Let [tex]X[/tex] be the random variable for the part's diameter, and [tex]Z[/tex] the same variable transformed into one that follows the standard normal distribution. You're looking for the proportion of the parts that lie 2 standard deviations away from the mean, which translates to
[tex]\mathbb P(|X|>15.000+2\times0.030)=\mathbb P(|X|>15.060)=\mathbb P(|Z|>2)[/tex]
Recall the empirical rule, which says that approximately 95% of a normal distribution falls within 2 standard deviations of the mean, or [tex]\mathbb P(|Z|<2)\approx0.95[/tex]. This means
[tex]\mathbb P(|Z|>2)=1-\mathbb P(|Z|<2)\approx0.05=5\%[/tex]
So out of 15000 parts, you should expect about 5% of them, or about 750 parts to be rejected.
