# hw1sol - ME363 Homework#1 Solution 1. x = 0.0462 (a) x =...

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ME 363 Homework #1 Solution 1. R R R R R (a) 9.2 0.0462 R=0.24 0.0478 UCL 3 9.2 3 0.0462 9.3386 LCL 3 9.2 3 0.0462 9.0614 UCL R 3 0.24 3 0.0478 0.3834 LCL R-3 0.24 3 0.0478 0.0966 (b) U x xx x x x σ == = =+ = +× = =− = −× = =+ = = ×= 22 2 CL A R A 0.729 for n=4 =9.2+0.729 0.24=9.3750 LCL A R =9.2-0.729 0.24=9.0250 (c) See next page x x x x = × × {} pk Alternate Solution based on calculated in (a) UCL LCL = ˆ 6 9.3750 9.0250 = 1.263 60 . 0 4 6 2 min UCL , LCL C 3 min 9.3750 9.2 , 9.0250 9.2 = 30 . 0 4 6 2 0.175 = 3 p C = × −− = × × 1.263 0.0462 = 2 pk 0.24 ˆ (d) 0.1166 2.059 UCL LCL = ˆ 6 9.3750 9.0250 = 0.5 6 0.1166 min UCL , LCL C 3 min 9.3750 9.2 , 9.0250 9.2 = 3 0.1166 0.175 = 0.5 . 1 1 6 6 x p R d C = = × = × = ×

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(c 1) control charts for x 9.05 9.1 9.15 9.2 9.25 9.3 9.35 123456789 1 0 Data Mean UCL LCL (c 2) control chart for R 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 1 0 Data Mean UCL LCL
2. Shaft: mean = 1.983 variance = 0.0042 holes: mean = 2.003 variance = 0.0039

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hw1sol - ME363 Homework#1 Solution 1. x = 0.0462 (a) x =...

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