Answer :
Final answer:
The probability of the tire wearing out before 60000 miles is about 7.64%, whereas the probability of it lasting more than 73000 miles is approximately 33.36%.
Explanation:
This problem is a good example of how to solve mathematical probability problems pertaining to normally distributed data. Given that the lifespan of the tires is represented by a normal distribution with a mean (μ) of 70000 and a standard deviation (σ) of 7000, we can apply the rules of a standard normal distribution to find the desired probabilities.
A. If we want to find the probability that the tire wears out before 60000 miles, we have to calculate the Z-score for 60000. The Z-score formula is Z = (X - μ) / σ. So, Z = (60000 - 70000) / 7000 = -1.43 approximately. The area ('probability') corresponding to the Z-score -1.43 is 0.0764, meaning there is about a 7.64% chance the tire wears out before 60000 miles.
B. To find the probability that a tire lasts more than 73000 miles, we calculate the Z-score for 73000, i.e., Z = (73000 - 70000) / 7000 = 0.43. The corresponding 'probability' for Z=0.43 is 0.3336, which means the area to the right of it (representing a lifespan greater than 73000 miles) is 1 - 0.3336 = 0.6664 or about a 66.64% chance. However, this result is incorrect. We must note that a Z-score of 0.43 represents the area up to 0.43 (about 0.6664) from the left tail. Thus, the area on the right (representing the lifespan greater than 73000 miles) is 1 - 0.6664 = 0.3336 or 33.36% approximately.
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