PROPERTIES OF POISSON DISTRIBUTION

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). 

Practice Problems

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

Problem 3 :

If the random variable X follows a Poisson distribution with mean 3.4, find P(X = 6).

Solution : 

Mean = 3.4

P(X = x) = (e^-λ λ^x)/x!

P(X = 6) = (e^-3.4 3.4⁶)/6!

= 0.072

Problem 4 :

The number of industrial injuries per working week in a particular factory is known to follow a Poisson Distribution with mean 0.5.

Find the probability that,

a) In a particular week there will be 

i) less than 2 accidents

ii)  more than 2 accidents.

b) in a three week period there will be no accidents.

Let A be the number of accidents in one week, so A ~ P₀(0.5)

i) P(x < 2) 

Number of accidents should be lesser than 2.

P(x < 2) = P(x = 0) + P(x = 1)

= (e^-0.5 0.5⁰)/0! + (e^-0.5 0.5¹)/1!

= e^-0.5 + e^-0.5 0.5

= e^-0.5(1 + 0.5)

= e^-0.5(1.5)

= 0.9098

ii) P(x > 2) = 1 - P(X ≤ 2)

= 1 - 0.9856

= 0.0144

b)  P(0 in 3 weeks)

= (e^-0.5)³

= 0.223

Problem 5 :

If X~P(λ) and P(X = 4) = 3P(X = 3), find λ and P(X = 5).

Solution :

P(X = 4) = 3P(X = 3)

P(X = x) = (e^-λ λ^x)/x!

(e^-λ λ⁴)/4! = 3(e^-λ λ³)/3!

λ/4 = 3

λ = 12

P(X = 5) = (e^-12 12⁵)/5!

= 6.15(248832) / 120

= 12752.6

Problem 6 :

If the mean of a Poisson variable X is 1, what is the P(x = takes the value at least 1) ?

a) 0.456   b)  0.821     c) 0.632    d)  0.254

Solution :

P(X = x) = (e^-λ λ^x)/x!

Mean = λ = 1

At least 1, then x ≥ 1

P(x ≥ 1) = 1 - P(x < 1)

So P(x < 1) = 1 - P(X = 0)

= 1 - (e^-1 1⁰)/0!

= 1 - 1/e

= 1 - 1/2.718

= 1 - 0.36

= 0.63

so, option c is correct.

Problem 7 :

IF X ~ P(m) and its coefficient of variation is 50, what is the probability that X would assume only non zero values ?

a)  0.018   b)  0.982    c) 0.989    d)  0.976

Solution :

Coefficient of variation = (Standard deviation / mean) x 100%

Standard deviation = √m

50 = (√m/m) x 100

50m = 100√m

50m/100 = √m

m/2 = √m

m²/4 = m

m² = 4m

m² - 4m = 0

m(m - 4) = 0

m = 0 and m = 4

At least 1, then x ≥ 1

P(x ≥ 1) = 1 - P(x < 1)

So P(x < 1) = 1 - P(X = 0)

= 1 - (e^-4 4⁰)/0!

= 1 - e^-4

= 1 - (1/e⁴)

1 - 1/54.6

= 1 - 0.018

= 0.982

So, option b is correct.

Problem 8 :

For a Poisson variate X, P(x = 1) = P(x = 2), what is the value of x?

a)  1    b) 1.50    c) 2     d) 2.5

Solution :

P(x = 1) = P(x = 2)

P(X = x) = (e^-λ λ^x)/x!

(e^-λ λ¹)/1! = (e^-λ λ²)/2!

1 = λ/2

λ = 2

Mean = variance

So, 2 is the answer option c.

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