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![]() Statistical Techniques and 6 Sigma
![]() Exponential Distributions
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| Author | Topic: Exponential Distributions |
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TickTax unregistered |
I would be grateful for help with the following problem. I know how to calculate DPMO for a Normally distributed process, but how do I calculate DPMO for an exponential distribution? What DPMO figures do 3sigma, 4sigma, 5sigma and 6sigma processes produce? IP: Logged |
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dWizard Forum Contributor Posts: 20 |
Basically, I ainât gotta clue. I have some more references at home I will check tonight, but I hold little hope. Try these links. Maybe they will help. http://www.itl.nist.gov/div898/handbook/eda/section3/eda3667.htm http://www.itl.nist.gov/div898/handbook/apr/section1/apr161.htm Regards, ------------------ IP: Logged |
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Rick Goodson Forum Wizard Posts: 102 |
I don't have a clue either but will take a look in some of my stat books. I am curious though, why do you need DPMO for an exponential distribution? What is the process your are trying to measure? IP: Logged |
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Laura M Forum Contributor Posts: 299 |
It's late and its been a long day. But I'm thinking its the area under the curve from the point (spec limit?) to infinity. Which is the integral, isn't it? Old brain cells are getting stirred up, but I once did the probability of getting a part out of a "leak spec." Leak distributions are typically exponential. Lots of 0's, and low values, a few creeping up to the spec. I was able to superimposed an "e to the x" distribution over real data. Then from the upper spec forward is the probability, convert to PPM. I excluded "major failures" because we were looking for those that were borderline and subject to measurement error. I probably should have consulted books rather than shooting from the hip on this one, but you guys that did can correct me! Laura IP: Logged |
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TickTax unregistered |
Thanks to those of you who have responded so far. Perhaps I should explain my problem a little more. I'm looking at a queuing process which has an exponential p.d.f. I am aware that for an exponential process with p.d.f lamda*exp(-lambta*t) the mean is 1/lambda and the variance is 1/(lambda^2) If I integrate from 4 sigma(mean plus 3sigma) to infinity (a 3sigma process?) I get 18316 DPMO. Other DPMO's calculated this way are as follows: 4 sigma - 6738 However, Motorola's DPMO figures assume a mean which drifts +- 1.5 sigma. What I would like to be able to do is to calculate the figures assuming a similar drift in the mean of my exponential distribution. IP: Logged |
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