A mechanic sells a brand of automobile tire that has a life expectancy that is normally distributed, with a mean life of 31,000 miles and a standard deviation of 2100 miles. He wants to give a guarantee for free replacement of tires that don't wear well. How should he word his guarantee if he is willing to replace approximately 10% of the tires?

#### Solution

Let $X$ denote the life expectancy of automobile tire. Given that $X\sim N(31000,2100^2)$ distribution.

That is $\mu = 31000$ and $\sigma = 2100$.

Let the 10th percentile of $X$ is $a$.

` $$ \begin{aligned} & P(X < a) =0.1\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma} < \frac{a-31000}{2100}\big)=0.1\\ &\Rightarrow P(Z < \frac{a-31000}{2100}\big)=0.1\\ &\Rightarrow \frac{a-31000}{2100}= -1.282\\ & \qquad \text{(From normal statistical table)}\\ &\Rightarrow a = 31000 + -1.282* 2100\\ &\Rightarrow a = 28307.8\\ &\Rightarrow a \approx 28308 \end{aligned} $$ `

Thus tires that wear out by $a \approx 28308$ miles will be replaced free of charge.

#### Further Reading

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