Answer to Question #104477 in Statistics and Probability for BARUN YADAV

Question

If X is the number scored in a throw of a fair die, show that the Chebychev’s inequality gives , P{|X-µ|2.5}<0.47, where µis the mean of X, while the actual probability is zero.

Answer 1

When a die is thrown, $X$ denotes the number that turns up. We find that the mean of $X$ is

$\mu=E(X)={7 \over 2}, Var(X)=\sigma^2={35 \over 12}$

By Chebyshev's inequality we have

$P(|X-\mu|\geq C)\leq{\sigma^2 \over C^2}$

Taking $C=2.5,$ we have

$P(|X-\mu|>2.5)<{{35 \over 12} \over 2.5^2}$

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

Or

$P(|X-3.5|>2.5)<0.47$