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The Poisson Distribution
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The Poisson Probability Distribution
Simeon Denis “Fish”!
• "Researches on the probability ofcriminal civil verdicts" 1837
• Looked at the form of the binomial
distribution
When the Number of
Trials is Large.
• He derived the cumulative Poissondistribution as the
Limiting case of the
Binomial When theChance of Success
Tends to Zero.
Simeon DenisPoisson
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Other phenomena that often follow a Poisson
distribution are
•death of infants,
•the number of misprints in a book,
•the number of customers arriving, and t
•he number of activations of a Geiger counter.
The distribution was derived by the Frenchmathematician Siméon Poisson in 1837, and the
first application was the description of the number of
deaths by horse kicking in the Prussian army.
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• Poisson Distribution: An approximation to the binomial
distribution for the SPECIAL CASE when the average
number (mean µ ) of successes is very much smaller than
the possible number n. i.e. µ << n because p << 1.
• This distribution is important for the study of such phenomena asradioactive decay. This distribution is NOT necessarily symmetric!
Data are usually bounded on one side & not the other.
An advantage of this distribution is that σ2
= μ
The Poisson Distribution
µ = 1.67
σ = 1.29
µ = 10.0
σ =
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The Poisson Distribution Models Counts
• If events happen at a constant rate over time, the PoissonDistribution gives
The Probability of X Number of
Events Occurring in a time T.• This distribution tells us the
Probability of All Possible Numbers of
Counts, from 0 to Infinity.• If X= # of counts per second, then the Poisson probability that
X = k (a particular count) is:
l
λ ≡ the average number of counts per second.
!)(
k
ek X p
k
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Mean and Variance for the
Poisson Distribution• It’s easy to show that for this distribution,
The Mean is:
• Also, it’s easy to show that The Variance is:l
L
The Standard Deviation is:
2
For a Poisson Distribution, the
variance and mean are equal!
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Terminology: A “Poisson Process” • The Poisson parameter can be given as the mean
number of events that occur in a defined time period OR,
equivalently, can be given as a rate, such as
= 2
events per month. must often be multiplied by a time t
in a physical process
(called a “Poisson Process” )
!
)(
)( k
et
k X P
t k
μ =
t σ =
t
More on the Poisson Distribution
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Example
1. If calls to your cell phone are a Poisson processwith a constant rate = 2 calls per hour, what
is the probability that, if you forget to turn your
phone off in a 1.5 hour class, your phone ringsduring that time?
Answer: If X = # calls in 1.5 hours, we want
P(X ≥ 1) = 1 – P(X = 0)
p
P(X ≥ 1) = 1 – .05 = 95% chance
05.!0
)3(
!0
)5.1*2()0( 3
30)5.1(20
eee
X P
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Example Continued
2. How many phone calls do you expectto get during the class?
<X> =
t = 2(1.5) = 3
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10
Conditions Required for the
Poisson Distribution to hold:l
1. The rate is a constant, independent of time.
2. Two events never occur at exactly the same
time.3. Each event is independent. That is, the
occurrence of one event does not make thenext event more or less likely to happen.
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Example
λ = (5 defects/hour)*(0.25 hour) =
λ = 1.25
p(x) = (
xe-)/(x!)x = given number of defects
P(x = 0) = (1.25)0e-1.25)/(0!)
= e-1.25 = 0.287
= 28.7%
• A production line produces 600 parts per hour withan average of 5 defective parts an hour. If you test
every part that comes off the line in 15 minutes,what
is the probability of finding no defective parts (andincorrectly concluding that your process is perfect)?
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0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
P r o b a b i l i t y
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
m
binomial
poisson
N=10,p=0.1
0
0.1
0.2
0.3
0.4
0.5
P r o b a b i l i t y
0 1 2 3 4 5m
poisson
binomial
N=3, p=1/3
Comparison of the Binomial & Poisson
Distributions with Mean μ = 1
Clearly, there is not much difference between them!
For N Large & m Fixed:
Binomial
Poisson
N N
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Poisson Distribution: As λ (Average # Counts)
gets large, this also approaches a Gaussian
l
λ = 5 λ = 15
λ = 25 λ = 35
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14
1. The probability an event occurs inthe interval is proportional to thelength of the interval.
2. An infinite number of occurrencesare possible.
3. Events occur independently at a
rate .
Poisson DistributionX=number of occurrences of event in a given time
period
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15
Poisson Distribution
Source: http://en.wikipedia.org/wiki/Poisson_distribution
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16
lλ= np
= 10,000 x 0.00024 = 2.4
e.g. Probability of an accident in a year is
0.00024. So in a town of 10,000, the rate
Poisson Distribution - Example
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17
Poisson with =2.4
Poisson Distribution
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Examples of possible Poisson distributions
1) Number of messages arriving at a telecommunications system in a day
2) Number of flaws in a metre of fibre optic cable
3) Number of radio-active particles detected in a given time
4) Number of photons arriving at a CCD pixel in some exposure time
(e.g. astronomy observations)
Sum of Poisson variables
The probability of events per unit time does not have to be constant for the total number
of events to be Poisson – can split up the total into a sum of the number of events in
smaller intervals.
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Assuming these are independent random events, the number
of people killed in a given year therefore has a Poisson
distribution:
Answer :
Poisson distribution
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1 2 3 4
0% 0%0%0%
Poisson distribution
Suppose that trucks arrive at a
receiving dock with an average
arrival rate of 3 per hour. What
is the probability exactly 5 trucks
will arrive in a two-hour period?
Question from Derek Bruff
Countdown
20
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Poisson distribution
Suppose that trucks arrive at a
receiving dock with an average
arrival rate of 3 per hour. What is
the probability exactly 5 trucks
will arrive in a two-hour period?
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Mean and variance
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Example: Telecommunications
Messages arrive at a switching centre at random and at an average rate of 1.2 persecond.
(a) Find the probability of 5 messages arriving in a 2-sec interval.
(b) For how long can the operation of the centre be interrupted, if the probability oflosing one or more messages is to be no more than 0.05?
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Question: (b) For how long can the operation of the centre be interrupted, if the
probability of losing one or more messages is to be no more than 0.05?
(b) Let the required time = t seconds. Average rate of arrival is 1.2/second.
Answer:
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1 2 3 4 5
0% 0%0%0%0%
Poisson or not?
Which of the following are
likely to be well modelled by a
Poisson distribution?
(can click more than one)
Can you tell what is fish, or will you flounder?
1. Number of duds found when I test
four components2. The number of heart attacks in
Brighton each year
3. The number of planes landing at
Heathrow between 8 and 9am4. The number of cars getting
punctures on the M1 each year
5. Number of people in the UK flooded
out of their home in July Countdown
60
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Number of duds found when I test four components
Number of people in the UK flooded out of their home in July
- NO: this is Binomial(it is not the number of independent random events in a continuous interval)
The number of heart attacks in Brighton each year
- YES: large population, no obvious correlations between heart attacks in
different people
The number of planes landing at Heathrow between 8 and 9am
- NO: 8-9am is rush hour, planes land regularly to land as many as possible
(1-2 a minute) – they do not land at random times or they would hit each other!
The number of cars getting punctures on the M1 each year
- YES (roughly): If punctures are due to tires randomly wearing thin, then expect
punctures to happen independently at random But: may not all be independent, e.g. if there is broken glass in one lane
Are they Poisson? Answers:
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Example
The probability of a certain part failing within ten years is 10-6
. Five million of the parts have been sold so far.
What is the probability that three or more will fail within ten years?
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Poisson Distribution Summary
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An example is the improvement of trafficsafety, where the government wants to know
whether seat belts reduce the number of
death in car accidents. Here, the Poissondistribution can be a useful tool to answer
questions about benefits of seat belt use.
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Example
Arrivals at a bus-stop follow a
Poisson distribution with an averageof 4.5 every quarter of an hour.
Obtain a barplot of the distribution(assume a maximum of 20 arrivals in
a quarter of an hour) and calculate
the probability of fewer than 3 arrivalsin a quarter of an hour.
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The probabilities of 0 up to 2 arrivals canbe calculated directly from the formula
( )
!
xe p x
x
4.5 0
4.5(0)0!
e p
with =4.5
So p(0) = 0.01111
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Similarly p(1)=0.04999 and p(2)=0.11248
So the probability of fewer than 3 arrivalsis 0.01111+ 0.04999 + 0.11248 =0.17358
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Consider a collection of graphs for
different values of
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=3
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=4
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=5
6
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=6
10
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=10
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In the last case, the probability of 20
arrivals is no longer negligible, sovalues up to, say, 30 would have to be
considered.
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