Normal No Spacin 4) Tolerances for a new part require weights between 243.9 and

Question

Normal No Spacin 4) Tolerances for a new part require weights between 243.9 and 260.1 pounds. The part is made using a process that has a mean of 256.0 pounds with a standard deviation of 3.8 pounds. The process population is normally distributed. What is the sigma capability for this process? (use four decimal places) 5) Tolerances for a new part require weights between 249.6 and 268.4 pounds. The part is made using a process that has a mean of 256.8 pounds with a standard deviation of 2.8 pounds. The process population is normally distributed. If the process can be adjusted so that it is centered, what will be the sigma capability for the adjusted process? (use four decimal places) kindle

Explanation / Answer

4)

Given that,

USL = 260.1

LSL= 243.9

Process mean, m = 256

Standard deviation, s = 3.8

Process capacity index, Cpk = MIN((USL-m)/3s, (m-LSL)/3s)

= MIN((260.1-256)/(3*3.8), (256-243.9)/(3*3.8))

= MIN(0.3596, 1.0614)

= 0.3596

Cpk is less than 1.33, therefore process is not capable of producing within the specifications.

4)

Given that,

USL = 268.4

LSL= 249.6

Adjusted process mean, m = (268.4+249.6)/2 = 259

Standard deviation, s = 2.8

Process capacity index, Cpk = MIN((USL-m)/3s, (m-LSL)/3s)

= MIN((268.4-259)/(3*2.8), (259-249.6)/(3*2.8))

= MIN(1.1190, 1.1190)

= 1.1190

Sigma capability of the adjusted process can be simply calculated as Cp = (USL-LSL)/6s = (268.4-249.6)/(6*2.8) = 1.1190

Cpk (and Cp) is less than 1.33, therefore process is not capable of producing within the specifications.

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