I beg to disagree. The Binomial Distribution is done with replacement.
The 80/80 rule is supposed to be done without replacement. Therefore, the only discrete distribution which satisfies the question - What is the probability that M samples of a sample population S will pass/fail when S is taken from a total population N with an expected pass/fail rate of P/F? without replacement is the HyperGeometric Distribution. In other words, the HyperGeometric Distribution answers the question - if your company makes a total of 100 products per month and you expect 20 of them to fail, what's the probability that 2 samples of a random sample of 10 being tested without replacement each month will also fail? The HyperGeometric Distribution says in such a scenerio that the chances 2 samples will fail are 31.8%. Also, the probability that 3 samples in the above given conditions will fail is 20.9%. Which makes for some odd interpretations. But in any event, even though you think you're working with an 80% success rate, you have only a 70% confidence level. Now, I'm certainly no statistical expert, but I believe the Binomial and Hypergeometric distributions become nearly equal in results only with large numbers. You can play around with this on Excel since the various distributions are built-in spreadsheet functions. - Doug ------------------------------------------- This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. Visit our web site at: http://www.ewh.ieee.org/soc/emcs/pstc/ To cancel your subscription, send mail to: [email protected] with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Michael Garretson: [email protected] Dave Heald [email protected] For policy questions, send mail to: Richard Nute: [email protected] Jim Bacher: [email protected] All emc-pstc postings are archived and searchable on the web at: No longer online until our new server is brought online and the old messages are imported into the new server.
