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
