例2 测试一批零件外径尺寸的平均值=190101,S=0.0143,规 格要求93,试计算工序能力指数并估计不合格品率。 解:由题意: T=19.04T=18.97 T=0.07 T,+ 19.005≠x=190101 e 9005-190101=0005 计算CpFp T-2e0.07-2×0.0051 ≈0.70 6S 6×0.0143 19.005-19010l 0.145 0.07/2 0.07 0.816 6×0.0143 Cnk=(1-k)C,=(1-0.145)×0.816≈0.7 Fp=1-[中 1904-19010 18.97-19.0101 00145 0.0143 1-Φ209y-(2841=0021=21% 或由C=0.816,k=0145查表得不良品率估计约为21%23%9 例2 测试一批零件外径尺寸的平均值 =19.0101，S=0.0143，规 格要求为 解：由题意： 计算Cpk 0.04 19 0.03 +  − 19.005 19.0101 2 19.04 =  = + = = x T T T T U L m U TL =18.97 T = 0.07 (1 ) (1 0.145) 0.816 0.7 0.816 6 0.0143 0.07 0.145 0.07 2 19.005 19.0101 0.70 6 0.0143 0.07 2 0.0051 6 2 19.005 19.0101 0.0051 = − = −   =  = = − =   −  = − = = − = − = p k p p p k m C k C C k S T e C e T x 1 [ ] ) 0.0143 18.97 19.0101 ) ( 0.0145 19.04 19.0101 ( = − Φ − −Φ − p =1−[(2.093) −(−2.804) ] = 0.021 = 2.1% 或由Cp=0.816，k=0.145查表得不良品率估计约为2.1%～2.3% x