Question

Diameters of ball bearings are normally distributed with mean 0.498 cm and standard

deviation 0.002 cm. If specifications require 0.5 +- 0.004 cm, what percentage of

production will be rejected?

+- = plus - minus

Answer 1

$P(0.5-0.004<X<0.5+0.004)=P(0.496<X<0.504)=\\=P(\frac{0.496-0.5}{0.002}<Z<\frac{0.504--0.5}{0.002})= P(-2<Z<2)=$

$=P(Z<2)-P(Z<-2)=0.9772-0.0228=0.9544=95.44\%$ .

Thus, $1-0.9544=0.0456=4.56\%$ of production will be rejected.