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(3a). A process makes parts with a critical dimension that can be modeled as a N

ID: 420078 • Letter: #

Question

(3a). A process makes parts with a critical dimension that can be modeled as a Normal distribution with mean 39.98 and standard deviation 0.08 inches. If the specification for that dimension is 39.8 to 40.2 inches, what percentage of output would be defective? (b). A process makes parts with a critical dimension that can be modeled as a Normal distribution with mean 40.0 and standard deviation 0.08 inches. If the specification for that dimension is 39.8 to 40.2 inches, what should be the standard deviation so that it becomes a six-sigma process? (Provide three significant digits to the right of the decimal point)

Explanation / Answer

(a) process mean, m = 39.98

Standard deviation, s = 0.08

For lower specification, z = (39.8-39.98)/0.08 = -2.25, P(x<39.8) = NORMSDIST(-2.25) = 0.0122

For upper specification, z = (40.2-39.98)/0.08 = 2.75, P(x>40.2) = 1-NORMSDIST(2.75) = 0.0030

Percentage of defective output = 0.0122 + 0.0030 = 0.0152

(b) For 6 sigma process, process capability index Cpk needs to be equal to 2.

Process mean is the arithmatic mid point of specification limits (39.8+40.2)/2 = 40. Therefore, process is centered. So, Cpk = Cp

Cp = (USL-LSL)/6s = (40.2-39.8)/6s = 2

Solving for s, we get, s = (40.2-39.8)/(6*2) = 0.033

Therefore, standard deviation should be 0.033 or lesser