In 6 sigma with a nominal specification of 90 If Cp=0.23 , Cpk = -0.02 LSL = 80
ID: 3333176 • Letter: I
Question
In 6 sigma with a nominal specification of 90 If Cp=0.23 , Cpk = -0.02 LSL = 80 , USL = 100 , Mean = 79.1667 StDev overall = 15.5806, StDev within =14.3615 What is the percentage of data that falls higher than upper specification limit ? And how can I caulcate it. In 6 sigma with a nominal specification of 90 If Cp=0.23 , Cpk = -0.02 LSL = 80 , USL = 100 , Mean = 79.1667 StDev overall = 15.5806, StDev within =14.3615 What is the percentage of data that falls higher than upper specification limit ? And how can I caulcate it. If Cp=0.23 , Cpk = -0.02 LSL = 80 , USL = 100 , Mean = 79.1667 StDev overall = 15.5806, StDev within =14.3615 What is the percentage of data that falls higher than upper specification limit ? And how can I caulcate it.Explanation / Answer
probability of higher than upper limit is
p [X > 100]
= p [Z > 100 - 79.1667 / 15.5806]
= p [Z > 1.337]
= 0.0906 [it is obtained from Z table]