Download 4.5 Poisson Distribution

Poisson Distribution
Requirements
4.5 Poisson Distribution
Slide 1
! The random variable x is the number of occurrences
of an event over some unit interval.
The Poisson distribution is a discrete probability
distribution that applies to occurrences of some event
over a specified unit interval. The random variable x is the
number of occurrences of the event in an unit interval.
The interval can be time, distance, area, volume, or some
similar unit.
! The occurrences must be random.
! The occurrences must be independent of each other.
! The occurrences must be uniformly distributed over
the interval being used.
Formula
P(x) =
Slide 2
Parameters
µ x • e -µ where e ≈ 2.71828
! The mean is µ.
x!
! The standard deviation is
Difference from a
Binomial Distribution
The Poisson distribution differs from the binomial
distribution in these fundamental ways:
! In a binomial distribution the possible values of
the random variable are x are 0, 1, . . . n, but a
Poisson distribution has possible x values of
0, 1, . . . , with no upper limit.
Example
Slide 5
µ.
Example
Slide 3
! The binomial distribution is affected by the
sample size n and the probability p, whereas
the Poisson distribution is affected only by
the mean µ.
σ=
1
Slide 4
World War II Bombs In analyzing hits by V-1 buzz bombs
in World War II, South London was subdivided into 576
regions, each with an area of 0.25 km2. A total of 535
bombs hit the combined area of 576 regions
If a region is randomly selected, find the probability
that it was hit exactly twice.
The Poisson distribution applies because we are
dealing with occurrences of an event (bomb hits)
over an unit interval (a region with area of 0.25 km2).
Poisson as Approximation
to Binomial
Slide 6
The mean number of hits per region is
µ=
P( 2) =
number of bomb hits
number of regions
=
535
= 0.929
576
(0.929) 2 ( 2.71828) −0.929 (.863)(.395)
=
= 0.170
2!
2
The probability of a particular region being hit exactly
twice is P(2) = 0.170.
The Poisson distribution is sometimes used
to approximate the binomial distribution
when n is large and p is small.
Rule of Thumb:
assumptions
! n≥ 100
! np ≤ 10
Value for µ
µ = n •p
Key Principle
Example Illinois Pick 3 Game
Slide 7
Pay 50 cent to select a sequence of three digits, such as
911. If you play this game once every day, find the
probability of winning exactly once in 365 days.
To find the probability of at least one of
something, calculate the probability of
none, then subtract that result from 1.
That is,
Solution: There are 1000 possible numbers: 000-999.
• Let X=# of winnings in a year~Bin(365, 1/100), where
p=P(winning)=1/1000 and n=365.
• We are looking for P(2).
• Use Poisson approximation: µ=np=365/1000=.365
P(1) =
Slide 8
P(at least one) = 1 – P(none)
(.365)1 e −.365 (.365)(2.71828) −.365
=
= 0.253.
1!
1
PS. The exact P(1)=365(.001)(.999)364=0.2536.
Example
Slide 9
Solution (cont’d)
Example
Step 3: Find the probability of the complement.
Gender of Children Find the probability of a couple
having at least 1 girl among 3 children. Assume that
boys and girls are equally likely and that the gender of
a child is independent of the gender of any brothers or
sisters.
Solution
Step 1: Let A = {at least 1 of the 3 children is a girl}
Step 2: Ā ={not child among 3 children is girl}
= {all 3 children are boys}
2
= P(boy) • P(boy) • P(boy)
=
1 1 1 1
• • =
2 2 2 8
Step 4: Find P(A) by evaluating 1 – P(Ā).
P ( A) = 1 − P ( A) = 1 −
1 7
=
8 8
Interpretation There is a 7/8 probability that if a
couple has 3 children, at least 1 of them is a girl.
= {boy and boy and boy}
Example
P(Ā) = P(boy and boy and boy)
Slide 11
The expected number of cancer deaths in a
community is 63 per year. Assuming a Poisson
distribution, what would be the probability of
observing
(a) Exact one death in a year?
(b)At least one deaths in a year?
Let X = # of cancer deaths in a year, µ = 63
631 (2.71828) −63
= 2.7465 *10 −26 ≈ 0.0000.
(a) P(X = 1) =
1!
630 (2.71828) −63
≈ 1.0000.
(b) P ( X ≥ 1) = 1 − P( X = 0) = 1 −
0!
Slide 10