A tire company has developed a new type of steel-belted radial tire. Extensive testing indicates the population of mileages obtained by all tires of this new type is normally distributed with a mean of 37,402 miles and a standard deviation of 4,135 miles. The company wishes to offer a guarantee providing a discount on a new set of tires if the original tires purchased do not exceed the mileage stated in the guarantee. What should the guaranteed mileage be if the tire company desires that no more than 2 percent of the tires will fail to meet the guaranteed mileage?

To find the guaranteed mileage such that no more than 2% of the tires will fail to meet it, we need to find the value \(x\) such that 2% (or 0.02) of the tires have mileages below \(x\). This involves determining a particular percentile of the normal distribution.

Given:
- Mean (\(\mu\)) = 37,402 miles
- Standard deviation (\(\sigma\)) = 4,135 miles
- We want the mileage \(x\) such that 2% of the tires are below this value. This corresponds to the 2nd percentile of the normal distribution.

### Step 1: Find the Z-score corresponding to 2%
Using the standard normal distribution, the Z-score corresponding to the 2nd percentile is approximately \(-2.054\). This value can be found using a Z-table or a statistical calculator.

### Step 2: Use the Z-score formula
The Z-score formula is:
\[
Z = \frac{x - \mu}{\sigma}
\]
Where:
- \(Z\) is the Z-score corresponding to the percentile (in this case, \(-2.054\)),
- \(x\) is the unknown guaranteed mileage,
- \(\mu = 37,402\) is the mean mileage,
- \(\sigma = 4,135\) is the standard deviation.

Now we solve for \(x\):
\[
-2.054 = \frac{x - 37,402}{4,135}
\]

### Step 3: Solve for \(x\)
\[
x = -2.054 \times 4,135 + 37,402
\]
\[
x \approx -8,494.49 + 37,402
\]
\[
x \approx 28,907.51
\]

### Conclusion:
The company should set the guaranteed mileage to approximately **28,908 miles** to ensure that no more than 2% of the tires fail to meet the guaranteed mileage.

Would you like further clarification on any of the steps or additional details?

Here are some related questions for further practice:
1. What would the guaranteed mileage be if only 5% of the tires were allowed to fail?
2. How would the guarantee change if the standard deviation of the tire's mileage increased to 5,000 miles?
3. What is the probability that a randomly selected tire lasts between 30,000 and 40,000 miles?
4. What would the company guarantee if they wanted to increase the failure threshold to 10%?
5. How does changing the mean mileage affect the guaranteed mileage for the same failure percentage?

**Tip:** When working with normal distributions, the Z-score helps to standardize data, making it easier to find probabilities and percentiles.

Learn how to calculate the guaranteed mileage for steel-belted radial tires using normal distribution and Z-scores. This problem involves determining the 2nd percentile of tire mileages with a mean of 37,402 miles and a standard deviation of 4,135 miles. A detailed solution is provided, making it suitable for high school students in Grades 10-12.

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