTION:
Used to calculate probability of a number of successes that
take place in a certain interval of time or space
Poisson Rule for Probability
e 
P( X  r ) 
r!

HUH !
Where did that come from?
What is Lambda?
Why use e?
r
The Quest for the Meaning of Lambda!
This requires a bit of lateral thinking so hang on tight!
Question
What is the only function , f(x) that has
1. f(x) = f’(x)
2. f(0) = 1
ANSWER : f(x) = ex
Okay try and find f’(x) of this function
2
3
n
x x
x
x
1     ..........   ......
1! 2! 3!
n!
What about f(0)?
It has the same properties as f(x) = e^x
and if they were both graphed you can confirm that
are in fact equivalent in every way
GREAT BUT HOW DOES THIS HELP TO EXPLAIN WHAT LAMBDA IS ?
BE PATIENT – WE’LL GET THERE
WE KNOW
2
3
n
x x
x
x
e  1     ..........   ......
1! 2! 3!
n!
x
AND
e 
P( X  r ) 
r!


AND
r
 P( X  r )  1
r 0

 P( X  r ) 
r 0
e
0 
0!

e
1 
1!

e
2 
2!
 ..........
 1  2
 
 1  
 ..........  e
 1! 2!

 
e e
1
AS IT SHOULD!
So what about Lambda?
Okay I’m coming to that – I said there would be some lateral thinking involved
A Poisson distribution is a discrete distribution so the expected
value rule still holds.
X
Po ( )

E ( X )   r.P( X  r )
r 0
e   0
e  1
e   2
e   3
E ( X )  0.
 1.
 2.
 3.
 ...
0!
1!
2!
3!
Keep going. What do you find?
E(X) = LAMBDA!
While we’re at it lets check variance as well

Var ( X )   r 2 .P( X  r )  ( E ( X )) 2
r 0
What do you find this time?
Variance = Expected Value (Mean) = Lamdba!
So if you are doing a problem which fits a
Poisson Distribution the problem would define
lambda and it could be given as the mean or
as the variance (or even standard deviation).
Example
Top Car Rentals rents to tourists. They have 4 cars which are
hired out on a daily basis. The number of requests each
day occurs randomly with a mean of 3. Determine the
probability that
(a)None of the cars are rented
(b)At least 3 of the cars are rented
(c)Some requests are refused
NOTE: poissonpdf and cdf are in your TI83’s DISTR Menu
Yipee!
ANS:a) 0.0498, b) 0.577,c) 0.185
So where did the rule come from?
It came from Statisticians attempts to fit a curve to the
graphed distributions shown on page 242
Look here for further evidence
http://www.math.csusb.edu/faculty/stanton/m
262/poisson_distribution/Poisson_old.html