To solve this problem, we need to find the standard deviation ($\sigma$) of the lifetime of the tires so that only 1% of the tires fail before reaching the 60,000-mile warranty limit. The tire lifetimes are normally distributed with a mean of 74,500 miles.

Step-by-step Solution:

1. Set up the problem using the normal distribution formula:

   • Mean ($\mu$) = 74,500 miles
   • Standard deviation ($\sigma$) = unknown
   • $P(X < 60,000) = 0.01$ (1% of tires fail before 60,000 miles)

2. Find the z-score corresponding to a cumulative probability of 0.01: Using a standard normal distribution table or a calculator:

   • $z = -2.3263$ (negative because it's in the left tail)

3. Relate z-score to the normal distribution formula: The z-score formula is: $z = \frac{X - \mu}{\sigma}$ Substituting the values: $-2.3263 = \frac{60,000 - 74,500}{\sigma}$

4. Solve for $\sigma$: Rearrange the formula to isolate $\sigma$: $\sigma = \frac{60,000 - 74,500}{-2.3263}$ Calculate the numerator: $60,000 - 74,500 = -14,500$ Divide by $-2.3263$: $\sigma = \frac{-14,500}{-2.3263} = 6234.1632 \, \text{miles}$

Final Answer:

$\sigma \approx 6234.1632 \, \text{miles}$ The standard deviation should be set to approximately 6234.1632 miles.

Let me know if you'd like further clarification or additional assistance!

Follow-up Questions:

1. What does the z-score represent in a normal distribution?
2. How does adjusting the standard deviation affect the failure rate of tires?
3. What other factors could impact the lifetime of these tires?
4. Why is it important to round to multiple decimal places in such calculations?
5. How would the answer change if the warranty mileage increased?

Tip: Always double-check cumulative probabilities when working with z-scores to ensure you’re interpreting the left and right tails correctly.