The lengths of 3 inch nails manufactured on a machine are normally distributed with a mean of 3 and a standard deviation of .009.

The nails are either shorter than 2.98 inches or longer than 3.02 inches are unusable. what percentage of the nails produced by this machine are unusable?

#### Solution

Given that $\mu = 3$ and $\sigma = 0.009$.

The nails are either shorter than 2.98 inches or longer than 3.02 inches are unusable.

The probability that the nails produced by this machine are unusable is

` $$ \begin{aligned} P(X < 2.98 \text{ OR }X > 3.02)&=1- P(2.98 < X < 3.02)\\ &=1-\bigg[P\bigg(\frac{2.98-3}{0.009}\leq \frac{X-\mu}{\sigma} \leq \frac{3.02-3}{0.009}\bigg)\bigg]\\ &=1-\bigg[P\bigg(-2.222 \leq Z \leq 2.222\bigg)\bigg]\\ &= 1-\bigg[P(Z < 2.222) -P(Z < -2.222)\bigg]\\ &=1- \big[0.9869-0.0131\big]\\ &= 1-(0.9737)\\ &= 0.0263 \end{aligned} $$`

$2.63$ % of the nails are unusable.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators