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A Poisson distribution is the probability distribution that results from a Poisson experiment.

**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 (Î¼) 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.)

Î¼: 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; Î¼): The

**Poisson probability**that__exactly__x successes occur in a Poisson experiment, when the mean number of successes is Î¼.

**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 (Î¼) that occur in a specified region, we can compute the Poisson probability based on the following formula:

P(x; Î¼) = (e 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 Î¼ .

The variance is also equal to Î¼ .

**Poisson Distribution Example**

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:

Î¼ = 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; Î¼) = (e^{-Î¼}) (Î¼^{x}) / x!

P(3; 2) = (2.71828^{-2}) (2^{3}) / 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.

**Cumulative Poisson Example**

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:

Î¼ = 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})(5^{0}) / 0! ] + [ (e^{-5})(5^{1}) / 1! ] + [ (e^{-5})(5^{2}) / 2! ] + [ (e^{-5})(5^{3}) / 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.