Quote the part of an article:

1. "Approximately 2700 parts per million parts/steps will fall outside the normal variation of +/- 3 Sigma. This, by itself, does not appear disconcerting. However, when we build a product containing 1200 parts/steps, we can expect 3.24 defects per unit (1200 x .0027), on average. This would result in a rolled yield of less than 4%, which means fewer than 4 units out of every 100 would go through the entire manufacturing process without a defect."

Question 1,
how to calculate the 4%, and how to calcuate only 4 units of every 100 would go through the entire process without a defect.

2. "In the same case of a product containing 1200 parts/steps, we would now expect only 0.0041 defects per unit (1200 x 0.0000034). This would mean that 996 units out of 1000 would go through the entire manufacturing process without a defect. To quantify this, Capability Index (Cp) is used."

Queation 2,
how to calculate the 996 unit from 1000 would go through entire process without defect?
Is it (1-0.0041)*1000=996? if like this , how to calcuate above the 3 sigma no defect level, is it (1-0.04)*100? that should be 96 product without defect through whole process, not 4 ?

thanks

Michael

1. "Approximately 2700 parts per million parts/steps will fall outside the normal variation of +/- 3 Sigma. This, by itself, does not appear disconcerting. However, when we build a product containing 1200 parts/steps, we can expect 3.24 defects per unit (1200 x .0027), on average. This would result in a rolled yield of less than 4%, which means fewer than 4 units out of every 100 would go through the entire manufacturing process without a defect."

Just wanted to point out that the question is based on erroneous reasoning; just because .27% of the population may be expected to fall outside the +/- 3-sigma limits doesn't mean that any of those will be defective. Also, the .27% is included in "normal variation"--that's the whole idea. Where the limits are overlaid on the curve is a matter of choice. There's a very good chance that a point outside the limits is due to assignable cause, but it ain't necessarily so.

First of all, I agree with JSW05 that the article is poorly worded, in that outside 3 sigma is not necesarily defective.

If we assume that 0.27% (2700 ppm) are indeed defective, then with 1200 opportunities, the odds of no defects would be (1-0.0027)^1200 = 0.039 = 3.9%. You could also use the binomial distribution function - =BINOMDIST(0,1200,0.0027,TRUE) in Excel

With only 3.4 ppm, the probability of no defects becomes (1-0.0000034)^1200 = 0.996 = 99.6%

Does that get you on the right track?


Tim F

First of all, I agree with JSW05 that the article is poorly worded, in that outside 3 sigma is not necesarily defective.

If we assume that 0.27% (2700 ppm) are indeed defective, then with 1200 opportunities, the odds of no defects would be (1-0.0027)^1200 = 0.039 = 3.9%. You could also use the binomial distribution function - =BINOMDIST(0,1200,0.0027,TRUE) in Excel

With only 3.4 ppm, the probability of no defects becomes (1-0.0000034)^1200 = 0.996 = 99.6%

Does that get you on the right track?


Tim F

thanks Tim