# How to Calculate Sigma without UCL

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• #250668

weasle23
Participant

Can someone walk me through this problem?

Suppose that:
1) The lower specification limit (tolerance) for a particular output is 330mm
2) The specification width for this output is 10.3mm
3) This output’s process mean is “centered” within the specification interval
If our goal is a process that has a Cp that is at least 1.333,
what is the largest value for σ that would be acceptable in this situation?

• This topic was modified 1 year, 2 months ago by weasle23.
• This topic was modified 1 year, 2 months ago by Katie Barry.
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#250674

Robert Butler
Participant

Obviously the problem is assuming a normal distribution of the data.

The basic formula for Cp in this case is

Cp = (USL-LSL/(6*std)

So, LSL = 330, USL = 330+10.3 and you want a minimum of 1.33 for Cp – therefore plug in the numbers in the above equation, re-arrange and solve for std.

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#250675

weasle23
Participant

I was able to get that far. I wouldn’t even have to find USL because the specification width was given. Still, for the sake of learning I will fill this out and re-engineer this here to the best of my ability. I have the correct answer so I know i’m wrong anyways.

(340.3-330)/6SD = 1.33

10.3/6(SD) = 1.33

10.3/(SD) = 6(1.33)

10.3/(SD) = 7.98

And I do not know how to go any further than that.

• This reply was modified 1 year, 2 months ago by weasle23.
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#250677

weasle23
Participant

I see what i did wrong, I should have put in more 3’s into the CP value

1.333*6 = 7.99998

10.3/7.99998 = 1.2875

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#250679

Robert Butler
Participant

It sounds like you are taking some kind of exam or course or something and trying to match whatever is offered on a multiple choice problem – this is fine but there is something you should keep in mind when you are faced with questions like this in the real world and that is the issue of significant digits.

It is possible to get measurements out to the 100,000th place, or almost any other place for that matter, but when it comes to calculations the rule is the precision of your final result can be no more precise than the precision of the least precise term used in the calculation.  In this case (if we assume 330 mm has actually been measured to 330.0 mm) then when we combine this information with  10.3 this means the final value for sigma should be 1.3 – anything else is just empty precision.

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#250786

Chris Seider
Participant

It seems you do have a UCL. :)

hope all is well with you @rbutler !

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#251150

MBBinWI
Participant

@rbutler – a function of the computer generation.  If they had grown up with a slide rule, they would understand.  ;-)

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