Poisson Distribution Examples
Example 1: If 4% of the total items made by a factory are defective. Find the probability that less than 2 items are defective in the sample of 50 items.
Solution:
Here we have, n = 50, p = (4/100) = 0.04, q = (1-p) = 0.96, λ = 2
Using Poisson’s Distribution,
P(X = 0) = [Tex]\frac{2^0e^{-2}}{0!}[/Tex] = 1/e2 = 0.13534
P(X = 1) = [Tex]\frac{2^1e^{-2}}{1!}[/Tex] = 2/e2 = 0.27068
Hence the probability that less than 2 items are defective in sample of 50 items is given by:
P( X > 2 ) = P( X = 0 ) + P( X = 1 ) = 0.13534 + 0.27068 = 0.40602
Example 2: If the probability of a bad reaction from medicine is 0.002, determine the chance that out of 1000 persons more than 3 will suffer a bad reaction from medicine.
Solution:
Here we have, n = 1000, p = 0.002, λ = np = 2
X = Number of person suffer a bad reaction
Using Poisson’s Distribution
P(X > 3) = 1 – {P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)}
P(X = 0) = [Tex]\frac{2^0e^{-2}}{0!}[/Tex] = 1/e2
P(X = 1) = [Tex] \frac{2^1e^{-2}}{1!}[/Tex] = 2/e2
P(X = 2) = [Tex] \frac{2^2e^{-2}}{2!}[/Tex] = 2/e2
P(X = 3) = [Tex]\frac{2^3e^{-2}}{3!}[/Tex] = 4/3e2
P(X > 3) = 1 – [19/3e2] = 1 – 0.85712 = 0.1428
Example 3: If 1% of the total screws made by a factory are defective. Find the probability that less than 3 screws are defective in a sample of 100 screws.
Solution:
Here we have, n = 100, p = 0.01, λ = np = 1
X = Number of defective screws
Using Poisson’s Distribution
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
P(X = 0) = [Tex] \frac{1^0e^{-1}}{0!}[/Tex] = 1/e
P(X = 1) = [Tex]\frac{1^1e^{-1}}{1!}[/Tex] =1/e
P(X = 2) = [Tex]\frac{1^2e^{-1}}{2!}[/Tex] =1/2e
P(X < 3) = 1/e + 1/e +1/2e
= 2.5/e = 0.919698
Example 4: If in an industry there is a chance that 5% of the employees will suffer by corona. What is the probability that in a group of 20 employees, more than 3 employees will suffer from the corona?
Solution:
Here we have, n = 20, p = 0.05, λ = np = 1
X = Number of employees who will suffer corona
Using Poisson’s Distribution
P(X > 3) = 1-[P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]
P(X = 0) = [Tex]\frac{1^0e^{-1}}{0!}[/Tex] = 1/e
P(X = 1) = [Tex]\frac{1^1e^{-1}}{1!}[/Tex] = 1/e
P(X = 2) =[Tex]\frac{1^2e^{-1}}{2!}[/Tex] =1/2e
P(X = 3) =[Tex]\frac{1^3e^{-1}}{3!}[/Tex] =1/6e
P(X > 3) = 1 – [1/e + 1/e + 1/2e + 1/6e]
= 1 – [ 8/3e] = 0.018988
Poisson Distribution | Definition, Formula, Table and Examples
Poisson Distribution is one of the types of discrete probability distributions like binomial distribution in probability. It expresses the probability of a given number of events occurring in a fixed interval of time.
Poisson distribution is a type of discrete probability distribution that determines the likelihood of an event occurring a specific number of times (k) within a designated time or space interval. This distribution is characterized by a single parameter, λ (lambda), representing the average number of occurrences of the event.
In this article, we will discuss the Poisson Distribution including its definition, Poisson Distribution formula, Poisson Distribution examples, and properties of Poisson Distribution in detail.
Table of Content
- What is Poisson Distribution?
- Poisson Distribution Definition
- Poisson Distribution Formula
- Poisson Distribution Table
- Poisson Distribution Characteristics
- Poisson Distribution Graph
- Poisson Distribution Mean and Variance
- Poisson Distribution Mean
- Poisson Distribution Variance
- Standard Deviation of Poisson Distribution
- Probability Mass Function of Poisson Distribution
- Difference between Binomial and Poisson Distribution
- Poisson Distribution Examples
- Poisson Distribution Practice Problems