Poisson distribution
A random variable X taking on one of the values 0,1,2,.........is said to be poissons Random variable with parameter K if for some
K > 0 is given by \[P(X=x)=\frac{e^{-k}k^{x}}{x!}\].
For Poisson's distribution Mean =E[X]= K
Variance =V[X]=K
Therefore expected value and variance of a Poisson's Random variable are both equal to its parameter K.
Here K is average no of occurrences of event in an observation period t2-t1. so K=a(t2-t1.) where k is the number of occurrence of event per unit time.
Example - A certain airport receives on an average of a aircrafts per hour. What is the probability that no aircraft land in particular 2 hr period.
solution.
Given that a= rate of occurrence of event per unit time =4/hr.
K= a(t2-t1)=2a=2*4=8
Now we wish that no aircraft should land for 2 hrs. That means x=0.
Hence Using \[[P(X=0)=\frac{e^{-8}8^{0}}{0!}\]=0.00033546.
Example 2.
A certain company sells tractors which fail at a rate of 1 out of thousand . If 500 tractors are purchased from this company what is probability of 2 of them failing within first year.
solution. Here K=np=500*(1/1000)=0.5
X is 2 (since asked probability of 2)
Therefore \[P(X=2)=\frac{e^{-0.5}0.5^{2}}{2!}=0.07582\]
