The name of French mathematician Simeon Poisson, Poisson probabilities are useful when there are a large number of independent trials with a small probability of success on a single trial and the variables occur over a period of time. It can also be used when a density of items is distributed over a given area or volume.
\[p(x;\lambda )=\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!},x=0,1,2,...\].
Lambda in the formula is the mean number of occurrences. If you're approximating a binomial probability using the Poisson, then lambda is the same as mu or $n * p$.
Example:
If there are 500 customers per eight-hour day in a check-out lane, what is the probability that there will be exactly 3 in line during any five-minute period?
The expected value during any one five minute period would be $500 / 96 = 5.2083333$. The $96$ is because there are $96$ five-minute periods in eight hours. So, you expect about $5.2$ customers in $5$ minutes and want to know the probability of getting exactly $3$.
\[p(3;\frac{500}{96})=\frac{{{e}^{-\left( \frac{500}{96} \right)}}{{\left( \frac{500}{96} \right)}^{3}}}{3!}=0.1288(approx)\]
Attributes of a Poisson Experiment
A Poisson experiment is a statistical experiment that has the following properties:
- The experiment results in outcomes that can be classified as successes or failures.
- The average number of successes (μ) that occurs in a specified region is known.
- The probability that a success will occur is proportional to the size of the region.
- The probability that a success will occur in an extremely small region is virtually zero.
Note that the specified region could take many forms. For instance, it could be a length, an area, a volume, a period of time, etc.
Notation
The following notation is helpful, when we talk about the Poisson distribution.
e: A constant equal to approximately 2.71828. (Actually, e is the base of the natural logarithm system.)
μ: The mean number of successes that occur in a specified region.
x: The actual number of successes that occur in a specified region.
P(x; μ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is μ.
Poisson Distribution
A Poisson random variable is the number of successes that result from a Poisson experiment. The probability distribution of a Poisson random variable is called a Poisson distribution.
Given the mean number of successes (μ) that occur in a specified region, we can compute the Poisson probability based on the following formula:
Poisson Formula. Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is μ. Then, the Poisson probability is:
\[P(x; μ) = (e-μ) (μx) / x!\]
where x is the actual number of successes that result from the experiment, and e is approximately equal to 2.71828.
The Poisson distribution has the following properties:
The mean of the distribution is equal to μ .
The variance is also equal to μ .
Example 1
The average number of homes sold by the Acme Realty company is 2 homes per day. What is the probability that exactly 3 homes will be sold tomorrow?
Solution: This is a Poisson experiment in which we know the following:
μ = 2; since 2 homes are sold per day, on average.
x = 3; since we want to find the likelihood that 3 homes will be sold tomorrow.
e = 2.71828; since e is a constant equal to approximately 2.71828.
We plug these values into the Poisson formula as follows:
P(x; μ) = (e-μ) (μx) / x!
P(3; 2) = (2.71828-2) (23) / 3!
P(3; 2) = (0.13534) (8) / 6
P(3; 2) = 0.180
Thus, the probability of selling 3 homes tomorrow is 0.180 .
Cumulative Poisson Probability
A cumulative Poisson probability refers to the probability that the Poisson random variable is greater than some specified lower limit and less than some specified upper limit.
Example 1
Suppose the average number of lions seen on a 1-day safari is 5. What is the probability that tourists will see fewer than four lions on the next 1-day safari?
Solution: This is a Poisson experiment in which we know the following:
μ = 5; since 5 lions are seen per safari, on average.
x = 0, 1, 2, or 3; since we want to find the likelihood that tourists will see fewer than 4 lions; that is, we want the probability that they will see 0, 1, 2, or 3 lions.
e = 2.71828; since e is a constant equal to approximately 2.71828.
To solve this problem, we need to find the probability that tourists will see 0, 1, 2, or 3 lions. Thus, we need to calculate the sum of four probabilities: P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5). To compute this sum, we use the Poisson formula:
$P(x < 3, 5) = P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5)$
$P(x < 3, 5) = [ (e-5)(50) / 0! ] + [ (e-5)(51) / 1! ] + [ (e-5)(52) / 2! ] + [ (e-5)(53) / 3! ] $
$P(x < 3, 5) = [ (0.006738)(1) / 1 ] + [ (0.006738)(5) / 1 ] + [ (0.006738)(25) / 2 ] + [ (0.006738)(125) / 6 ] $
$P(x < 3, 5) = [ 0.0067 ] + [ 0.03369 ] + [ 0.084224 ] + [ 0.140375 ] $
$P(x < 3, 5) = 0.2650$
Thus, the probability of seeing at no more than 3 lions is 0.2650.
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