Binomial and Poisson Distribution

Description
Binomial and Poisson Distribution

Definitions: Bernoulli trial
Bernoulli trial: If there is only 1 trial with
probability of success p and probability of
failure 1-p, this is called a Bernoulli
distribution. (It is a special case of the
binomial with n=1)

Probability of success:
Probability of failure:


p p p X P = ÷ |
.
|

\
|
= =
÷1 1 1
1
1
) 1 ( ) 1 (
p p p X P ÷ = ÷ |
.
|

\
|
= =
÷
1 ) 1 ( ) 0 (
0 1 0
1
0
Characteristics of Bernoulli distribution
For Bernoulli (n=1)
E(X) = p
Var (X) = p(1-p)


Binomial Distribution
?What the binomial distribution is

?How to recognise situations where the binomial
distribution applies

?How to find probabilities for a given binomial
distribution, by calculation and from tables


When to use the binomial distribution
• Independent variables

The Binomial distribution is all about success
and failure.
When to use the Binomial Distribution

– A fixed number of trials
– Only two outcomes
– (true, false; heads tails; girl,boy; six, not six …..)
– Each trial is independent

IF the random variable X has Binomial distribution, then
we write X ? B(n,p)
X
Binomial distribution
n X p p
X n X
n
X
.... 2 , 1 , 0 , ) 1 ( = ÷
|
.
|

\
|
÷
1-p = probability
of failure
p =
probability of
success
X = #
successes
out of n
trials
n = number of trials
Note the general pattern emerging ? if you have only two possible
outcomes (call them 1/0 or yes/no or success/failure) in n independent
trials, then the probability of exactly X “successes” in n trials is given by
Binomial distribution
Ex - Eggs are packed in boxes of 12. The probability that each egg
is broken is 0.35.Find the probability that in a random box of
eggs there are 4 broken eggs.
Ans- [12 boxes are equivalent to 12 independent trials,i.e.n=12
and p=0.35, To find P(X=4) =p(4)]

figures t significan 3 to 235 . 0
65 . 0 35 . 0 495 65 . 0 35 . 0
4
12
) 4 (
8 4 ) 4 12 ( 4
=
× × = ×
|
|
.
|

\
|
= =
÷
X P
Ex 2 - Eggs are packed in boxes of 12. The probability that
each egg is broken is 0.35.Find the probability in a random
box of eggs there are less than 3 broken eggs.
0151 . 0 01346 . 0 1225 . 0 66 65 . 0 35 . 0 12 005688 . 0 1 1
65 . 0 35 . 0
2
12
65 . 0 35 . 0
1
12
65 . 0 35 . 0
0
12
) 2 ( ) 1 ( ) 0 ( ) 2 ( ) 1 ( ) 0 ( ) 3 (
11 1
) 10 ( 2 ) 11 ( 1 ) 12 ( 0
= × × + × × + × × =
×
|
|
.
|

\
|
+ ×
|
|
.
|

\
|
+ ×
|
|
.
|

\
|
=
+ + = = + = + = = < p p p X P X P X P X P
USING TABLES of the Binomial
distribution
An easier way to add up binomial
probabilities is to use the cumulative
binomial tables
Find the probability of getting 3 successes in 6 trials,
when n=6 and p=0.3
n=6 x 0 1 2 3 4 5 6
P=0.3 P(X=x) 0.1176 0.4202 0.7443 0.9295 0.9891 0.9993 1.000
n=6 x 0 1 2 3 4 5 6
P=0.3 P(X=x) 0.1176 0.4202 0.7443 0.9295 0.9891 0.9993 1.000
The probability of getting 3 or fewer successes is found by adding:
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1176 + 0.3026 + 0.3241 + 0.1852 =
0.9295 , This is a cumulative probability.

Mean, variance & standard deviation
• E(X)= ? = mean = ? x P(X=x) ... By Definition of E(X) for
discrete random variable
= ? x p(x)
On substituting value of p(x) and simplifying
? = mean = np
• V(X) = ?² = E(X
2
) – [E(X)]
2
... By Definition of variance
for discrete random variable
= ?x² P(X=x) – ?
2
= ?x² p(x) – ?
2
= np(1-p) = npq

? =

) 1 ( p np ÷
• Thus the mean and variance of a random
variable whose distribution is B(n,p) is given
as ? =np and V(X) = ?²= np(1-p) = npq
Thus V(X)< E(X) in Binomial distribution.
• Note –1) Binomial distribution is symmetric
for p=q= 0.5
2) Positively skewed for p <0.5
3) Negatively skewed for p >0.5


Poisson distribution
• Poisson distribution is for counts—if events
happen at a constant rate over time, the Poisson
distribution gives the probability of X number of
events occurring in time T.
• Poisson distribution is a limiting case of binomial
distribution when n is large and p is small such
that np is finite
• When n is large and p is small, the calculation of
binomial probabilities involve prohibitive amount
of work.





• A discrete random variable X is said to follow Poisson
distribution with parameter ? if its probability mass function
is given by
P(X=x)=P(x) = [e
-

?
?
x
] / x! , x = 0,1,2………
= 0 otherwise

where e =2.71828
and we write X ? P(?)

• It is discovered by Simeon Denis Poisson
(1781-1840). Its parameter is ? .
Parameter ? = np = expected no of occurrences of the event

• Poisson distribution occurs when there are events
which do not occur as outcomes of definite no.of
trials (Unlike that of Binomial) of an experiment but
which occur at random pts of time and space wherein
our interest lies only in the no. of occurrences of the
event and not in the nonoccurrences
Ex- 1 No of deaths from a disease.
2 No.of suicides reported in a particular
city
3 No of persons killed in a road accident .
4 No of mistakes per page in the final proof of
the book.


Poisson Mean and Variance
Mean

Variance and Standard Deviation
ì µ =
ì o =
2
ì o =
where ì is the parameter of the distribution =
expected number of ocurrences in a given time
period
Note – Poisson distribution is always positively
skewed and its skewness decreses as ì increases.

For a Poisson
random variable, the
variance and mean
are the same!
Example: Poisson distribution
1 Suppose that a rare disease has an incidence of 1 in 1000
person-years. Assuming that members of the population are
affected independently, find the probability of k cases in a
population of 10,000 (followed over 1 year) for k=0,1,2.
n =10,000 and p = 1/1000 = 0.001
The expected value (mean) =ì = .001*10,000 = 10
10 new cases expected in this population per year?
00227 .
! 2
) 10 (
) 2 (
000454 .
! 1
) 10 (
) 1 (
0000454 .
! 0
) 10 (
) 0 (
) 10 ( 2
) 10 ( 1
) 10 ( 0
= = =
= = =
= = =
÷
÷
÷
e
X P
e
X P
e
X P
Ex 2
• A factory produces blades in packets of 10 .The
probability that a blade is defective is 0.2 percent
Find the no of packets having two defective blades in
a consignment of 10,000 packets.
• Ans - n=10 ,p = probability of getting a defective
blade = 0.2% = 0.002
Therefore ?= n*p =10*0.002 = 0.02
P(X=2) = [e
-

?
?
x
] / x!
= [e
-

0.02
(0.02)
2
] / 2!
= 0.000196


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