Amachine 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?

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(x<0.496) and P(x>0.504) 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(x<0.496)+P(x>0.504) = 0.3707+0.0099 = 0.3806 or 38.06%

Logical equivalent of the give question is its contrapositive