Poisson Distribution Questions

Introduction :

Poisson distribution is mainly used for finding the average of the given functions by using the formulas. The x values and the `lambda` is used in the formula of Poisson distribution. Both the values are used for the event in success and the rate of change values. In this article, we are going to see about the Poisson distribution questions and the example problems.
Explanation to Poisson Distribution Questions

The explanation for Poisson distribution questions are given below the following section,
Formula:

Poisson distribution =   `((e^-lambda)(lambda^x))/(x!)`

where,

x = Poisson value

`lambda` = Rate of change
e = Log function

Example Problems to Poisson Distribution Questions

Question 1: Solve Poisson distribution where,`lambda` = 2, x = 4 and e = 2.718.

Solution:

Step 1: Given:

`lambda`  = 2

x = 4

Step 2: Formula:

Poisson distribution =   `((e^(-lambda))(lambda^x))/(x!)`

Step 3: To find e:

e-2 = (2.718)-2

= 0.1353

Step 4: Solve:

`lambda`  = 2

x = 4

`lambda^x` = (2)4 = 16

Step 4: Substitute:

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

= `2.1648/(4!)`

= `2.1648/24`

= 0.09

Result: Poisson Distribution = 0.09

Thus, this is the obtained answer for Poisson distribution questions.

Question 2: Solve Poisson distribution where,`lambda` = 2, x = 6 and e = 2.718.

Solution:

Step 1: Given:

`lambda`  = 2

x = 6

Step 2: Formula:

Poisson distribution =   `((e^(-lambda))(lambda^x))/(x!)`

Step 3: To find e:

e-2 = (2.718)-2

= 0.1353

Step 4: Solve:

`lambda`  = 2

x =  6

`lambda^x` = (2)6 = 64

Step 4: Substitute:

`((e^-lambda)(lambda^x))/(x!)`  = `(0.1353(64))/(6!)`

= `8.6952/(6!)`

= `8.6952/720`

= 0.012

Result: Poisson Distribution = 0.012

Thus, this is the obtained answer for Poisson distribution questions.
Practice Problems to Poisson Distribution Questions

Question 1: Solve Poisson distribution where,`lambda` = 42, x = 44 and e = 2.718.