This document provides information about the Poisson distribution. It begins with background on the mathematician Simeon Denis Poisson who developed the distribution in 1837. The Poisson distribution models the number of random events occurring in a fixed interval of time or space. The key properties are that the probability of an event is constant and events are independent. Several examples of real-world applications are given such as disease occurrences, mutations, and telephone calls. The document then provides the Poisson probability mass function equation and explains how to calculate probabilities for specific values. It also discusses Poisson processes and fitting observed data to a Poisson distribution. Finally, it compares the Poisson distribution to the binomial distribution and outlines their key differences.
4. What is Poisson distribution?
• The Poisson distribution is a discrete probability distribution for
the counts of events that occur randomly in a given interval of
time (or space).
• Many experimental situations occur in which we observe the
counts of events within a set unit of time, area, volume, length
etc.
Examples:
• The number of cases of a disease in different towns
• The number of mutations in set sized regions of a chromosome
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5. Equation
The probability of observing x events in a given interval is given
by,
.
e is a mathematical constant. e≈2.718282
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6. • The probability that there are r occurrences in a
given interval is given by
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7. Poisson Process
Poisson process is a random process which
counts the number of events and the time that
these events occur in a given time interval.
The time between each pair of consecutive
events has an exponential distribution with
parameter λ and each of these inter-arrival
times is assumed to be independent of other
inter-arrival times.
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8. Example
1. Births in a hospital occur randomly at an
average rate of 1.8 births per hour. What is the
probability of observing 4 births in a given hour at
the hospital?
2. If the random variable X follows a Poisson
distribution with mean 3.4 find P(X=6)?
3. Number of telephone calls in a week.
4. Number of people arriving at a checkout in a day.
5. Number of industrial accidents per month in a
manufacturing plant. .
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9. 6. A typist makes on average 2 mistakes per page. What is the
probability of a particular page having no errors on it?
Solution;
P0 = 2^0/0! * exp(-2) = 0.135
7. A computer crashes once every 2 days on average. What is the
probability of there being 2 crashes in one week?
Solution;
P2 = (3.5)^2/2! * exp (-3.5) = 0.185
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10. 8.The mean number of faults in a new house is 8. What is the
probability of buying a new house with exactly 1 fault?
so P1 = 8^1/1! * exp(-8) = 0.0027.
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11. Problem number 1:
Consider, in an office 2 customers arrived today. Calculate the
possibilities for exactly 3 customers to be arrived on tomorrow.
Solution
Find f(x).
P(X=x) = e-λ λ x / x!
P(X=3) = (0.135)(8) / 3! = 0.18.
Hence there are 18% possibilities for 3
customers to be arrived on tomorrow
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12. Problem number 2:
Births in a hospital occur randomly at an average rate of 1.8 births
per hour.
i) What is the probability of observing 4 births in a given hour at
the hospital?
ii) What about the probability of observing more than or equal to
2 births in a given hour at the hospital?
Solution:
Let X = No. of births in a given hour
i)
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13. ii) More than or equal to 2 births in a given hour
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14. Problem number 3:
Consider a computer system with Poisson job-arrival stream
at an average of 2 per minute. Determine the probability
that in any one-minute interval there will be
(i) 0 jobs;
(ii) exactly 2 jobs;
(iii) at most 3 arrivals.
Solution
(i) No job arrivals:
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15. (iii) At most 3 arrivals
(ii) Exactly 3 job arrivals:
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16. Fitting a Poisson distribution to observed
data.
The following tables gives a number of days in a 50 – day period during
which automobile accidents occurred in a city
Fit a Poisson distribution to the data :
No of accidents : 0 1 2 3 4
No. of days :21 18 7 3 1
solution
X F Fx
0 21 0
1 18 18
2 7 14
3 3 9
4 1 4
N=50 ∑fx=45
M = X = ∑fx/N = 45/50 = 0.9
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17. In order to fit Poisson distribution , we shall multiply each
probability by N i.e. 50
Hence the expected frequencies are :
X 0 1 2 3 4
F 0.4066
X 50
0.3659
X 50
0.1647
X 50
0.0494
X 50
0.0111
X 50
=20.33 =18.30 =8.24 =2.47 =0.56
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18. Mean and Variance for
the Poisson Distribution
It’s easy to show that for this
distribution,
The Mean is:
e.g. mean=1.8
The Variance is:
then Variance=1.8
So, The Standard Deviation is:
Standard Deviation=1.34
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20. Graph :
• Let’s continue to assume we have a continuous
variable x and graph the Poisson Distribution, it
will be a continuous curve, as follows:
Fig: Poison distribution graph.
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21. Applications of Poisson distribution
A practical application of this
distribution was made
by Ladislaus Bortkiewicz in
1898 when he was given the
task of investigating the
number of soldiers in the
Russian army killed
accidentally by horse kicks
this experiment introduced the
Poisson distribution to the
field of reliability engineering. Ladislaus Bortkiewicz
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22. • the number of deaths by horse kicking in the
Prussian army
• birth defects and genetic mutations
• rare diseases
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23. • traffic flow and ideal gap distance
• number of typing errors on a page
• hairs found in McDonald's
hamburgers
• spread of an endangered animal in
Africa
• failure of a machine in one month
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24. Binomial Distribution Poisson Distribution
• Binomial distributions are useful
to model events that arise in a
binomial experiment.
• If, on the other hand, an exact
probability of an event
happening is given, or implied, in
the question, and you are asked
to calculate the probability of
this event happening k times out
of n, then the Binomial
Distribution must be used
• Poisson distributions are useful
to model events that seem to
take place over and over again in
a completely haphazard way.
• If a mean or average probability
of an event happening per unit
time/per page/per mile cycled
etc., is given, and you are asked
to calculate a probability
of n events happening in a given
time/number of pages/number
of miles cycled, then the Poisson
Distribution is used.
Binomial vs Poisson
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25. Binomial Distribution Poisson Distribution
• The random variable of interest is the
total number of trials that ended in a
success.
• Every trial results in either a success,
with probability p, or a failure, with
probability 1-p. These must be the only
two outcomes for a trial.
• A fixed number of repeated, identical,
independent trials. n is usually the
parameter chosen to label the number
of trials.
• The probability mass function for the
binomial distribution is given by:
• Key assumptions for the Poisson model
include:
The random variable counts the number
of events that take place in a given
interval (usually of time or space)
• All events take place independently of
all other events
• The rate at which events take place is
constant usually denoted λ
• The probability mass function is given
by
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26. Binomial Distribution Poisson Distribution
• Infinite Number of Trials
• Unlimited Number of Outcomes
Possible
• Mean of the Distribution is the
Same for All Intervals (mu)
• Number of Occurrences in Any
Given Interval Independent of
Others
• Predicts Number of Occurrences
per Unit Time, Space, ...
• Can be Used to Test for
Independence
• Fixed Number of Trials (n)
[10 pie throws]
• Only 2 Possible Outcomes
[hit or miss]
• Probability of Success is Constant
(p)
[0.4 success rate]
• Each Trial is Independent
[throw 1 has no effect on throw 2]
• Predicts Number of Successes
within a Set Number of Trials
• Can be Used to Test for
Independence
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