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 } < 47.0 , where µ is the mean of X, while the actual
probability is zero.

Answer 1

Chebyshev's inequality:

$P\{|X-\mu|\geq k\sigma\}\leq\frac{1}{k^2}, k\in R.$

Let X is our random variable.

$MX=\frac{1}{6}(1+2+3+4+5+6)=3.5.\\ MX^2=\frac{1}{6}(1^2+2^2+3^2+4^2+5^2+6^2)=\frac{91}{6}.\\ DX=MX^2-(MX)^2=\frac{91}{6}-(3.5)^2.\\ \sigma X=\sqrt{\frac{91}{6}-(3.5)^2}.\\ k\sigma X=2.5.\\ k=\frac{2.5}{\sqrt{\frac{91}{6}-(3.5)^2}}\approx 1.46385.\\ \frac{1}{k^2}\approx 0.4666.$