1. 1 Lecture 9: The Poisson Random Variable and its PMF Devore, Ch. 3.6

2. Topics I.Poisson Failure (or”Arrival”) Processes II.Poisson Distribution III.Poisson Time Process IV.Poisson Industry Applications

3. 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

4. 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!)

5. 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 0)

6. 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 PMF: ?

7. 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?

8. 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):

9. Cumulative Poisson Table ~ F(x,l)

10. 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. 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.

12. Interesting Poisson Special Case 1 What is the equation for k = 0? –Process has no errors / no defects / no flaws. –Assume = . t

13. 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.

14. 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.

15. 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?

16. 16 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)

17. 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

18. 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

19. Which Distribution to Use When?