How to find the following integration Let $X_1, \cdots, X_n$ be $iid$ normal random variables with unknown mean $\mu$ and known variance $\sigma^2$. How to find $E[\Phi(\bar X)]$, where $\bar X:=\frac{\sum_{i=1}^nX_i}{n}$, please? I guess the answer should be $\Phi(\mu)$. Here is how I started. Note that $Y:= \bar X$ is also normal with mean $\mu$ and variance $\frac{\sigma^2}{n}$.
$$E[\Phi(\bar X)] = \int_{-\infty}^\infty \Phi(y) f_Y(y) dy.$$
Let $z=\frac{y-\mu}{\sigma/\sqrt{n}}$ and the above integration becomes
$$\int_{-\infty}^\infty \Phi(\mu+\frac{\sigma z}{\sqrt{n}}) \phi(z) dz$$
I could not proceed any further from here. Does anyone know what to do, please? Thank you!

Find $E[\Phi(c\bar X)]$ where $X_i \sim N(\mu, 1)$ and $c$ is a constant. Then, proceed following Dilip but replace $X$ with $\bar Xc.$
Let $Z\sim N(0,1)$ be independent of all $X_i.$ Then
$$\begin{align}
E[\Phi(c\bar X)]
&= \int_{-\infty}^\infty P\{Z \leq cy\mid \bar X = y\}f_\bar X(y)\,\mathrm dy\\
&= P\{Z \leq c\bar X\}\\
&= P\{Z-c\bar X \leq 0\} \\
&= \Phi\left(\frac{c\mu}{\sqrt{1 + c^2/n}}\right)
\end{align}$$
If we then choose $c$ to make this last expression equal to $\Phi(\mu),$ then we will obtain its MVUE. In this case $c=\sqrt\frac n{n-1}.$ If  $\sigma^2\ne1$, then $c=\sqrt\frac n{n-\sigma^2}$ for $n$ sufficiently large.
