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
Answer
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 \( \boxed{65075} \)
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 \( \boxed{65075} \)
