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Hello,

I am having a hard time with one of the chapter questions regarding process performance: I have followed the formula for Pp and I am getting it wrong:  The problem is:

A process produces a part that is required to measure between 79.75 and 80.25 inches wide. As part of the supplier qualification process, samples were collected and yielded a process mean of 79.5 inches, and the long-term standard deviation was 0.625 inches.   What is the Pp for this process?

I am using the formula Pp= USL-LSL/ 6(sigma).  I am assuming the USL= 80.25 and LSL =79.75.  The answer is .333 but I am not getting that with following this formula.

80.25 - 79.75 = 0.5

6 x .625 = 3.75 (You multiple the six by your standard deviation)

0.5 / 3.75 = .133

That suppliers process is not meeting expectations, as demonstrated by the sample size average being below the LSL. 

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Shannon Rice
Quality Assurance Evaluator
Hot Springs AR
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Thank you so much! That is what I did, but the book answer is giving me an answer of .333;  so I thought I was doing it wrong?  

There is also a question in the study guide similar this where I would think you would use the same equation as above to find Cp but it says to use this equation instead:

Cp=USL-LSL/ 6(R bar/d2)

Are you familiar with this equation and why we would use it to find Pp or Cp?

Original Message:
Sent: 01/27/24 08:09 PM
From: Shannon Rice
Subject: Process Performance Pp- please help

80.25 - 79.75 = 0.5

6 x .625 = 3.75 (You multiple the six by your standard deviation)

0.5 / 3.75 = .133

That suppliers process is not meeting expectations, as demonstrated by the sample size average being below the LSL. 

------------------------------
Shannon Rice
Quality Assurance Evaluator
Hot Springs AR
------------------------------