1. n the number of trials is indefinitely large. That is,
n → ∞
2. p the constant probability of success in each trial is very small. That is,
p → 0
3. Poisson distribution is known as a uni-parametric distribution as it is characterized by only one parameter m.
4. The mean of Poisson distribution is given by m. That is,
μ = m
5. The variance of the Poisson distribution is given by
σ2 = m
6. Like binomial distribution, Poisson distribution could be also uni-modal or bi-modal depending upon the value of the parameter m.
m is a non integer ----> Uni-modal
Here, the mode = the largest integer contained in m
m is a integer ----> Bi-modal
Here, the mode = m, m - 1
7. Poisson approximation to Binomial distribution :
If n, the number of independent trials of a binomial distribution, tends to infinity and p, the probability of a success, tends to zero, so that m = np remains finite, then a binomial distribution with parameters n and p can be approximated by a Poisson distribution with parameter m (= np).
In other words when n is rather large and p is rather small so that m = np is moderate then
(n, p) ≅ P(m)
8. Additive property of binomial distribution.
Let X and Y be the two independent Poisson variables.
X is having the parameter m1
and
Y is having the parameter m2
Then (X + Y) will also be a Poisson variable with the parameter (m1 + m2).
Problem 1 :
If the mean of a Poisson distribution is 2.7, find its mode.
Solution :
Given : Mean = 2.7
That is, m = 2.7
Since the mean 2.7 is a non integer, the given Poisson distribution is uni-modal.
Therefore, the mode of the given Poisson distribution is
= Largest integer contained in m
= Largest integer contained in 2.7
= 2
Problem 2 :
If the mean of a Poisson distribution is 2.25, find its standard deviation.
Solution :
Given : Mean = 2.25
That is, m = 2.25
Standard deviation of the Poisson distribution is given by
σ = √m
= √2.25
= 1.5
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