Introduction to Poisson distribution table:

Theoretical distributions are classified into many types. They are binomial distribution, normal distribution, Poisson distribution etc. In this article we can learn Poisson distribution, which figure most significantly in statistical theory and in application. Poisson distribution is also known the discrete probability distribution. Let us see the Poisson distribution table in this article.

The Poisson distribution is the limited form of the binomial distribution. Examples for the Poisson distribution are the number of cars passing through a certain street in the time period t and the number of printing mistakes at each page of the book.

Some more examples are as follows:

(1) The quantity of alpha particles emitted by a radioactive source in a known time interval.

(2) The number of phone calls acknowledged at a telephone exchange in a known time interval.

(3) The amount of flawed research paper in a packet of 100, produced by a good industry.

(4) The amount of printing mistakes at each page of a book by a good publication.

(5) The number of road accidents reports in a city at a particular junction at a particular time.

Definition of Poisson Distribution and Poisson Distribution Table:

A random variable X is said to have a Poisson distribution if the probability mass function of X is

P(X = x) = ‘(e^(-lambda) lambda^(x))/(x!)’ , x = 0,1,2, … for some λ > 0 The mean of the Poisson Distribution is λ, and the variance is also λ. The parameter of the Poisson distribution is denoted by λ. The mean value is λ = n p

Where n is the number of trails and p is the possibilities for the event.

The above given are the Poisson distribution table. By using such table values we can find the solution for various Poisson distribution problems.

Problems Using Poisson Distribution Table:

Ex 1: A manufacturer of cotton pins knows that 5% of his product is defective. If he sells pins in boxes of 100 and guarantees that not more than 2 pins will be defective. Determine the probability for a box will fail to meet the guaranteed quality.

Sol:

The value of p is p = 5% = 5/100 , n = 100

The mean value is = n p = (5/100 ) X (100) = 5

By the Poisson distribution

P[X = x] = ‘(e^(-lambda)lambda^x)/(x!)’

Probability for a box will to meet the guaranteed quality = P[X > 2]

P[X > 2] = 1- P[X ≤ 2]

P[X > 2] = 1- (P (0) +P (1) +P (2))

P[X > 2] = 1- (1 + 5 + 25/2)

P[X > 2] = 1- (1+ 5 + 12.5)

P[X > 2] = 1- (18.5)

P[X > 2] = 1- 0.0067(18.5)

P[X > 2] = 1- 0.12395

P[X > 3] = 0.87605

The probability for the box will fail to meet the guaranteed quality is 0.87605.

Ex 2: A car hire firm has three cars. The number of demands for a car as a Poisson distribution with mean of 2.3. Calculate the proportion of days on which neither car is used and the proportion of days on which some demand is refused.

Sol:

Let X denote the number of demands for a car.

The given mean value is 2.3.

By the Poisson distribution

P[X = x] = ‘(e^(-lambda)lambda^x)/(x!)’

Proportion of the days on which neither car is used = P[X = 0] = = 0.1003

Proportion of days on which some demand is refused = P[X > 3]

P[X > 3] = 1- P[X ≤ 3]

P[X > 3] = 1- [P (0) +P (1) + P (2) +P (3)]

P[X > 3] = 1- (1+ 2.3+ 2.645+ 2.0278)

P[X > 3] = 1 – (0.1003) (7.9728)

P[X > 3] = 1 – 0.7997

P[X > 3] = 0.2003

The proportion of days on which neither the car used is 0.1003. The proportion of days on which some demand refused is 0.2003.