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Question 6.S6.1: A manufacturer of precision machine parts produces round sha...

A manufacturer of precision machine parts produces round shafts for use in the construction of drill presses. The average diameter of a shaft is .56 inch. Inspection samples contain 6 shafts each. The average range of these samples is .006 inch. Determine the upper and lower \bar{x} control chart limits.

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The mean factor A_2 from Table S6.1, where the sample size is 6, is seen to be .483. With this factor, you can obtain the upper and lower control limits:
UCL_{\bar{x}} = .56 + (.483) (.006)
= .56 + .0029
= .5629 inch
LCL_{\bar{x}} = .56 – .0029
= .5571 inch

TABLE S6.1 Factors for Computing Control Chart Limits (3 sigma)
SAMPLE SIZE, n MEAN FACTOR, A_2 UPPER RANGE, D_4 LOWER RANGE, D_3
2 1.880 3.268 0
3 1.023 2.574 0
4 .729 2.282 0
5 .577 2.115 0
6 .483 2.004 0
7 .419 1.924 0.076
8 .373 1.864 0.136
9 .337 1.816 0.184
10 .308 1.777 0.223
12 .266 1.716 0.284

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