## Poisson Distribution Table

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.

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## The Poisson Distribution

In this video, I briefly discuss a situation that can be modeled by a Poisson Distribution, give the formula, and do a simple example illustrating the Poisson Distribution.
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## Poisson Distribution Conditions

Introduction to poisson distribution conditions

In probability theory and statistics, the Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate and independently of the time since the last event. The Poisson distribution can also be used for the number of events in other specified intervals such as distance, area or volume.

The Poisson distribution can be defined as the distribution of the number of events within a fixed time interval, offered to the events that happen random, in a separate time and on a constant rate. The occurrence rate, λ, is the number of proceedings per unit time. When λ is huge, the form of a poisson distribution is extremely related to that of the standard normal distribution. Let us see about the poisson distribution conditions.

Poisson Distribution Conditions

The probability of x events occurrence within unit time through an event rate of λ is:

P(x) = ‘(e^-lambda lambda^x)/(|__x)’

Where

x = 0, 1, 2, 3, 4…

e = 2.71828

λ = mean number of achievements in the known time interval or region of space

The poisson distribution has the following conditions

The number of achievements within the two disjoint time intervals is independent.

The probability of a success through a little time interval is proportional to the whole length of the time interval.

The probability of two events happening within the equal narrow interval is insignificant.

The probability of an event contained through a confident interval do not modify over dissimilar intervals.

Example

Find the poisson distribution If ‘lambda’= 2, x = 4 and e = 2.718

Solution

Step 1: Given

‘lambda’ = 2

x = 4

Step 2: Find e

e^-2 = (2.718)^-2

= 0.1353

Step 3: Find ‘lambda^x’

‘lambda’ = 2

x = 4

‘lambda^x’ = (2)^4= 16

Step 4: Substitute the value of poisson distribution formula

‘((e^-lambda)(lambda^x))/(x!)’ = ‘((0.1353)(16))/(4!)’

='((0.1353)(16))/(24)’

= 0.0902

Therefore the poisson distribution = 0.0902

Examples for Poisson Distribution Conditions

Example 1 for poisson distribution

Find the poisson distribution If ‘lambda’= 3, x = 5 and e = 2.718

Solution

Step 1: Given

‘lambda’ = 3

x = 5

Step 2: Find e

e^-3 = (2.718)^-3

= 0.0498

Step 3: Find ‘lambda^x’

‘lambda’ = 3

x = 5

‘lambda^x’ = (3)^5= 243

Step 4: Substitute the value of poisson distribution formula

‘((e^-lambda)(lambda^x))/(x!)’ = ‘((0.0498)(243))/(5!)’

='((0.0498)(243))/(120)’

= 0.1008

Therefore the poisson distribution = 0.1008

Example 2 for poisson distribution

Find the poisson distribution If ‘lambda’= 4, x = 7 and e = 2.718

Solution

Step 1: Given

‘lambda’ = 4

x = 7

Step 2: Find e

e^-4 = (2.718)^-4

= 0.0183

Step 3: Find ‘lambda^x’

‘lambda’ = 4

x = 7

‘lambda^x’ = (4)^7= 16384

Step 4: Substitute the value of poisson distribution formula

‘((e^-lambda)(lambda^x))/(x!)’ = ‘((0.0183)(16384))/(7!)’

='((0.0183)(16384))/(5040)’

= 0.0594

Therefore the poisson distribution = 0.0594

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