Poisson distribution is a discrete probability distribution which is described by a single parameter λ, the mean of the distribution. Poisson distribution is the limiting form of Binomial distribution as n -> $\infty$. It serves as an approximation to Binomial distribution when the number of trials n is very large.

Poisson distribution is used to find probabilities when

  1. n is very large and p is very small when compared to n.
  2. the discrete variable measures the density, that is when the events are counted within an interval of time, length, area , volume etc.

Poisson distribution is also used as an approximation to binomial distribution when np < 5 and for situations when only the mean of the distribution is known.

A Poisson variable is used to represent

  1. The number of natural calamities like earth quakes happening during the course of an year.
  2. The number of deaths of the insurance policy holders before the maturity period.
  3. Number of incoming calls during a period of time.

Poisson distribution is the discrete probability distribution that gives the probability of occurrence of events within a given interval of time or any n dimensional space. It is defined completely by a single parameter λ the mean number of occurrences.
P( X : λ) = $\frac{e^{-\lambda }\lambda ^{X}}{X!}$ where λ is the mean of the distribution.
In a binomial distribution as n -> ∞ (n is very large) and p -> 0 (p is very small) the product np is a finite quantity = λ, which is the parameter that describes the Poisson distribution for such a situation.

As the mean of the binomial distribution = np, λ is the mean of the Poisson distribution, which is also the expected value E(x) of the distribution.
Poisson distribution has the special property that both the mean and variance of the distribution = λ, the parameter that defines the distribution.
Var (X) = λ.
The cumulative probability distribution F(X : λ) of a random variable X is the summation of probabilities less than or equal to P(X).
Cumulative probability distribution of the Poisson distribution P(X : λ) is given by

F(X : λ) = $\sum_{i=0}^{x}\frac{e^{-\lambda }\lambda ^{i}}{i!}$
Specifically when X = 3,

F(3) = $\frac{e^{-\lambda }\lambda ^{0}}{0!}$ + $\frac{e^{-\lambda }\lambda ^{1}}{1!}$ + $\frac{e^{-\lambda }\lambda ^{2}}{2!}$ + $\frac{e^{-\lambda }\lambda ^{3}}{3!}$
Evaluating cumulative probabilities for a Poisson variable is quite tedious when X assumes larger numbers, as the number of probabilities to compute are quite high.  Poisson distribution table is handy for this purpose. The table lists the cumulative probabilities for different λ values against the variable x values.

Poisson Table

 For example to find F(8 : 6) the reading along x = 8 under
λ = 6 is taken as  0.8472.
Poisson distribution graphs exhibit the following properties.
  1. They are uni modal
  2. The graphs are positively skewed.
  3. The mean is somewhat close to the median.
  4. The variance increases as λ increases and the graph becomes wider.
  5. The skewness decreases as λ increases and the Poisson graph approaches the normal symmetrical shape.
Integration is the technique applied to find the probability or cumulative probability which is equivalent to summation used for discrete random variables. Based on this a continuous version of Poisson distribution can be defined using advanced techniques in terms of Gamma functions.

Solved Examples

Question 1: Rain storms can hit a hypothetical island on an average once a week during rainy season.
    1) Find the probability at least 2 rain storms hit the island in the first two weeks of a rainy season.
    2) Find the probability distribution of time , starting from a day after a storm , until the next storm strikes.
Solution:
 
The average number of rain storms per week =1.  Hence the average number of rain storms per two weeks λ = 2.
    If X is the random variable representing the number of rain storms in two weeks we need to determine P(X > 2).
    P( X > 2) = 1 - P(X ≤ 2)
                   = 1 - [P(x =0) + P(x=1) + P(x=2)]

    Using Poisson distribution, P( X : λ) = $\frac{e^{-\lambda }\lambda ^{X}}{X!}$

    P(X=0) = $\frac{e^{-2}2^{0}}{0!}$ = e-2 = 0.1353

    P(X=1) = $\frac{e^{-2}2^{1}}{1!}$ = 2e-2 = 0.2707

    P(X=2) = $\frac{e^{-2}2^{2}}{2!}$ = 2e-2 = 0.2707

    P(X > 2) = 1 - [0.1353 + 0.2707 + 0.2707} = 1 - 0.6767 = 0.3233

    Let Y denote the time in week until the next storm strikes. As the storm hits the island on an average once a week,
    If no rains occur during a time interval t, then Y > t.
   P (Y > t) = probability of zero rain storm striking in t weeks = e-t.
   Hence the probability distribution of variable Y is
   F(t) = P(Y ≤ t) = 1 - P(Y>t) = 1 - e-t.
 

Question 2: A customer care executive handles on an average 3 calls in 30 minutes. Find the probability
    1.  He handles not more than 5 calls in an hour
    2. More than five calls in an hour
Solution:
 
The average number of calls attended in an hour λ = 6.
    1.   P(X ≤ 5) = P(X=0) + P(X=1) + P(X=2) + P(x=3) + P(x=4) + P(X = 5) = 0.4457
    2. P( X > 5) = 1 - P(X ≤ 5) = 1 - 0.4457 = 0.5543