Module 4: Discrete Random Variables
Poisson Distribution
Barbara Illowsky & OpenStax et al.
There are two main characteristics of a Poisson experiment.
- The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages.
- The Poisson distribution may be used to approximate the binomial if the probability of success is “small” (such as 0.01) and the number of trials is “large” (such as 1,000). You will verify the relationship in the homework exercises. n is the number of trials, and p is the probability of a “success.”
The random variable [latex]X=[/latex] the number of occurrences in the interval of interest.
Example
The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of interest is the number of loaves of bread put on the shelf in five minutes. The time interval of interest is five minutes. What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three?
[reveal-answer q=”869903″]Show Solution[/reveal-answer]
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Let [latex]X=[/latex] the number of loaves of bread put on the shelf in five minutes. If the average number of loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on the shelf in five minutes is [latex]left(frac{5}{30}right)left(12right)=2[/latex] loaves of bread.
The probability question asks you to find [latex]P(x=3)[/latex].
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Notation for the Poisson: [latex]P=[/latex] Poisson Probability Distribution Function
[latex]X{sim}P(mu)[/latex]
Read this as “X is a random variable with a Poisson distribution.” The parameter is [latex]mu[/latex] (or [latex]lambda[/latex]); [latex]mu[/latex] (or [latex]lambda[/latex])[latex]=[/latex] the mean for the interval of interest.
Example
Leah’s answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes?
[reveal-answer q=”877222″]Show Solution[/reveal-answer]
[hidden-answer a=”877222″]
Let [latex]X=[/latex] the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or [latex]frac{1}{4}[/latex] hour.)
[latex]x=0,1,2,3,...[/latex]
If Leah receives, on the average, six telephone calls in two hours, and there are eight 15-minute intervals in two hours, then Leah receives [latex]left(frac{1}{8}right)left(6right)=0.75[/latex] calls in 15 minutes, on average. So, [latex]mu=0.75[/latex] for this problem.
[latex]X{sim}P(0.75)[/latex]
Find [latex]P(x>1)[/latex]. [latex]P(x>1)=0.1734[/latex] (calculator or computer)
- Press