A brand of automobile tire has a life expectancy that is normally distributed, with a mean life of 40,000 miles and a standard deviation of 2,500 miles. The lifespan of three randomly selected tires are 35,000 miles, 40,000 miles, and 45,000 miles. Find the z-score that corresponds to each lifespan. Then find the probability of attaining a value larger than (positive z-scores) or smaller (negative z-scores) than the original lifespan. Hint: What is the proportion remaining in the tail for each z-score?

MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
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A brand of automobile tire has a life expectancy that is normally distributed, with a mean life of 40,000 miles and a standard deviation of 2,500 miles. The lifespan of three randomly selected tires are 35,000 miles, 40,000 miles, and 45,000 miles. Find the z-score that corresponds to each lifespan. Then find the probability of attaining a value larger than (positive z-scores) or smaller (negative z-scores) than the original lifespan. Hint: What is the proportion remaining in the tail for each z-score?

 

The lifespan of a species of fruit fly is normally distributed with a mean of 36 days and a standard deviation of 4 days. The lifespan of three randomly selected flies are 28 days, 34 days, 36 days, and 42 days. Find the z-score that corresponds to each lifespan. Then find the probability of attaining a value larger than (positive z-scores) or smaller (negative z-scores) than the original lifespan.

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