Answer to Question #105572 in Statistics and Probability for Anu

Question

If X is the number scored in a throw of a fair die, show that the Chebychev’s
inequality gives {P X −µ > }5.2 < 47.0 , where µis the mean of X, while the actual
probability is zero.

Answer 1

$$P\{|X-\mu|>2.5\}<0.47$$

Chebyshev's inequality:

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

Let $X$ is our random variable.

$$MX={1 \over 6}(1+2+3+4+5+6)={7 \over 2}$$

$$MX^2={1 \over 6}(1^2+2^2+3^2+4^2+5^2+6^2)={91 \over 6}$$

$$DX=MX^2-(MX)^2$$

$$DX={91 \over 6}-({7 \over 2})^2={35 \over 12}$$

$$\sigma_x=\sqrt{DX}=\sqrt{{35 \over 12}}$$

$$k\sigma_X=2.5=>k={2.5 \over \sqrt{{35 \over 12}}}=\sqrt{{15 \over 7}}\approx1.463850$$

$${1 \over k^2}={7 \over 15}\approx0.4667<0.47$$

Hence if $X$ is the number scored in a throw of a fair die, then the Chebychev’s inequality gives 

$$P\{|X-\mu|>2.5\}<0.47$$