1

Lecture 9: The Poisson Random Variable and its

PMF

Devore, Ch. 3.6

Topics

I. Poisson Failure (or”Arrival”) Processes

II. Poisson Distribution

III. Poisson Time Process

IV. Poisson Industry Applications

• Multiple failures possible for one unit

• Separate failures are independent

• Probability of failure increase with exposure (e.g. time, area, volume)

• Examples:– Part with many features– # flaws in rolls of textiles (# snags in carpet)– # paint chips on an automotive body– # calls to a telephone exchange– # bad sectors in a disk space

I. Poisson Failure Processes

Under What Conditions is a discrete RV a Poisson RV?

• Example: Suppose flaws or defects occur at random along the length of a thin copper wire. – X = Random variable representing the number of

flaws across L millimeters of wire.• This will be a Poisson RV, if: • For any small interval Dx, and for a constant

failure rate ,• 1. prob(exactly one flaw)= *Dx• 2. Prob (no flaws)=1- *Dx, and therefore,• 3. Prob (more than one flaw in Dx) =0 (Zero!)

Poisson and Binomial PMFs

• The Poisson distribution approaches the Binominal PMF under the following conditions.– As n infinity and p 0 (large # of subintervals with a small

probability of occurrence),– np approaches a value > 0.

– Then: b(x; n, p) p(x: )

– General rules for using the Poisson over the Binomial:• For any binomial experiment with large n and small p, we may

approximate using Poisson. In general, if• n (# of subintervals) >= 100• p <= 0.01 • and np <= 20. (and np ~ > 0)

II. The Poisson PMF

= rate (number of defects per unit time or unit area or..)

X = The Poisson RV, (x = 0,1,2, … )where x is the number of defects

0 !

);(

someforx

exp

x

PMF:

?

Poisson Example

• Copper wire example: suppose flaws occur at random along the length of a thin copper wire such that the probability of a flaw in any 1 millimeter of wire is p=0.005. – Let X – # flaws in 1 mm of wire.– What is the prob of having exactly 2 flaws

in 200 mm of wire?

– What is the prob of having at most 2 flaws in 200 mm of wire?

Poisson Distribution Tables

• Similar to binomial, cumulative distribution tables are available for poisson, F(x; l).

• Use CDF table for the following.– What is P(X = 1) = ? if l = 2– Use table on next slide (also on page

739, Appendix A.2 of Devore textbook):

Cumulative Poisson Table ~ F(x,l)

x 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 15.0 20.00 0.135 0.050 0.018 0.007 0.002 0.001 0.000 0.000 0.000 0.000 0.0001 0.406 0.199 0.092 0.040 0.017 0.007 0.003 0.001 0.000 0.000 0.0002 0.677 0.423 0.238 0.125 0.062 0.030 0.014 0.006 0.003 0.000 0.0003 0.857 0.647 0.433 0.265 0.151 0.082 0.042 0.021 0.010 0.000 0.0004 0.947 0.815 0.629 0.440 0.285 0.173 0.100 0.055 0.029 0.001 0.0005 0.983 0.916 0.785 0.616 0.446 0.301 0.191 0.116 0.067 0.003 0.0006 0.995 0.966 0.889 0.762 0.606 0.450 0.313 0.207 0.130 0.008 0.0007 0.999 0.988 0.949 0.867 0.744 0.599 0.453 0.324 0.220 0.018 0.0018 1.000 0.996 0.979 0.932 0.847 0.729 0.593 0.456 0.333 0.037 0.0029 0.999 0.992 0.968 0.916 0.830 0.717 0.587 0.458 0.070 0.00510 1.000 0.997 0.986 0.957 0.901 0.816 0.706 0.583 0.118 0.01111 0.999 0.995 0.980 0.947 0.888 0.803 0.697 0.185 0.02112 1.000 0.998 0.991 0.973 0.936 0.876 0.792 0.268 0.03913 0.999 0.996 0.987 0.966 0.926 0.864 0.363 0.06614 1.000 0.999 0.994 0.983 0.959 0.917 0.466 0.10515 0.999 0.998 0.992 0.978 0.951 0.568 0.15716 1.000 0.999 0.996 0.989 0.973 0.664 0.22117 1.000 0.998 0.995 0.986 0.749 0.29718 0.999 0.998 0.993 0.819 0.38119 1.000 0.999 0.997 0.875 0.47020 1.000 0.998 0.917 0.55921 0.999 0.947 0.64422 1.000 0.967 0.721…36 1.000

Mean, Variance of Poisson RV

• If X has a Poisson Distribution with parameter

– b(x;n,p) p (x;)– As n infinity and p 0, then np

• Therefore, mean =

– How about the variance of the binomial• If b(x;n,p), then Var(X) = np(1-p)• What happens to the variance for p(x;) as n large, p 0?

• V(X) = E(X) =

11

III. Poisson Time Process

• If the following conditions are reasonably satisfied, then a process may be deemed a Poisson process.

• Given time intervals with random counts of X across them, the intervals may be partitioned such that:– Prob. of more than 1 event in a subinterval is ~0.– Prob. of a count of 1 in a subinterval is the same for

all intervals and proportional to the length of intervals.– Count of each subinterval is independent of other

subintervals. !

)(k

tetP

kt

k

Interesting Poisson Special Case 1

• What is the equation for k = 0?– Process has no errors / no defects / no flaws.– Assume = . t

tt

kt

k

ete

tP

k

tetP

!0)(

!)(

0

0

IV. Poisson Process Applications

• Poisson processes are widely used in fields of quality and reliability, particularly time dependent analysis.

• Examples– Reliability (warranty analysis)

• Predicting the likelihood of failure or # of warranty returns per some time interval.

– Rolled Throughput Yield Analysis• Predicting the likelihood that a part (which can have 1 or

more defects occur at multiple assembly operations) will pass through all assembly operations without a defect.

Reliability

• Review: Reliability is the probability that a system or product will perform (working vs. not working) in a satisfactory manner for a given period of time, when operated under specified conditions.

• Some Uses of Reliability: – predicting manufacturing process

uptime.– predicting warranty returns.

Example: Spare Parts Problem

• Suppose you have a machine that contains a non-repairable motor. – Let X = # of spare parts needed in 10 year period– Machine breakdown rate () = 0.05 failures per year

(assume failures follow an exponential distribution and occur independent of other failures)

– Design life of the machine is 10 years.

• What is the probability that you will not need a spare part in the next 10 years?

• What is the probability that 2 spares will be adequate?

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Interesting Case

• Given Poisson process, if X = and t= 1/?– example, = 0.05 failures per year and t = 1/ 0.05 =

20 yrs– E(X) = t =1

• Example: so for time t = 1/, only 37% of units will not fail (63% of units will fail)

368.0!0

1

!

)()0XPr(

01

e

x

te xt

Defect Example

• Suppose you have a process with 4 operations.• At each one, a part is either good, or has 1 or more defects.

• Given the following reject rates, what is the probability of a part passing through each and every operation without being scrapped or reworked? (known as the the rolled

throughput yield problem)

= 0.07

?

= 0.11 = 0.16 = 0.13

Assume independent operations that follow exponential distribution for failures

Summary : Discrete Random Variables and PMFs

• Random variables (rv’s) help us measure characteristics of a sample– An rv is a rule that associates a number

(value) with each outcome in a sample– What for? to focus on numerical aspects

of a sample such as mean, proportion, standard deviation

• Probability Distributions: Binomial, hypergeometric, geometric, negative binomial, Poisson

Which Distribution to Use When?

Discrete Probability Distribution Trials Sample Size Random VariableType of

OutcomesProbabilty of

Success

Binomialn independent trials, fixed in

advance nX is # successes

among n trialssuccess or

failure constant (p)

HypergeometricN objects in the population (M successes among the N

objects)n

X is # successes among the n objects in the

sample

success or failure

p=M/N

GeometricIndependent trials, n is not

known in advanceUntil first success

X is # of trials until first success

success or failure

constant (p)

Negative BinomialIndependent trials, until r

successes, n is not known in advance

Until r successesX is # failures that

precede the rth success

success or failure constant (p)

Poisson large n, small p large n, small p

X is # of occurrences (S) in an interval or time

period

success or failure, but

multiple possible per

part or object

Rate per unit (lambda) =

np