The following solution to this problem was found by @Wolfies:
(E^(-((-2 z + \[Mu]1 + \[Mu]2)^2/(2 (\[Sigma]1^2 + \[Sigma]2^2)))) Sqrt[2/\[Pi]] (-1 + Erf[((z - \[Mu]2) \[Sigma]1^2 + (-z + \[Mu]1) \[Sigma]2^2)/(Sqrt[2] \[Sigma]1 \[Sigma]2 Sqrt[\[Sigma]1^2 + [Sigma]2^2])]))/(-1 + Erf[(\[Mu]1 - \[Mu]2)/(Sqrt[2] Sqrt[\[Sigma]1^2 + \[Sigma]2^2])] Sqrt[\[Sigma]1^2 +[Sigma]2^2])
Note that $Z = \frac{X + Y}{2}$ where $X$ and $Y$ are independent random variables that both follow the Normal distribution, with possibly different means and variances (denoted by subscripts $1$ for $X$ and $2$ for $Y$).
I am now trying to plot the two normal distributions together with the random variable $Z$ described by Wolfies' solution, on the same axis. What gets me is that $Z$ does not have the size I would expect relative the two Normal distributions; the distribution for $Z$ is too large.
Below is a plot of $Z$ and $X$ on two different axis, and the size problem is clear:
(E^(-((-2 z + \[Mu]1 + \[Mu]2)^2/(2 (\[Sigma]1^2 + \[Sigma]2^2)))) \ Sqrt[2/\[Pi]] (-1 + Erf[((z - \[Mu]2) \[Sigma]1^2 + (-z + \[Mu]1) \[Sigma]2^2)/(Sqrt[ 2] \[Sigma]1 \[Sigma]2 Sqrt[\[Sigma]1^2 + \ \[Sigma]2^2])]))/((-1 + Erf[(\[Mu]1 - \[Mu]2)/(Sqrt[ 2] Sqrt[\[Sigma]1^2 + \[Sigma]2^2])]) Sqrt[\[Sigma]1^2 + \ \[Sigma]2^2])
. $\endgroup$