A machine operation produces bearings whose diameters are normally distributed, with a mean of 0.497 inch and a standard deviation of 0.003 inch. suppose that specifications require that the bearing diameter be 0.500 inch plus or minus 0.004 inch. normal distribution: $/mu = 0.497$, $/sigma = 0.003$. specifications: (0.496, 0.504) what proportion of the production will be unacceptable?

Question

Mathematics

Kasen Lynn

7 January, 22:04

A machine operation produces bearings whose diameters are normally distributed, with a mean of 0.497 inch and a standard deviation of 0.003 inch. suppose that specifications require that the bearing diameter be 0.500 inch plus or minus 0.004 inch. normal distribution: $/mu = 0.497$, $/sigma = 0.003$. specifications: (0.496, 0.504) what proportion of the production will be unacceptable?

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Alberto Lambert · Answer 1

Mean = 0.497 in, SD = 0.003 in

Required diameter ranges between 0.496 in and 0.504 in

Anything other diameter obtained is not acceptable.

That is;

P (x0.504) are not acceptable.

Now,

P (x<0.496) = P (Z< (0.496-0.497) / 0.003)) = P (Z<-0.33)

From Z tables, P (Z<-0.33) = 0.3707

Similarly,

P (x>0.504) = P (Z> (0.504-0.497) / 0.003)) = P (Z>2.33)

From Z tables, P (Z>2.33) = 1-0.9901 = 0.0099

Therefore, unacceptable proportion = P (x0.504) = 0.3707+0.0099 = 0.3806 or 38.06%