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Poisson Distribution

I. When           (i) the number of trials is indefinitely large i.e., n ->∞

(ii) Constant probability of success for each trial is very small i.e., p ->0,

and                  (iii) np =  a finite value.

Poisson distribution is obtained as a limiting case of binomial distribution.

The probability mass function is

where A. is called the parameter of this distribution.

Proof.

Let np = ð  P= /n,  q=1-/n

when n->∞             

Therefore,

II. Properties

 

 

(i)               

(ii)         Distribution function

(i)                 First two moments about origin.

When  is not an integer,

Mode = integral part of

When is an integer,

Mode = - 1 and

 

(i)                 In queuing theory (to be discussed in part B), it can be shown under certain assumptions that the probability of arriving x customers in time t is

which is a Poisson distribution with parameter A.t.? This is also called Poisson process i.e., the number of customers generated (arriving) until any specific time has a Poisson distribution.

 

Example 1. There are 150 misprints in a book of 520 pages. What is the probability that a given

page will contain at most 2 misprints ?

 

Solution: Here            =150/520     and let the misprints follow Poisson distribution.

Required probability= P[X ≤ 2]

=  P[X=0] + P[X=1] + P[X=2]

=0.9968.

Example 2. Let the probability that an individual suffers a bad reaction from an injection is 0.001. What is the probability that out of 3000 individuals (a) exactly 3, (b) more than 2 individuals will suffer a bad reaction?

Solution.   Here                       =3000 * 0.001 = 3

(a)    Required probability= P[X = 3]

(b) Required probability = P[X > 2] = 1 - P[X ≤ 2]

=1-(P[X = 0] + P[X = 1] + P[X = 21])

Example 3. A controlled manufacturing process is 0.2% defective. What is the probability of taking 2 or more defectives from a lot of 100 pieces? (a) By using binomial distribution. (b) By using Poisson approximation.

Solution. (a) p = Probability of defective = 0.002, q = 1 - p = 0.998

n = 100

Probability of finding 2 or more defective

=1 - [Probability of zero and one defective]

=1- (P[X = 0] + P[X = 1])

=1-[(0.998)100 + 100 (0.002) (0.998)99]

=1- 0.983 = 0.017

(b) Here                        = np = 100 x 0.002 = 0.2

Probability of finding 2 or more defective

=1-[P[X=O] + P[X=1]]

=1 - [e-0.2 + (0.2) e-0.2]

=1 - 0.982 = 0.018.

 

Example 4. Fit a Poisson distribution to the set of observations:

Solution.        

Mean= =125/135=  0.926

The mean of the Poisson distribution is  = 0.926

Hence the expected frequencies are given by

Therefore the fitted distribution is given by

PROBLEMS

 

1. If X be a Poisson distributed random variable and P[X = 1] = 3P[X = 2], then find P[X > 2].

2. A medicine was supplied in 100 batches (each batch containing a fairly large number of items). A total of 50 items in all the batches were found to be defective. Find the probability that (a) a batch has no defective item; (b) a batch has at least three defective items.

3. If 2 per cent of electric bulbs manufactured by a certain company are defective, find the probability that in a sample of 200 bulbs (a) less than 2 bulbs are defective; (b) more than 3 bulbs are defective.

4. Let X follows Poisson distribution, find the value of the mean of the distribution of P[X = 1]= 3P[X = 2].

5. In a certain factory, blades are manufactured in packets of  10. There is a 0.1% probability for any blade to be defective. Using Poisson distribution calculates approximately the number of packets containing two defective blades in a consignment of 10000 packets.

6. Fit a Poisson distribution to the following :

7. Fit a Poisson distribution to the following :

8. Records show that the probability is 0.00002 that a car will have a flat tire while driving over a certain bridge. Use the Poisson distribution to determine the probability that among 20000 cars driven over this bridge, not more than one will have a flat tire.

9. A typ1st kept a record of mistakes made per day during 300 working days of a year :

Fit an appropriate Poisson distribution to the data.

10. The probability that a Poisson variate X takes a positive value is ( 1 – e-1.5). Find the variance and also the probability that X lies between -1.5 and 1.5.

11. A manufacturer, who produces medicine bottles, finds that 0.1% of the bottles are defective. The bottles are packed in boxes containing 500 bottles. A drug manufacturer buys 1 00 boxes from the producer of bottles. Using Poisson distribution, find how many boxes will contain (i) no defectives, (ii} at least two defectives.

12. What is the probability that in a company of 1000 people only one person will have birth day on New Year’s Day? (Assume that a year has 365 days).

 

 

ANSWERS

 

1. 0.0302                                 2. (a) 0.61                   (b) 0.01

3. (a) 0.09                                (b) 0.58                        4. 2/3

5. Two packets

6.

(The EF corresponding to x = 1, has been adjusted to make ∑ EF = 60)

8. 0.9380

 

10. Variance = 1.5, P(-1.5 <. X < 1.5) = 2.5 e- l.5

11. (i) 61                      (ii) 9                                                                12. 0.18.