Suppose that the lifetimes of tires of a certain brand are normally distributed with a mean of 74,000 miles and a standard deviation of σ miles. These tires come with a 55,000-mile warranty. The manufacturer of the tires can adjust σ during the production process, but the adjustment of σ is quite costly. The manufacturer wants to set σ once and for all so that only 5% of the tires will fail before warranty expires. Find the standard deviation to be set. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.miles
Question
Suppose that the lifetimes of tires of a certain brand are normally distributed with a mean of 74,000 miles and a standard deviation of σ miles. These tires come with a 55,000-mile warranty. The manufacturer of the tires can adjust σ during the production process, but the adjustment of σ is quite costly. The manufacturer wants to set σ once and for all so that only 5% of the tires will fail before warranty expires. Find the standard deviation to be set. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.miles
Solution
To solve this problem, we need to use the properties of the normal distribution.
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First, we need to find the z-score that corresponds to the top 5% of the distribution. This is because we want only 5% of the tires to fail before the warranty expires. Using a standard normal distribution table or a z-score calculator, we find that the z-score corresponding to the top 5% of the distribution is approximately 1.645.
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Next, we use the formula for the z-score, which is (X - μ) / σ, where X is the value we're interested in (in this case, the warranty mileage of 55,000 miles), μ is the mean (74,000 miles), and σ is the standard deviation, which we're trying to find.
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We plug in the values we know into the z-score formula and solve for σ:
1.645 = (55000 - 74000) / σ
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Solving for σ gives us:
σ = (55000 - 74000) / 1.645
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This gives us σ = -19000 / 1.645 = -11550.15 miles.
However, the standard deviation cannot be negative. This negative value indicates that the warranty mileage of 55,000 miles is less than the mean lifetime of the tires (74,000 miles). This makes sense because we want only 5% of the tires to fail before the warranty expires.
So, the standard deviation that the manufacturer should set is approximately 11550.2 miles (rounded to one decimal place).
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