AaBbCcDdEe AaBbCcDdEe Normal No Spacing 6) Requirements for the width of a tract

Question

AaBbCcDdEe AaBbCcDdEe Normal No Spacing 6) Requirements for the width of a tractor engine component are 23.43+/-0.637 millimeters. The current process produces components with an average width of 23.536 and a population standard deviation of 0.126. The process is normally distributed. What is the sigma capability of this process? (Use four decimal places) 7 Requirements for the width of a tractor engine component are 23.07+0.634 millimeters. The current process produces components with an average width of 23.500 and a population standard deviation of 0.079. The process is normally ributed. What is the Cpk of this process? (Use four decimal places) 8) Requirements for the width of a tractor engine component are 23.61 +-0.639 millimeters. The current process produces components with an average width of 23.538 and a population standard deviation of 0.053. 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 Cok of this process? (Use four decimal places) 9) Compare the processes described in the folowing tabie. What is the Cpk ot the process that will generate the fewest defects? Process Mean 197 50 Tolerance 197 +/-51 50 +7- 24 1159 + 346 870 4/-259 18.3 1159 870 114.5 86.2

Explanation / Answer

6)

UCL = 23.43+0.637 = 24.067

LCL = 23.43-0.637 =22.793

Mean = 23.536

standard deviation (s) = 0.126

Process capability = Min ((UCL-Mean)/(3*s), (Mean-LCL)/(3*s))

= Min ((24.067-23.536)/(3*0.126), (23.536-22.793)/(3*0.126))

= Min (1.4048, 1.9656)

= 1.4048

For sigma calculation, sigma = 3*Process capability = 3*1.4048 = 4.2143

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.