POISSON DISTRIBUTION Poisson Distribution

Structure
10.1 Introduction
Objectives

10.2 Poisson Distribution

10.3 Moments of Poisson Distribution
10.4 Fitting of Poisson Distribution
10.5 Summary
10.6 Solutions/Answers

10.1 INTRODUCTION
In Unit 9, you have studied binomial distribution which is applied in the cases
where the probability of success and that of failure do not differ much from
each other and the number of trials in a random experiment is finite. However,
there may be practical situations where the probability of success is very small,
that is, there may be situations where the event occurs rarely and the number of
trials may not be known. For instance, the number of accidents occurring at a
particular spot on a road everyday is a rare event. For such rare events, we
cannot apply the binomial distribution. To these situations, we apply Poisson
distribution. The concept of Poisson distribution was developed by a French
mathematician, Simeon Denis Poisson (1781-1840) in the year 1837.
In this unit, we define and explain Poisson distribution in Sec. 10.2. Moments
of Poisson distribution are described in Sec. 10.3 and the process of fitting a
Poisson distribution is explained in Sec. 10.4.
Objectives
After studing this unit, you would be able to:
• know the situations where Poisson distribution is applied;
• define and explain Poisson distribution;
• know the conditions under which binomial distribution tends to Poisson
distribution;
• compute the mean, variance and other central moments of Poisson
distribution;
• obtain recurrence relation for finding probabilities of this distribution;
and
• know as to how a Poisson distribution is fitted to the observed data.

27
Discrete Probability
Distributions 10.2 POISSON DISTRIBUTION
In case of binomial distributions, as discussed in the last unit, we deal with
events whose occurrences and non-occurrences are almost equally important.
However, there may be events which do not occur as outcomes of a definite
number of trials of an experiment but occur rarely at random points of time and
for such events our interest lies only in the number of occurrences and not in
its non-occurrences. Examples of such events are:
i) Our interest may lie in how many printing mistakes are there on each page
of a book but we are not interested in counting the number of words
without any printing mistake.
ii) In production where control of quality is the major concern, it often
requires counting the number of defects (and not the non-defects) per item.
iii) One may intend to know the number of accidents during a particular time
interval.
Under such situations, binomial distribution cannot be applied as the value of n
is not definite and the probability of occurrence is very small. Other such
situations can be thought of yourself. Poisson distribution discovered by S.D.
Poisson (1781-1840) in 1837 can be applied to study these situations.
Poisson distribution is a limiting case of binomial distribution under the
following conditions:
i) n, the number of trials is indefinitely large, i.e. n → .
ii) p, the constant probability of success for each trial is very small, i.e. p → 0.
iii) np is a finite quantity say ‘’.

Definition: A random variable X is said to follow Poisson distribution if it

assumes indefinite number of non-negative integer values and its probability
mass function is given by:
e− .x

p ( x ) = P ( X = x ) =  x ; x = 0, 1, 2, 3, ... and   0.

 0; elsewhere
where e = base of natural logarithm, whose value is approximately equal to
2.7183 corrected to four decimal places. Value of e− can be written from the
table given in the Appendix at the end of this unit, or, can be seen from any
book of log tables.

Remark 1
i) If X follows Poisson distribution with parameter  then we shall use the
notation X  P().
ii) If X and Y are two independent Poisson variates with parameters 1 and 2
repectively, then X + Y is also a Poisson variate with parameter 1+2. This
is known as additive property of Poisson distribution.

28
10.3 MOMENTS OF POISSON DISTRIBUTION Poisson Distribution

rth order moment about origin of Poisson variate is

r e 
 − x

' =
( )
r r
r E X = x p(x) =  x
x=0 x=0


e−x 
e−x 
e− .x
 = x
'
1 = x =
x=0 x x=1 x x −1 x=1 x −1

− 
1
2 3 
=e  + + + ...
 0 1 2 
   1 2

=  −
e 1 + + +...
 1 2 
  2 3 
= −  
e e  e = 1+ + + + ...(see Unit 2 of MST-001 )
1 2 3
 
=
 Mean = 

e−x
2 =  x
2

x=0 x

=  e− .x
 
x (x −1) + x
x=0 x
[As done in Unit 9 of this Course]


 e − x e −  x 
=  x (x −1) +x 
x=0  x x 

= x (x −1) x (x −1) x − 2 + x

e − x 
e−x
x=2 x=0 x

x 

e−x
= e x −2 
−+ x
x
x=2 x=0

2 3 4 
= − '
e  + + + ... + 1
 0 1 2 
  2 
= − 2 '
e  1+ + + ... + 1
 1 2 

= e−2e + '1

= 2 + 
( )
2
 Variance of X is given as V(X) =  = ' − '
2 2 1

= 2 +  − ()
2

=
29
Discrete Probability 3

3 =  x p ( x )
' 3
Distributions
x=0

Writing x3 as x ( x −1 )( x − 2) + 3x ( x −1) + x, we have

See
 Unit 9 of this course where 
the expression of ' is obtained
 3 

e 
− x
= x (x −1)( x − 2) + 3x (x −1) + x 
 x=0
 x


e−x 
e−x 
e−x
= x (x −1)(x − 2) + 3x (x −1) + x
x=3 x x=2 x x=1 x

x
= e− x (x −1)(x − 2) ( )
+ 3  2 + ( )
x=3 x (x −1)(x − 2) x − 3
x

 x − 3 + 3 +
2
=e −

x=3
 3 4 5 
= −
e  + + +...  + 3 2 + 
 0 1 2 
  2 
= − 3 2
e  1+ + + ... + 3 + 
 1 2 

= e− 3 e + 32 + 

= 3 + 32 + 
Third order central moment is

 3 = '3 − 3'2'1 + 2 (1' )

3

= [On simplification]

e−x
 =
' x4.
4
x=3 x

Now writing x4 = x (x −1)(x − 2)(x − 3) + 6x (x −1)(x − 2) + 7x (x −1) + x,

and proceeding in the similar fashion as done in case of ' 3, we have

 '4 = 4 + 63 + 72 + 

 Fourth order central moment is

( ) ( )
2 4
 = ' − 4' ' + 6' ' − 3 '
4 4 3 1 2 1 1

= 32 +  [On simplification]

30
Therefore, measures of skewness and kurtosis are given by Poisson Distribution
23 2
1 1
1 = = = , 1 =  = ; and
 32 3  1

4 3 + 
2
1 1
2 = = = 3+ , 2 =  2 − 3 = .
 22 2  

Now as 1 is positive, therefore the Poisson distribution is always positively

skewed distribution. Also as 2 > (  > 0), the curve of the distribution is
Leptokurtic.
Remark 2
i) Mean and variance of Poisson distribution are always equal. In fact this is
the only discrete distribution for which Mean = Variance = the third
central moment.
ii) Moments of the Poisson distribution can be deduced from those of the
binomial distribution also as explained below:
For a binomial distribution,
Mean = np
Variance = npq

3 = npq (q − p)

4 = npq 1+ 3pq(n − 2) = npq1+ 3npq − 6pq

Now as the Poisson distribution is a limiting form of binomial distribution

under the conditions:
(i) n → , (ii) p → 0 i.e. q → 1, and (iii) np =  (a finite quantity);
 Mean, Variance and other moments of the Poisson distribution are given as:
Mean = Limiting value of np = 
Variance = Limiting value of npq
= Limiting value of (np) (q)
= () (1) = 
3 = Limiting value of npq (q – p)
= Limiting value of (npq) (q – p)
= () (1 – 0)
=
4 = Limiting value of npq [1 + 3npq – 6pq]
= Limiting value of (npq) [1 + 3(npq) –6 (p)(q)]
= () [1 + 3() – 6 (0)(1)]
= [1 + 3] = 32 + 
Now let’s give some examples of Poisson distribution.
31
Discrete Probability
Distributions
Example 1: It is known that the number of heavy trucks arriving at a railway
station follows the Poisson distribution. If the average number of truck arrivals
during a specified period of an hour is 2, find the probabilities that during a
given hour
a) no heavy truck arrive,
b) at least two trucks will arrive.
Solution: Here, the average number of truck arrivals is 2
i.e. mean = 2
 =2
Let X be the number of trucks arrive during a given hour,
 by Poisson distribution, we have
e− 2 (2)
x
e−x
PX = x = = ; x = 0, 1, 2, ...
x x
Thus, the desired probabilities are:
(a) P[arrival of no heavy truck] = P[X = 0]
e−2 20
=
0

= e−2

See the table given 

in the Appendix at 
= 0.1353
 
the end of this unit 

(b) P[arrival of at least two trucks] = P  X  2

= P  X = 2  + P  X = 3+ ...

= 1− PX = 1+ PX = 0

 sum of all the 

probabilities is 1 
 

e−2 20 e−2 21 
= 1−  + 
 0 1 

−2  2
0
21
= 1− e  +  = 1− e −2 (1+ 2)
0 1

= 1− (0.1353)(3) =1− 0.4059 = 0.5941

Note: In most of the cases for Poisson distribution, if we are to compute the
probabilities of the type P  X  a  or P  X  a  , we write them as
PX  a = 1− PX  a and

32
P  X  a = 1− P  X  a  , because n may not be definite and hence we cannot Poisson Distribution

go up to the last value and hence the probability is written in terms of its
complementary probability.
Example 2: If the probability that an individual suffers a bad reaction from an
injection of a given serum is 0.001, determine the probability that out of 500
individuals
i) exactly 3,
ii) more than 2
individuals suffer from bad reaction
Solution: Let X be the Poisson variate, “Number of individuals suffering from
bad reaction”. Then,
n = 1500, p = 0.001,
  = np = (1500) (0.001) = 1.5
 By Poisson distribution,
e−x
PX = x = , x = 0, 1, 2, ...
x

e−1.5.(1.5)
x

= ; x = 0, 1, 2, ...
x
Thus,
i) The desired probability = P[X = 3]
e−1.5.(1.5)
3

=
3

=
(0.2231)(3.375)
= 0.1255
6

 e−0.5 = 0.6065, e−1 = 0.3679, so 

 −1.5 
e = e−1  e−0.5 = (0.3679) (0.6065) = 0.2231)
 
See the table given in the Appendix
 
 at the end of this unit 

ii) The desired probability = P  X  2

= 1− P  X  2

= 1 − PX = 2+ PX = 1+ PX = 0

 e−1.5. 1.5 2 e−1.5. 1.5 1 e−1.5. 1.5 0 

( ) ( ) ( )
= 1−  + + 
 2 1 0 

33
  2.25 +1.5 +  = 1− (3.625) e−1.5
Discrete Probability
Distributions = 1− e−1.5  1
 2 

= 1− (3.625)(0.2231) = 1 – 0.8087 = 0.1913

Example 3: If the mean of a Poisson distribution is 1.44, find the values of

variance and the central moments of order 3 and 4.
Solution: Here, mean = 1.44
  = 1.44
Hence, Variance =  = 1.44
3 =  = 1.44
4 = 32 +  = 3 (1.44)2 + 1.44 = 7.66.
Example 4: If a Poisson variate X is such that P[X = 1] = 2P[X = 2], find the
mean and variance of the distribution.
Solution: Let  be the mean of the distribution, hence by Poisson distribution,
e−x
PX = x = ; x = 0, 1, 2, ...
x

Now, P  X = 1 = 2PX = 2

e− .1 e− .2

 =2
1 2

  = 2  2 −  = 0  ( − 1) = 0   = 0, 1
But  = 0 is rejected
[ if  = 0 then either n = 0 or p = 0 which implies that Poisson distribution
does not exist in this case.]
=1
Hence mean =  = 1, and
Variance =  = 1.
Example 5: If X and Y be two independent Poisson variates having means 1
and 2 respectively, find P[X + Y < 2].
Solution: As X ~ P(1), Y ~ P(2), therefore,
X + Y follows Poisson distribution with mean = 1 + 2 = 3.
Let X + Y = W. Hence, probability function of W is
e−3.3w
PW = w = ; w = 0, 1, 2, ... .
w

Thus, the required probability= P[X + Y < 2]

= P[W < 2]
= P[W = 0] + P[W = 1]
34
 e−3.30 e−3.31 Poisson Distribution
= +
0 1

= (0.0498)(1 + 3) [From Table, e−3 = 0.0498]

= 0.1992.

You may now try these exercises.

E1) Assume that the chance of an individual coal miner being killed in a mine
1
accident during a year is . Use the Poisson distribution to calculate
1400
the probability that in a mine employing 350 miners, there will be at least
one fatal accident in a year. (use e−0.25 = 0.78 )
E2) The mean and standard deviation of a Poisson distribution are 6 and 2
respectively. Test the validity of this statement.
E3) For a Poisson distribution, it is given that P[X = 1] = P[X = 2], find the
value of mean of distribution. Hence find P[X = 0] and P[X = 4].

We now explain as to how the Poisson distribution is fitted to the observed

data.

10.4 FITTING OF POISSON DISTRIBUTION

To fit a Poisson distribution to the observed data, we find the theoretical (or
expected) frequencies corresponding to each value of the Poisson variate.
Process of finding the probabilities corresponding to each value of the Poisson
variate becomes easy if we use the recurrence relation for the probabilities of
Poisson distribution. So, in this section, we will first establish the recurrence
relation for probabilities and then define the Poisson frequency distribution
followed by the process of fitting a Poisson distribution.
Recurrence Formula for the Probabilities of Poisson Distribution
For a Poisson distribution with parameter , we have
− x
e 
p(x) = … (1)
x
Changing x to x + 1, we have
− x+1
e 
p ( x +1) = … (2)
x +1
Dividing (2) by (1), we have

(e  x+1
−
)
p(x +1) x +1 
= =
p(x) x +1
(e − x
 )
x

35
Discrete Probability

Distributions  p ( x +1) = p(x) … (3)
x +1
This is the recurrence relation for probabilities of Poisson distribution. After
obtaining the value of p(0) using Poisson probability function i.e.
e−0 −
p(0) = = e , we can obtain p(1), p(2), p(3),…, on putting
(0 )
x = 0, 1, 2, …. successively in (3).

Poisson Frequency Distribution

If an experiment, satisfying the requirements of Poisson distribution, is
repeated N times, then the expected frequency of getting x successes is given
by
−
e x
f ( x ) = N.PX = x = N. ; x = 0, 1, 2,...

Example 5: 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 100 boxes from the producer of bottles.
Using Poisson distribution, find how many boxes will contain at least two
defective bottles.
Solution: Let X be the Poisson variate, “the number of defective bottles in a
box”. Here, number of bottles in a box (n) = 500, therefore, the probability (p)
of a bottle being defective is
0.1
p = 0.1% = = 0.001
100
Number of boxes (N) = 100
 = np = 500 .001 = 0.5
Using Poisson distribution, we have
e−x
PX = x = ; x = 0, 1, 2, ...
x

e−0.5 (0.5)
x

= ; x = 0, 1, 2, ...
x

 Probability that a box contain at least two defective bottles

= P  X  2

= 1− P  X  2

= 1– PX = 0 + P  X = 1

 e−0.5 ( 0.5) 0 e−0.5 (0.5)1 

= 1−  +  = 1− e−0.5 1+ 0.5
 0 1 

= 1 – (0.6065) (1.5) = 1 – 0.90975 = 0.09025.

36
Hence, the expected number of boxes containing at least two defective bottles Poisson Distribution

= N.P[X  2]
= (100) (0.09025)
= 9.025

Process of Fitting a Poison Distribution

For fitting a Poisson distribution to the observed data, you are to proceed as
described in the following steps.

• First we obtain mean of the given distribution i.e. fx , being mean,
f
take this as the value of .
• Next we obtain p(0) = e− [Use table given in Appendix at the end of this
unit.]

• The recurrence relation p ( x +1) = p ( x ) is then used to compute the
x +1
values of p(1), p(2), p(3), …
• The probabilities obtained in the preceding two steps are then multiplied
with N to get expected/theoretical frequencies i.e.
f ( x ) = N.PX = x; x = 0, 1, 2, ...

Example 6: The following data give frequencies of aircraft accidents

experienced by 2480 pilots during a certain period:

Number of Accidents 0 1 2 3 4 5
Frequencies 1970 422 71 13 3 1

Fit a Poisson distribution and calculate the theoretical frequencies.

Solution: Let X be the number of accidents of the pilots. Let us first obtain the
mean number of accidents as follows:
Number of Frequency ( f ) fX
Accidents
(X)
0 1970 0
1 422 422
2 71 142
3 13 39
4 3 12
5 1 5
Total 2480 620
37
Discrete Probability
Distributions
 Mean =  = fx = 620
f 2480
  = 0.25

 by Poisson distribution,

p(0) = e− = e−0.25

See table given in the Appendix 

at the end of this unit
= 0.7788
 

Now, using the recurrence relation for probabilities of Poisson distribution i.e.

p ( x +1) = p ( x ) and then multiplying each probability with N, we get the
x +1
expected frequencies as shown in the following table

Number of 
=
0.25 p(x) = PX = x Expected/
Accidents x +1 x +1 Theoretical
(X) frequency
f(x) = 2480p(x)
(1) (2) (3) (4)
0 0.25 p(0) = 0.7788 1931. 4 □ 1931
= 0.25
0 +1
1 0.25 p(1) = 0.25  0.7788 482. 9 □ 483
= 0.125
1+1 = 0.1947
0.25
2 = 0.0833 p(2) = 0.125  0.1947 60.3 □ 60
2 +1 = 0.0243

0.25 4.96 □ 5
3 = 0.0625 p(3)= 0.0833  0.0243
3 +1 = 0.0020

0.25
4 = 0.05 p(4)= 0.0625  0.0020 0.248 □ 0
4 +1 = 0.0001
0.25
5 = 0.0417 p(5)= 0.05  0.0001 0
5 +1 = 0.000005

You can now try the following exercises

E4) In a certain factory turning out fountain pens, there is a small chance,
1
, for any pen to be defective. The pens are supplied in packets of
500
10. Calculate the approximate number of packets containing (i) one
defective (ii) two defective pens in a consignment of 20000 packets.

38
E5) A typist commits the following mistakes per page in typing 100 pages. Poisson Distribution
Fit a Poisson distribution and calculate the theoretical frequencies.
Mistakes per 0 1 2 3 4 5
page(X)
Frequency 42 33 14 6 4 1
(f)

We now conclude this unit by giving a summary of what we have covered in it.

10.5 SUMMARY
The following main points have been covered in this unit:
1. A random variable X is said to follow Poisson distribution if it
assumes indefinite number of non-negative integer values and its
probability mass function is given by:
 e − x

p ( x ) = P ( X = x ) =  x ; x = 0, 1, 2, 3,... and   0.
0; elsewhere


2. For Poisson distribution, Mean = Variance = μ3 =  , μ4 = 32 + 

1 1 1 1
3.  = ,  = ,  = 3 + ,  = for this distribution .
1 1 2 2
   
4. Recurrence relation for probabilities of Poisson distribution is

p ( x +1) = .p ( x ) , x = 0, 1, 2, 3,...
x +1
5. Expected frequencies for a Poisson distribution are given by
−
e x
f ( x ) = N.PX = x = N. ; x = 0, 1, 2, ...
x
If you want to see what our solutions/answers to the exercises in the unit are,
we have given them in the following section.

10.6 SOLUTIONS/ANSWERS

E1) Let X be the Poisson variable “Number of fatal accidents in a year”.

1
Here n = 350, p =
1400
 1 
  = np = (350 ) = 0.25 .
 
 1400 
By Poisson distribution,

39
Discrete Probability
e− .x
Distributions
PX = x = , x = 0, 1, 2, ...
x

e−0.25 (0.25)
x

= , x = 0, 1, 2, ...
x
Therefore, P [at least one fatal accident]

= P  X  1 = 1 – P[X < 1] = 1 – P[X = 0]

e−0.25 ( 0.25 )
0

= 1− = 1 – e−0.25 = 1 – 0.78 = 0.22

0

E2) As mean = 6, therefore,  = 6.

As standard deviation is 2, therefore, variance = 4   = 4.
We get two different values of , which is impossible. Hence, the
statement is invalid.

E3) Let  be the mean of the distribution,

 by Poisson distribution, we have
e−x
PX = x = ; x = 0, 1, 2, 3, ...
x
Given that P[X = 1] = P[X = 2],
e−1 e−2
 =

2
 =  2 – 2 = 0  ( – 2) = 0
2
  = 0, 2.
 = 0 is rejected,
 =2
Hence, Mean = 2.
e−0
Now, P[X = 0] = = e− = e−2 = 0.1353,
0

[See table given in the Appendix at the end of this unit.]

e−2 (2) e−2 (16) 2
4
e−4
and P  X = 4 = = = = (0.1353)
4 4 24 3
= 2(0.0451)
= 0.0902.

40
 1 Poisson Distribution
E4) Here p = , n = 10, N = 20000,
500
1
  = np = 10  = 0.02
500
By Poisson frequency distribution
f ( x ) = N.PX = x

e−x
= (20000) ; x = 0, 1, 2,...
x
Now,
i) The number of packets containing one defective
= f(1)

e−0.02 .(0.02)
1

= (20000)
1
= (20000) (0.9802) (0.02) See
 the table given 
in the Appendix
 
= 392.08 □ 392; and
ii) The number of packets containing two defectives
e−0.02 (0.02)
2

= f(2) = 20000
2
(0.9802)(0.0004)
= ( 20000 ) = 3.9208 □ 4
2
E5) The mean of the given distribution is computed as follows
X f fX
0 42 0
1 33 33
2 14 28
3 6 18
4 4 16
5 1 5
Total 100 100

 Mean  = fx = 100 = 1

f 100
 p ( 0 ) = e− = e−1 = 0.3679.

41
Discrete Probability
Distributions
Now, we obtain p(1), p(2), p(3), p(4), p(5) using the recurrence relation for
probabilities of Poisson distribution i.e.

p ( x +1) = p ( x ) ; x = 0, 1, 2, 3, 4 and then obtain the expected frequencies
x +1
as shown in the following table:

X 
=
1 p ( x) Expected/Theoretical
x +1 x +1 frequency
f ( x ) = N.P (X = x )
= 100.P(X = x)

0 1
=1 p(0) = 0.3679 36.79 □ 37
0 +1
1 1 p(1) = 1 0.3679 = 0.3679 36.79 □ 37
= 0.5
1+1
2 1 p ( 2 ) = 0.5 0.3679 = 0.184 18.4 □ 18
= 0.3333
2 +1
3 1 p(3) = 0.3333  0.184 = 0.0613 6.13 □ 6
= 0.25
3 +1
4 1 p(4)=0.25  0.0613 = 0.0153 1.53 □ 2
= 0.2
4 +1
5 1 p(5)=0.2  0.0153 = 0.0031 0.3 □ 0
= 0.1667
5 +1

42
 Appendix Poisson Distribution

Value of e−λ (For Computing Poisson Probabilities)

(0 < λ < I)

 0 1 2 3 4 5 6 7 8 9

0.0 1.0000 0.9900 0.9802 0.9704 0.9608 0.9512 0.9418 0.9324 0.9231 0.9139

0.1 0.9048 0.8958 0.8860 0.8781 0.8694 0.8607 0.8521 0.8437 0.8353 0.8270

0.2 0.7187 0.8106 0.8025 0.7945 0.7866 0.7788 0.7711 0.7634 0.7558 0.7483

0.3 0.7408 0.7334 0.7261 0.7189 0.7118 0.7047 0.6970 0.6907 0.6839 0.6771

0.4 06703 0.6636 0.6570 0.6505 0.6440 0.6376 0.6313 0.6250 0.6188 0.6125
0.5 0.6065 0.6005 0.5945 0.5886 0.5827 0.5770 0.5712 0.5655 0.5599 0.5543
0.6 0.5448 0.5434 0.5379 0.5326 0.5278 0.5220 0.5160 0.5113 0.5066 0.5016
0.7 0.4966 0.4916 0.4868 0.4810 0.4771 0.4724 0.4670 0.4630 0.4584 0.4538
0.8 0.4493 0.4449 0.4404 0.4360 0.4317 0.4274 0.4232 0.4190 0.4148 0.4107
0.9 0.4066 0.4026 0.3985 0.3946 0.3906 0.3867 0.3829 0.3791 0.3753 0.3716

(=1, 2, 3, ...,10)

 1 2 3 4 5 6 7 8 9 10

e− 0.3679 0.1353 0.0498 0.0183 0.0070 0.0028 0.0009 0.0004 0.0001 0.00004

Note: To obtain values of e− for other values of , use the laws of exponents i.e.

e−(a+b) = e−a .e−b e. g. e−2.25 = e−2 .e−0.25 = (0.1353)(0.7788) = 0.1054.

43