I don't understand what the question is actually asking for. I've never seen the notation for what I'm being asked to find. Any help would be greatly appreciated.
could it be some wierd notation for the identity function?
That is:
$\displaystyle 1(X>1.96)=\left\{\begin{array}{cc}0,&\mbox{ if }
x\leq 1.96\\1, & \mbox{ if } x>1.96\end{array} \right.$
if so
$\displaystyle E \left( 1(X>1.96) \right) = P(X>1.96)$
You can recognize this as the 97.5 percentile of the distribution, so the answer would be 0.025
It's the indicator function : $\displaystyle \bold 1_A(x)=\begin{cases} 1 \text{ if } x\in A \\ 0 \text{ otherwise}\end{cases}$
Why do you find it "weird" ?
And while talking probabilities, if A is an event, $\displaystyle \mathbb{E}[\bold 1_A]=\mathbb{P}(A)$
So it's as SpringFan25 wrote, I'm just confirming it (I've already used this notation tons of times )