Question: PART C ONLY PLS! The table below shows the total area under the normal curve for a point that is z standard deviations to the

PART C ONLY PLS!
PART C ONLY PLS! The table below shows the total
PART C ONLY PLS! The table below shows the total
PART C ONLY PLS! The table below shows the total
PART C ONLY PLS! The table below shows the total
The table below shows the total area under the normal curve for a point that is z standard deviations to the right of the mean. Z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 0.00 0.01 0.02 0,03 0.04 0.05 0.06 0.07 0.08 0.09 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5754 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.6554 0.6591 0.6628 0.66640.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.7258 0.72910.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.75180.7549 0.7580 0.7612 0.7642 0.7673 0.77040.7734 0.7764 0.7794 0.7823 0.7852 0.7881 0.79100.7939 0.7967 0.7996 0.8023 0.8051 0.8079 0.8106 0.8133 0.8159 0.8186 0.8212 0.8238 0.82640.8289 0.8315 0.8340 0.8365 0.8389 0.84130.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.87T0 0.8790 0.8810 0.8830 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.89620.8980 0.8997 0.9015 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 027 non 2011 ANINA 10 Innl OVAC An automatic lathe produces roller for for bearings, and the process is monitored by statistical process control charts. The centralne of the chart for the sample means is set at 8.00 and for the meant 0.34 The process is in controla stablished by samples of the upper and lower specification for the diameter of the role 8.00 +0.25) and 360-0.25), respectively Click the icon to view the table of factors for calculating to soma limits for the chart and chart Calculate the controllimits for the mean and range charts The Ultim and the L.Cheque omm (Enter your response rounded to two decimo pice) The UCL,X71m and the LC 348 m. (Enter your responses rounded to two decina pame) te standard deviation of the procederbution it and to be 0.11 nm na procent of any specification? Antune tour digna qully CA Y because than the value of 133 O Y Decause is greater than the critical of 133 We No. because is less than the entical value of 1.33 O No. because a greater than the oral value of 133 c. If the process is not capable what percent of the actuall outside the specification mit? Mint Refer to the roa) 10:40 (ar your response rounded to two decimal place Typen the 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 1.023 0.729 5 0.577 6 0.483 0.419 0.373 9 0.337 10 0.308 OVOU AWN 0 0 0 0 o 0.076 0.136 0.184 0.223 Factor for UCL for R-Chart (D) 3.267 2.575 2.282 2.115 2.004 1.924 1.864 1.816 1.777 An automatic same produces roller for roter Deanngs, and the process is monitored by statical process control charts. Ine centraline of the chart for the sample means is set to and for the mean tange at 0.34 mm. The process is in control, as established by samples of size 6. The upper and lower specifications for the diameter of the roles aro (8.60 025) 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 1. Calculate the control limits for the mean and range charts The Uche equals 68 mm and the L. CLx quails 0 mm. (Enter your responses rounded to two decimal places) The UCL, equals 8.71 mm and the LCL, equals 8,49 mm. (Enter your responses rounded to two decimal places) bilt the standard deviation of the process distribution is estimated to be 0.11 mm, in the process capable of meeting specifications? Assume four-sigma quality. O Yos, because he less than the ortical value of 133 Yes, because greater than the critical value of 133 No, because he is less than the critical value of 1.33 CD No, because it is greater than the critical value of 133 t. If the process is not capable, what percent of the output will fall outde the specification imito? Hint Refer to the standard normal sabat) 18:40 %. Enter your reasons rounded to nec

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