12.2. Consider shafts with specification limits 2.0002 and 2.0019 in. The matching bearings also carry specification limits of 2.0009 and 2.0033 in. Note the sizable overlapping of the two sets of limits. Suppose that these limits are met in a

3s sense with normal distributions. Make a guess as to the proportion of random choices of bearing and shaft, which will have negative clearance, that is, will not go together. To illustrate, throw three dice for the number of 0.0001 in. above 2.0000 in. for a shaft, for example, 12 total is 2.0012 in. Likewise, throw six dice (or three, twice) for the inside diameter of a bearing. The difference then is the diametral clearance. You can tabulate the differences, for example, 2012¼8. Obtain 50 clearances by dice throws. For three-dice totals, m ¼ 10.5, s ¼ 2.96, and for six-dice totals m¼21, s¼4.18. Use (12.7)–

(12.9) to find m and s for the clearance, then use (2.11) and Table A to approximate the percentage of clearances below zero. Is this less than you expected?