Download The Poisson distribution

The Poisson
distribution
(Session 07)
SADC Course in Statistics
Learning Objectives
At the end of this session, you will be able to:
• describe the Poisson probability distribution
including the underlying assumptions
• calculate Poisson probabilities using a
calculator, or Excel software
• apply the Poisson model in appropriate
practical situations
To put your footer here go to View > Header and Footer
2
Examples of data on counts
A common form of data occurring in practice
are data in the form of counts, e.g.
• number of road accidents per year at
different locations in a country
• number of children in different families
• number of persons visiting a given website
across different days
• number of cars stolen in the city each month
An appropriate probability distribution for this
type of random variable is the Poisson
distribution.
To put your footer here go to View > Header and Footer
3
The Poisson distribution
• The Poisson is a discrete probability
distribution named after a French
mathematician Siméon-Denis Poisson,
1781-1840.
• A Poisson random variable is one that
counts the number of events occurring
within fixed space or time interval.
• The occurrence of individual outcomes are
assumed to be independent of each other.
To put your footer here go to View > Header and Footer
4
Poisson Distribution Function
• While the number of successes in the
binomial distribution has n as the maximum,
there is no maximum in the case of Poisson.
• This distribution has just one unknown
parameter, usually denoted by  (lambda).
• The Poisson probabilities are determined by
the formula:
P( X  k ) 
k 
e
k!
,
for k  0,1,2,3,
To put your footer here go to View > Header and Footer
5
Example: Number of cars stolen
• Suppose the number of cars stolen per
month follows a Poisson distribution with
parameter  = 3
What is the probability that in a given month
• Exactly 2 cars will be stolen?
• No cars will be stolen?
• 3 or more cars will be stolen?
To put your footer here go to View > Header and Footer
6
Example: Number of cars stolen
For the first two questions, you will need:
λ 2e 
=
P(X = 2) =
2!
λ 0e 
=
P(X = 0) =
0!
The 3rd is computed as
= 1 – P(X=0) – P(X=1) – P(X=2)
=
To put your footer here go to View > Header and Footer
7
Graph of Poisson with  = 15
0.12
Probability
0.10
0.08
0.06
0.04
0.02
0.00
0
4
8
12
16
20
24
28
X
To put your footer here go to View > Header and Footer
8
Graph of Poisson with  = 10
0.14
0.12
Probability
0.10
0.08
0.06
0.04
0.02
0.00
0
4
8
12
16
20
24
28
X
To put your footer here go to View > Header and Footer
9
Graph of Poisson with  = 7
0.16
0.14
Probability
0.12
0.10
0.08
0.06
0.04
0.02
0.00
0
4
8
12
16
20
24
28
X
To put your footer here go to View > Header and Footer
10
Graph of Poisson with  = 4
0.25
Probability
0.20
0.15
0.10
0.05
0.00
0
4
8
12
16
20
24
28
X
To put your footer here go to View > Header and Footer
11
Graph of Poisson with  = 1
0.40
0.35
Probability
0.30
0.25
0.20
0.15
0.10
0.05
0.00
0
4
8
12
16
20
24
28
X
To put your footer here go to View > Header and Footer
12
Practical quiz
• What do you observe about the shapes of
the Poisson distribution as the value of
the Poisson parameter  increases?
• Approximately where does the peak of
the distribution occur?
To put your footer here go to View > Header and Footer
13
Properties of the Poisson distribution
• The mean of the Poisson distribution is the
parameter .
• The standard deviation of the Poisson
distribution is the square root of . This
implies that the variance of a Poisson
random variable = .
• The Poisson distribution tends to be more
symmetric as its mean (or variance)
increases.
To put your footer here go to View > Header and Footer
14
Expected value of a Poisson r.v.
• The expected value of the Poisson random
variable (r.v.) with parameter  is equal to

E( X )   x
x 0

x
x!
e

 .
Note that, since Poisson is a probability
distribution,
x


 x!e

 1.
x 0
To put your footer here go to View > Header and Footer
15
Variance of a Poisson r.v.
• The second moment, E(X2) can be shown
to be:

E( X )   x
2
x 0
2

x
x!
e 
  .
2
Hence
Var( X )    E( X )    
2
2
2
• The standard deviation of a Poisson
random variable is therefore  .
To put your footer here go to View > Header and Footer
16
Cumulative probability distribution
Poisson cumulative distribution with
mean = 5
1.2
Probability
1.0
0.8
0.6
0.4
0.2
30
27
24
21
18
15
12
9
6
3
0
0.0
X
To put your footer here go to View > Header and Footer
17
Interpreting the cumulative distn
• Note that for X larger than about 12, the
cumulative probability is almost equal to 1.
• In applications this means that, if say, the
family size follows a Poisson distribution
with mean 5, then it is almost certain that
every family will have less than 12
members.
• Of course there is still the possibility of rare
exceptions.
To put your footer here go to View > Header and Footer
18
Class Exercise
In example above, we assumed X=family size,
has a Poisson distribution with =5.
Thus P(X=x) = 5x e-5/x! , x=0, 1, 2, …etc.
(a)What is the chance that X=15?
Answer: P(X=15) = 515 e-5/15!
= 0.000157
This is very close to zero. So it would be
reasonable to assume that a family size of 15
was highly unlikely!
To put your footer here go to View > Header and Footer
19
Class Exercise – continued…
(b) What is the chance that a randomly
selected household will have family size < 2 ?
To answer this, note that
P(X < 2) = P(X = 0) + P(X = 1)
=
(c) What is the chance that family size will be
3 or more?
To put your footer here go to View > Header and Footer
20
Further practical examples follow…
To put your footer here go to View > Header and Footer
21