QBM117
Business Statistics
Probability Distributions
Poisson Distribution
1
Objectives
•
To introduce the Poisson distribution
•
Learn how to recognize an experiment whose
outcomes follow a Poisson probability distribution
•
Learn how to calculate Poisson probabilities
•
Find the mean and variance of a Poisson distribution
•
Learn how to use Excel to find Poisson probabilities
2
Poisson Distribution
• The binomial distribution can be used to calculate the
probability of getting a specified number of successes
for a given number of repeated trials.
• The Poisson distribution can be used to calculate the
probability that there will be a specified number of
occurrences within a unit of time or space.
3
Poisson Experiment
A Poisson experiment possesses the following
properties:
1. The experiment consists of counting the number
of times a certain event occurs during a given
unit of time or over a unit of space.
2. The probability that an event occurs in an interval
is the same for all intervals of equal size, and is
proportional to the size of the interval.
3. The number of events that occur in any interval is
independent of the number of events that occur
in any other interval.
4. The probability of more than one occurrence in a
very small interval is close to zero.
4
Example 1
A study was carried out to examine the number of
emails received by employees of a large company.
The study found that on average an employee
receives 110 emails per week.
– The experiment consists of counting the number
of emails received per week.
– If an average of 110 emails are received per
week then an average of 110/5=22 will be
received per day.
– The number of emails that arrive in a day is
independent of the number of emails that arrive
in any other day.
– Assume that emails arrive one at a time.
5
Poisson Random Variable
• The random variable in a Poisson experiment is the
number of occurrences of an event, within a specified
time or space.
• It is called the Poisson random variable.
• It is a discrete random variable as the possible values
of the random variable can be listed: 0, 1, 2, …
6
Examples of Poisson Random Variables
• The number of telephone calls received at a
switchboard in a minute.
• The number of machine breakdowns during a night
shift.
• The number of particles of a particular pollutant in a
cubic meter of air emitted from a factory.
• The number of customers arriving at a bank in an
hour.
• The number of flaws in a pane of glass.
7
Poisson Distribution
• If X is a Poisson random variable, the probability
distribution is given by
e   x
P( X  x) 
x!
where  is the average number of occurrences in a
given time interval or region.
• Note that e is a constant value of 2.71828… that can
be found on your calculator.
8
• Hence the probability of observing exactly x
occurrences per unit of time or space is given by
e   x
P( X  x) 
x!
9
Example 2
A paint factory uses “Agent A” in the paint
manufacturing process. There is an average of 3
particles of Agent A in a cubic meter of air emitted
during the production process.
The number of Agent A particles has a Poisson
distribution with mean 3 particles per cubic meter of
air emitted from the factory.
Let X = the number of particles of Agent A in a cubic
meter of air emitted from the factory.
 = 3 particles per cubic meter.
10
a. What is the probability that there will be 5 particles of
Agent A in a cubic meter of air emitted from the
factory?
e 3 35
P( X  5) 
5!
 0.1008
11
b. What is the probability that there will be no Agent A
particles in a cubic meter of air emission?
e 3 30
P( X  0) 
0!
 0.0498
12
c. What is the probability that there will be less than 2
particles of agent A in a cubic meter of air emitted
from the factory?
P( X  2)  P( X  0)  P ( X  1)
3 1
e 3
 0.0498 
1!
 0.0498  0.1494
 0.1992
13
Example 3
We are interested in the number of arrivals at the
drive-thru window of a fast-food restaurant during a 5
minute period. If we can assume that the probability
of a car arriving is the same for any two periods of
equal length and that the number of arrivals in any
period is independent of the number of arrivals in any
other period, then the number of arrivals can be
modelled as a Poisson random variable. Suppose
that we are only interested in the busy lunch hour
period and these assumptions are satisfied. The
average number of cars arriving in a 5 minute period
is 4.
14
Let X = the number of cars arriving in a 5 minute
period.
 = 4 cars per 5 minute period
X has a Poisson distribution with a mean of 4 cars
per 5 minute period.
15
a. What is the probability of exactly 5 cars arrivals
during a 5 minute period?
4
5
e 4
P( X  5) 
5!
 0.1563
16
b. What is the probability of 5 arrivals during a 10
minute period?
We are now interested in a 10 minute period rather
than a 5 minute period.
= 4 cars per 5 minute period, is equivalent to
= 8 cars per 10 minute period.
8 5
e 8
P( X  5) 
5!
 0.0916
17
Poisson Tables
• An alternative to calculating Poisson probabilities
using the formula is to use Table 2 of Appendix C of
the text.
• The probabilities given in this table are cumulative
probabilities.
P( X  k )  P( X  0)  P( X  1)  ...  P( X  k )
k
  P( X  x)
x 0
18
Example 4
The number of accidents that occur at a busy
intersection is Poisson distributed with a mean of 3.5
per week. Find the probability of the following
events:
a. Less than three accidents in a week
b. Five or more accidents in a week
c. No accidents today
19
Let X = the number of accidents per week
= 3.5 accidents per week
a. The probability that there less than three accidents
in a week is
P( X  3)  P( X  2)
 0.321
20
b. The probability that there are five or more accidents
in a week is
P( X  5)  1  P( X  4)
 1  0.725
 0.275
21
c. We want to find the probability that there are no
accidents today.
The average number of accidents per week is 3.5.
Therefore the average number of accidents per day
is 3.5/7=0.5.
Hence
= 0.5 accidents per day
The probability that there are no accidents today is
P( X  1)  0.607
22
Expected Value and Variance of
Poisson Random Variables
• If X is a Poisson random variable, the mean and
variance of X are
E( X )  
V (X )  

where
is the average number of occurrences in a
given interval of time or space.
23
Example 4 revisited
The number of accidents that occur at a busy
intersection is Poisson distributed with a mean of 3.5
per week.
a. What is the expected number of accidents per
week?
b. What is the variance of the number of accidents
per week?
24
The expected number of accidents per week:
E ( X )  3.5
The variance of the number of accidents per week:
V ( X )  3.5
25
Calculating Binomial and Poisson
Probabilities in Excel
• There are instructions on page 204 of the text (pg
202 abridged).
• You will be asked to calculate binomial and Poisson
probabilities in Excel in Tutorial 5.
26
Reading for next lecture
• Chapter 5 Section 5.6 – 5.7
Exercises
• 5.38
• 5.45
• 5.75
27