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A fair (unbiased) die is thrown. Let X be the number shown up.

(c) Find $\Bbb P (\lvert X − \Bbb E(X)\rvert \geqslant 2.5)$.

(d) Apply Chebyshev’s inequality to find an upper bound to the probability $\Bbb P (\lvert X − \Bbb E(X)\rvert \geqslant 2.5)$

Okay so Chebychev's Inequality is $\Bbb P(\lvert X−\Bbb E(X)\rvert > k~σ)~\leqslant~\frac 1{k^2}$ for $k>0$, where $σ^2$ is the variance of $X$.

I'm not sure how to do this question, so far I've only found the expected value and variance!   My probability test is tomorrow so help is much appreciated! Descriptive answers would be great.

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  • $\begingroup$ $\Bbb P (\lvert X − \Bbb E(X)\rvert \geqslant 2.5)$ is the probability of throwing a $1$ or a $6$ $\endgroup$ – Henry Feb 25 '16 at 23:40
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  • $X$ is the result of a fair (unbiased) die roll.   The first step is to find the mean and variance of this random variable.

    • Well, just the variance; we do not care about the mean (which, anyway, is $3.5$).
    • So, do you know the variance of a discrete uniform distribution?   Otherwise can you obtain it from first principles?
  • When you have the variance, $\sigma^2$, you need to find $k$ so that $k\sigma=2.5$.   Then you can apply Chebychev's inequality.

    • OR simply find $k^2$ so that $k^2\sigma^2 = 2.5^2$ and save some needless squaring of square roots.
  • Then you have $k^{-2}$ and are done; as that is the upper bound according to the inequality.

$$\Bbb P\big(\lvert X-\Bbb E(X)\rvert \geqslant 2.5\big) ~ \leqslant~ \dfrac {\sigma^2}{6.25}$$

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