MT2Sol2010Sec15B

MT2Sol2010Sec15B - C p = (US-LS)/6 = (100.5-99.5)/(6x0.414)...

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DS 412 Sec 15 Midterm 2 Solution (Blue) Spring 2010 1. Let V = annual volume P,Q: (50-35)V – 480,000 = (48-36)V – 360,000, so 3V = 120,000 or V = 40,000 P: (50-35)V – 480,000 = 0, so 15V = 480,000 or V = 32,000 Q: (48-36)V – 360,000 = 0, so 12V = 360,000 or V = 30,000 Location Volume P ≥ 40,000 Q 30,000 – 40,000 2. Warehouse at P: Cost/year = (90x90x0.15) + (110x120x0.124) = $2,799 Warehouse at Q: Cost/year = (120x90x0.15) + (110x130x0.13) = $3,479 Warehouse at R: Cost/year = (120x120x0.12) + (90x130x0.13) = $3,249 Choose P 3a. = 99.765 gms., = 1.077 gms. n = 5, so A 2 = 0.58, D 3 = 0, D 4 = 2.11 Mean chart : UCL = + A 2 = 99.765 + (0.58x1.077) = 100.39 gms. LCL = - A 2 = 99.765 - (0.58x1.077) = 99.14 gms. Range chart : UCL = D 4 = 2.11x1.077 = 2.272 gms. LCL = D 3 = 0x1.077 = 0 gms. b. Sample mean = (100.2+99.5+99.7+99.8+100.1)/5 = 99.86 gms. . Sample range = 100.2 – 99.5 = 0.7 gms. Yes, the sample satisfies control limits for both charts. c. = 99.765 gms., σ = 0.414 gms., US = 100.5 gms., LS = 99.5 gms.
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Unformatted text preview: C p = (US-LS)/6 = (100.5-99.5)/(6x0.414) = 0.40; Not capable because C p < 1. C pk = Min{(US-)/3 , (-LS)/3 } = Min{(100.5-99.765)/(3x0.414), (99.765-99.5)/(3x0.414)} = 0.725; Not capable because C pk < 1. Since C p < 1, the variability (std. dev.) must be reduced. Further, the mean must be increased, since C pk < 1. 4a. A p-chart should be used because control limits are required for the proportion of wrong reports. Also, the sample size and # defective are given. b. = 20/2000 = 0.01, n = 200, z = 3 = = 0.007 UCL = + z = 0.01 + (3x0.007) = 0.031 LCL = - z = 0.01 (3x0.007) = -0.011 0 c. Sample p 1 1/200 = 0.005 2 3/200 = 0.015 3 3/200 = 0.015 4 5/200 = 0.025 5 2/200 = 0.01 All 5 samples (100%) lie between control limits. We need at least 99.7% with 3-sigma limits, so the process is in control....
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MT2Sol2010Sec15B - C p = (US-LS)/6 = (100.5-99.5)/(6x0.414)...

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