# S3 2010 8 OCR

(i)

The CDF of S is given by:

\begin{align*}F(s) &= \int_1^s \frac 8 {3x^3} \mathrm dx \\ & = \left[-\frac 4 {3x^2}\right]_1^s \\ & = -\frac 4 {3s^2} + \frac 4 3\end{align*}

Note that:

\displaystyle X = \frac 1 2 \times S \times S \times \sin 30^\circ = \frac 1 4 S^2

Since 1 \le S \le 2, we have 1 \le S^2 \le 4 and hence \dfrac 1 4 \le X \le 1.

We therefore have for \dfrac 1 4 \le x \le 1:

\begin{align*}\mathrm P\left(X \le x\right) & = \mathrm P \left(\frac 1 4 S^2 \le x\right) \\ & = \mathrm P \left(S^2 \le 4x\right) \\ & = \mathrm P(S \le 2 \sqrt x) \\ & = F(2 \sqrt x) \\ & = -\frac 4 {3 \times (2 \sqrt x)^2} + \frac 4 3\\ & = -\frac 1 {3 x} + \frac 4 3\end{align*}

the CDF of X.

Differentiating we get the PDF of X:

\displaystyle f_X(x) = \frac 1 {3x^2}

over \dfrac 1 4 \le x \le 1 and 0 otherwise.

(ii)

At the median we have:

\displaystyle \mathrm P(X \le m) = -\frac 1 {3 m} + \frac 4 3 = \frac 1 2

so that:

\displaystyle -\frac 1 {3m} = -\frac 5 6

the median then being given:

\displaystyle m = \frac 6 5 \times \frac 1 3 = \frac 2 5