3.2.4 Poisson Distribution
Definition Let X be the number of events per basic unit: For example,
• Number of rain drops in one minute.
• Number of cars passing by you for an hour.
• Number of chocolate particles in one ChoCoChip cookie.
• Number of typos in one page.
• Number of defaults in one cm2 area.
• Number of visitors of a certain web site between 10:00-11:00pm.
• Number of characters in Page 256 of the text book.
The Poisson distribution has a single parameter λ, sometimes called the intensity parameter.
A random variable X, taking values in the nonnegative integers, has a Poisson(λ) distribution if
P (X = x|λ) = e−λλx
x! , x = 0, 1, . . . The mean of X is
EX = X∞ x=0
xe−λλx x! =
X∞ x=1
xe−λλx x!
= λ X∞
x=1
e−λλx−1 (x − 1)!
= λ X∞
y=0
e−λλy y! = λ A similar calculation will show that
VarX = λ.
The mgf is
MX(t) = eλ(et−1).
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Example 3.2.4 (Waiting time)
As an example of a waiting-for-occurrence application, consider a telephone operator who, on the average, handles five calls every 3 minutes. What is the probability that there will be no calls in the next minute? At least two calls?
If we let X =number of calls in a minute, then X has a Poisson distribution with EX = λ = 5/3. So
P (no calls in the next minute) = P (X = 0)
= e−5/3(53)0
0! = e−5/3 = 0.189
P (at least two calls in the next minute) = P (X ≥ 2)
= 1 − P (X = 0) − P (X = 1) = 1 − .189 − e−5/3(5/3)1
1! = 0.496.
Example 3.2.5 (Poisson approximation)
A typesetter, on the average, makes one error in every 500 words typeset. A typical page contains 300 words. What is the probability that there will be no more than two errors in five pages?
If we assume that setting a word is a Bernoulli trial with success probability p = 5001 , then X =number of errors in five pages (1500 words) is binomial(1500, 5001 ). Thus,
P (no more than two errors) = P (X ≤ 2)
= X2 x=0
µ1500 x
¶ ( 1
500)x(499
500)1500−x
= .4230.
If we use the Poisson approximation with λ = 1500/500 = 3, we have P (X ≤ 2) ≈ e−3(1 + 3 +32
2) = 0.4232.
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