\begin{problem}[15]
Two factories produce the same product.  The number, $R$, of items
produced per day by the first factory fluctuates randomly with mean 80 and
standard deviation 2.  The number, $T$, of items produced per day by the
second factory fluctuates independently of the first factory's output, and
$T$ is observed to have mean 90 and standard deviation 3.  \iffalse Use
the Chebychev Bound to\fi Show that there is at least a 96\% chance that
tomorrow, the excess output of the second factory over the first factory
will differ from what it is expected to be by less than a factor of two.

\solution{
Let $D \eqdef T-R$ be the excess output in a day.  So
\begin{align}
\expect{D} & = \expect{T} - \expect{R} = 90-80=10,\label{E}\\
\variance{D} & = \variance{T} + \variance{R} = 2^2+3^2=13,\label{V}\\
\pr{\abs{D-\expect{D}} \geq 2\expect{D}} & \leq \frac{13}{20^2}
&\text{(by~(\ref{E}), (\ref{V}) \& Chebychev)},\notag\\
\pr{\abs{D-\expect{D}} <  2\expect{D}} & \geq 1 - \frac{13}{20^2} =
   \frac{387}{400} > 0.96. \notag
\end{align}
}

\end{problem}

