# Chebyshev's inequality exercise

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 ?

• "...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). Commented Jan 8, 2017 at 15:35

Note $P[|X-55|>15]\leq \dfrac{E(X-55)^2}{15^2}=\dfrac{Var(X)+(E(X)-55)^2}{15^2}$
• 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?
• Google markov inequality and note $|x-55|>15$ iff $(x-55)^2>15^2$ Commented Jan 8, 2017 at 15:47