Question

The length of 5-inch nails manufactured on a machine are normally distributed with a mean of 5.0 inches and a standard deviation of 0.009 inch. The nails that are either shorter than 4.98 inches or longer than 5.02 inches are unusable. What percentage of all the nails produced by this machine are unusable? 

Answer 1

Let $X=$ the length of nail: $X\sim N(\mu, \sigma^2).$

Given $\mu=5.0 \ in, \sigma=0.009\ in.$

$$P(X<4.98\ or\ X>5.02)$$

$$=P(X<4.98)+1-P(X\leq 5.02)$$

$$=P(Z<\dfrac{4.98-5.0}{0.009})+1-P(Z\leq \dfrac{5.02-5.0}{0.009})$$

$$\approx P(Z<-2.222)+1-P(Z\leq 2.222)$$

$$\approx0.013134+0.013134\approx0.026268$$

2.627 % of all the nails produced by this machine are unusable.