Power transformation (Y = X^n, n≥2)

5 The continuous random variable \(X\) has a triangular distribution with probability density function given by $$f ( x ) = \left\{ \begin{array} { l r } 1 + x & - 1 \leqslant x \leqslant 0
1 - x & 0 \leqslant x \leqslant 1
0 & \text { otherwise } \end{array} \right.$$

1. Show that, for \(0 \leqslant a \leqslant 1\), $$\mathrm { P } ( | X | \leqslant a ) = 2 a - a ^ { 2 } .$$ The random variable \(Y\) is given by \(Y = X ^ { 2 }\).
2. Express \(\mathrm { P } ( Y \leqslant y )\) in terms of \(y\), for \(0 \leqslant y \leqslant 1\), and hence show that the probability density function of \(Y\) is given by $$g ( y ) = \frac { 1 } { \sqrt { } y } - 1 , \quad \text { for } 0 < y \leqslant 1 .$$
3. Use the probability density function of \(Y\) to find \(\mathrm { E } ( Y )\), and show how the value of \(\mathrm { E } ( Y )\) may also be obtained directly using the probability density function of \(X\).
4. Find \(\mathrm { E } ( \sqrt { } Y )\).

8 The continuous random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 6 } x & 2 \leqslant x \leqslant 4
0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is defined by \(Y = X ^ { 3 }\). Show that \(Y\) has probability density function g given by $$g ( y ) = \begin{cases} \frac { 1 } { 18 } y ^ { - \frac { 1 } { 3 } } & 8 \leqslant y \leqslant 64
0 & \text { otherwise } \end{cases}$$ Find \(\mathrm { E } ( Y )\).

8 The continuous random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 4 } ( x - 1 ) & 2 \leqslant x \leqslant 4
0 & \text { otherwise } \end{cases}$$

1. Find the distribution function of \(X\).
   The random variable \(Y\) is defined by \(Y = ( X - 1 ) ^ { 3 }\).
2. Find the probability density function of \(Y\).
3. Find the median value of \(Y\).

10 The continuous random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 2 } & 1 \leqslant x \leqslant 3
0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is defined by \(Y = X ^ { 3 }\). Find the distribution function of \(Y\). Sketch the graph of the probability density function of \(Y\). Find the probability that \(Y\) lies between its median value and its mean value.

7 The continuous random variable \(X\) has probability density function given by $$\mathrm { f } ( x ) = \begin{cases} \frac { 1 } { 21 } x ^ { 2 } & 1 \leqslant x \leqslant 4
0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is defined by \(Y = X ^ { 2 }\). Show that \(Y\) has probability density function given by $$\operatorname { g } ( y ) = \begin{cases} \frac { 1 } { 42 } y ^ { \frac { 1 } { 2 } } & 1 \leqslant y \leqslant 16
0 & \text { otherwise } \end{cases}$$ Find

1. the median value of \(Y\),
2. the expected value of \(Y\).

7 The random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 6 } x & 2 \leqslant x \leqslant 4
0 & \text { otherwise } \end{cases}$$

1. Find the distribution function of \(X\). The random variable \(Y\) is defined by \(Y = X ^ { 3 }\). Find
2. the probability density function of \(Y\),
3. the value of \(k\) for which \(\mathrm { P } ( Y \geqslant k ) = \frac { 7 } { 12 }\).

10 The random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 30 } \left( \frac { 8 } { x ^ { 2 } } + 3 x ^ { 2 } - 14 \right) & 2 \leqslant x \leqslant 4
0 & \text { otherwise } \end{cases}$$

1. Find the distribution function of \(X\).
   The random variable \(Y\) is defined by \(Y = X ^ { 2 }\).
2. Find the probability density function of \(Y\).
3. Find the value of \(y\) such that \(\mathrm { P } ( Y < y ) = 0.8\).

7 The continuous random variable \(X\) has probability density function given by $$f ( x ) = \begin{cases} \frac { 1 } { 21 } x ^ { 2 } & 1 \leqslant x \leqslant 4
0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is defined by \(Y = X ^ { 2 }\).

1. Show that \(Y\) has probability density function given by $$g ( y ) = \begin{cases} \frac { 1 } { 42 } y ^ { \frac { 1 } { 2 } } & 1 \leqslant y \leqslant 16
   0 & \text { otherwise } \end{cases}$$
2. Find the median value of \(Y\).
3. Find the expected value of \(Y\).