# Math 370 HW 10 - Homework 10 Solution 8-18, 8-19, 8-28,...

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1Homework 10 Solution8-18, 8-19, 8-28, 8-29, 8-31, 8-328-18.a) If the process uses 66.7% of the specification band, then 6= 0.667(USL-LSL). Therefore, 1/0.667 = (USLLSL)/6s =1/0.667 = 1.5Because the process is centered)(667.0)(667.03LSLUSLTherefore (USLm)/3s = 1/0.667 = 1.5 and (mUSL)/3s = 1/0.667 = 1.5 and Cpk = 1.5Because Cpand Cpkexceeds unity, the natural tolerance limits lie inside the specification limits and few defective unitswill be produced.b) Assuming a normal distribution with 6= 0.667(USLLSL) and a centered process, then 3= 0.667(USL).Consequently, USL= 4.5andLSL = 4.5011)5.4(1)5.4(5.4)(ZPZPZPUSLXPBy symmetry, the fraction defective is 2[P(X > USL)] = 0.8-19.a)0002922.0ˆor922.2693.1947.4drˆ2141.1)0002922.0(64020.04040.0)ˆ(6LSLUSLCP6525.0]629.1,6525.0min[)0002922.0(34020.04030428.0,)0002922.0(3403428.04040.0minˆ3,ˆ3min