Now, if your question what is the formula, then the formula you need is the one to calculate the cummulative probability in a binomial distribution.
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If you find 50 in 10,000, the point estimation remains 0.005 (5%), but the 90% confidence interval narrows to 0.4% ~ 0.6% (compare with 0.26% ~ 1.0%). It is interesting to note the effect of the sample size. Compare with the point estimation of 0.5%. Computers can certainly make your live easier here.
#HOW TO CALCULATE PPM IF NUMBER IS REPEATING TRIAL#
These p have to be found by trial and error (or an iterative method). The 90% confidnece interval that that leaves 5% of chances on each side. I said the figures were the 90% confidence interval. Let’s work with P(X P(X P(X<=X1)=P(X5 is impossible.Ĭlearly the result of these probabilities are too extremme. Let’s call p the actual defectives rate in the lot (that you don’t know and will not know either), n the sample size, and X1 the number of defecives found in the sample. In all these cases the lot is many times larger than the sample, so the “infinitely large lot” is a very good model. The defects in a lot are binomially distributed IF the defects were produced at random OR the sample is taken at random (or both), as long as the sample is taken with repetition (a sampled part is returned to the lot after sampling, existing a probability to take the same part twice) OR the lot is infinitely large. the range where you expect with a 90% of confidence that the ture lot PPM will be) for the three lots based what was found in the sample:Īs you see, there is a wide overlapping, so it is even possible that all lots have the same PPM.įor more info about sampling variation and PPM, try this link to a previos post from today: For example, these are the 90% confidence intervals (i.e. With so few defectives found in the sample, the uncertainty due to sampling variation is high. As the total number of units in lots 1+2 is 1700k+500k=2200k=2.2M, the PPM for lots 1+2 is 5199/2.2M*1M=2363 PPM.Īnother problem is the reliability of the sample PPM as an estimate of the population PPM. Then, the number of defectives in lots 1+2 is 4048+1071=5199. Making the same reasoning for the second lot, you can estimate the number of defectives to be 500k*2143/1M=1071. If you take that value as the estimation of the PPM in the lot, then you can estimate 1700k*2381/1M=4048 as the number of defectives in the lot. PPM of the sample: 10/4200*1M=2381PPM (the same figure you got) Further more, with your reasoning if you had 3 lots of 400,000PPM each the total PPM would be 1.2 millions fefective per million parts! (you cannot have more defective parts than parts)
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You are not taking into account the population sizes.