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Diano4ka-milaya [45]
3 years ago
6

A machining operation produces bearings with diameters that are normally distributed with mean 3.0005 inches and standard deviat

ion .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?
Mathematics
1 answer:
padilas [110]3 years ago
7 0

Answer:

5%

Step-by-step explanation:

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 <em>around 5% of total production will be scrap</em>.

 

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