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Problem Text

The prevailing winds blow in a constant southerly direction from an enchanted castle. Each year, according to an ancient tradition, a princess releases 96 magic seeds from the castle, which are carried south by the wind before falling to rest. South of the castle lies one league of grassy parkland, then one league of lake, then one league of farmland, and finally the sea. If a seed falls on land it will immediately grow into a fever tree. (Fever trees do not grow in water).
Seeds are blown independently of each other. The random variable $L$ is the distance in leagues south of the castle at which a seed falls to rest (either on land or water). It is known that the probability
density function $\mathrm{f}$ of $L$ is given by 
\[
\mathrm{f}(x)=\begin{cases}
\frac{1}{2}-\frac{1}{8}x & \mbox{ for }0\leqslant x\leqslant4,\\
0 & \mbox{ otherwise.}
\end{cases}
\]
What is the mean number of fever trees which begin to grow each year?
\begin{questionparts}
\item The random variable $Y$ is defined as the distance in leagues south of the castle at which a new fever tree grows from a seed carried by the wind. Sketch the probability density function of $Y$, and
find the mean of $Y$. 
\item One year messengers bring the king the news that 23 new fever trees have grown in the farmland. The wind never varies, and so the king suspects that the ancient tradition have not been followed properly.
Is he justified in his suspicions?
\end{questionparts}

Solution (Optional)

\begin{align*}
\mathbb{P}(\text{fever tree grows}) &= \mathbb{P}(0 \leq L \leq 1) + \mathbb{P}(2 \leq L \leq 3) \\
&= \int_0^1 \frac12 -\frac18 x \d x + \int_2^3 \frac12 - \frac18 x \d x \\
&= \left [\frac12 x - \frac1{16}x^2 \right]_0^1+ \left [\frac12 x - \frac1{16}x^2 \right]_2^3 \\
&= \frac12 - \frac1{16}+\frac32-\frac9{16} - 1 + \frac{4}{16} \\
&= \frac58
\end{align*}

The expected number of fever trees is just $96 \cdot \frac58 = 60$.

\begin{questionparts}
\item $f_Y(t)$ must match the distribution for $L$, but limited to the points we care about, therefore it should be:

$f_Y(t) = \begin{cases}  ( \frac45 - \frac15t ) & \text{if } t \in [0,1]\cup[2,3] \\
0 & \text{otherwise} \end{cases}$

\begin{center}
    \begin{tikzpicture}
        \draw[->] (0,0) -- (4,0) node [right] {$d$};
        \draw[->] (0,0) -- (0,1);

\draw (0, {4/5}) -- (1, {3/5}) -- (1, 0);;
        \draw (2,0) -- (2, {2/5}) -- (3, {1/5}) -- (3,0);
    \end{tikzpicture}
\end{center}

\begin{align*}
\mathbb{E}(Y) &= \frac12 \cdot \frac15 (4 - \frac12)+\frac52 \cdot (1 -  \frac15 (4 - \frac12)) \\
&= \frac12 \cdot \frac7{10} + \frac52 \cdot \frac3{10} \\
&= \frac{22}{20} \\
&= \frac{11}{10}
\end{align*}

\item Given the seeds are blown independently and the wind hasn't changed, it is reasonable to model the number of fever trees as $B(96, \frac{5}{8})$, it is also acceptable to approximate this using a Normal distribution, ie $N(60, 22.5)$, $23$ is $\frac{23-60}{\sqrt{22.5}}$ is a very negative number, so he should be extremely suspicious.

\end{questionparts}

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