Question: please awnser C An automatic lathe produces rollers for roller bearings, and the process is monitored by statistical process control charts. The central line of
please awnser C
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 8.60 and for the mean range at 0.25 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 (8,60 +0.25) and (8.60 -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 50 mm and the LCLg equais 0 mm. (Enter your responses rounded to two decimal places) The UCL; equals 8.72 mm and the LCL equals 8,48 mm. (Enter your responses rounded to two decimal places.) b. If the standard deviation of the process distribution is estimated to be 0.15 mm, is the process capable of meeting specifications? Assume four-sigma quality. DA. Yes, because Cox is greater than the critical value of 1.33 B. No, because Cox is less than the critical value of 1.33. C.C. Yos, because Cpk is less than the critical value of 1.33 D. No, because Cok is greater than the critical value of 1.33 c. If the process is not capable, what percent of the output will fall outside the specification limits? (Hint. Refer to the standard normal ab% (Enter your response rounded to two decimal places Type Nif there is no solution) 3 Factors for calculating three-sigma limits for the x-chart and R-chart Size of Sample Factor for UCL and LCL Factor for LCL for Factor for UCL for (n) for X-chart (A2) R-Chart (D3) R-Chart (DA) 2 1.880 0 3.267 1.023 0 2.575 4 0.729 0 2.282 5 0.577 0 2.115 6 0.483 0 2.004 7 0.419 0.076 1.924 0.373 0.136 1.864 9 0.337 0.184 1.816 10 0.308 0.223 1.777 8

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