Question: i am trying to find C? please help An automatic lathe produces rollers for roller bearings, and the process is monitored by statistical process control
i am trying to find C? please help
An automatic lathe produces rollers for roller bearings, and the process is monitored by statistical process control charts. The central line of the chart for the sample means is set at 8 10 and for the mean range at 031 mm. The process is in control , as established by samples of size 6. The upper and lower specifications for the diameter of the rollers are (8 10 +0.25) and (8.10 - 0 25) mm, respectively Click the icon to view the table of factors for calculating three-sigma limits for the x-chart and R-chart a. Calculate the control limits for the mean and range charts The UCLA equals 0.62 mm and the LC e quals 0 mm. (Enter your responses rounded to two decimal places) The UCL equals 625 mm and the LCL ; equals 795 mm (Enter your responses rounded to two decimal places) b. If the standard deviation of the process distribution is estimated to be 0.19 mm, is the process capable of meeting specifications? Assume four-sigma quality On No, because Cpk is greater than the critical value of 1.33 8. Yes, because Cpk is less than the critical value of 133 C. No, because Cyk is less than the critical value of 1.33 OD Yes, because Cox is greater than the critical value of 133 c. If the process is not capable, what percent of the output will fall outside the specification limits? (Mint. Refer to the standard pomal table.) % (Enter your response rounded to huo decimal places Type Nil there is no solution) Factors for calculating three-sigma limits for the x-chart and R-chart Factor for LCL for R-Chart (D3) Size of Sample Factor for UCL and LCL (n) for x-chart (A2) 2 1.880 3 1.023 4 0.729 5 0.577 6 0.483 7 0.419 8 0.373 9 0.337 10 0.308 0 0 0 0 0 0.076 0.136 0.184 0.223 Factor for UCL for R-Chart (DA) 3.267 2.575 2.282 2. 115 2.004 1.924 1.864 1.816 1.777

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