The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000 miles and a standard deviation of 2900 miles. What warranty should the company use if they want the top $96 \%$ of the tires to last longer than the warranty? 62,900 miles 65,075 miles 54,925 miles 57,100 miles

Step 1 ：Calculate the z-score corresponding to the top 96% of the normal distribution using the formula: \( z = \frac{x - \mu}{\sigma} \)

Step 2 ：Find the z-score that corresponds to a cumulative probability of 0.96, which is approximately 1.75

Step 3 ：Use the z-score to find the warranty mileage using the formula: \( x = z \cdot \sigma + \mu \)

Step 4 ：Plug in the values: \( x = 1.75 \cdot 2900 + 60000 \)

Step 5 ：Calculate: \( x \approx 5075 + 60000 \)

Step 6 ：The warranty mileage that the company should use if they want the top 96% of the tires to last longer than the warranty is approximately 65,075 miles. Therefore, the answer is \( \boxed{65075} \)