June 20, 2012- Lesson on Poisson Distribution and Hypergeometric Distribution

Poisson Distribution

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 (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) (μ^x) / 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) (μ^x) / 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 .

Poisson Calculator

Clearly, the Poisson formula requires many time-consuming computations. The Stat Trek Poisson Calculator can do this work for you - quickly, easily, and error-free. Use the Poisson Calculator to compute Poisson probabilities and cumulative Poisson probabilities. The calculator is free. It can be found under the Stat Tables tab, which appears in the header of every Stat Trek web page.

Poisson Calculator

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.

Hypergeometric Distribution

The probability distribution of a hypergeometric random variable is called a hypergeometric distribution. This lesson describes how hypergeometric random variables, hypergeometric experiments, hypergeometric probability, and the hypergeometric distribution are all related.

Notation

The following notation is helpful, when we talk about hypergeometric distributions and hypergeometric probability.

• N: The number of items in the population.
• k: The number of items in the population that are classified as successes.
• n: The number of items in the sample.
• x: The number of items in the sample that are classified as successes.
• _kC_x: The number of combinations of k things, taken x at a time.
• h(x; N, n, k): hypergeometric probability - the probability that an n-trial hypergeometric experiment results in exactly x successes, when the population consists of N items, k of which are classified as successes.

Hypergeometric Experiments

A hypergeometric experiment is a statistical experiment that has the following properties:

• A sample of size n is randomly selected without replacement from a population of N items.
• In the population, k items can be classified as successes, and N - k items can be classified as failures.

Consider the following statistical experiment. You have an urn of 10 marbles - 5 red and 5 green. You randomly select 2 marbles without replacement and count the number of red marbles you have selected. This would be a hypergeometric experiment.
Note that it would not be a binomial experiment. A binomial experiment requires that the probability of success be constant on every trial. With the above experiment, the probability of a success changes on every trial. In the beginning, the probability of selecting a red marble is 5/10. If you select a red marble on the first trial, the probability of selecting a red marble on the second trial is 4/9. And if you select a green marble on the first trial, the probability of selecting a red marble on the second trial is 5/9.
Note further that if you selected the marbles with replacement, the probability of success would not change. It would be 5/10 on every trial. Then, this would be a binomial experiment.

Hypergeometric Distribution

A hypergeometric random variable is the number of successes that result from a hypergeometric experiment. The probability distribution of a hypergeometric random variable is called a hypergeometric distribution.
Given x, N, n, and k, we can compute the hypergeometric probability based on the following formula:

Hypergeometric Formula. Suppose a population consists of N items, k of which are successes. And a random sample drawn from that population consists of n items, x of which are successes. Then the hypergeometric probability is:

h(x; N, n, k) = [ _kC_x ] [ _N-kC_n-x ] / [ _NC_n ]

The hypergeometric distribution has the following properties:

• The mean of the distribution is equal to n * k / N .
• The variance is n * k * ( N - k ) * ( N - n ) / [ N² * ( N - 1 ) ] .

Example 1

Suppose we randomly select 5 cards without replacement from an ordinary deck of playing cards. What is the probability of getting exactly 2 red cards (i.e., hearts or diamonds)?
Solution: This is a hypergeometric experiment in which we know the following:

• N = 52; since there are 52 cards in a deck.
• k = 26; since there are 26 red cards in a deck.
• n = 5; since we randomly select 5 cards from the deck.
• x = 2; since 2 of the cards we select are red.

We plug these values into the hypergeometric formula as follows:

h(x; N, n, k) = [ _kC_x ] [ _N-kC_n-x ] / [ _NC_n ]
h(2; 52, 5, 26) = [ ₂₆C₂ ] [ ₂₆C₃ ] / [ ₅₂C₅ ]
h(2; 52, 5, 26) = [ 325 ] [ 2600 ] / [ 2,598,960 ] = 0.32513

Thus, the probability of randomly selecting 2 red cards is 0.32513.

Hypergeometric Calculator

As you surely noticed, the hypergeometric formula requires many time-consuming computations. The Stat Trek Hypergeometric Calculator can do this work for you - quickly, easily, and error-free. Use the Hypergeometric Calculator to compute hypergeometric probabilities and cumulative hypergeometric probabilities. The calculator is free. It can be found under the Stat Tables tab, which appears in the header of every Stat Trek web page.

Hypergeometric Calculator

Cumulative Hypergeometric Probability

A cumulative hypergeometric probability refers to the probability that the hypergeometric random variable is greater than or equal to some specified lower limit and less than or equal to some specified upper limit.
For example, suppose we randomly select five cards from an ordinary deck of playing cards. We might be interested in the cumulative hypergeometric probability of obtaining 2 or fewer hearts. This would be the probability of obtaining 0 hearts plus the probability of obtaining 1 heart plus the probability of obtaining 2 hearts, as shown in the example below.
Example 1

Suppose we select 5 cards from an ordinary deck of playing cards. What is the probability of obtaining 2 or fewer hearts?
Solution: This is a hypergeometric experiment in which we know the following:

• N = 52; since there are 52 cards in a deck.
• k = 13; since there are 13 hearts in a deck.
• n = 5; since we randomly select 5 cards from the deck.
• x = 0 to 2; since our selection includes 0, 1, or 2 hearts.

We plug these values into the hypergeometric formula as follows:

h(x < x; N, n, k) = h(x < 2; 52, 5, 13)
h(x < 2; 52, 5, 13) = h(x = 0; 52, 5, 13) + h(x = 1; 52, 5, 13) + h(x = 2; 52, 5, 13)
h(x < 2; 52, 5, 13) = [ (₁₃C₀) (₃₉C₅) / (₅₂C₅) ] + [ (₁₃C₁) (₃₉C₄) / (₅₂C₅) ] + [ (₁₃C₂) (₃₉C₃) / (₅₂C₅) ]
h(x < 2; 52, 5, 13) = [ (1)(575,757)/(2,598,960) ] + [ (13)(82,251)/(2,598,960) ] + [ (78)(9139)/(2,598,960) ]
h(x < 2; 52, 5, 13) = [ 0.2215 ] + [ 0.4114 ] + [ 0.2743 ]
h(x < 2; 52, 5, 13) = 0.9072

Thus, the probability of randomly selecting at most 2 hearts is 0.9072.



From: 

StatTrek

 By: Kent Spencer Manalo Mendez
       & Tito Nuevacobita