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Topic History of : Six Sigma

Max. showing the last 6 posts - (Last post first)
6 years 3 months ago #12550

MAHJABEEN MUSTOFA

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your question says 6 sigma is the target so it should be SDX6. = 6
6 years 4 months ago #12463

Anonymous

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Dear Fresh,
I am sure the standard deviation was given or not described as pessimist or optimist viewpoint.. Probably they have not given as 3 point PERT estimation. That's why your standard deviation was 16-10=6
9 years 2 months ago #5024

Steve Sandoval

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Michael,

I agree with your calculation that using the PERT beta distribution you would come up with a 13 day mean and a standard deviation of 1 day. I appears, however, that the question intends you to take +/- 6 standard deviations (six sigma) from the mean. This produces the answer of 7 to 19 days.

However, I personally would have found this question a bit frustrating. If I see "a quality requirement of Six Sigma", then because it is capitalized then I typically would interpret it to mean that they are following the whole Six Sigma quality philosophy / framework -- and not necessarily that they intend you to take +/- 6 standard deviations in your calculations.

The bottom line is that I would view that question as poorly worded and a little unclear. Remember that the PMP exam is multiple choice. Since it can be interpreted two ways, I would do both methods and check to see if both answers appear as options. If only one answer is there, then you know which to pick. If both answers appear as options, then you'll just have to choose whichever interpretation makes most sense to you.
9 years 2 months ago #5019

Michael Waserman

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Steve, thank you for your input.
Just to clarify the question I am struggling with...

The activity has an optimistic estimate of 10 days, pessimistic estimate of 16 days and most likely estimate of 13 days. If your company has a quality requirement of Six Sigma what is the duration within which task must be completed?

I calculated 13 days mean.. and standard deviation = 1. Correct answer is 7 - 19 days. I don’t understand why it is +- 6 days from the mean.

Thanks.
9 years 2 months ago #5016

Steve Sandoval

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Hi Michael,

I think that you're keying in on the word "sigma" and that it is perhaps the cause of some confusion.

Generally when you hear "Six Simga" mentioned, it is as Cornelius says -- it is a quality managmenet methodology. Think of it as just one more method in a family of quality managment approaches over they years. Other examples would be TQM (Total Quality Management), or ZD (Zero Defects).

The idea behind Six Sigma is that they wanted to drive their defect rate down to near zero. The goal was to achieve a rate of approximagtely one defect per million items. If you look at a normal distribution (a bell curve), to achieve a probability at such a low level you need to go approximately six standard deviations away from the mean. That is where the name Six Sigma comes from. It's just a shorthand way of referencing their goal of defects being at a rate comparable with one in a million.


WIth respect to the PERT task duration, the beta distribution version of the formula basically is taking your date estimates and forcing them onto a simple probability function that looks like a bell curve.
The beta distribution PERT formula is: duration = (optimistic + pessimitic + 4*most likely) / 6
When using this simple model, the standard deviation (sigma) is equal to (optimistic - pessimistic) / 6.


You'll need to look at the context to know whether they're talking about the quality methodology or whether they are actually referencing six standard deviations away from the mean. However, if they intend the quality methodology then Six Sigma is generally capitalized.

With respect to your specific question, I'm afraid that I'm unable to speak to the details without seeing the details of the question. However, I hope that the above discussion helps you understand Six Sigma vs. PERT.
9 years 2 months ago #4995

Michael Waserman

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Would it be possible to elaborate a little more on the relation between six sigma quality requirement and process duration?

I have a question that results 13 days of the PERT task duration estimate and 1 day as a standard deviation. The correct answer reads that the task can be completed in 7 to 19 days period. I don't really understand where 6 days duration deviation came from...

Any input is greatly appreciated.

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