Examples of Poisson Distribution

Definition: A random variable X is  a Poisson distribution if the probability mass function of X is P(X = x) =e−λ  λx / x!,                                       x = 0,1,2, …for some λ > 0

The mean of  Poisson Distribution denoted by λ, and the variance is denoted by λ.
The parameter of Poisson distribution is λ.

The Poisson distribution is a restrictive case of Binomial distribution under the following conditions.
(i)   Number of trials(n) is indefinitely huge(large), that is n → ∞.
(ii)  The constant probability(p) of success in each trial is very less.
ie., p → 0.
(iii) np = λ is finite where λ is a positive real number. When an event occurs rarely, the distribution of such event may be assumed to follow a Poisson distribution. The example problems of poisson distribution is   given below

Examples of Poisson distribution:

Some Examples of poisson distributions are given below

(1) The number of gamma particles emitted by a radio active source in a given time interval.
(2) The number of phone calls received at a telephone exchange in a given time interval.
(3) The number of defective articles in a packet of 250, produced by a good industries limited.
(4) The number of printing errors at each page of a book by a good publication centre.
(5) The number of road accidents reported in a city at a particular time.

Example Problems for Poission distribution:

Example problem 1: If a publisher of  technical books takes a great pain to ensure that his books are free of typological errors, so that the probability of any given page containing atleast one such error is 0.005 and errors are independent from page to page

(i) what is the probability of its 400 page novels will contain exactly one page with error.

(ii) atmost three pages with errors.
[e−2 = 0.1353 ; e−0.2. = 0.819].


Solution :

n = 400 , p = 0.005
np = 2 = λ
(i)  P(one page with error) = P(X = 1) = e−λ λ1/1! = e-2 21/1!
= 0.1363 × 2 = 0.2726
(ii)  P(atmost 3 pages with error) = P(X ≤ 3)
= Σ e−λ λx / x!    [limits 0 to 3]
= `sum` e−2 (2)x  /  x!
= e2 [1 +2/1! + 22/2! + 23/3!]
= e−2 (19/3 )= 0.8569