So X~Bin(3,0.4)
$P(X=k)$ = $3 \choose k$ $0.4^{k} (1-0.4)^{3-k}$, k = 0,1,2,3
I tried using the CDF technique
F(y) = P(Y < y) = P$((X-2)^{2} < y)$ = $ (-\sqrt y + 2 < X <\sqrt y + 2)$
= $\sum$ $3\choose y $ $0.4^{y} 0.6^{3-y}$ y=1,2,3,4 and the sum if from 1 to 4
Is this correct?
The cdf technique is awkward for random variables with discrete distribution. Here are a couple of approaches:
Approach 1: We find the distribution of $Y$. Note that $X$ takes on the values $0$, $1$, $2$, and $3$, with probabilities you know how to compute. We have $\Pr(X=0)=(0.6)^3$, $\Pr(X=1)=\binom{3}{1}(0.4)(0.6)^2$, and so on.
Now $Y=(X-2)^2$. So $Y$ takes on the value $4$ when $X=0$, the value $1$ when $X=1$ or $X=3$, and the value $0$ when $X=2$.
Now we can completely describe the distribution of $Y$. We get $\Pr(Y=0)=\Pr(X=2)=\binom{3}{2}(0.4)^2(0.6)$.
We have $\Pr(Y=1)=\binom{3}{1}(0.4)(0.6)^2+(0.4)^2$.
And finally we have $\Pr(Y=4)=(0.6)^2$.
Now that you know the distribution of $Y$, you can find $E(Y)$ in the usual way.
Approach 2: The expectation of $(X-2)^2$ is equal to $$\sum_{x=0}^3 (x-2)^2 \Pr(X=x).\tag{1}$$ But $\Pr(X=x)=\binom{3}{x}(0.4)^x (0.6)^{3-x}$. Now use (1).