In an attempt to put the language to help us, $E(X\mid Y>2)$ is a "conditional expected value given a specific event", and it is a number, while $E(X\mid Y)$ is shorthand for $E(X\mid \sigma(Y))$ - a conditioning on the whole sigma algebra generated by $Y$, and hence a function of $Y$, called the "conditional expectation of $X$ given $Y$".
Conditional Expectations are defined for random variables $X$ such that $E(|X|) < \infty$, and they, too, satisfy $E(|E(X\mid Y)|) < \infty$ (this is provable).
Half- formally, the defining property of the function "Conditional Expectation of $X$ given $Y$", write $Z = E(X\mid Y)$ is
$$E(Z\cdot I_{\{B_j\}}) = E(X\cdot I_{\{B_j\}})\;\; \forall B_j $$
where $B_j = \{Y=y_i\}$ is an event (so we have numbers in both side of the equation), and $I_{\{\cdot \}}$ is the indicator function.
Since the indicator function takes the values $0$ or $1$, then $$E(X\cdot I_{\{B_j\}})=\cases {0 \\ \\
E(X)}$$
Since $E(X) < \infty$, we have $$E(X\cdot I_{\{B_j\}}) < \infty \Rightarrow E(Z\cdot I_{\{B_j\}})< \infty$$
and the known result
$$E(Z) = E\big[E(X\mid Y)\big] = E(X)$$
So, if the Conditional Expectation has a finite expected value, is it possible that itself would be infinite in some sense?
This seems to be a qqplot of the data compared with a standard normal distribution, so I would have thought the $x$ values should the typical values of the population quantiles of a standard normal distribution
So with $105$ observations I would have thought the extreme left $x$ value should be not far away from $\Phi^{-1}\left(\dfrac{0.5}{105}\right) \approx -2.59$ and the one next to it near $\Phi^{-1}\left(\dfrac{1.5}{105}\right) \approx -2.19$, with the extreme right values being the corresponding $\Phi^{-1}\left(\dfrac{104.5}{105}\right) \approx +2.59$ and $\Phi^{-1}\left(\dfrac{103.5}{105}\right) \approx +2.19$. Visually, this seems to be close to what you have in the charts
Best Answer
You almost got there.
Your 0.32 would more accurately be 0.315 (for number of sd's below the mean). This gives a central area of about 25% and a left tail area of about 38%. More accurately, (100-24.8)/2=37.6 (Your 24.8 is 100-?-? = 24.8 and you want to figure out what that ? value would be)