Poisson Distribution Examples
Poisson Distribution Calculator
Poisson distribution calculator helps you to determine the probability and cumulative probabilities for Poisson random variable given the mean number of successes ($\lambda$).
Poisson Distribution Calculator | |
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Average rate of success ($\lambda$): | |
Number of success (x): | |
Result | |
Probability : P(X = x) | |
Cumulative Probability : P(X ≤ x) | |
Cumulative Probability : P(X < x) | |
Cumulative Probability : P(X ≥ x) | |
Cumulative Probability : P(X > x) | |
How to find Poisson Distribution Probabilities?
Step 1 - Enter the average rate of sucess $\lambda$
Step 2 - Enter the value of $x$
Step 3 - Click on "Calculate" button to get Poisson distribution probabilities
Step 4 - Gives the output probability at $x$ for Poisson distribution
Step 5 - Gives the output cumulative probabilities for Poisson distribution
Poisson distribution
A discrete random variable $X$ is said to have Poisson distribution with parameter $\lambda$ if its probability mass function is
$$ \begin{align*} P(X=x)&= \begin{cases} \frac{e^{-\lambda}\lambda^x}{x!} , & x=0,1,2,\cdots; \lambda>0; \\ 0, & Otherwise. \end{cases} \end{align*} $$
Mean and variance of Poisson Distribution
The mean and variance of Poisson distribution is $E(X)=\lambda$ and $V(X)=\lambda$ rexpectively.
Poisson Distribution Example 1
A book contains 500 pages. If there are 200 typing errors randomly distributed throughout the book, use the Poisson distribution to determine the probability that a page contains
a. exactly 3 errors,
b. at least 3 errors,
c. at most 2 errors,
d. 2 or more errors but less than 5 errors.
Solution
A book contains 500 pages and there are 200 typing errors randomly distributed throughout the book.
The average no. of typing errors per page $=\lambda =\frac{200}{500}= 0.4$.
The random variable $X$ is no. of typing errors per page. $X\sim P(0.4)$.
The probability mass function of Poisson distribution with $\lambda =0.4$ is
$$ \begin{aligned} P(X=x) &= \frac{e^{-0.4}(0.4)^x}{x!},\; x=0,1,2,\cdots \end{aligned} $$
a. The probability that a page contains exactly 3 errors is
$$ \begin{aligned} P(X=3) &= \frac{e^{-0.4}0.4^{3}}{3!}\\ &= 0.0072 \end{aligned} $$
b. The probability that a page contains at least 3 errors is
$$ \begin{aligned} P(X\geq3) &= 1- P(X\leq 2)\\ &= 1- \sum_{x=0}^{2}P(X=x)\\ &= 1- \big[P(X=0) + P(X=1) + P(X=2)\big]\\ &= 1- \bigg[ \frac{e^{-0.4}0.4^{0}}{0!}+ \frac{e^{-0.4}0.4^{1}}{1!}+ \frac{e^{-0.4}0.4^{2}}{2!}\bigg]\\ &= 1-\big(0.6703+0.2681+0.0536\big)\\ &= 1-0.992\\ &= 0.008 \end{aligned} $$
c. The probability that a page contains at most 2 errors is
$$ \begin{aligned} P(X\leq2) &= \sum_{x=0}^{2}P(X=x)\\ &= P(X=0) + P(X=1) + P(X=2)\\ &= \frac{e^{-0.4}0.4^{0}}{0!}+ \frac{e^{-0.4}0.4^{1}}{1!}+ \frac{e^{-0.4}0.4^{2}}{2!}\\ &= 0.6703+0.2681+0.0536\\ &= 0.992 \end{aligned} $$
d. The probability that a page contains 2 or more errors but less than 5 errors is
$$ \begin{aligned} P(2\leq X< 5) &=P(2\leq X\leq 4)\\ &=P(X=2)+P(X=3)+P(X=4)\\ &= \frac{e^{-0.4}0.4^{2}}{2!}+\frac{e^{-0.4}0.4^{3}}{3!}+\frac{e^{-0.4}0.4^{4}}{4!}\\ &= 0.0536+0.0072+0.0007\\ &= 0.0615 \end{aligned} $$
Poisson Distribution Example 2
Suppose that in a certain area there are on average 5 traffic accidents per month. Find the probability of
a. 4 accidents in a given month,
b. at least 2 accidents in a given month,
c. at most 2 accidents in a given month.
Solution
The average no. of traffic accidents per month $=\lambda = 5$.
The random variable $X$ is no. of traffic accidents per month. $X\sim P(5)$.
The probability mass function of Poisson distribution with $\lambda =5$ is
$$ \begin{aligned} P(X=x) &= \frac{e^{-5}(5)^x}{x!},\; x=0,1,2,\cdots \end{aligned} $$
a. The probability of $4$ accidents in a given month is
$$ \begin{aligned} P(X=4) &= \frac{e^{-5}5^{4}}{4!}\\ &= 0.1755 \end{aligned} $$
b. The probability of at least 2 accidents in a given month is
$$ \begin{aligned} P(X\geq2) &= 1- P(X\leq 1)\\ &= 1- \sum_{x=0}^{1}P(X=x)\\ &= 1- \big[P(X=0) + P(X=1)\big]\\ &= 1- \bigg[ \frac{e^{-5}5^{0}}{0!}+ \frac{e^{-5}5^{1}}{1!}\bigg]\\ &= 1-\big(0.0067+0.0337\big)\\ &= 1-0.0404\\ &= 0.9596 \end{aligned} $$
c. The probability of at most 2 traffic accidents is
$$ \begin{aligned} P(X\leq2) &= \sum_{x=0}^{2}P(X=x)\\ &= P(X=0) + P(X=1) + P(X=2)\\ &= \frac{e^{-5}5^{0}}{0!}+ \frac{e^{-5}5^{1}}{1!}+ \frac{e^{-5}5^{2}}{2!}\\ &= 0.0067+0.0337+0.0842\\ &= 0.1246 \end{aligned} $$
Conclusion
In this tutorial, you learned about how to calculate mean, variance and probabilities of Poisson distribution. You also learned about how to solve numerical problems based on Poisson distribution.
To read more about the step by step tutorial on Poisson distribution refer the link Poisson Distribution. This tutorial will help you to understand Poisson distribution and you will learn how to derive mean of Poisson distribution, variance of Poisson distribution, moment generating function and other properties of Poisson distribution.
To learn more about other discrete probability distributions, please refer to the following tutorial:
Let me know in the comments if you have any questions on Poisson Distribution Examples and your thought on this article.