Examples: Poison Distribution | Applied Mathematics for Class 12 - Commerce PDF Download

Example 1: A random variable X has a Poisson distribution with parameter λ such that P (X = 1) = (0.2) P (X = 2). Find P (X = 0).

For the Poisson distribution, the probability function is defined as:
P (X =x) = (e^{– λ} λ^x)/x!, where λ is a parameter.
Given that, P (x = 1) = (0.2) P (X = 2)
(e^{– λ} λ¹)/1! = (0.2)(e^{– λ} λ²)/2!
⇒ λ = λ²/ 10
⇒ λ = 10
Now, substitute λ = 10, in the formula, we get:
P (X =0 ) = (e^{– λ} λ⁰)/0!
P (X =0) = e^-10 = 0.0000454
Thus, P (X= 0) = 0.0000454

Example 2: Telephone calls arrive at an exchange according to the Poisson process at a rate λ= 2/min. Calculate the probability that exactly two calls will be received during each of the first 5 minutes of the hour.

Assume that “N” be the number of calls received during a 1 minute period.
Therefore,
P(N= 2) = (e^-2. 2²)/2!
P(N = 2) = 2e^-2.
Now, “M” be the number of minutes among 5 minutes considered, during which exactly 2 calls will be received. Thus “M” follows a binomial distribution with parameters n = 5 and p = 2e^-2.
P(M = 5) = 32 x e^-10
P(M = 5) = 0.00145, where “e” is a constant, which is approximately equal to 2.718.