POISSON DISTRIBUTION Poisson distribution may be expected in

POISSON DISTRIBUTION Poisson distribution may be expected in cases where the chance of any individual event being a success is small and n tends to infinity. This distribution is used to describe the behavior of rare events such as the number of accidents on road , number of printing mistakes in a book etc.

DEFINITION

Assumptions for applying the Poisson distribution: Poisson distribution can be applied under the following assumptions: i). The outcome of trial/experiment must be of dichotomous nature. ii). The probability of success must remain the same for trials. iii). The trials should be conducted under identical conditions. iv). The trials should be statistically independent. v). The probability of success should be very small and ‘n’ should be large such that ‘np’ is a constant ‘m’.

Characteristics of Poisson Distribution: •

Physical situations in daily life when Poisson distribution is applicable: Ø Number of telephone calls received per minute at a particular switch board. Ø Number of accidents in a city per unit time. Ø Number of defective parts produced in a factory per unit time. ØNumber of printing mistakes in each page of a book.

This means there is a 14% chance of receiving no phone calls in 5 minute period. This means there is 86% chance of receiving at least one phone call within this 5 minute time interval , given that the rate is 4 calls /10 minute. So this is not in the best interest of the receptionist to step out for 5 minutes.

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Poisson Distribution as an Approximation of the Binomial Distribution: • To avoid the tedious job of calculating binomial probability distributions, we can use the Poisson instead. • The Poisson distribution can be reasonable approximation of the binomial, but only under certain conditions: • These conditions occur when: • n is large • p is small • The rule most often used by statisticians is that the Poisson is a good approximation of the binomial when n is greater than or equal to 20 and p is less than or equal to 0. 05.

Problem No. 10: Poisson distribution as an approximation to Binomial distribution 2 percent of the fuses manufactured by a firm are expected to be defective. Find the probability that a box containing 200 fuses contains i). Defective fuses. ii). 3 or more defective fuses. [ at least 3 defective fuses] iii). At most 2 fuses are defective. Ans: P( a fuse is defective ) = p = 2/100 = 0. 02, n = 200 X is B( n= 200, p = 0. 02) i. e. , binomial with parameters n and p Here, p is very small and n is very large. Therefore X can be treated as Poisson variate with parameters m =np = 200 x 0. 02 = 4

Problem number 11 • Suppose a life insurance company insures the lives of 5000 persons aged 42. If studies show the probability that any 42 -years old person will die in a given year to be 0. 001, find the probability that the company will have to pay at least two claims during a given year. • [ Ans: 0. 9598 ]

Problem number 12 • A manufacturer who produces medicine bottles finds that 0. 1 per cent of the bottles are defective. The bottles are packed in boxes containing 500 bottles. A drug manufacturer buys 100 boxes from the producer of bottles. Using Poisson distribution, find how many boxes will contain: • i). No defectives. [Ans: 0. 6065 X 100 = 61 (approx. )] • Ii). At least two defectives. [ Ans: 100 X • 0. 09025 =10 ( approx. )]