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Statistics Homework Solutions 235

Statistics Homework Solutions 235 - 235 SUPPLEMENTARY...

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SUPPLEMENTARY EXERCISES FOR CHAPTER 10 235 Therefore σ 0 002171. Since μ is closer to USL than to LSL , C pk USL μ 3 σ 2 303. (b) Yes. Since C pk 1, the process capability is acceptable. 3. (a) The capability is maximized when the process mean is equidistant from the specification limits. Therefore the process mean should be set to 0.20. (b) LSL 0 18, USL 0 22, σ 0 002171. If μ 0 20, then C pk 0 22 0 20 3 0 002171 3 071. 5. (a) Let μ be the optimal setting for the process mean. Then C p USL μ 3 σ μ LSL 3 σ , so 1 2 USL μ 3 σ μ LSL 3 σ . Solving for LSL and USL yields LSL μ 3 6 σ and USL μ 3 6 σ . (b) The z -scores for the upper and lower specification limits are z 3 60. Therefore, using the normal curve, the proportion of units that are non-conformingis the sum of the areas under
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