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------------------------------------------------ The life expectancy of a particular brand of tire is normally distributed with a mean of 36,000 and a standard deviation of 4000 miles. refer to exhibit 6-6. what is the probability that a randomly selected tire will have a life of at least 45,000 miles?

To solve this problem, we need to find the probability that a randomly selected tire will have a life of at least 45,000 miles. We can use the standard normal distribution to solve this problem.

First, we need to calculate the z-score for 45,000 miles using the formula z = (x - µ) / σ where x is the value, µ is the mean, and σ is the standard deviation. In this case, the mean is 36,000 miles and the standard deviation is 4000 miles. So, the z-score is z = (45,000 - 36,000) / 4000 = 2.25.
Next, we need to find the probability of a z-score greater than or equal to 2.25. We can use a standard normal distribution table or a calculator to find this probability. From the table, we can determine that the probability of a z-score greater than or equal to 2.25 is approximately 0.9878.
Therefore, the probability that a randomly selected tire will have a life of at least 45,000 miles is approximately 0.9878.

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