Question

1) Requirements for the width of a tractor engine component are 23.57 +/- 0.659 millimeters. The current process produces components with an average width of 23.500 and a population standard deviation of 0.099. The process is normally distributed. What is the sigma capability of this process? (Use four decimal places)

2) Requirements for the width of a tractor engine component are 23.32 +/- 0.638 millimeters. The current process produces components with an average width of 23.458 and a population standard deviation of 0.071. The process is normally distributed.

What is the Cpk of this process?

(Use four decimal places)

3) Requirements for the width of a tractor engine component are 23.37 +/- 0.602 millimeters. The current process produces components with an average width of 23.512 and a population standard deviation of 0.061. The process is normally distributed.

If the control limits are set at +/- 2 standard deviations instead of +/- 3 standard deviations from the mean, what is the Cpk of this process?

(Use four decimal places)

Answer 1

1)

UCL = 23.57 + 0.659 = 24.229

LCL = 23.57 - 0.659 = 22.911

Mean = 23.500

SD = 0.099

Sigma Capability = Min [(UCL-Mean)/Sd, (Mean-LCL)/SD]

Sigma Capability = Min [(24.229-23.500)/0.099 , (23.500-22.911)/0.099]

Sigma Capability = Min (7.3636,5.9495)

Sigma Capability = 5.9495

2)

UCL = 23.32 + 0.638 = 23.958

LCL = 23.32 - 0.638 = 22.682

Mean = 23.458

SD = 0.071

Cpk = Min [(UCL-Mean)/(3*Sd) , (Mean-LCL)/(3*SD)]

Cpk = Min [(23.958-23.458)/(3*0.071) , (23.458-22.682)/(3*0.071)]

Cpk = Min (2.3474,3.6432)

Cpk = 2.3474

3)

UCL = 23.37 + 0.602 = 23.972

LCL = 23.37 - 0.602 = 22.768

Mean = 23.512

SD = 0.061

Cpk = Min [(UCL-Mean)/(3*Sd) , (Mean-LCL)/(3*SD)]

Cpk = Min [(23.972-23.512)/(3*0.061) , (23.512-22.768)/(3*0.061)]

Cpk = Min (2.5137,4.0656)

Cpk = 2.5137