A certain type of light bulb has an average lifetime of 10,000 hours. The SD of bulb lifetimes is 470 hours. What fraction of bulbs could last more than 10,705 hours? I think the correct answer should be 2/9. My reasoning: k = 705/470 = 3/2. So upper bound is 1 / (3/2) ^ 2 = 4/9. But then I divide it into 2 because it is the the fraction of elements that are k*SD from the mean. So I believe that 4/9 is the fraction of bulbs that could last below than 9295 (10000-470*3/2) or more than 10705 (10000+470*3/2) hours. But the book shows 4/9 as the correct answer. Why I'm wrong?
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1$\begingroup$ The reason why you can just halve it is well-explained in the answer, but you can actually do better than 4/9 -- see en.wikipedia.org/wiki/…, which yields the bound 4/13. $\endgroup$– Glen_bCommented Aug 29, 2017 at 10:09
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$\begingroup$ Hint: one lifetime distribution that is consistent with these data assigns a chance of $9/13$ to a lifetime of $10000 - 940/3$ hours and a chance of $4/13$ to a lifetime of $10000 + 705$ hours. $\endgroup$– whuber ♦Commented Aug 29, 2017 at 15:11
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1 Answer
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When you divide 4/9 by two, you assume that distribution of bulb's lifetime is symmetric.
Indeed
4/9 is the fraction of bulbs that could last below than 9295 (10000-470*3/2) or more than 10705 (10000+470*3/2) hours
but since you don't know anything about relation between $P(X<9295)$ and $P(X>10705)$, all you can say is that $P(X>10705) \leq \frac 49$.
answered Aug 29, 2017 at 7:18