Given $E(X)=100,\sigma(X)=10$ and $P(X\le 10k)\ge 4/5$, how do we find a lower bound on $k$ using the Chebyshev's inequality
Let $X$ be any r.v. with finite mean, $μ$, and finite variance. Then $∀a > 0,$
$$ P\bigg(|X-\mu|\ge a\bigg)\le\frac{\sigma^2(X)}{a^2} $$ which is the Chebyshev’s inequality.
$$ P(X\le 10k)\ge 4/5\implies P(X> 10k)\le 1/5\implies P(X-\mu>10k-\mu)\le 1/5\\ $$ And from Chebyshev’s inequality $$ P(|X-\mu|\ge 10k-\mu)=P\Big(X-\mu\ge 10k-\mu\text{ or }X-\mu\le -(10k-\mu)\Big)\le \frac{100}{(10k-\mu)^2}\\ P(X-\mu\ge 10k-\mu)+P(X-\mu\le -(10k-\mu))\le \frac{100}{(10k-\mu)^2} $$ Is there a way to proceed further?
Note that as long as $10k-\mu > 0$, $$P(X-\mu > 10k-\mu) \leq P(|X-\mu| \geq 10k-\mu) \leq \frac{\sigma^2(X)}{(10k-\mu)^2}$$ so we'll have the left-most quantity $\leq 1/5$ if the right-most quantity is $ \leq 1/5$. Using the given values, $$\frac{100}{(10k-100)^2} \leq \frac15 \\ (k-10)^2 \geq 5$$
This last inequality will hold if $k \geq 10 + \sqrt{5}$, and for any such $k$, we can check that $10k-\mu > 0$, so $$k \geq 10+\sqrt{5}$$ is sufficient for $P(X\leq 10k) \geq \frac45$.