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.
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.
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