Poisson distribution used in cases where the chance of any individual event being a success is very small. The distribution is used to describe the behavior of rare events. It is limiting case of binomial distribution.

1. POISSON DISTRIBUTION
Presented By:
Javaid Ahmad Wani
20903113

2. History
The distribution was derived by the
French mathematician Siméon Poisson
in 1837, and the first application was
the description of the number of deaths
by horse kicking in the Prussian army.

3. Poisson distribution used in cases where the chance of any
individual event being a success is very small. The
distribution is used to describe the behavior of rare events.
It is limiting case of binomial distribution.
Examples;
The number of defective screws per box of 5000 screws.
The number of printing mistakes in each page of the first proof of
book.
The number of air accidents in India in one year.
WHY WE NEED POISSON
DISTRIBUTION!!

4.  The random variable X should be discrete.
 Happening of the event must be of two alternatives
such as success & failure.
 Applicable in those cases where the number of trials n is
very large and the probability of success p is very small
but the mean np = λ is finite.
 Statistical independence is assumed.
CONDITION UNDER WHICH POISSON
DISTRIBUTION IS USED

5. EQUATION
If X = The number of events in a given interval,
Then, if the mean number of events per interval is λ.
The probability of observing x events in a given interval is
given by,
e is a mathematical constant. e≈2.718282.

6. Consider, in an office 2 customers arrived today. Calculate the possibilities for
exactly 3 customers to be arrived on tomorrow.
Solution
Step1: Find e-λ.
Where , λ=2 and e=2.718 e-λ = (2.718)-2 =
0.135.
Step2:Find λ x. where, λ=2 and x=3. λ x
= 23 = 8.
Step3: 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.
PROBLEM 1

7. Give that 2% of fuses manufactured by firm are defective find the probability that a
box contain 200 fuses has at least 1 defective fuses.
Solution
Here n = 200 , p = 0.02
λ = n*p = 4
P(X=x) = e-λ λ x / x!
p(x>=1) = p( 1- p(x<1)) = (1 - p(0))
= 1 - e-4 4 0/ 0! = 1 - e-4
= 0.98
PROBLEM 2

8. THANK YOU

13. PROBLEM 3
The average number of accidents at a particular intersection every year is 18.
(a) Calculate the probability that there are exactly 2 accidents there this month.
Solution
There are 12 months in a year, so  = 18
12
= 1.5 accidents per month
P(X = 2) =
e
x
x!
1.5 2

e 1.5
2!
= 0.2510

14. MEAN AND VARIANCE FOR THE
POISSON DISTRIBUTION
It’s easy to show that for this
distribution,
The Mean is:
The Variance is:
So, The Standard Deviation is:
  
  2
  
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4

15. 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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5

16. 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.
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BINOMIAL VS POISSON