(a) A machine spindle has a specification (tolerance) on its diameter of 1.500 ± 0.009 inches. If the Process Capability Ratio Cp = 1.0, what is the standard deviation of the spindles being produced by the cylindrical grinder?
(b) What would the standard deviation have to be to achieve a process capability index of 1.33?
(c) If Cp = 1.33 and the process mean is centered within the tolerance range, how many oversized parts would be expected in grinding the spindle described in Problem 1 (b)?
(d) The process mean has moved 1.5 times the center of the tolerance range. From Example 1 (b), σ = 0.00226 inches. The shift k = 1.5(0.00226) = 0.003 inches toward the USL. Now μ = 1.500 + 0.003 = 1.503. Determine Cpkl and Cpku and the effect of the shift in the process mean on the defect rate by comparing the percentage of good parts produced and defective parts produced ppm for different sigma levels of tolerance by filling the following table (Show the related calculations for the probability of parts within the tolerance range and falling outside the tolerance range for each sigma level).
Process Centered | Process Mean 1.5 Sigma from Center
--- | ---
Tolerance Range (± σ level) | Cp | Percent good parts | Defective parts ppm | Percent good parts | Defective parts ppm
±3 sigma | 1.00 | | | |
±4 sigma | 1.33 | | | |
±6 sigma | 2.00 | | | |