Suppose that the lifetimes of old-fashioned TV tubes are normally distributed with a standard deviation of 1.2 years. Suppose also that exactly 35% of the TV tubes die before 4.5 years. Find the mean lifetime of TV tubes. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.
Question
Suppose that the lifetimes of old-fashioned TV tubes are normally distributed with a standard deviation of 1.2 years. Suppose also that exactly 35% of the TV tubes die before 4.5 years. Find the mean lifetime of TV tubes. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.
Solution
To solve this problem, we need to use the properties of the normal distribution.
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First, we know that 35% of the TV tubes die before 4.5 years. This means that the z-score corresponding to 35% in the standard normal distribution table is the z-score for 4.5 years.
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Looking up 35% (or 0.3500) in a standard normal distribution table, we find that the closest z-score is -0.3853.
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The z-score is calculated using the formula: Z = (X - μ) / σ, where X is the value from the dataset (in this case, 4.5 years), μ is the mean we're trying to find, and σ is the standard deviation (which is 1.2 years).
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We can rearrange the formula to solve for μ: μ = X - Zσ.
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Substituting the known values into the formula, we get: μ = 4.5 - (-0.3853 * 1.2) = 4.5 + 0.46236 = 4.96236.
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Rounding this to at least one decimal place, we find that the mean lifetime of TV tubes is approximately 5.0 years.
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