Average of two incomes, taken from a normal distribution In a large corporation, people over age thirty have an annual income whose distribution can be approximated by a normal distribution with mean 60,000 and standard deviation 10,000. Two people are selected at random. What is the chance that the average of their two incomes is over 65,000? 
 A: Hint: Let $X$ and $Y$ be the incomes of the first selected person, and the second. Let $W=\frac{X+Y}{2}$. 
Then $W$ has normal distribution, with mean $60000$ and standard deviation $\frac{10000}{\sqrt{2}}$.
You want to find $\Pr(W\gt 65000)$. This is probably a familiar sort of calculation. 
A: $\newcommand{\+}{^{\dagger}}%
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Lets $\ds{{\rm P}\pars{x} \equiv {1 \over \root{2\pi}\sigma}\,
          \exp\,\pars{\bracks{x - \overline{x}}^{2} \over 2\sigma^{2}}}$ where
$\ds{\overline{x} = 60000}$ and $\ds{\sigma \equiv 10000}.\quad$ $\ds{\pp\pars{t}: {\large ?}}.\quad$ $\ds{t \equiv 65000}$.

\begin{align}
\pp\pars{t}&=
\left.\int_{-\infty}^{\infty}\dd x\int_{-\infty}^{\infty}\dd y\,
{\rm P}\pars{x}{\rm P}\pars{y}\right\vert_{\pars{x + y}/2\ >\ t}
=
\left.\int_{-\infty}^{\infty}\dd x\int_{-\infty}^{\infty}\dd y\,
{\rm P}\pars{x + \overline{x}}{\rm P}\pars{y + \overline{x}}
\right\vert_{\pars{x + y}/2 + \overline{x}\ >\ t}
\\[3mm]&=\left.%
{1 \over 2\pi\sigma^{2}}\int_{0}^{2\pi}\dd\theta\int_{0}^{\infty}
\exp\,\pars{-\,{r^{2} \over 2\sigma^{2}}}r\,\dd r
\right\vert_{\cos\pars{\theta} + \sin\pars{\theta}\ >\
2\pars{t - \overline{x}}/r}
\end{align}

The angular integration is given by
$\pars{~\mbox{with}\ \tilde{t} \equiv {2\bracks{t - \overline{x}} \over r}~}$:
\begin{align}
&\int_{0}^{2\pi}\Theta\pars{\cos\pars{\theta} + \sin\pars{\theta} - \tilde{t}}
\,\dd\theta
\\[3mm]&=
\int_{0}^{\pi}\Theta\pars{\cos\pars{\theta} + \sin\pars{\theta} - \tilde{t}}
\,\dd\theta
+
\int_{-\pi}^{0}\Theta\pars{\cos\pars{\theta} + \sin\pars{\theta} - \tilde{t}}
\,\dd\theta
\\[3mm]&=
\int_{0}^{\pi}\sum_{\mu = \pm}
\Theta\pars{\cos\pars{\theta} + \mu\sin\pars{\theta} - \tilde{t}}\,\dd\theta
\\[3mm]&=
\int_{0}^{\pi/2}\sum_{\mu = \pm}
\Theta\pars{\cos\pars{\theta} + \mu\sin\pars{\theta} - \tilde{t}}\,\dd\theta
+
\int_{-\pi/2}^{0}\sum_{\mu = \pm}
\Theta\pars{-\cos\pars{\theta} - \mu\sin\pars{\theta} - \tilde{t}}\,\dd\theta
\\[3mm]&=
\int_{0}^{\pi/2}\sum_{\mu = \pm}\sum_{\mu' = \pm}
\Theta\pars{\mu'\cos\pars{\theta} + \mu\mu'\sin\pars{\theta} - \tilde{t}}\,\dd\theta
\end{align}
Since $\tilde{t} > 0$, the angular integration is reduced to:
\begin{align}
&\int_{0}^{2\pi}\Theta\pars{\cos\pars{\theta} + \sin\pars{\theta} - \tilde{t}}
\,\dd\theta
\\[3mm]&=
\int_{0}^{\pi/2}\Theta\pars{-\cos\pars{\theta} + \sin\pars{\theta} - \tilde{t}}
\,\dd\theta
+
\int_{0}^{\pi/2}\Theta\pars{\cos\pars{\theta} - \sin\pars{\theta} - \tilde{t}}
\,\dd\theta
\\[3mm]&+
\int_{0}^{\pi/2}\Theta\pars{\cos\pars{\theta} + \sin\pars{\theta} - \tilde{t}}
\,\dd\theta
\end{align}
Can you take it from here ?. Maybe, there should be some modification since incomes are positive numbers !!!.
