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I have a pretty simple question about how to use the Chebyshev's inequality in this case:

First of all we know that for expected value $\mu$ and variance $\sigma^2 \leq \infty$ and a random variable $X$, it holds for $k > 0$:

$$P\left[|X - \mu| \geq k \right] \leq \frac{\sigma^2}{k^2}$$

In a case where we have a random variable Y with $\mu = 50$ and $\sigma^2 = 25$ what is the probability of the random variable to have a value between 40 and 70 ?

I know that is possible to find this probability if instead of 70 we had 60 since: $$P[40 < X < 60] = P[-10 < X - 50 < 10] = P[|X - 50| < 10]$$ So we compute $1 - P[|X - 50| \geq 10]$ to find what we search for.

But it's not possible to use the same approach for a value between 40 and 70 right? Cause we would have $P[|X - 55| < 15]$ and we couldn't use the Chebyshev's inequality am I right? Which other ways would we have to solve the problem, if some ?

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    _"...it's possible to find this probability if..."_ The Chebyshev's inequality does not allow to _find_ the probability, only to _bound_ it (and the bounds are not typically tight). – 2017-01-08

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Note $P[|X-55|>15]\leq \dfrac{E(X-55)^2}{15^2}=\dfrac{Var(X)+(E(X)-55)^2}{15^2}$

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    Oh I didn't knew that, so I could solve this with $\frac{25 + (50-55)^2}{15^2}$? Is there a reference to that equation? Or could you show me how the transofrmations you made to reach that equality? – 2017-01-08
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    Google markov inequality and note $|x-55|>15$ iff $(x-55)^2>15^2$ – 2017-01-08
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    Oh thanks I got it :) – 2017-01-09