Question

Lean Six Sigma
The Lower Specification Limit is 8.0 seconds

36. Representative data gathered from a process that follows a normal distribution shows a mean of 8.2 seconds with a standard deviation of 0.3 seconds. The process has 8.5 seconds as an upper specification limit. What is the process yield? a. Calculate: Cp and Cpk b. How many defects per million opportunities the process produces? (hint: use the Excel spreadsheet given in class) c. What is the Sigma of the process?

Answer 1

a) Calculate Cp and Cpk

Lower specification Limit, LSL 8.0 = Upper specification Limit, USL 8.5 Process mean, i 8.2 Process Standard Deviation, o 0.3 Process capability index, Cpk Lower, Cpl (H-LSL)/30 0.222 Upper, Cu (USL-H)/30 0.333 Process, Cpk min ( Cpu, Cpl) =0.222 Process capability ratio, Cp Cp (USL-LSL)/6a = 0.278 LD

b)

z value for LTL = (8-8.2)/0.3 = -0.667, F(z) = NORMSDIST(-0.667) = 0.2525

z value for UTL = (8.5-8.2)/0.3 = 1, F(z) = NORMSDIST(1) = 0.8413

Fraction defective lower than LTL = 0.2525

Fraction defective greater than UTL = 1 - 0.8413 = 0.1587

Total fraction defective = 0.2525+0.1587 = 0.4112

Defects per million opportunities = 0.4112*10^6 = 411,200

c) Sigma of the process with 1.5 sigma shift = NORMSINV(1-411200/10^6)+1.5 = 1.7245