Example of Process Capability

The specifications for one characteristic of a part call for its width to be 3.000 + 0.008 centimetres. The procedure has been run under controlled conditions so that no assignable causes of variation have been introduced, & samples have been taken. The standard deviation of the process was calculated to be 0.003 centimeter.

A.    What is C_p for this process? What does this value say regarding the process capability?

B.     What percentage of the units would be outside the specifications for the width of the part if the procedure were operated and centered at 3.000 centimeters? Suppose a normal distribution of measurements.

Solution

(a)     C p   = USL - LSL/6 σ = 3.008 - 2.992/6 (0.003)

                   = 0.016/0.018

                    = 0.889

As the index is less than 1, the process is not capable of generating all the items within the specifications limits, even if the process is operated so that its mean is centered exactly among the specification limits.

 (b)       As the normal distribution is symmetrical and centered among the specification limits, the portion outside the upper & lower limits shall be twice the portion outside one limit. The portion outside the upper limit might be found by converting the upper limit to a Z value and referring to the normal distribution curve  as following:

Z =  x - μ / σ  =  3.008 - 3.000/0.003 =  0.008/0.003 = 2.667 or 2.67

The area to the left of a Z of 2.67 is 0.9962. The area to the right of this value is 1 - 0.9962, or 0.0038. As the curve is symmetrical, an equal amount shall be to the left of the lower specification limit. The overall portion outside the specifications shall be 2 (0.0038) = 0.0076, that is 0.76%.