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Six Sigma and Process Development/Measurement

The problem I am having trouble figuring out is the following:

Suppose that a process with a normally distributed output has a mean of 55.0 and a variance of 4.0.
a. If the specifications are 55.0 +/- 4.00, compute Cp, Cpk,Cpm.

b. Suppose the mean shifts to 53.0 but the variance remains unchanged. Recompute and interpret these process capability indexes.

c. If the variance can be reduced to 40 percent of its original value, how do the process capability indices change (using the original mean of 55.0) ?

Solution Preview

Please see the attached file.

Suppose that a process with a normally distributed output has a mean of 55.0 and a variance of 4.0.
a. If the specifications are 55.0 +/- 4.00, compute Cp, Cpk,Cpm.
Mean = 55
Standard Deviation = Variance^0.5=4^0.5=2.0
USL = 55+4=59
LSL=55-4=51
Cp = Tolerance / (6*standard deviation) = (USL-LSL)/standard deviation
Cp=(59-51)/(6*2)=0.667

Cpk = Minimum of (USL-mean)/(standard deviation*3) and (mean-LSL)/(standard deviation*3)
=Minimum of (59-55)/(3*2) and (55-51)/(3*2)
= Minimum of 0.667 and 0.667
=0.667

Cpm = Cp = Tolerance / (6*standard deviation from target) = (USL-LSL)/standard deviation from target
Since target here is same as mean, the standard ...

Solution Summary

This solution contains step-by-step calculations and full explanations on how to calculate the Cp, Cpk, and Cpm from using the mean and vairance values.

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