I have the following problem. Let $(X)_{n>0}$ be a sequence of $4000$ independent random variables (discrete), all of them have a Bernoulli distribution with $p = 0.8$. Let $X$ be the sum of these $4000$ random variables, that is $$X = \sum_{i=1}^{4000} X_i. $$ I've calculated the probability $P(3000 < X <3400)$ using Cebîshev inequality: $$ P(|X-3200|<200)\ge1-\frac{\text{Var}(X)}{200^2}.$$
Now, suppose I want to calculate $P(X\le3000)$. I've done it using the Central Limit Theorem but the answer is quite small: $0.00003$. My question is: can I calculate $P(X\le3000)$ using Cebîshev inequality?
Answering your final question: yes. Once your random variable is positive, you can elevate both sides of $X \le 3000$ to the power of 2. In terms of probability, we are saying there is a equality of events
$$ \{X \le 3.000 \} = \{ X^2 \le (3.000)^2 \} $$
Because $X \ge 0$.
In fact, the result is more general.
Let $X$ a random variable, $\phi : \mathbb{R} \longrightarrow [0, \infty)$, $A \subset \mathbb{R}$, measurable, and $i_A = \inf\{ \phi(y) : y \in A \}$ then
$$ P(X \in A) \le \frac{\mathbb{E}[\phi(X)]}{i_A} $$
If you put $\phi(x) = x^2$ and $A = [3.000, \infty)$ you can determine $P(X \ge 3.000)$ and finally, $P(X < 3.000)$