Page 121 - Applied statistics and probability for engineers

Section 3-9/Poisson Distribution 99

Example 3-30 can be generalized to include a broad array of random experiments. The
interval that was partitioned was a length of wire. However, the same reasoning can be applied
to an interval of time, an area, or a volume. For example, counts of (1) particles of contami-
nation in semiconductor manufacturing, (2) laws in rolls of textiles, (3) calls to a telephone

exchange, (4) power outages, and (5) atomic particles emitted from a specimen have all been
successfully modeled by the probability mass function in the following dei nition.
In general, consider subintervals of small length Δt and assume that as Δt tends to zero,

1. The probability of more than one event in a subinterval tends to zero.
2. The probability of one event in a subinterval tends to λΔt.
3. The event in each subinterval is independent of other subintervals.

A random experiment with these properties is called a Poisson process.
These assumptions imply that the subintervals can be thought of as approximate inde-
pendent Bernoulli trials with the number of trials equal to n = /Δ t and success probability
T
p = λΔ t = λ T n. This leads to the following result.
/
Poisson
Distribution The random variable X that equals the number of events in a Poisson process is a
Poisson random variable with parameter 0 < λ, and
f x ( ) = e −λ T (λ T) x x = 0 , , ,…1 2
x!

The sum of the probabilities is 1 because
∞ e −λ T λ ( T) x ∞ λ ( T) x
∑ = e −λ T ∑
x=0 x! x=0 x!
and the summation on the right-hand side of the previous equation is recognized to be Taylor’s
λ
T
x
expansion of e evaluated at λT. Therefore, the summation equals e and the right-hand side
equals 1.
Historically, the term process has been used to suggest the observation of a system over
time. In our example with the copper wire, we showed that the Poisson distribution can also
apply to intervals such as lengths, and a following example uses areas.
The parameter λ is the mean number of events per unit length. It is important to use con-
sistent units for λ and T. For example, if λ = 2 3 l aws per millimeter, then T should be
.
expressed in millimeters. If λ = 7 1 square centimeters, then an area of 4.5 square inches
.
.
should be expressed as T = 4 5 2 54. ( . 2 ) = 29 03 square centimeters (Figure 3-14).

Example 3-31 Calculations for Wire Flaws For the case of the thin copper wire, suppose that the number of

laws follows a Poisson distribution with a mean of 2.3 laws per millimeter. Determine the prob-

ability of exactly two laws in 1 millimeter of wire.
.
Let X denote the number of laws in 1 millimeter of wire. Then, λT = 2 3 l aws and

2
(
P X = ) = e − .3 2 3 . 2 = 0 265
2
.
2!
Determine the probability of 10 laws in 5 millimeters of wire. Let X denote the number of l aws in 5 millimeters of

wire. Then, X has a Poisson distribution with
λT = 2 3 flaws/mm × 5mm = 11 5 flaws
.
.