The Poisson Distribution

The discrete probability distribution \(P(X) = \frac{{\lambda^x}{e}^{-\lambda}}{X!}\) X = 0,1,2,3

where e = natural log = 2.7182818 and \(\lambda\) is a given constant.

Some important Formulas for Poisson Distribution

Mean \(\mu = \lambda\)

Variance \(\sigma^2 = \lambda\)

Standard deviation \(\sigma = \sqrt{\lambda}\)

Moment coefficient of skewness \(\alpha_3 = \frac{1}{\sqrt{\lambda}}\)

Moment coefficient of kurtosis \(\alpha_3 = 3 + \frac{1}{\lambda}\)

Lets try to plot poisson distribution

 y <- c()
 p <- c()
 lambda = 5
 y =  (lambda^0)*exp(-lambda)*(1/factorial(0))
 x <- seq(0,10,by =1)
 i = 0
  for (i in 0:10) {
  p[i] = (lambda^i)*exp(-lambda)*(1/factorial(i))}
  y <- c(y,p)
  barplot (y,x, col = "red",width=1, space =0)

Examples

Prob1. Suppose a man recieves 3 call per day from different tele-marketing agencies on an average.

Sol.
Then the proability of receiving a one call at most in a day \(\frac{{3^0}{e}^{-3}}{0!} + \frac{{3^1}{e}^{-3}}{1!}\) = 0.1991483

Here’s the R syntax
ppois(1, lambda=3, lower = TRUE) = 0.1991483

Problem2. Admissions to emergency room has a poison distribution and mean 5 patient per day.Find the probability of having atmost 3 admission per day and probability of atleast 8 admissions. Also find probability of having exactly 2 admissions per day.

solution

The probability of at most 3 admission = \(\frac{{5^0}{e}^{-5}}{0!} + \frac{{5^1}{e}^{-5}}{1!} + \frac{{5^1}{e}^{-5}}{2!}\) = 0.2650259

The probability of atleast 8 admissions = 1- probability of less than 8 admission = 0.1333717

The probability of exactly 2 admissions =\(\frac{{5^2}{e}^{-5}}{2!}\) = 0.0842243
R syntax

Probability of atmost 3 admissions = ppois( 3,lambda =5, lower = TRUE) =0.2650259
Probability of atleast 8 admissions = ppois( 7,lambda =5, lower = FALSE) =0.1333717
probability of having exactly 2 admissions = dpois( 2,lambda =5) =0.0842243