Poisson Distribution

7/21/2019 Poisson Distribution

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The Poisson distribution

The Poisson probability Mass functionPoisson distribution is a limiting case of the Binomialdistribution when

Then probability mass function of Poisson is given

( )

( )ondistributi Binomial of averagenpiii

pei pii

neini

==

≤ →

≥∞ →

λ .

05.0..0.

20...

( )!

xe p x

x

λ

λ −

= ∞= ,........4,3,2,1,0 x



Another Approach to Poisson Distribution

The Poisson distribution is useful when dealing with the

number of occurrences of a particular event over a

specified interval, where the interval can be time or space.

A random variable must have the following properties to be

classed as a Poisson random variable:1. the probability of an occurrence of the event

in any interval is the same as for any other

interval of equal length

. the occurrence of the event in any interval is

independent of the occurrence in any other

interval.



The following random variables can possess these properties:

i. The number of telephone calls arriving at a switch board in a onehour period

ii. traffic flow and ideal gap distance



iii. The number of customers arriving at a cash desk in a shop

iv. Number of typing errors on a page.



v. hairs found in McDonald's burgers

v. Failure of a machine in one month



!ny e"periment meeting the following conditions is a Poisson Process

#Poisson e"periment$

The Poisson probability mass function, for any Poisson process with

parameter The probability of x occurrences in an interval of si%e t .This is

&here

t = no. of units of time

x = no. of occurrences in t units of time.

= no of occurrences per unit of time.

'ote: the Poisson random variable has no limit on the number of

occurrences.

∞==

−

,........4,3,2,1,0;!

)( )x(P x for

x

et t x λ

λ

λ



()!MP*( + 1 ! company maes electric motors. The probability an electric motor is

defective is -.-1. &hat is the probability that a sample of -- electric

motors will contain e"actly / defective motors0

olutionThe average number of defectives in -- motors is

2 -.-1 3 -- 2

The probability of getting / defectives is:λ

10082.0!5

3 )5x(P

......,3,2,1,0;!

)x(P

35

===

==

−

−

e

x for x

e x λ λ



("ample 4

5f electricity power failures occur according to a Poisson distribution

with an average of failures every twenty wees, calculate the

probability that there will not be more than one failure during a

particular wee

olution

The average number of failures per wee is:

6'ot more than one failure6 means we need to include the probabilities

for 6- failures6 plus 61 failure6.

15.020

3==λ

( ) ( ) 989.0!1

15.0

!0

15.0

10

115.0015.0

=×

+×

==+=

−−

ee

x P x P



("ample 4

There were 78 aircraft hi9acings worldwide. The mean number of

hi9acings per day is estimated as 788/ 2 -.18.there is need to

now about the chances of multiple hi9acing in one day. ;se2 -.18 and find the probability that the number of hi9acings # " $

in one day is - , 1

olution

The Poisson distribution applies because we are dealing with the

occurrences of hi9acing event over a time interval of one day. The

probability of - and 1 is

calculated as

λ

( )

( ) 111.0!1

126.01

882.0!0

126.00

1126.0

0126.0

=×

==

=

×

==

−

−

e x P

e x P



("ample 4 7

The probability that a person dies from a certain infection is

0.003 .<ind the probability that e"actly 4 of the ne"t 3000 so infected will die

olution

2 n " p 2 --- " -.-- 2 =λ

( ) 0337.0!4

94

49

=×

==

−

e x P



("ample 4 /

>iven that the witch board of an office receives on the average 0.9 calls

per minute, find the probability that

i. in a given minute there will be at least one incoming call.

ii. Between 9:00 A.M. and 9:02 A.M . there will be e"actly 2 incomingcalls

5ii. ?uring an interval of 7 minutes there will be at most 2 incomingcalls

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Poisson Distribution