Random variable is uniformly distributed on interval $[a, 4]$ and mean of this variable is $3$. Estimate $P(|X| < 3)$. Use Chebyshev’s inequality!
It is obvious that $a = 2$, as we have a uniform distribution. I have no idea how to solve it using Chebyshev’s inequality. I got stuck here: $$P(|X - 3| < k) \ge 1 - \frac{\sigma^2}{k^2}$$.
Since you have $P(|X|<3)$, and the interval is $[2,4]$, then you hace $$ P(|X|<3) = P(X<3) = P(X \leq 3) \tag{because of equalty P-ctp} $$ then, because of X is uniformly distributed on $[2,4]$, the probability is just $P(X\leq3)= \frac{1}{4-2}=\frac{1}{2}$ wich has (for intuition) sense.
In the first equation, notice that X is a random variable positive!