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Problem 3-18 a. UCL x = x + A3s = 11.22 + (1.7665)(1.427) = 13.72 LCL x = x − A3s = 11.2 − (1.7665)(1.427) = 8.68 All plot within limits, so no instabilities UCLs = B 4s = 2.089(1.7665) = 3.69 LCLs = B3s = None All plot within limits, so no instabilities b. σ(hat) = s / c 4 = 1.776 / .94 = 1.879 c. Pr[ z p 0 − µ/σ ] = [ z p (0 − 11.2) / 1.879] = [ z p −5.957] ≈ 0 0% will be below nominal in length d. Each 25” above nominal = $100,000/year loss…if we take our µ from 11.2 to 5.6277 this would give us a (11.2-5.6377)(1/64) = .0869 inch reduction Savings will be .0869 inches (1/.25inches)($100,000) = $134,764 e. x = 5.6377 n=3 CL = 5.6377 UCL x = µ + 3(1.879 / 3 ) = 8.89 LCL x = µ − 3(1.879 / 3 ) = 2.38 UCLs = B6σ = 4.277 LCLs = B5σ = none σ(hat) = 1.879 (from part b) Problem 3-22 a. UCL x = x + A2 R = 1.18097 + (.0004)(.729) = 1.181265 LCL x = x − A2 R = 1.18097 − (.0004)(.729) = 1.180675 UCLr = D 4 R = (.0004)(2.282) = .0009128 LCLr = D3R = (.0004)(0) = none x chart indicates the alarm bell ringing 8 times R chart indicates no instabilities b. 1st rule: A single point is outside 3σ controls limits. This alarm bell rings 5 times with 5 points being outside the 3σ limits. The first time it rings is at x = 4. 2nd rule: 2 out of any 3 consecutive points outside 2σ limits on one side of the center. This alarm rings at x = 3, 4, 5 for the first time with 4 and 5 being outside the 2σ limits. 3rd rule: 4 out of any 5 consecutive points outside 1σ limit on one side of the center. This alarm rings for the first time with x = 2, 3, 4, 5, 6 with 2, 4, 5, 6 being out of the1σ limit. 4th rule: 8 consecutive points on one side of the center. This alarm bell doesn’t ring! c. σ = R / d 2 = .0004 / 2.059 = .0001943 d. 1.1809 + / − .005" 1 – P[# of diameters not meeting specifications] 1 – P[-25.99 p z p 24.99] = 1-0 = 100% of the diameters meet specifications e. σ = .0001943 Es = C 4σ = (.0001943)(.94) = .0001826 n=5 UCL = B 4 s = (.0001826)(2.089) = .000381 LCL = B3s = (.0001826)(0) = None