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
