Suppose you work as a quality control official for a tire company. From actual r

Question

Suppose you work as a quality control official for a tire company. From actual road tests with the tires, your company's engineering group determines that the mean tire life is normally distributed with mu = 36, 500 miles with standard deviation sigma = 5000 What is the probability that a tire lasts greater than 40,000 miles? Suppose your company is considering a guarantee that it will give a discount on replacement tires if the original tires don't last for some guaranteed mileage. If your company sets this guarantee at the 10th percentile, below what mileage would the tire have to fail to qualify for the discount on the next tire?

Explanation / Answer

= 36500, = 5000

(a) z = (x - )/ = (40000 - 36500)/5000 = 0.70

P(x > 40000) = P(z > 0.7) = 0.242

(b) Area under the standard normal curve is 10% (0.10) to the left of z = -1.2816

x = + z * = 36500 - 1.2816(5000) = 30092 miles

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