Poisson Distribution

Distribution

The following are the two types of Theoretical distributions :

1. Discrete distribution  2. Continous distribution

Discrete distribution

The binomial and Poisson distributions are the most useful theoretical distributions for discrete variables.

POISSON DISTRIBUTION

Poisson distribution was derived in 1837 by a French Mathematician Simeon D. Poisson. If n is large,, the evaluation of the binomial probabilities can involve complex computations, in such a case, a simple approximation to the binomial probabilities could be use. Such approximation of binomial when n is large and p is close to zero is called the Poisson distribution.

Poisson distribution occurs when there are events which do not occur as a definite number on trials but an events occurs rarely and the following examples may be analysed:

(i) Number of bacteria in one cubic centimeter.

(ii) Number of printing mistakes per page in a text book

(iii) the number of alpha particles emitted by a radioactive substance in a fraction of a second.

(iv) Number of road accidents occurring at a particular interval of time per day.

(v) Number of lightnings per second.

Poisson distribution is a limiting case of binomial distribution under the following conditions :

i.             n, the number of trials is indefinitely large i.e n → ∞ .

ii.             p, the constant probability of success in each trial is very small, i.e. p → 0

iii.            np = λ is finite. Thus p = λ/n = l and q=1-( λ/n) where λ is a positive real number.

Definition 7.2

A random variable X is said to follow a Poission distribution with parameter λ if it assumes only non-negative values and its probability mass function is given by

Derivation of Mean and variance of Poisson distribution

Variance (X) = E(X²) – E(X)²

= λ² + λ – (λ)²

= λ

Properties of Poisson distribution :

1. Poisson distribution is the only distribution in which the mean and variance are equal .

Example 7.14

In a Poisson distribution the first probability term is 0.2725. Find the next Probability term

Solution :

Example 7.15

In a book of 520 pages, 390 typo-graphical errors occur. Assuming Poisson law for the number of errors per page, find the probability that a random sample of 5 pages will contain no error.

Solution :

The average number of typographical errors per page in the book is given by λ = (390/520) = 0.75.

Hence using Poisson probability law, the probability of x errors per page is given by

The required probability that a random sample of 5 pages will contain no error is  given by : [P(X=0)]⁵ = (e^-0.75)⁵ = e^-3.75

Example 7.16

An insurance company has discovered that only about 0.1 per cent of the population is involved in a certain type of accident each year. If its 10,000 policy holders were randomly selected from the population, what is the probability that not more than 5 of its clients are involved in such an accident next year? ( e⁻¹⁰ = .000045)

Solution :

p = probability that a person will involve in an accident in a year

 = 0.1/100 = 1/1000

 given n = 10,000

so,  λ = np = 10000 (1/1000) = 10

Probability that not more than 5 will involve in such an accident in a year

 P(X < 5)= P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)

 = 0.06651

Example 7.17

One fifth percent of the the blades produced by a blade manufacturing factory turn out to be defective. The blades are supplied in packets of 10. Use Poisson distribution to calculate the approximate number of packets containing no defective, one defective and two defective blades respectively in a consignment of 1,00,000 packets (e^–0.2 =.9802)

Solution :

P = 1/5/100 = 1/500 =0.002

n = 10 .

λ = np = 0.02

(i) Number of packets containing no defective = N p(o) = 1,00,000 × e^–0.02

 = 98020

(ii) Number of packets containing one defective = N p(1) = 1,00,000 × 0.9802 × 0.02

 = 1960

(iii) Number of packets containing 2 defectives = N p(2) = 20

Example 7.18

If the probability that an individual suffers a bad reaction from injection of a given serum is 0.001, determines the probability that out of 2,000 individuals (a) exactly 3, and (b) more than 2 individuals will suffer a bad reaction.

Solution :

Consider a 2,000 individuals getting injection of a given serum , n = 2000

Let X be the number of individuals suffering a bad reaction.

Let p be the probability that an individual suffers a bad reaction = 0.001

 and q = 1– p = 1– 0.001 = 0.999

Since n is large and p is small, Binomial Distribtuion approximated to poisson distribution

So, λ = np = 2000 × 0.001 = 2

(i) Probability out of 2000, exactly 3 will suffer a bad reaction is

(ii) Probability out of 2000, more than 2 individuals will suffer a bad reaction

 = P(X > 2)

 = 1– [( P X( ≤ 2)]

 = 1 – [P(x = 0) + P(x = 1) + P(x = 2)]

 = 0.323

Example 7.19

When counting red blood cells, a square grid is used, over which a drop of blood is evenly distributed. Under the microscope an average of 8 erythrocytes are observed per single square. What is the probability that exactly 5 erythrocytes are found in one square?

Solution :

Let X be a random variable follows poisson distribution with number of erythrocytes.

Hence, Mean λ = 8 erythrocytes/single square

P(exactly 5 erythrocytes are in one square) = P(X = 5) =

The probability that exactly 5 erythrocytes are found in one square is 0.0916. i.e there are 9.16% chances that exactly 5 erythrocytes are found in one square.

Example 7.20

Assuming one in 80 births is a case of twins, calculate the probability of 2 or more sets of twins on a day when 30 births occur.

Solution :

Let x devotes the set of twins on a day

P(twin birth) = p = 1/80 = 0.0125 and n = 30

The value of mean λ = np = 30 × 0.0125 = 0.375

Hence, X follows poisson distribution with p(x) = 

The probability is

P(2 or more) = 1 – [p (x = 0) + p (x = )] =