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| Poisson Distribution |
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| Poisson distribution is a limiting process of binomial distribution. Poisson distribution occurs when there are events which do not occur as outcomes of a definite number of outcomes. |
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| Poisson distribution is used under the following conditions: |
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 Number of trials n tends to infinity |
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| Probability of success p tends to zero and |
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| np = l is finite. |
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| We shall now deduce the Poisson distribution from the binomial
distribution by assuming that n ® ¥ and p
® 0 such that the product np always remains
finite, say l. |
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| We shall now use a very important result of limits in Calculus. We state this result without proof: |
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| where e is a constant lying between the number 2 and 3 and ex is defined by |
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| with x a real number. From equation (1), we observe that each of |
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| (r - 1) factors, |
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 that |
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| Thus |
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| where l is a finite number and is equal to np. |
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| The sum of the probabilities P(X = r) or simply P(r) for r = 0, 1, 2, … is 1. This can be seen by putting r = 0, 1, 2, … in (4) and adding all the probabilities. |
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| Also, each of the probabilities is a non-negative fraction. This leads to the distribution defined below: |
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| A random variable X taking values 0, 1, 2, … is said to have a Poisson distribution with parameter l (finite), if its probability distribution is given by |
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| There are many daily life situations where n is very large and p is very small. In such situations, the Poisson distribution can be more conveniently used as an approximation to binomial distriburtion which may prove cumbersome for large values of n. This is called Poisson approximation to binomial distribution. The Poisson approximation to binomial distribution is easier to compute directly and easier to tabulate than the binomial distribution, since the values of e-l for various values of l are found in standard tables. |
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| Some examples of such situations are |
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| i) telephone trunk lines with a large number of subscribers and the probability of telephone lines being available is very small, |
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| ii) traffic problems with repeated occurrence of events such as accidents whose probability is very small, |
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| iii) many industrial processes undergoing mass scale production with probability of events as 'faults' or 'breakdowns' being very small, etc. |
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| The probability mass function of the Poisson distribution given by |
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| Note 1: |
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| = e-l el |
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| = 1 |
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| Example: |
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| A random variable X has a Poisson distribution with parameter l such that P (X = 1) = (0.2) P (X = 2). Find P (X = 0). |
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| Suggested answer: |
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| For the Poisson distribution, the probability function is given by |
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| Given P (x = 1) = (0.2) P (X = 2) |
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| Mean and Variance |
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| = le-l el |
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| = l |
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| Example: |
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| If the variance of the Poisson distribution is 2, find the probabilities for r = 1, 2, 3, 4 and 5 from the recurrence relation of the Poisson distribution. |
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| Suggested answer: |
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| The variance of the Poisson distribution = l = 2 |
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| Recurrence relation is given by |
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