Question

A key joint in a precision machining process has a lower specification limit of a width of 0.99 mm and an upper specification limit of 1.01 mm. The standard deviation is 0.005 mm and the mean is 1 mm. The company wants to reduce its defect probability and operate a "Six Sigma process." To what level would this company have to reduce the standard deviation in the process to meet this target? Note: A "Six Sigma process" has a process capability index Cp value of 2.0.

A. 0.000001

B. 0.00001

C. 0.0017

D. 0.07

Answer 1

Process capability Index, Cp is defined as :

Cp =   ( USL – LSL) / 3 X Sd

Where,

USL = Upper Specification Limit =    1.01 mm

LSL = Lower Specification Limit = 0.99 mm

Sd = Corresponding Standard deviation of the process

Since . Cp = 2 for the “six sigma process”.

Cp = ( USL – LSL) / 6 x Sd

Or, 2 = ( 1.01 – 0.99) / 6xSd

Or, Sd = ( 1.01 – 0.99) / 12 =0.02 /12 = 0.00166 ( or 0.0017 by rounding to 4 decimal places)

Therefore, Standard deviation has to be reduced to = 0.0017 mm

COMPANY HAS TO REDUCE THE STANDADRD DEVIATION IN THE PROCESS TO THE LEVEL OF 0.0017 MM