I'm trying to solve the following exercise but I'm not sure if what I'm doing is right.

"Let $X$ be an r.v. distributed as $\chi_{40}^{2}$. Use Tchebichev’s inequality in order to find a lower bound for the probability $P(|(X/40) − 1| ≤ 0.5)$, and compare this bound with the exact value found from the $\chi^{2}$ Distribution Table."

Considering that $\mu=40$ and $\sigma=\sqrt{2\times40}$ my approach was turning the inequality into:

$P(-20\leq|X-40|\leq 20)\geq 1-\frac{1}{k^{2}}$

In order to obtain:

$P(|X-40| ≤ 20)\geq 1-\frac{1}{k^{2}}$

$P(|X-40| ≤ 20)\geq 1-\frac{1}{2.236^{2}}=0.8$

But this result doesn't match with the Distribution Table.


1 Answer 1


Chebyshev's inequality is an inequality. It tells you that $P(|X-40| ≤ 20)\geq 0.8$. So if the probability that $20\leq X\leq 60$ is at least $0.8$, the inequality is satisfied. It is:

> pchisq(60,40)-pchisq(20,40)
[1] 0.9746722

which is greater than $0.8$, so the lower bound given by the inequality works just as it should here.


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