Question

The specification limits for the width of a part produced in a new machine is 3 ? 0.008 centimeters. The standard deviation of the process is estimated to be 0.003 centimeters. a. What is the value of process capability index (Cp) for this process? (2 points) b. What percentage of the parts would be outside the specifications for the width of the part if the process were operated and centered at 3.000 centimeters? Assume normal distribution of measurements. (4 points)

Answer 1

USL = 3+0.008 = 3.008 cm

LSL = 3-0.008 = 2.992 cm

Standard deviation, s = 0.003

a) Cp = (USL-LSL)/6s = (3.008-2.992)/(6*0.003) = 0.889

b) Process mean, m = 3.000

For USL, z = (USL-m)/s = (3.008-3.000)/0.003 = 2.67, P(x>3.008) = 1-NORMSDIST(2.67) = 1-0.9962 = 0.0038

For LSL, z = (USL-m)/s = (2.992-3.000)/0.003 = -2.67, P(x<2.992) = NORMSDIST(-2.67) = 0.0038

Percentage of parts outside the specification = P(x>3.008) + P(x<2.992) = 0.0038+0.0038 = 0.0076 or 0.76 %