# Use Chebyshev's inequality to ﬁnd a lower bound of a Chi-Square Distribution

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.

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)

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