If X is the number scored in a throw of a fair die,show that the Chebyshev's inequality gives

$\displaystyle P(\mid X-\mu\mid >2.5)<0.47$

Where $\displaystyle \mu$ is the mean of X,while the actual probability is zero.

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- Jul 17th 2009, 08:27 PMroshanheroLast question on chebyshev's inequality
If X is the number scored in a throw of a fair die,show that the Chebyshev's inequality gives

$\displaystyle P(\mid X-\mu\mid >2.5)<0.47$

Where $\displaystyle \mu$ is the mean of X,while the actual probability is zero. - Jul 18th 2009, 12:01 AMmr fantastic
I don't see where the trouble could be here. Where are you stuck?

You should know how to calculate the standard deviation $\displaystyle \sigma$ of the number scored.

So solve $\displaystyle k \sigma = 2.5$ for k and then substitute this value of k into the right hand side of the Chebyshev inequality given here: Chebyshev's inequality - Wikipedia, the free encyclopedia