Download Poisson Distribution

Poisson Distribution
This occurs when we are counting the number of successes in some given interval (time, space).
The properties of the Poisson Distribution are:
(a) The average number of successes (λ,μ) occurring in the interval is known.
(b) Occurrences are independent.
(c) Occurrences are random and the probability of a single success is proportional to the interval
length.
(d) The probability of more than one occurrence is small compared to the probability of one success
in a small interval.
If X is a random variable representing the number of successes in a given interval and λ is the average
e   x
number of successes then P(X = x) =
x = 0, 1, 2, …
x!
[ Note: the mean for the Poisson R.V. = λ and the variance is also λ (s.d. =  ) ]
Examples when the Poisson Distribution arises:
1. The number of phone calls arriving at an exchange in a given time interval.
2. Typing errors on a page of an author’s manuscript.
3. The number of faults in a computer’s circuitry over a period of a month.
4. The number of deaths from horsekicks in an army corps over a period of one year.
Problems Using the Poisson Distribution
Example 1: The average number of days school is closed due to snow during the winter in a certain city
is 4. What is the probability that the schools in this city will close for 6 days during a winter?
Ans: Let X be a random variable representing the number of days schools will be closed. Then X is a
Poisson R.V. with λ = 4. We want P(X = 6)
e 4 6
P(X = 6) =
6!
Σ Pg 81
= 0.1042 (check tables)
Example 2: During weekdays a garage owner counts the number of cars and finds that on average there
are 10 every hour. What is the probability that during a particular ¼ hour there will be some cars on his
driveway?
Ans: Let X be a random variable representing the number of cars on the driveway in a ¼ hour interval.
Then X is Poisson with λ = 10 ÷ 4
= 2.5
We want P(X > 0)
P(X > 0) = 1 – P(X = 0) or
e 2.5 2.5 0
= 1 – 0.0821 (tables)
or P(X = 0) =
0!
= 0.9179
= e-2.5
Example 3: (Inverse Problems)
In samples of milk taken from a bulk transportation vehicle, 40% proved to have no bacterial spores.
Estimate the mean number of spores per sample, and hence find the probability of a randomly selected
sample containing 2 spores.
Ans: Let X be a random variable representing the number of spores per sample. Then X is Poisson with
λ unknown.
(a) P(X = 0) = 0.4
e-λ = 0.4
e   0
= 0.4
-λln e = ln 0.4
0!
e-λ = 0.4
-λ = ln 0.4
λ = 0.92 (looking backwards in tables)
or
λ = 0.92
e   2
2!
0.4  0.846
=
2
= 0.17
(b) P(X = 2) =
Poisson Approximation to the Binomial
If in a binomial problem we have π very small (or very close to 1 – some texts also say n large), then we
can approximate the distribution with a Poisson random variable.
Example: 2% of items produced by a factory are defective. What is the probability that in a sample of
200 we get: (a) 3 defectives
(b) At least one defective
Ans: Let X be a R.V. representing the number of defectives. Then X is Binomial with n = 100, π = 0.02.
Since π is small we can approximate by a Poisson R.V., Y with:
λ = nπ
= 200 0.02
=4
(i) P(X = 3) ≈ P(Y = 3)
e   x
=
,x=3
x!
= 0.1954 (tables)
(ii) P(X ≥ 1) = 1 – P(X = 0)
≈ 1 – P(Y = 0)
= 1 – 0.0183
= 0.9817
Σ Pg 119