A machining operation produces bearings with diameters that are normally distributed with mean 3.0005 inches and standard deviation .0010 inch. Specifications require the bearing diameters to lie in the interval 3.000±.0020 inches. Those outside the interval are considered scrap and must be remachined. With the existing machine setting, what fraction of total production will be scrap?

The 68-95-99.7 rule for the Normal distribution is an empirical rule that remind us the percentages of data that falls between the mean ± 1, ± 2 and ± 3 standard deviations.

That is to say, if the mean is m and the standard deviation s, roughly speaking 68% of the data falls between [m-s, m+s], 95% between [m-2s, m+2s] and 99.7% between [m-3s, m+3s].

Since the mean is 3.0005 and the standard deviation is s=0.0010, 2s=0.0020, 95% of the data should fall between [3.0005-0.0020, 3.0005+0.0020] and 5% outside this interval. So around 5% of total production will be scrap.