A mechanic sells a brand of automobile tire that has a life expectancy that is normally​ distributed, with a mean life of 

25 comma 00025,000 

miles and a standard deviation of 

28002800 

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?

To determine how the mechanic should word his guarantee, we need to find the mileage threshold below which approximately 10% of the tires will fail. This involves using the properties of the normal distribution.

Given:
- Mean (μ) = 25,000 miles
- Standard deviation (σ) = 2,800 miles
- Desired proportion of tires to replace = 10%

Step-by-step solution:

1. **Determine the z-score corresponding to the 10th percentile of the normal distribution:**

   The 10th percentile corresponds to a cumulative probability of 0.10. We can use the standard normal distribution table or a calculator to find the z-score for this probability.

   \[
   P(Z \leq z) = 0.10
   \]

   From the standard normal distribution table, the z-score that corresponds to a cumulative probability of 0.10 is approximately:

   \[
   z \approx -1.28
   \]

2. **Convert the z-score to the corresponding mileage:**

   Use the formula for converting a z-score to an x-value in a normal distribution:

   \[
   x = \mu + z\sigma
   \]

   Substitute the given values into the formula:

   \[
   x = 25,000 + (-1.28 \times 2,800)
   \]

   Calculate the value:

   \[
   x = 25,000 - 3,584
   \]

   \[
   x \approx 21,416
   \]

Therefore, the mechanic should word his guarantee such that he will replace tires that wear out before they reach approximately 21,416 miles. This ensures that approximately 10% of the tires, which are expected to wear out sooner than this mileage, will be covered under the guarantee.

### Suggested Wording for the Guarantee:
"The tires are guaranteed for 21,416 miles. If the tire wears out before reaching 21,416 miles, it will be replaced free of charge."

Would you like more details or have any questions about this solution? Here are some related questions you might consider:

1. How is the z-score determined for different percentiles?
2. What is the relationship between standard deviation and the spread of the data?
3. How does the normal distribution apply to real-world scenarios?
4. What other factors might affect the life expectancy of tires?
5. How can one calculate the probability for other percentages of tire replacements?

**Tip:** When working with normal distributions, remember to check if the data set follows a normal distribution before applying related statistical methods.

Learn how a mechanic determines a tire guarantee based on normal distribution and probability. Includes step-by-step calculation using mean and standard deviation. Suitable for high school students studying statistics and probability.

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