Poisson Distribution

The Poisson distribution The Poisson probability Mass function Poisson distribution is a limiting case of the Binomial distribution when i.

n     → ∞

i.e.

( n ≥ 20)

ii.

 → 0  p   

i.e.

(  p ≤ 0.05)

iii.

np = λ  = average of   Binomial  distribution

Then probability mass function of Poisson is given

  p ( x )

e =

−

λ 

λ 

 x  x

 x !

=

0 , 1 , 2  , 3 , 4 ,........

∞

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 one hour 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

P(x)

=

( λ  t  )  x e

 x !

− λ 

t 

;  for  x = 0 ,1, 2 , 3 , 4 ,........∞

&here t x

λ 

= = =

no. of units of time 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.

()!MP*( + 1  ! company maes 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

olution The average number of defectives in -- motors is

λ  2

-.-1 3 -- 2 

The probability of getting / defectives is :

P(x)

λ  =

P( x =5)

 x

e

− λ 

;  for   x = 0 ,1 , 2 , 3 , ......

 x ! =

35 e 5!

−

3

=

0.10082

("ample 4  5f electricity power failures occur according to a Poisson distribution with an average of  failures every twenty wees, 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.

λ  =

3

= 0.15

20

 P  ( x = 0

)

+

 P  (  x = 1 )

=

e

−

0.15

× 0.15

0!

0 +

e

−

0.15

×

1!

0.15

1 =

0.989

("ample 4  There were 78 aircraft hi9acings worldwide. The mean number of hi9acings per day is estimated as 788/ 2 -.18.there is need to now about the chances of multiple hi9acing in one day. ;se λ  2 -.18 and find the probability that the number of hi9acings # " $ in one day is - , 1 olution The Poisson distribution applies because we are dealing with the occurrences of hi9acing event over a time interval of one day. The probability of - and 1 is calculated as  P  ( x = 0

)

 P  (  x = 1 )

=

=

e

−

0.126

×

0.126

0

0.126

1

0!

e

−

0.126

×

1!

=

0.882

=

0.111

("ample 4 7 The probability that a person dies from a certain infection is 0.003 .
λ 

 P 

(

 x

=

4

)

e =

−

9

×

  4!

9

4

=

0.0337

("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  incoming calls 5ii. ?uring an interval of 7 minutes there will be at most 2  incoming calls