6
\includegraphics[max width=\textwidth, alt={}, center]{8f9e5f25-05c1-4a4e-9094-2a1b42416588-08_490_1195_255_475} The diagram shows the graph of the probability density function of a random variable \(X\) that takes values between - 1 and 3 only. It is given that the graph is symmetrical about the line \(x = 1\). Between \(x = - 1\) and \(x = 3\) the graph is a quadratic curve. The random variable \(S\) is such that \(\mathrm { E } ( S ) = 2 \times \mathrm { E } ( X )\) and \(\operatorname { Var } ( S ) = \operatorname { Var } ( X )\).

1. On the grid below, sketch a quadratic graph for the probability density function of \(S\).
   \includegraphics[max width=\textwidth, alt={}, center]{8f9e5f25-05c1-4a4e-9094-2a1b42416588-08_490_1191_1169_479} The random variable \(T\) is such that \(\mathrm { E } ( T ) = \mathrm { E } ( X )\) and \(\operatorname { Var } ( T ) = \frac { 1 } { 4 } \operatorname { Var } ( X )\).
2. On the grid below, sketch a quadratic graph for the probability density function of \(T\).
   \includegraphics[max width=\textwidth, alt={}, center]{8f9e5f25-05c1-4a4e-9094-2a1b42416588-08_488_1187_1996_479} It is now given that $$f ( x ) = \begin{cases} \frac { 3 } { 32 } \left( 3 + 2 x - x ^ { 2 } \right) & - 1 \leqslant x \leqslant 3
   0 & \text { otherwise } \end{cases}$$
3. Given that \(\mathrm { P } ( 1 - a < X < 1 + a ) = 0.5\), show that \(a ^ { 3 } - 12 a + 8 = 0\).
4. Hence verify that \(0.69 < a < 0.70\).