Introduces the Poisson Probability Distribution

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Poisson Distribution 2009. Max Chipulu, University of Southampton
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The Poisson DistributionThe Poisson Distribution
1. To introduce the Poisson distribution as the situation
where a discrete number of successes can be observed
in a finite interval
2. To calculate Poisson distribution probabilities using the
recurrence formula
3. To recognize and use the Poisson distribution as an
approximation of the Binomial distribution in the
situation where the number of trials is very large but
probability of a success is small, i.e. when the expected
number of successes in the Binomial is small.
Poisson Distribution 2009. Max Chipulu, University of Southampton
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This wasThis was SimSimééonon--Denis PoissonDenis Poisson
Born 21Born 21
JuneJune
17811781
Died 25Died 25
AprilApril
18401840

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Poisson Distribution 2009. Max Chipulu, University of Southampton
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He is the Frenchman that said,
‘Life is good for only two things: to study
Mathematics and to teach it.’
Can you imagine that?Can you imagine that?
A life good only for studying MathematicsA life good only for studying Mathematics……
Poisson Distribution 2009. Max Chipulu, University of Southampton
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Poisson was a great Mathematician. HePoisson was a great Mathematician. He
achieved many great Mathematical featsachieved many great Mathematical feats
For example,For example, ‘‘What are the odds that a horseWhat are the odds that a horse
kicks a soldier to death?kicks a soldier to death?’’
He oftenHe often agonisedagonised over Mathematical issuesover Mathematical issues

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Poisson Distribution 2009. Max Chipulu, University of Southampton
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He found that the probability of a soldier being
killed by horse kick on any given day was very
small.
Poisson observed ten army corps over twenty
years.
However, because he had observed each army
corps over many days, the number of
opportunities of this event happening was
very large… so that the event actually
happened a few times.
Poisson Distribution 2009. Max Chipulu, University of Southampton
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We don’t know this!
But since Poisson was great, lets assume he had
considerable powers of Statistical Reasoning
Lets also assume he summoned them to ponder
this:
But how to calculate the probability. . .
To die or not to die by horse kickTo die or not to die by horse kick

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Poisson Distribution 2009. Max Chipulu, University of Southampton
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My experiment is this: Does a soldier get killed by
horse kick?
He proceeded, thus:He proceeded, thus:
Let each day that I observe be an
experiment
If a soldier gets killed by horse kick, then
the experiment is a success. If not, the
experiment has failed.
Poisson Distribution 2009. Max Chipulu, University of Southampton
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Clearly, a horse deciding to kick a soldier toClearly, a horse deciding to kick a soldier to
death is a random event.death is a random event.
And so, observing each day is aAnd so, observing each day is a BernouliBernouli TrialTrial
But I know about Bernoulli trials: I can work outBut I know about Bernoulli trials: I can work out
probabilities using the Binomial distribution.probabilities using the Binomial distribution.

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Poisson Distribution 2009. Max Chipulu, University of Southampton
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But if I observe a corps over a year, this is 365But if I observe a corps over a year, this is 365
trials!!trials!!
I mustn't forget my resources are limited:I mustn't forget my resources are limited:
I have never heard of computers or electronicI have never heard of computers or electronic
calculators.calculators.
How can I work this out?How can I work this out?
THIS IS IMPOSSIBLETHIS IS IMPOSSIBLE
Poisson Distribution 2009. Max Chipulu, University of Southampton
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And so, Poisson found a way toAnd so, Poisson found a way to
approximate the binomial probabilities:approximate the binomial probabilities:
!
)(
)( )(
i
eixP
i
µµ−
==
We knowWe know ee is Euleris Euler’’s number. Its value iss number. Its value is
2.71828. It is more commonly known as the2.71828. It is more commonly known as the
exponential.exponential.
We also know what the productWe also know what the product is: This is theis: This is the
expected number of successes in a Binomialexpected number of successes in a Binomial
DistributionDistribution

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Poisson Distribution 2009. Max Chipulu, University of Southampton
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For Zero Successes, Poisson Formula SimplifiesFor Zero Successes, Poisson Formula Simplifies
toto
)(
0
)(
!0
)(
)0( µµ µ −−
=== eexP
Poisson Distribution 2009. Max Chipulu, University of Southampton
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A little more Mathematics simplifies things as follows:A little more Mathematics simplifies things as follows:
)!1()!1(
)(
)1(
1
)(
)1(
)(
−
=
−
=−=
−
−
−
−
i
e
i
eixP
ii
µµµ µµ
!
)(
)(since,But )(
i
eixP
i
µµ−
==
)1()(Therefore
)(
!)!1(
)1(Then )(
1
)(
−===
===
−
=−= −
−
−
ixP
i
ixP
ixP
i
i
e
i
i
eixP
ii
µ
µ
µ
µ
µµ µµ
i
i
iiiii
!
)!1(:grearrangin;)!1(*!:!Expand =−−=

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Poisson Distribution 2009. Max Chipulu, University of Southampton
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Therefore, once we know the probability of zeroTherefore, once we know the probability of zero
successes, we can calculate other probabilitiessuccesses, we can calculate other probabilities
successively as:successively as:
)1()( −=== ixP
i
ixP
µ
Number ofNumber of
successessuccesses
minus 1minus 1
ExpectedExpected
number ofnumber of
successsuccess
Number ofNumber of
successsuccess
Poisson Distribution 2009. Max Chipulu, University of Southampton
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Poisson collected data on the number ofPoisson collected data on the number of
men that died in each the 10 corps over themen that died in each the 10 corps over the
twenty years.twenty years.
This is equivalent to observing 200 samplesThis is equivalent to observing 200 samples
of the same distribution.of the same distribution.
Poisson found that the average number ofPoisson found that the average number of
men that died in each corps was 0.61 permen that died in each corps was 0.61 per
yearyear

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Poisson Distribution 2009. Max Chipulu, University of Southampton
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:iesprobabilitthe
outworktoablewasPoisson,61.0withThus, =µ
5433.0)0( )61.0(
=== −
exP
3314.05433.0*61.0)0(
1
)1( ===== xPxP
µ
1011.03314.0*
2
61.0
)1(
2
)2( ===== xPxP
µ
etc;0021.01011.0*
3
61.0
)2(
3
)3( ===== xPxP
µ
Calculating Poisson ProbabilitiesCalculating Poisson Probabilities
Poisson Distribution 2009. Max Chipulu, University of Southampton
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Poisson Distribution 2009. Max Chipulu, University of Southampton
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These are the probabilities of observing 0, 1,These are the probabilities of observing 0, 1,
2 deaths etc in one corps over a year.2 deaths etc in one corps over a year.
To find the number of corps out of the 200To find the number of corps out of the 200
sampled in which 0, 1, 2 deaths etc weresampled in which 0, 1, 2 deaths etc were
expected, Poisson multiplied theseexpected, Poisson multiplied these
probabilities by 200.probabilities by 200.
He then compared the numbers resultingHe then compared the numbers resulting
from his theory with those he had collected.from his theory with those he had collected.
Poisson Distribution 2009. Max Chipulu, University of Southampton
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Deaths in Army over 20yrs: Actual VsDeaths in Army over 20yrs: Actual Vs
Poisson RatesPoisson Rates
0
20
40
60
80
100
120
Frequency
0 1 2 3 4 5
Deaths per year
Actual Poisson

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Poisson Distribution 2009. Max
Chipulu, University of Southampton
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Besides being suitable for approximating the binomial,Besides being suitable for approximating the binomial,
the Poisson distribution is generally suitable forthe Poisson distribution is generally suitable for
modelling discrete events that can occur within amodelling discrete events that can occur within a
constrained interval of time or spaceconstrained interval of time or space
• SPACE
• e.g. a square piece of
metal can develop a
micro crack at any
point within its
surface. The Poisson
distribution can be
used to model
number of cracks
observed
• TIME
• e.g. the number of
arrivals at a queue
within a specific
period of time such
as 1 minute are
usually modelled by
a Poisson
distribution.
Poisson Distribution 2009. Max Chipulu, University of Southampton
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Example: Call CentreExample: Call Centre
A call centre receives, on average, three calls per minute. What is
the probability that there will be, at most, 1 calls in a half
minute?
Suggested Steps for Calculation:
1. This is a Poisson Distribution: Discuss with the person nearest
you.
2. Take out your calculators and work out
-the expected number of calls
-the probability of 0 calls
-probability of 1 call
-the required probability

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Poisson Distribution 2009. Max Chipulu, University of Southampton
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Example: Call Centre SolutionExample: Call Centre Solution
Poisson Distribution 2009. Max Chipulu, University of Southampton
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Example: Call Centre SolutionExample: Call Centre Solution

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Poisson Distribution 2009. Max Chipulu, University of Southampton
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Example:Example: GosportGosport && FarehamFareham Leukemia RatesLeukemia Rates
Altogether, there are 170, 000 inhabitants in the townsAltogether, there are 170, 000 inhabitants in the towns GosportGosport andand
FarehamFareham. A few years ago, local residents were alarmed to discover that. A few years ago, local residents were alarmed to discover that
13 children in families that lived within a 513 children in families that lived within a 5--mile radius of each other hadmile radius of each other had
suffered with leukemia.suffered with leukemia.
However, according to a medical doctor from Southampton GeneralHowever, according to a medical doctor from Southampton General hospital,hospital,
the alarm was unwarranted as there are fluctuations in any procethe alarm was unwarranted as there are fluctuations in any process thatss that
is subject to the rates of chance. He stated that he did not finis subject to the rates of chance. He stated that he did not find 13 casesd 13 cases
particularly high given that the incidence of the disease is 8.3particularly high given that the incidence of the disease is 8.3 per million.per million.
Not surprisingly, the residents were unconvinced. But lacking exNot surprisingly, the residents were unconvinced. But lacking expertise, theypertise, they
were unable to present a coherent rebuttal to the doctorwere unable to present a coherent rebuttal to the doctor’’s statement.s statement.
Are the rates of Leukemia inAre the rates of Leukemia in GosportGosport andand FarehamFareham consistent with a randomconsistent with a random
causal process?causal process?
Poisson Distribution 2009. Max Chipulu, University of Southampton
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GosportGosport andand FarehamFareham Leukemia Rates SolutionLeukemia Rates Solution

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Poisson Distribution 2009. Max Chipulu, University of Southampton
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GosportGosport andand FarehamFareham Leukemia Rates SolutionLeukemia Rates Solution
Poisson Distribution 2009. Max Chipulu, University of Southampton
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GosportGosport andand FarehamFareham Leukemia Rates SolutionLeukemia Rates Solution