During one of my lectures, we went over a problem that I'm still trying to figure out. Let X be random variable with expectation μ. If $P[|X-μ|≥2]=0.5$ then what is the value of $P[(X-μ)2≥4]$?
At first, I simply assumed that since everything within the probability mass function is being squared, that squaring the answer should suffice. However, when I brought this up to the attention of my professor, I was told that my answer and reasoning was incorrect, and was left with one hint: Note that for any outcome o, $|X(o)-μ|≥2$ if and only if $(X(o)-μ)^2≥4$
I've tried to read into the interpretation of this hint just to see where to even start; yet, I find myself lost. I've tried backtracking and obtaining the variance since we know that the answer will be equivalent to $\frac{var[x]}{2^2}$.
Here is the background for your problem. Markov's Inequality is as follows: Let $Y$ be a random variable for which $\mu_Y = E(Y) < \infty$ exists, and which takes only non-negative values, so that $P(Y \ge 0) = 1.$ Then $P(Y \ge a) \le \mu_Y/a,$ for any $a > 0.$ (In some probability textbooks, Chebyshev's Inequality is stated as a corrollary of Markov's Inequality. Essentially, this exercise illustrates the key step in that relationship.)
In your case let $Y = (X - \mu_X)^2$ and $a = 4.$ Thus, $$P(Y \ge a) = P[(X-\mu_X)^2 \ge 4] = P(|X - \mu_X| \ge 2) \le \frac{E[(X-\mu_x)^2]}{4} =\frac{\sigma_X^2}{4}.$$
By definition, $\sigma_X^2 = Var(X) = E[(X-\mu_x)^2].$
Now it is important to separate the background of your problem from the problem itself: The events $\{(X-\mu_X)^2 \ge 4\}$ and $\{|X - \mu_x| \ge 2\}$ are identical. They have the same probability. That statement is the essence of your problem. So you can say, "It is also true that $P[(X-\mu)^2 \ge 4] = 0.5.$"