THE POISON DISTRIBUTION

Having discussed the binomial distribution in the preceding section, we now turn to

Poisson distribution, which is also a discrete probability distribution. It was discovered by

a French mathematician. Simeon Dennis Poison (1781 –1840) and hence named after

him. Poisson distribution is also a discrete distribution.

Poisson distribution is a limiting case of Binomial distribution under the following

conditions.

(i) n the number of trials is indefinitely large ie., n → ∞.

(ii) P the constant probability of success in each trial is very small

i.e., p → 0.

(iii) np = λ is finite where λ is a positive real number. When an event occurs rarely, the

distribution of such an event may be assumed to follow a Poisson distribution.

Definition: A random variable X is said to have a Poisson distribution if the probability

mass function of X is P(X = x) =

e−λ λx

x

, x = 0, 1, 2 …for some λ > 0

The mean of the Poisson distribution is λ, and the variance is also λ.

The parameter of the Poisson distribution is λ.

A Poisson distribution is often a good model for many naturally occurring discrete

random variables which are counts, e.g.

X = number of events occurring in a unit of time (or unit of space)

X = number of cases of a rare disease in a city per month (or a country per day)

X = number of cars passing a point on a road per minute

X = number of telephone calls to a telesales company per minute

X = number of alpha particles emitted by a (slowly decaying) radioactive source per

second

X = number of plants of a particular species per square meter of land

It is also used in quality statistics to count the number of defective items or in insurance

problems to count the number of casualties or in waiting time problems to count the

number incoming customers or the number of patients arriving to consult a doctor in a

given time period, radioactive particle emitted by a sample over a set time period, flaws

in a long manufactured tape or wire, breakdowns , breakdowns or failures of a piece of

equipment or component over a given period of time, counts from a Geiger counter,

collisions of cars at a specific intersection under specific conditions, flaws in a casting,

and telephone calls to a particular telephone and so forth. All these are examples of

Poisson experiments (subject to some conditions). In a Poisson experiment, the number

of times an outcome occurs in an interval is counted. The interval is usually an interval of

time but it can be a distance or an area. In the specification below, success refers to the

occurrence of the outcome of interest, length refers to the size of an interval and sub-

interval refers to a part of the interval.

In the first experiment listed above a success is the machine breaking down, the length of

the interval is a day and a sub-interval is a minute (or second) in the day.

The characteristics of the Poisson distribution are:

1. The events occur independently. This means that the occurrence of a subsequent

event is not all influenced by the occurrence of an earlier event.

2. Theoretically, there is no upper limit with the number of occurrence of an event

during a specified time period.

3. The probability of a single occurrence of an event within a specified time period is

proportional to the length of the time period of intervals.

4. In an extremely small portion of the time period, the probability of two or more

occurrence of an event is negligible.

5. The mean number of event in each unit will be denoted by the Greek letter A.

The Poisson experiment

1. There is an interval of known fixed length.

2. There is an average of μ successes in the interval.

3. The number of successes in a sub-interval is independent of the number of successes in

any other non –overlapping sub-interval.

4. The probability of a success occurring in a sub-interval is proportional to the length of

the interval.

5. As a sub-interval becomes smaller and smaller so the probability of two or more

successes occurring in the sub-interval approaches zero.

6. The number of successes X observed in the interval of known fixed length is counted.

The random variable X defined in the Poisson experiment is said to have a Poisson

distribution.

The Poisson distribution has one parameter:

μ = the mean number of successes in the interval of known fixed length

If the random variable X has a Poisson probability distribution with parameter μ, we

write X~ Po ( )

It can be shown that if X~ Po ( ) then

, for x = 0, 1, 2, 3 ...n

Note that in evaluating the above expression:

i. 0! is defined to be 1.

ii. k0 is 1 for any non-zero value of k.

iii. The number e that appears in the formula is one of the fundamental constants of

mathematics. Its value is approximately 2.71828.

You can calculate the value of p(x) for different values of μ and x quite easily using your

calculator.

A random variable with the above probability function is called a poison random variable

with parameter t. If we consider a poisson random variable over an interval of unit length,

then writing X for Xt, we get

P(x = k) e - t ( t) k , k=0,1,2,3….

k!

Note:

i. P(x = k) e - t ( t) k , > 0,K=0,1,….

k!

ii. e - t ( t) k = e-t ( t) k

k! k!

e-t. e-t = 1

The Mean and Variance of Poisson Distribution

The Mean: =. The is Greek letter (Lamba) is the parameter of the distribution and

it is the average number of times the random events occurs in the time period.

The Variance: 2 =.

Note that instead of time, the Poisson random variable may be considered in the experiment

of counting the number X of times a particular event occurs during a given unit area, volume

etc.

Example 1

On average, six people per hour use an electronic teller machine during the prime shopping

hours in a shopping more. What is the probability that:

i. Exactly six people will use the machine during the randomly selected

hour?

ii. Fewer than five people will use the machine during a randomly selected

hour?

iii. No one will use the facility during a 10 minutes interval?

iv. No one will use the facility during a 5 minutes interval?

Solution:

Using Poisson probability distribution

Where = 6

(i)

(ii)

k = o

k = o

(iii)

(iv)

Example 3

An engineering Company manufactured a specific component for a tuner amplifier. If it

was that out of every 1000 component produced, 8 are defective. If the components are

packed in boxes 250, find

(i) The probability of obtaining 0, 1, 2 and 3 defective in a box.

(ii) The probability that the box contained at least 3 defectives

(iii) The probability of at most 3 defective not in the box

Number of components = 1000, 8 defectives, boxes = 250

P (1 defective) = P = = 0.06

X = µ P – 250 x 0.08 = 2.

P (x) =

(1) P (x = 0) = = 0.1353

P (x =1) = = 0.270

P (x =2) = = 0.2707

P (x= 3) = = 0.1804

(ii) P (x >3) = 1. (P (0) + P (1) + P (2))

= 1 – (0.1353 + 0.270 + 0.2707).

= 1- 0.6767

= 0.3233

(iii) P (x >3) = P (0) + P (1) + P (2) + P (3)

N = 1000, not parked in the box.

P = 0.008

= NP = 1000 x 0.008 = 8

P (0) = = 0.0003

P (2) = = 0.0107

P (3) = = 0.0286

P ( at most three defectives not in the box) = P (x <3)

= P( 0) + P (1) + P(2) + (3)

= 0.0003 + 0.002 + 0.017 + 0.0286

= = 0.0416

Example(1)

Suppose that the average number of telephone calls arriving at the switchboard of a small

corporation is 30 calls per hour. What is the probability that:

(i) No call will arrive in a 3 minutes period?

(ii) More than 5 calls will arrive in a 5-minutes period?

(iii) At least 3 calls will arrive in a 6-minute period?

(iv) At most 5 calls will arrive in a half hour period?

Solution:

b(i) = 30 per hour or t = ½ per minute, P(X=x) =

P(X = 0) =

= = 0.223

(ii) P(X>5) = 1 – P (x< 5). t =

=1– (P (x = 0) + P (x = 1)+ P (x = 2) + P(x= 3) + P(x = 4)+ P (x = 5)

=1-

= 1 – (0.0821+ 0.2052 + 0.2565 + 0.2138 + 0.1336 +0.0668)

= 1 – 0.958

= 0.042.

(iii) P (x >3) = 1 – P (x < 2), = = 3

= 1- (P (x = 0) + P (x = 1) + P (x = 2) )

=1-

= 1 – (0.0498 + 0.1444 + 0.2240)

= 1 – 0.4232

= 0.5768

t = = 15

(iv) P (x<5) = P (X = 0) + P (x= 1) + P (x = 2) + P (x = 3) + P (x = 4) + P (x = 5)

=

= 0.0000 + 0.0000 + 0.0001 + 0.0010 + 0.0155 + 0.2323

= 0.2489

Example (2)

A special machine is used in recording radioactive event coming to a counter. Each event

arriving at the counter. Shows as a number. The number of radioactive event recorded in

a room is on the average of 3 events in 2 seconds.

Find the probability of:

(i) Exactly 3 events in 2 seconds

(ii) At least 3 events in 2 second

(iii) At most 3 events in 2 seconds.

Solution

= 2 P(x = x) =

(i) P (x = 3) = = 0.2244

(ii) P (x >3) = 1 – P (x < 2).

= 1 –

= 1 = 0.2244

= 1- (0.0448 + 0.1494 + 0.2240).

= 1-0.4232 = 0.3768.

(iii) P (X<3) = P (x=0) + P (x= 1) + P (x = 2) + P (x=3)

= = 0.2244

= 0.0498 + 0.1494 + 0.224 + 0.224

= 0.6472

Example (3)

A machine producing metal strip is estimated to average 2 minor flows per 24 metres. If

flaws occur at random, find the probability of:

(i) 2 flaws in a 4 meters

(ii) 2 flaws in a 12 metre length

(iii) At most 2 flaws in a 24 meter length

Solution

= 2 per 24m, P (x =x) =

(i) t =

P(x = 2) = = 0.0398

(ii) t =

P(x = 2) = = 0.2707

(iii) = 2

P(x < 2) = P (x = 0) + (x = 1) + P (x =2)

=

= 0.1353 + 0.2707 + 0.2707

= 0.6767.

Example 2:

Suppose that we are investigating the safety of a dangerous intersection. Past road safety

records shows a mean of 5 accidents per month at this intersection. Suppose the number

of accidents is distributed according to Poisson distribution. Calculate the probability in

any month of exactly 0, 1, 2, 3, or 4 accidents.

Solution

Since the number of accidents is distributed according to a Poisson distribution and the

mean number of accidents per month is 5, we have the probability of happenings

accidents in any month by this formula we can calculate.

Poisson distribution as an Approximation of Binomial Distribution

When n is becoming large the problem of computing n! in binomial distribution involves

tedious calculations, the Poisson distribution can be used in its place. However, this holds

well under certain condition such as when the number of trials is large while the binomial

probability of success is small. When n = > 20 and P < 0.05, then calculations of

binomial probabilities becomes much simpler by using Poison distribution.

The formula for calculating Poisson probabilities is.

If we change by np in the above formula, then the Poisson distribution becomes an

approximation of the binomial

Example 1:

Past experience shows that 2 percent defective is the process capability of producing

fuses in a factory. Obtain the probability that there will be more than 5 defective fuses in

a box of 200 fuses. Arrive at the probability using Binomial and Poisson distribution and

comments on the result.

Solution

Here P = 0.02, n = 200, np = 200 x 0.02 = 4 = mean

Using Poisson distribution, the probabilities of getting X defective fuses

=

= 1 – 0.785 = 0.215

Using Binomial distribution

:.

=

= 1 – 0.788

= 0.212

NOTE: One of the main uses of the Poisson distribution is to

calculate approximate binomial probabilities. Binomial tables are

available only for small values of n but the Poisson tables can be used

to approximate binomial probabilities or large n provided the value of p

is small.

Here, we see that the probabilities calculated in both distributions are more or less equal.

Hence the Poisson approximation to Binomial is very good here.

Application Questions Poisson probability distribution

1. The number of cars entering a small parking lot is a random variable having a Poisson

distribution with a mean of 1.5 per hour. The lot holds only 12 cars.

a) Find the probability that the lot fills up in the first hour (assuming that all cars stay in

the lot longer than one hour).

b) Find the probability that more than 3 cars arrive between 9 am and 11 am.

2. Customers arrive at a checkout counter at an average rate of 1.5 per minute. What

distribution will apply if reasonable assumptions are made? List those assumptions.

Find the probabilities that

a) Exactly two will arrive in any given minute;

b) At least three will arrive during an interval of two minutes;

c) At most 13 will arrive during an interval of six minutes.

3. Records of an electrical distribution system in a particular area indicate that over the

past twenty years there have been just six years in which lightning has not hit a

transformer. Assume that the factors affecting lightning hits on transformers have not

changed over that time, and that hits occur at random and independently.

a) Then what would be the best estimate of the average number of hits on transformers

per year?

b) In how many of the next ten years would we expect to have more than two hits on

transformers in a year?

4. In a plant manufacturing light bulbs, 1% of the production is known to be defective

under normal conditions. A sample of 30 bulbs is drawn at random.

Assume defective bulbs occur randomly and independently. What is the probability that:

a) the sample contains no defective bulbs;

b) more than 3 defective bulbs are in the sample.

Do this problem both (1) using the binomial distribution, and (2) using the Poisson

distribution.

5. Fifteen percent of piglets raised in total confinement under certain conditions will live

less than three weeks after birth. Assume that deaths occur randomly and independently.

Consider a group of eight newborn piglets.

a) What probability distribution applies without any approximation to the number of

piglets which will live less than three weeks?

b) What is the expected mean number of deaths?

c) What is the probability that exactly three piglets will die within three weeks of birth?

Use the binomial distribution.

d) Calculate the probability that exactly three piglets will die within three weeks of birth,

but now use the Poisson distribution.

e) Use the binomial distribution to calculate the probability that fewer than three piglets

will die within three weeks of birth.

f) Use the Poisson distribution to calculate the probabilities that exactly 0, 1, and 2 piglets

will die within three weeks of birth, and then that fewer than 3 piglets will die within

three weeks of birth.

(4) Alpha particles are emitted by a radio active source at an average rate of 5 in a 20

minutes interval. Using Poisson distribution finds the probability that there will be (i) 2

emissions (ii) at least 2 emission in a particular 20 minutes interval. [e−5 = 0.0067].

10. In a certain factory, blades are manufactured in packets of 10. There is a 0.2%

probability for any blade to be defective. Calculate approximately the number of packets

containing two defective blades in a consignment of 20,000 packets.

14. Assuming that on an average 2% of the output in a factory manufacturing certain

bolts is defective and that 200 units are in a package, what is the probability

that (i) None is defective (ii)At most 3 defective bolts may be found in

that package?

15. One –fifth percent of the blades produced by a blade manufacturing factory turn out

to be defective. The blades are supplied in packets of 10.

Use Poisson distribution to calculate the approximate number of packets containing no

defective, one defective and two defective blades respectively in a consignment of

1,00,000 packets.

LECTURE NOTE: ON POISSON PROBABILITY DISTRIBUTION

BY AJIBADE, F. BRIGHT

DEPT: GENERAL STUDIES PETROLEUM TRAINING INSTITUTE,

NIGERIA.