# 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 editor