Statistics Homework Solutions 235

Statistics Homework Solutions 235 - 235 SUPPLEMENTARY...

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Unformatted text preview: 235 SUPPLEMENTARY EXERCISES FOR CHAPTER 10 Therefore σ ¡ 0 002171. ¨ 3σ ¡ ¡ ¤ ©¥ £ µ 2 303. ¡¥ ¡ ¤ ¡ ¡ 1, the process capability is acceptable. ¡ (b) Yes. Since C pk USL ¨ ¡ Since µ is closer to USL than to LSL, C pk 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. 0 22, σ ¡ 3 071. ¡ ¢¥ ¡ 3 0 002171 ¨ ¨ ¤ £ ¥ 0 20 ¨ ¨ ¡ 0 22 ¨ ¤ ¡ ¨ ¡ 0 20, then C pk 0 002171. ¡ ¡ ¨ ¡ If µ 0 18, USL ¨ (b) LSL 5. (a) Let µ be the optimal setting for the process mean. µ µ LSL 3σ . 3 60. ¨ ¡ ¡ ¨ ¥ ¨ ¤ ¥ £ ¡ ¡ ¤ ¡¥ ¤ ¥ £ ¡ ¤ ¡ ¡ (b) The z-scores for the upper and lower specification limits are z 3σ 3 6σ. ¢ ¡ 3 6σ and USL µ ¨ ¤ µ USL ¡ Solving for LSL and USL yields LSL 3σ , so 1 2 ¤ ¥ £ LSL ¡¥ µ ¤ 3σ ¡ µ ¤ ¥ £ USL ¥ Then C p Therefore, using the normal curve, the proportion of units that are non-conforming is the sum of the areas under 3 60. the normal curve to the right of z 3 60 and to the left of z ¡ ¡ ¨ ¡ 0 0004. ¨ ¡ 0 0002 ¨ ¨ ¢ The proportion is 0 0002 ¨ (c) Likely. The normal approximation is likely to be inaccurate in the tails. Supplementary Exercises for Chapter 10 ¡ ¨ 0 1045. ¨ ¡ £ ¥ p 250 0 0147. ¡ £ ©¥ ¡ ¤ p1 p 250 ¨ 3. The upper and lower limits for the R-chart are D 3 R and D4 R, respectively. 0 and D4 ¡ ¡ ¤ ¡ ¢ ¡ From the control chart table, D3 2 575. ¨ ¡ p1 3. (a) The sample size is n 0 110. Therefore LCL = 0 and UCL = 0.283. The variance is in control. ¨ ¡ R 0 0596. 0 0596 ¡ 3 ¡ The UCL is p 3 ¨ The LCL is p 2 98 50 £ The centerline is p 250. p ¡ The sample size is n ¨ 1. ...
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