Poisson Distribution with Formulas and Examples

Definition Of Poisson Distribution

Poisson Distribution is one of the types of Discrete Distribution. A Random Variable X is said to follow Poisson Distribution with parameter λ, if its probability mass function is given by:

This is the Poisson Distribution formula.

where λ is the average occurrence per unit time or space over a certain period to time or space.
where λ = np

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Applications of  Poisson Distribution:

Here are some applications of Poisson Distribution:

• the arrival of customers at a service station.
• The number of telephone calls to a mail order service.
• Typographical errors in manuscripts.
• The number of defects in a fabric.
• The number of machine breakdown in a day.
• The number of accidents in a highway.

It is used for counting the number of events which occurs over time or space.

What are the conditions to use Poisson Distribution?

• The number of the trial (n) is large.
• The probability of success in a trial is small. i.e p is small

Mean of Poisson Distribution:

The mean of the Binomial Distribution is denoted by μ or E(X)

E(X) = λ

The variance of Binomial Distribution:

The variance of the Binomial Distribution is denoted by σ² or V(X).

V(X) = λ

Use of New Parameter in Poisson Distribution:

α = λ.T

Here α  is the new parameter and T is the Time period. α gives the average occurrence per unit over a time period T.

Poisson Distribution Examples:

Solved Problem 1: It is known that 5% of the books bound at a certain bindery have defective bindings. Find the probability that 2 of the 100 books bound by this bindery will have defective bindings using Poisson Distribution?

= Solution:

here given, 

number of trials (n) = 100

Probability of success (p) = 5% = 0.05

so, q = 1 - p = 0.95

Let X be the random variable for defective bindings.

We know that,

 λ = np = 100*0.05 = 5

we have, 

so,
P(X = 2) = (e^-5.λ²) / 2! = 0.084

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Solved Problem 2: The arrival of a truck at a receiving dock is a Poisson Process with a mean average rate of 2 per hour.
a. Find the probability that exactly 5 trucks arrive in a two hour time period?
b. Find the probability that 8 or more trucks arrive in two hour period?

= Solution:
here given,
average arrival rate = λ = 2 per hour
arrival rate in 2 hour time period (T = 2) is
α = λ.T = 2*2 = 4 trucks per 2 hour.

a.   the probability that exactly 5 trucks arrive in a two hour time period is:

P(X = 5) =  (e^-α.α^x) / x!  
           = (e^-4.4⁵) / 5!
= 0.15

b. the probability that 8 or more trucks arrive in two hour period is:

P(X ≥ 8) = 1 - P(X < 8)
               = 1 -[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)]
               = 1 - [(e^-4.4⁰) / 0! + (e^-4.4¹) / 1! + (e^-4.4²) / 2! + (e^-4.4³) / 3! + (e^-4.4⁴) / 4! + (e^-4.4⁵) / 5! + (e^-4.4⁶) / 6! + (e^-4.4⁷) / 7!]