Order Statistic Expectation / Probability Let $Y_1<Y_2$ be order statistics from a random sample of size $2$ from a normal distribution, $\mathcal{N}(\mu,\sigma^2)$, where $\sigma^2$ is known. Show that $P(Y_1<\mu<Y_2)=\frac12$ and find $E(Y_1-Y_2)$. 
I am not exactly sure how to solve the question above. Any help would be appreciated. Thanks.

Let $X_1,X_2$ be the i.i.d. sample; then $Y_2 =\max\{X_1,X_2\}$ and $Y_1=\min\{X_1,X_2\}$ (I'm regarding "${<}$" as a typo in the question, in view of the result to be proved).
Then either $Y_1<Y_2<\mu$ or $Y_1<\mu<Y_2$ or $\mu<Y_1<Y_2$.  (I'm discounting the event of probability $0$ that two or more of these are equal.)
The first happens if and only if both $X_1$ and $X_2$ are less than $\mu$; the second if and only if one (either one) is less than $\mu$ and the other greater; the third if and only if both are greater than $\mu$.
The probability that $X_1>\mu$ is $1/2$; similarly for $X_2$.
So the event $Y_1<\mu<Y_2$ is the event of exactly one success in two independent trials, with probability $1/2$ of success on each trial.  Therefore its probability is $1/2$.
Now notice that $E(Y_2-Y_1) = E(|X_2-X_1|)$, and $X_1-X_2 \sim \mathcal{N}(\mu-\mu,\sigma^2+\sigma^2)=\mathcal{N}(0,2\sigma^2)$.  So $E(|X_2-X_1|)= \sqrt{2}\sigma E\left(\dfrac{|X_2-X_1|}{\sqrt{2}\sigma}\right)$ and $Z=\dfrac{X_2-X_1}{\sqrt{2}\sigma}\sim\mathcal{N}(0,1)$.  So we want $\sqrt{2}\sigma E(|Z|)$.
So
$$
\begin{align}
E(|Z|) & = \int_{-\infty}^\infty |z| \varphi(z)\;dz = 2\int_0^\infty z \varphi(z)\;dz = 2\int_0^\infty z \frac{1}{\sqrt{2\pi}} e^{-z^2/2} \; dz \\  \\
& = \sqrt{\frac{2}{\pi}} \int_0^\infty ze^{-z^2/2} \; dz = \sqrt{\frac{2}{\pi}} \int_0^\infty e^{-u} \; du = \sqrt{\frac{2}{\pi}}.
\end{align}
$$
Multiplying that by $\sqrt{2}\;\sigma$, we get $\dfrac{2\sigma}{\sqrt{\pi}}$.
