The Poisson 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.


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This was This was SimSimééonon--Denis PoissonDenis Poisson

Born 21 Born 21 June June 17811781

Died 25 Died 25 AprilApril

18401840

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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……


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Poisson was a great Mathematician. He Poisson was a great Mathematician. He achieved many great Mathematical featsachieved many great Mathematical feats

For example, For example, ‘‘What are the odds that a horse What are the odds that a horse

kicks a soldier to death?kicks a soldier to death?’’

He often He often agonisedagonised over Mathematical issuesover Mathematical issues

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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.


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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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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.


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Clearly, a horse deciding to kick a soldier to Clearly, a horse deciding to kick a soldier to death is a random event.death is a random event.

And so, observing each day is a And so, observing each day is a BernouliBernouli TrialTrial

But I know about Bernoulli trials: I can work out But I know about Bernoulli trials: I can work out probabilities using the Binomial distribution.probabilities using the Binomial distribution.

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But if I observe a corps over a year, this is 365 But 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 electronic I 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


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And so, Poisson found a way to And so, Poisson found a way to approximate the binomial probabilities:approximate the binomial probabilities:

!

)()( )(

ieixP

iµµ−

==

We know We know ee is Euleris Euler’’s number. Its value is s number. Its value is 2.71828. It is more commonly known as the 2.71828. It is more commonly known as the

exponential.exponential.

We also know what the product We also know what the product µµ is: This is the is: This is the expected number of successes in a Binomial expected number of successes in a Binomial

DistributionDistribution

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For Zero Successes, Poisson Formula Simplifies For Zero Successes, Poisson Formula Simplifies toto

)(0

)(

!0

)()0( µµ µ −−

=== eexP


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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(

)(

−=

−=−=

−

−

−

−

ie

ieixP

iiµµµ µµ

!

)()( since,But )(

ieixP

iµµ−

==

)1()( Therefore

)(!)!1(

)1(Then )(1

)(

−===

===−

=−=−

−

−

ixPi

ixP

ixPi

iei

ieixP

ii

µ

µ

µ

µ

µµ µµ

i

iiiiii

!)!1( :grearrangin ;)!1(*! :! Expand =−−=

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Therefore, once we know the probability of zero Therefore, once we know the probability of zero successes, we can calculate other probabilities successes, we can calculate other probabilities

successively as:successively as:

)1()( −=== ixPi

ixPµ

Number of Number of successes successes minus 1minus 1

Expected Expected number of number of successsuccess

Number of Number of successsuccess


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Poisson collected data on the number of Poisson collected data on the number of men that died in each the 10 corps over the men that died in each the 10 corps over the

twenty years. twenty years.

This is equivalent to observing 200 samples This is equivalent to observing 200 samples of the same distribution. of the same distribution.

Poisson found that the average number of Poisson found that the average number of men that died in each corps was 0.61 per men that died in each corps was 0.61 per

yearyear

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:iesprobabilit the

outwork to able was Poisson ,61.0 with Thus, =µ

5433.0)0( )61.0(===

−exP

3314.05433.0*61.0)0(1

)1( ===== xPxPµ

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


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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 200 To find the number of corps out of the 200 sampled in which 0, 1, 2 deaths etc were sampled in which 0, 1, 2 deaths etc were

expected, Poisson multiplied these expected, Poisson multiplied these probabilities by 200. probabilities by 200.

He then compared the numbers resulting He then compared the numbers resulting from his theory with those he had collected.from his theory with those he had collected.


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Deaths in Army over 20yrs: Actual Vs Deaths 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

1010

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 for the Poisson distribution is generally suitable for

modelling discrete events that can occur within a modelling 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.


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Example: Call CentreExample: Call CentreA 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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Example: Call Centre SolutionExample: Call Centre Solution


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Example: Call Centre SolutionExample: Call Centre Solution

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Example: Example: GosportGosport & & FarehamFareham Leukemia RatesLeukemia RatesAltogether, there are 170, 000 inhabitants in the towns Altogether, there are 170, 000 inhabitants in the towns GosportGosport and and

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 had mile radius of each other had

suffered with leukemia. suffered with leukemia.

However, according to a medical doctor from Southampton General However, 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 that ss 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 cases d 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, they pertise, 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 in Are the rates of Leukemia in GosportGosport and and FarehamFareham consistent with a random consistent with a random

causal process?causal process?


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GosportGosport and and FarehamFareham Leukemia Rates SolutionLeukemia Rates Solution

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The Poisson Distribution

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