Answer to Question #220203 in Statistics and Probability for Annon

Question

Nylon strands are manufactured to a mean diameter of 2.5 mm with a standard deviation of 0.4 mm. Quality control accepts strands that are between 1.8 mm and 3.2 mm. Assuming that lengths of the Nylon strands are normally distributed, what percentage of strands will be rejected?

Answer 1

$P(X<1.8)=P(Z<\frac{1.8-2.5}{0.4})=P(Z<-1.75)=0.0401.$

$P(X>3.2)=P(Z>\frac{3.2-2.5}{0.4})=P(Z>1.75)=0.0401.$

Percentage of rejected strands: $0.0401+0.0401=0.0802=8.02\%.$