Six Sigma – Bell Curve
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 This topic has 9 replies, 5 voices, and was last updated 14 years, 9 months ago by Fake Stan Alert.

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November 15, 2007 at 2:48 pm #48681
Anoop NairParticipant@AnoopNair Include @AnoopNair in your post and this person will
be notified via email.Is a Six Sigma process +/ 3 Sigma (Total of 6 standard deviations) on a bell curve or +/ 6 Sigma (total of 12 standard deviations) on a bell curve?
0November 15, 2007 at 3:07 pm #164833
Dr. ScottParticipant@Dr.Scott Include @Dr.Scott in your post and this person will
be notified via email.Anoop Nair,
Six sigma is +/ 6 standard deviations between the mean of your bell curve and the nearest specification(s) for the process. I am not sure why you are referring to the bell curve. Your sigma level is simply the minimum of USLMean/S or MeanLSL/S.
I hope that makes sense.
Regards,
Dr. Scott0November 15, 2007 at 3:35 pm #164835
Anoop NairParticipant@AnoopNair Include @AnoopNair in your post and this person will
be notified via email.Dear Dr. Scott – Thank you for your reply. I think I did not articulate my question very well. Assuming I have a Six Sigma process and I have an LSL and a USL, how many standard deviations would I have on either side of my mean? In other words, what would be the Zvalue on each side of the mean. Thanks again for your help.
0November 15, 2007 at 4:00 pm #164838
Dr. ScottParticipant@Dr.Scott Include @Dr.Scott in your post and this person will
be notified via email.Anoop Nair,
Apparently I articulated even less effectively than you did. :)
You would need at least 6 standard deviations between your mean and either the USL or LSL. In other words, if your process is perfectly centered on the target (i.e., you process mean is the same as your target), then you would have to have 6 standard deviations of space between your mean and the USL and 6 standard deviations space between your mean and the LSL.
Does that clarify things?
Regards,
Dr. Scott0November 15, 2007 at 9:29 pm #164867
Anoop NairParticipant@AnoopNair Include @AnoopNair in your post and this person will
be notified via email.Perfect . Thank you!
0November 15, 2007 at 9:43 pm #164872
Dr. ScottParticipant@Dr.Scott Include @Dr.Scott in your post and this person will
be notified via email.Anoop,
You are very welcome. Hope I helped.
Cordially,
Dr. Scott0November 15, 2007 at 10:12 pm #164873Anoop,
While the technical definition for the sigma level of a process is the number of standard deviations that exist between the processes average and the closest of either the upper/lower specification (assuming that we can count of 6 increments before the limit) the projection of the VOP is still the normal curve at +/3 standard deviations from the average (explaining 99.73% of the expected outcomes).
Sigma levels of a process provide a capability assessment and are integrating both VOC and VOP. Standard deviations are used to portray the expected variation of the process and are a reflection of the VOP.
Regards,
Erik0November 16, 2007 at 6:45 am #164888
Fake Stan AlertParticipant@FakeStanAlert Include @FakeStanAlert in your post and this person will
be notified via email.Good question:the second one
0November 16, 2007 at 11:04 am #164897Dear Eric – Thanks for your note. I think I understand using Sigma as representing process capability and standard deviation explaining variation in the process. The question that I had was that if we look at the bell curve having the mean equivalent to the target, how many standard deviations on either side needs to be fit to depict a Six Sigma process. I understand from Dr. Scott reply that it has to be six standard deviations on both sides (ie, a total of 12 standard deviations). Put another way, if I have the classical bell curve with mean=target, with 3 standard deviations on either side of the mean with Z=3 coinciding with the LSL and Z=3 coinciding with the USL, explains a 3 Sigma process and not a Six Sigma process.
Thanks and Regards,
Anoop
PS: The reason for posting this question was that I had somebody looking at this classical bell curve and telling me that it represents a 6 Sigma process and my argument was that it represents a 3 Sigma process. I hope with the answers that I got from you and Dr. Scott, my understanding is right!0November 16, 2007 at 12:12 pm #164899
Fake Stan AlertParticipant@FakeStanAlert Include @FakeStanAlert in your post and this person will
be notified via email.Usually both can present the 6 sigma process,but the 12 sd curve is more accurate
0 
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