Average of two incomes, taken from a normal distribution

Question

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?

Answer 1

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.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ 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 !!!.