Lecture 6
Dan Piett
STAT 211-019
West Virginia University
Last Week
 Expected Value of Probability Distributions
 Binomial Distributions
 Probabilities
 Mean
 Standard Deviation
Overview
 Poisson Distribution
 Poisson Probabilities
Poisson Distribution
 Suppose an experiment possesses the following properties:
The random variable X counts the number of occurrences of
some event of interest in a unit of time or space (or both)
2. The events occur randomly
3. The mean number events per unit of space/time is constant
4. The random variable X has no fixed upper limit
(This will usually be false, but assume that if there is an upper
limit, it is reasonably large)
1.
 This is a Poisson experiment (NOT pronounced Poison)
 Note that Poisson Distributions are Discrete (You cannot have
1.9976 occurrences)
Example: Number of Pieces of Mail in a Day
(Mean of 5 per day)
Requirements
1. Random variable X counts
the number of occurrences
of some event of interest in
a unit of time or space (or
both)
2. The events occur randomly
3. The mean number events
per unit of space/time is
constant (lambda)
4. The random variable X has
no fixed upper limit
This Experiment
1.
2.
3.
4.
Random variable X counts the
number of pieces of mail
occurring in 1 day.
We can assume that the pieces of
mail occur randomly
Mean number of events per day is
constant ( We can assume this as
long as we assume that this mean
holds true for the times we are
interested in)
There is no upper-limit to the
mail you can receive in 1 day
(Debatable, but immaterial)
General Poisson Distribution
 Suppose X counts the number of occurrences in a Poisson
experiment. Then X follows a Poisson Distribution
 Notation:
 Pois stands for Poisson distribution
 lambda stands for the mean number of occurrences per unit
time/space
 For the previous example X~Pois(5)
Formula for a Poisson Distribution
 Probability of Success in a Poisson Distribution
Problem on Board
 Assume the mean number of people who visit the emergency
room of a particular hospital is 6 per hour.
 Does this constitute a Poisson Experiment?
 Find
 The prob that 0 people will visit the emergency room in 1 hour.
 The prob that 7 people will visit the emergency room in1 hour.
Cumulative Poisson Probabilities
 The previous formula can be used to find the probability that
X equal to exactly some value
 What about other probabilities of interest?
 X equal to less than some value?
 X equal to more than some value?
 X is between two values?
 How do we do this?
 EXACTLY like Binomial Probabilities
Back to the Previous Example
 What is the probability that at most 3 people visit the
emergency room in 1 hour?
 At most = less than or equal to
 At most 3 people= {0, 1, 2, 3, 4,…}
 P(At most 3 people) = P(X=0)+P(X=1) + P(X=2)+P(X=3)
 Note: The probability of this event is defined as the sums of
the probabilities.
 Remember that this only works because Poisson Probabilities
are discrete
 Looking pretty familiar? I’m sure you can guess an easier
way.
New Example
 Suppose that an archaeologist finds artifacts at a dig site at an
average rate of 2 per day. What is the probability that the
archaeologist finds fewer than 4 artifacts in a day.
 Fewer than 4= {0, 1, 2, 3, 4, 5, …}
 P(3 or less) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
 We would need to compute 4 probabilities to solve this.
 Is there a better way?
 Unlike Binomial, we only have 1 alternative method
 Using cumulative probability tables
 Why doesn’t the complementary rule work for less than
probabilities?
Cumulative Probability Tables
 Because of the difficulty of calculating these probabilities
(and how common the poisson distribution is). Cumulative
probabilities for specific values of lambda and x have
been tabulated.
 Note: These tables will be provided on exams.
 How to read the table:
 Find the appropriate lambda value, look for x
 This is the probability that X is less than or equal to that
value
Back to our Example
 We have our archaeologist finding 2 artifacts on average per
day. What is the probability that he:
 Finds at most 1 artifact?
 Finds less than 7 artifacts
Greater than Probabilities
 So we now know how to calculate the probability that X is
equal to exactly some value or the probability that X is less
than/less than or equal to some value.
 What about the probability that X is greater than/greater
than or equal to some value?
 Think back to complementary probabilities
Headed back to our Example
 Suppose the archaeologist moves to a new dig site and can
find an average of 4 artifacts per day now. What is the
probability that he finds 5 or more artifacts?
 5 or more = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, …}
 P(5 or more) = P(5) + P(6) + … P(infinity)
 Remember back to how we handled this with binomial
probabilities
 P(5 or more) and P(4 or less) are complementary events
 What does this mean?
 P(5 or more) = 1 – P(4 or less)
Greater than Probabilities
 Remember back to our use of the tables for calculating less
than or equal to probabilities
 We can likewise calculate greater than/greater than or
equal to probabilities using the table.
 Watch the =
 We want to get our greater than probabilities in terms of less
than or equal to
 P(X>3) = 1 – P(X<=3)
 {1, 2, 3, 4, 5, …)
 P(X>=3) = 1 – P(X<3) = P(X<=2)
 {1,2 ,3, 4, 5, 6}
In-between Probabilities
 So far we’ve done
 P(X=x), P(X<x), P(X>x)
 One more to go (The probability the X is between 2 values)




P(a < =x <= b)
Example: P(X is between 2 and 6)
Between 2 and 6 = {0, 1, 2, 3, 4, 5, 6, 7, … )
P(X is between 2 and 6) = P(X<=6) – P(X<=1)
 Why?
 P(X<=6) = P(0) + P(1) + P(2)+P(3)+P(4)+P(5)+P(6)
 P(X<=1) = P(0) + P(1)
 Subtract these and the 0 and 1 cancel leaving:
 P(2)+P(3)+P(4)+P(5)+P(6)
 This is what we want
Coming back to Exact Probabilities
 We can use the cumulative table to find exact probability as
well
 P(X=2) = P(X<=2) – P(X<=1)
 Same logic as the previous examples
 P(X<=2) = P(X=0) + P(X=1) + P(X=2)
 P(X<=1) = P(X=0) + P(X=1)
 Subtract and you are left with P(X=2)
Mean and Standard Deviation of a
Poisson Distribution
 Mean of a Poisson Distribution
 lambda
 Standard Deviation of a Poisson Distribution
 Sqrt(lambda)