Principal Investigator

(1)

Paper.15- Quantitative Techniques for Management Decisions Module:13 Discrete Probability Distribution : Poisson Distribution

-

Dr. Nidhi Handa

Department of Mathematics & Statistics.

Dr. Pankaj Madan

Professor, School of Management, Maharaha Agrasen University, Baddi, HP

Prof YoginderVerma

Pro–Vice Chancellor

Central University of Himachal Pradesh. Kangra. H.P.

Prof. S P Bansal

Vice Chancellor

Maharaja Agrasen University, Baddi

Content Writer

Co-Principal Investigator

Paper Coordinator Principal Investigator

Dr. Nidhi Handa Assistt. Prof.

Department of Mathematics & Statistics, Gurukul Kangri Viswavidhalaya, Haridwar

Prof. Pankaj Madan

Dean FMS, Gurukul Kangri viswavidhalaya, Haridwar

Prof. YoginderVerma

Pro –

Vice Chancellor

Central University of Himachal Pradesh. Kangra. H.P.

Prof. S P Bansal

Vice Chancellor

Maharaja AgrasenUniversity, Baddi

Content Writer

Co -

Principal Investigator

Paper Coordinator

Principal Investigator

(2)

QUADRANT:-1

Module-13 Discrete Probability Distribution : Poisson Distribution 1. Introduction

2. Poisson Probability Distribution

3. Conditions to apply Poisson Probability Distribution 4. Approaches To Formula of Poisson Probability

5. Problems related to Poisson Distribution 6. Summary

THE POISSON PROBABILITY DISTRIBUTION

The Poisson probability distribution, named after the French mathematician Simeon D.

Poisson, is another important probability distribution of a discrete random variable that has a large number of applications. Suppose a washing machine in a Laundromat breaks down an average of three times a month. We may want to find the probability of exactly two break downs during the next month. This is an example of Poisson probability distribution problem.

Each breakdown is called an occurrence in Poisson probability distribution terminology.

The Poisoon probability distribution is applied to experiments with random and independent occurrences. The occurances are random in the sense that they do not follow any pattern and, hence, they are unpredictable. Independence of occurrences means that one occurrence(or non occurrence) of an event does not influence the successive occurrences or non occurrences of that event. The occurrences are always considered with respect to an interval. In the example of the washing machine, the interval represents one month. The interval may be a time interval, a space interval, or a volume interval. The actual number or occurrences with Items Description of Module

Subject Name Management

Paper Name QUANTITATIVE TECHNIQUES FOR MANAGEMENT DECISIONS Module Title . Discrete Probability Distribution : Poisson Distribution.

Module Id Module No.-13

Pre- Requisites Basic knowledge of what Probability and its mean, variance are.

Objectives To study the probability distribution when occurrences are random and independent

Keywords Mean , Variance , Standard Variance ,Moments , Kurtosis ,Skewness.

(3)

in an interval is random and independent. If the average number of occurrences for a given interval is known, then by using the Poisson probability distribution we can compute the probability of a certain number of occurrences x in that interval. Note that the number of actual occurrences in an interval is denoted by x.

CONDITIONS TO APPLY POISSON PROBABILITY DISTRIBUTION :

The following three conditions must be satisfied to apply the Poisson probability distribution.

1. X is a discrete random variable.

2. The occurrences are random.

3. The occurrences are independent.

The following are a few examples of discrete random variables for which the occurrences are random and independent. Hence, these are examples to which the Poisson probability distribution can be applied.

1 Consider the number of patients arriving at the emergency ward of a hospital during a one-hour interval. In this example, an occurrences is the arrival of a patient at the emergency ward, the intervals is one hour (an interval of time), and the occurrences are random. The total number of patients who may arrive at this emergency ward during a one-hour interval may be 0,1,2,3,4,……. The independence of occurrences in this example means that patient arrive individually and the arrival of any two ( or more) patients is not related.

2 Consider the number of defective items in the next 100 items manufactured on a machine. In this case, the interval is a volume interval (100times). The occurrences (number of defective items) are random because there may be 0,1,2,3,……, 100 defective items in 100 items. We can assume the occurrence of defective items to be independent of one another.

3 Consider the number of defects in a 5-foot long iron rod. The interval, in this example, is a space interval(5feet). The occurrences (defects) are random because there may be any number of defects in a 5-foot iron rod. We can assume that these defects are independent of one another.

The following examples also qualify for the application of the Poisson probability distribution.

1) The number of accidents that occur on a given highway during a one-week period.

2)The number of customers coming to a grocery store during a one-hour interval.

3)The number of television sets sold at a department store during a given week.

On the other hand, consider the arrival of patients at a physician’s office. These arrivals will be non random if the patients have to make appointments to see the doctor. The arrival of commercial airplanes at an airport is non random because all planes are scheduled to arrive at certain times, and airport authorities know the exact number of arrivals for any period (although this number may change slightly because of late or early arrivals and cancellations). The Poisson probability distribution cannot be applied to these examples.

(4)

In the Poisson probability distribution terminology, the average number of occurrences in an interval is denoted by  (Greek letter lambda). The actual number of occurrences in the interval is denoted by x. The, using the Poisson probability distribution, we find the probability of x occurrences during an interval given the mean occurrences are  during that interval.

POISSON PROBABILITY DISTRIBUTION FORMULA

According to the Poisson probability distribution, the probability of x occurrences in an interval is -

P(x) =  ^xe

x!

Where  (pronounced lambda ) is the mean number of occurrences in that interval and the value of e is approximately 2.71828.

The mean number of occurrences in an interval, denoted by  , is called the parameter of the Poisson probability distribution or the Poisson parameter. As is obvious from the Poisson probability distribution formula, we need to know only the value of  to compute the probability of any given value of x. We can read the value of e for a given  from table.

EXAMPLE 1: A washing machine in a Laundromat breaks down an average of three times per month. Using the Poisson probability distribution formula, find the probability that during the next month this machine will have

(a) Exactly two breakdowns (b) at most one breakdown

Solution let  be the mean number of breakdowns per month and x be the actual number of breakdowns observed during he next month for this machine. Then,

 = 3

(a) The probability that exactly two breakdowns will be observed during the next month is

P(x=2) =

 ^xe

x! = ⁽³²^)𝑒⁻³

2! = (9)(.049787)

2 = .2240

(b) The Probability that at most one breakdown will be observed during the next month is given by the sum of the probabilities of zero and one breakdown. Thus,

P(at most 1 breakdown) = P( 0 or 1 breakdown) = P(x = 0) + P(x=1) = ⁽³

0)𝑒⁻³

0! + ⁽³

1)𝑒⁻³ 1!

= (1)(.049787)

1 + (3)(.049787) 1 = 0.0498 + 0.1494

(5)

= 0.1992

One important point to remember about the Poisson probability distribution is that the intervals for  and x must be equal.

POISSON DISTRIBUTION :

We come across several situations in which the probability 𝜑 of an event is very small but the number of trials n is no large that the event occurs several times as can be seen from the following cases :

 The number of typographical errors by an expert typist in typing the manuscript of a book is fairly high.

 Individually being born blind is a rare event and hence its probability is very small, but the no. of persons born blind in a big city in a year is quite high.

 The probability of a person suffering form a disease such as blood cancer is very small, but in a big city in a year the no. of persons suffering from the disease is sufficiently high.

Thus, clearly Poisson Distribution is a limiting case of Binomial Distribution when p (or q) is very small and n is large so that average no. of successes np (or nq) is a finite constant 𝛾 (say).

In the Binomial Distribution, the prob. Of r successes is given by

B(r, n, p) = ⁿCrp^r, p+q=1,

𝑛→∞lim𝑛𝑝 =m

MEAN AND VARIANCE :

If X is Poisson variate i.e. X ~ P(  ) then 𝜇 = mean = E (X)

Variance = 𝜎² = E(X – 𝜇)² = E(X²) – 𝜇² = m² + m – m²

= m

Note : The mean number of occurrences in an interval, denoted by  , is called the parameter of Poisson Distribution or the Poisson parameter.

(6)

ANOTHER APPROACH TO POISSON DISTRIBUTION :

Let the random variable X denote the number of telephone calls during time t at a telephone switch boards. We assume that (1) the calls are independent and (2)  denotes mean arrival rate of telephone cells per second.

Further examples of Poisson varicates :

1. Number of buses reaching bus depot in time t.

2. Number of suicides in a city in a year.

3. Number of Persons joining a queue at a counter in time t.

So Here  is mean (m).

Example 2: Experience shows that a box of 450 components of a company has 1 percent defective components. Find the probability that such a box has (1) no defective (2) one component defective and (3) at most 3 components defective. Given e^-4 =0.0183.

Solution : Let selection of a component from the box be regarded as a trial and getting a defective component as the success. If p denotes the probability of success, then 𝑝 = ¹

100= 0.01 which is small, as the number of trials n= 400 is large and so the problem can be solved by poisson distribution.

The parameter m= np=400 ×0.01=4

Required Poisson distribution 𝑝(𝑥 = 𝑟) = ^𝑒^−𝑚

𝑟! 𝑚^𝑟 𝑟 = 0,1,2 … …

=𝑒⁻⁴

𝑟! 4^𝑟 𝑟 = 0,1,2 … … (1) P(0 defective)= p(x=0)= 𝑒^−𝑚 = 𝑒⁻⁴=0.0183

(2) P(1 defective)= p(x=1)= 𝑒^−𝑚 = 4𝑒⁻⁴=4×0.0183 =0.0732 (3) P(at most 3 defective)=p(x≤ 3)

=∑ ^𝑒⁻⁴

𝑟! 4^𝑟

30

= 𝑒⁻⁴(1 + 4 +⁴²

2!+⁴³

3!)

=0.0183(1+4+8+³²

3)

=0.0183×⁷¹

3 = 0.0183× 23.6667

=0.4331

Example 3: If the probability of a bad reaction from certain vaccine is 0.001. Determine the probability that one of 2000 individuals (1) none (2) 1 (3) more than 1; (4) more than 2; (5) exactly 3 will get a bad reaction given 𝑒⁻²= 0.1353

(7)

Solution : let getting a bad reaction be regarded as success, then p=0.001 and the selection of an individual being regarded as a trial, then the number of trials n =2000. Hence , parameter of poisson distribution m= 2000 × 0.001 = 2

𝑝(𝑟 = 𝑠𝑢𝑐𝑐𝑒𝑠𝑠) =𝑒⁻²

𝑟! 2^𝑟 𝑟 = 0,1,2 … … (1) p(0 success)= 𝑒⁻²= 0.1353

(2) p(1 success)= 2𝑒⁻²= 2 × 0.1353 = 0.2706

(3) p(more than 1 success)= 𝑝(2 success)+ 𝑝(3 success)+… … ….

= 1 − 𝑝(0 success) - 𝑝(1 success) +… … ….

= 1 − 𝑒⁻² -2𝑒⁻²

=1-3𝑒⁻²=1-3×0.1353

=0.5941

(4) p(more than 2 success)= 𝑝(3success)+ 𝑝(4 success)+… … ….

= 1 − 𝑝(0 success) - 𝑝(1 success) –p(2 successes) … … ….

=1 − 𝑒⁻² −2𝑒⁻² −²²

2 𝑒²

=1- 5𝑒⁻²= 1- 5 × 0.1353 = 0.3235 (5) p(3 successes)= ²³

3!𝑒⁻² = 0.1804

Example 4: If the probability of a bad reaction from certain vaccine is 0.001. Determine the probability that one of 2000 individuals (1) none (2) 1 (3) more than 1; (4) more than 2; (5) exactly 3 will get a bad reaction given 𝑒⁻²= 0.1353

Solution : let getting a bad reaction be regarded as success, then p=0.001 and the selection of an individual being regarded as a trial, then the number of trials n =2000. Hence , parameter of poisson distribution m= 2000 × 0.001 = 2

𝑝(𝑟 = 𝑠𝑢𝑐𝑐𝑒𝑠𝑠) =𝑒⁻²

𝑟! 2^𝑟 𝑟 = 0,1,2 … … (1) p(0 success)= 𝑒⁻²= 0.1353

(2) p(1 success)= 2𝑒⁻²= 2 × 0.1353 = 0.2706

(3) p(more than 1 success)= 𝑝(2 success)+ 𝑝(3 success)+… … ….

= 1 − 𝑝(0 success) - 𝑝(1 success) +… … ….

= 1 − 𝑒⁻² -2𝑒⁻²

=1-3𝑒⁻²=1-3×0.1353

=0.5941

(4) p(more than 2 success)= 𝑝(3success)+ 𝑝(4 success)+… … ….

= 1 − 𝑝(0 success) - 𝑝(1 success) –p(2 successes) … … ….

=1 − 𝑒⁻² −2𝑒⁻² −²²

2 𝑒²

=1- 5𝑒⁻²= 1- 5 × 0.1353 = 0.3235 (5) p(3 successes)= ²³

3!𝑒⁻² = 0.1804

(8)

Example 5: The probability of getting no misprint is e^-2. Find the probability that a page contains more than 2 errors. (Given e^-2=0.135)

Solution : The required poisson distribution is 𝑝(𝑟 𝑒𝑟𝑟𝑜𝑟𝑠) =𝑒⁻²

𝑟! 2^𝑟 𝑟 = 0,1,2 … …

Since p (0 error) = e^-2= e^-m ⇒ 𝑚 = 2, the parameter of poisson distribution.

Hence p (2 and more than 2 success)= 𝑝(2 errors)+ 𝑝(3 errors)+… … ….

= 1 − 𝑝(0 error) - 𝑝(1error) =1 − 𝑒⁻²− 2𝑒⁻²

=1 − 3𝑒⁻² =1-3× 0.1353 = 1- 0.4059= 0.5941.

Example 6: A manufacturer of cotter pins knows that 5 per cent of this product is defective if he sells cotter poins in boxes of 100 and guarantees that not more than 10 pins will be

defective.What is the probability that a box will fail to meet the guaranteed quality.

Solution: The number of trials n=100

The probability that a cotter pin is defective,i.e., the probability of success p=5/100

∴ Parameter for Poisson distribution m=np=100* ⁵

100=5

∴ Probability of r success = ^𝑒⁻⁵⁵^𝑟

𝑟! r=0,1,2,...

P(more than 10 defective) = 1 - ∑ ^𝑒⁻⁵⁵^𝑟

𝑟!

10𝑟=0

Example 7: Suppose the number of telephone calls an operator received from 8:30 a.m. to 8:40 a.m. follow Poisson distribution with mean 2.Find the probability that the operator will receive (1) no call; (2) 1 call in that time interval tomorrow, given 𝑒⁻² =0.1353.

Solution: Poisson distribution has mean m=2.

Hence, the required Poisson distribution is P(r calls in the given time) = ^𝑒^−𝑚^𝑚^𝑟

𝑟! = ^𝑒⁻²²^𝑟

𝑟! , r=0,1,2,...

(1) P(no call) = p(r=0) = 𝑒⁻² =0.1353

(2) P(1 call) = p(r=1) = 2 𝑒⁻²=2*0.1353=0.2706

Example 8: An auto salesperson sells an average of .9 cars per day. Let x be the number of cars sold by this salesperson on any given day. Using the poisson probability distribution table, write the probability distribution of x.

(9)

Solution : Let  be the mean number of cars sold per day by this salesperson. Hence,  = .9.

Using the portion of corresponding to  = .9 we write the probability distribution of x in table I.

Table I Probability Distribution of x for  =.9

X P(x)

0 .4066

1 .3659

2 .1647

3 .0494

4 .0111

5 .0020

6 .0003

Note that 6 is the largest value of x for  = .9 listed in table I for which the proability is greater than zero. However, this does not mean that this salesperson cannot sell more than six cars on a given day. What this means is that the probability of selling 7 or more cars is very small. Actually, the probability of x = 7 for  = .9 calculated by using the Poisson formula is .000039. When rounded to four decimal places, this probability is .0000, as listed in table I.

Summary

If the average number of occurrences for a given interval is known ,then by using the Poisson probability distribution we can compute the probability of a certain number of occurrences r (or x) in that interval. Note that the number of actual occurrences in an interval is denoted by r (or x ).The average

occurrences during an interval for a Poisson probability distribution is its parameter. In this distribution the number of occurrences is large and

probability is small. Mean and Variance are also discussed. Related problems

are also solved.

(10)

Table II Table of Poisson Probabilities

(11)