Question

For a process that is in statistical control calculate the Cp and Cpk index given the following data. The appropriate diagram is required if defective parts exist. If there are out of specification parts determine the percentage defective.

X doublebar = 3.432, R bar = .010, n =5, LSL = 3.405, USL = 3.460

Answer #1

Std deviation , **σ**= R-bar / d2

d2 = 2.326 for n= 5

Std deviation , σ = 0.010 / 2.326 = 0.004

Cp = (USL - LSL) / 6σ = (3.460 - 3.405) / 6 *0.004 = 2.29

Cpk = Min [ (X-doublebar - LSL) / 3σ . (USL - Xdoublebar) / 3σ]

= Min (3.432 - 3.405) / 3 *0.010 , (3.460 - 3.432) / 3 * 0.010)

= 0.90

Z-score upper = (USL - Xdouble bar) / σ = (3.460 - 3.432) / 0.010 = 2.8

When Z-score is 2.8 , the P-value from Standard normal distribution table = 0.9974

I.e. for Upper limit 3.460, 0.9744 are within the limts. Hence defective = 1- 0.9974 = 0.0026 or 0.26%

Z-score lower = (X-doublebar - LSL) / σ = (3.432 - 3.405) / 0.010 = 2.7

When Z-score is 2.7 , the P-value from Standard normal distribution table = 0.9965

Hence defective = 1- 0.9965 = 0.0035 or 0.35%

Total defective =0.26 + 0.35 = 0.61%

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