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A number of lightbulbs have lifetimes X that are iid and exponentially distributed with a mean of 1/4. As a lightbulb fails, it is replaced with another until the bulbs run out. Using Chebyshev Inequality, estimate the number of lightbulbs we will need such that the probability that we have a functioning bulb at t = 10 is at least 0.8.

My attempt so far:

I have a random variable Y which is a sum of N X's. This is because as they burn out I replace them and so the mean of Y is N/4 and since these are iid the variance of Y is N/16.

Now I try to use Chebyshev Inequality, since I'm interested in the point y = 10 I have $\ |Y-mu|=|10-N/4| $, and the standard form is $\ P(|Y-mu|>=k)<=var(Y)/k^2 $. This implies that the inequality puts an upper bound but I'm looking for "at least 0.8", a lower bound. My question is can I then do the following to transform it into an lower bound, $\ 1-P(|Y-mu|<=k)>=var(Y)/k^2 $?

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Suppose you use $k$ lightbulbs.

  • The mean is $\frac{k}{4}$.

  • The variance is $\frac{k}{16}$.

Note that

$$ P\left( X < 10\right) = P\left( \frac{k}{4} - X > \frac{k}{4} - 10\right) \leq P\left( \left| \frac{k}{4} - X \right| > \left| \frac{k}{4} - 10\right|\right) . $$

Also

$$ \left| \frac{k}{4} - 10\right| = \frac{\sqrt{k}}{4} \frac{\left| \frac{k}{4} - 10\right|}{\frac{\sqrt{k}}{4}} . $$

By Chebyshev, therefore, the probability is less than

$$ \frac{\frac{k}{16}}{\left(\frac{k}{4} - 10\right)^2}. $$

It is simple to continue from here.

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