# Pp and Ppk for Weibull data with Physical lower limit 0.

Six Sigma – iSixSigma Forums General Forums Methodology Pp and Ppk for Weibull data with Physical lower limit 0.

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• #254786 Roman Hodik
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Hi all. Im currently developing a tolerance analysis tool, where one of the steps is to get Pp and Ppk values from descriptive statistics. And Im struggling to get this done with Weibull data.

Example:
I assume data with Shape=2 and Scale=6.
LSL = 0 and USL = 15.5

X99.865% = 99.865% percentile = 15.423
X0.135% = 0.135% percentile = 0.221
X50% = 50% percentile = 4.995

Pp = (USL-LSL) / (X99.865% – X0.435%) = 15.5 / (15.423-0.221) = 1.02
Pp via Minitab is not calculated …. why ?

Ppk = MIN[((X50%-LSL)/(X50% – X0.135%)) , ((USL-X50%)/(X99.865% – X50%))]
Ppk = MIN[4.995/(4.995 – 0.221) , (15.5-4.995)/(15.423-4.995)] = MIN[1.046 , 1.007] = 1.007
Ppk via Minitab is 1.00

Now Same example but I set LSL to 1.0 … and now see some big differences.
Pp = (USL-LSL) / (X99.865% – X0.435%) = (15.5-1) / (15.423-0.221) = 0.954
Pp via Minitab = 0.82 … why different ?

PpL = (4.995-1) / (4.995-0.221) = 0.837
PpU = (15.5-4.995) / (15.423-4.995) = 1.007
Ppk = 0.837
Ppk (and Ppl) via Minitab 0.64 … shy different ?

I struggle to understand why.

Also, just using common sense, for data with LSL = 0 and Physical Limit = 0, I would calculate the Ppk as Tolerance field divided by 99.73% data :
Ppk = T/X99.73% …. this makes to me more sense compared to (USL-X50%)/(X99.865% – X50%).

I would be happy for any comments and suggestions.

###### Attachments:
1. Weibull-Cp-Cpk.pptx