PDF of transformed variable

6 A rectangle of area \(Y \mathrm {~m} ^ { 2 }\) has a perimeter of 16 m and a side of length \(X \mathrm {~m}\), where \(X\) is a random variable with probability density function, f, given by $$f ( x ) = \begin{cases} \frac { 1 } { 2 } & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$

1. Obtain the cumulative distribution function, F , of \(X\).
2. Show that $$16 - Y = ( 4 - X ) ^ { 2 }$$ and deduce that the probability density function of the random variable \(Y\) is $$g ( y ) = \begin{cases} \frac { 1 } { 4 \sqrt { 16 - y } } & 0 \leqslant y \leqslant 12 \\ 0 & \text { otherwise } \end{cases}$$
3. Find the median of \(Y\).
4. Find \(\mathrm { E } ( Y )\).

4 The continuous random variable \(X\) has cumulative distribution function given by $$\mathrm { F } ( x ) = \begin{cases} 0 & x < 0 \\ \frac { 1 } { 8 } x ^ { 3 } & 0 \leqslant x \leqslant 2 \\ 1 & x > 2 \end{cases}$$

1. Find \(\mathrm { E } ( X )\).
2. Find the probability density function of \(Y\), where \(Y = \frac { 1 } { X ^ { 2 } }\).

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

1. Find the distribution function of \(X\). [3]

The random variable \(Y\) is defined by \(Y = (X - 1)^3\).

1. Find the probability density function of \(Y\). [4]
2. Find the median value of \(Y\). [3]

The continuous random variable \(X\) has probability density function \(f\) given by $$f(x) = \begin{cases} \frac{1}{80}\left(3\sqrt{x} - \frac{8}{\sqrt{x}}\right) & 4 \leqslant x \leqslant 16, \\ 0 & \text{otherwise}. \end{cases}$$

1. Find the distribution function of \(X\). [3]

The random variable \(Y\) is defined by \(Y = \sqrt{X}\).

1. Find the probability density function of \(Y\). [3]

The continuous random variable \(X\) has distribution function given by $$\text{F}(x) = \begin{cases} 0 & x < 0, \\ \frac{1}{90}(x^2 + x^4) & 0 \leqslant x \leqslant 3, \\ 1 & x > 3. \end{cases}$$ The random variable \(Y\) is defined by \(Y = X^2\).

1. Find the probability density function of \(Y\). [4]
2. Find the mean value of \(Y\). [2]

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

1. Find the distribution function of \(X\). [3]

The random variable \(Y\) is defined by \(Y = X^2\).

1. Find the probability density function of \(Y\). [4]
2. Find the value of \(y\) such that \(\mathrm{P}(Y < y) = 0.8\). [3]

The continuous random variable \(X\) has probability density function f given by $$f(x) = \begin{cases} \frac{1}{8} & 0 \leq x < 1, \\ \frac{1}{28}(8 - x) & 1 \leq x \leq 8, \\ 0 & \text{otherwise}. \end{cases}$$

1. Find the cumulative distribution function of \(X\). [3]

1. Find the value of the constant \(a\) such that P\((X \leq a) = \frac{5}{7}\). [3]

The random variable \(Y\) is given by \(Y = \sqrt[3]{X}\).

1. Find the probability density function of \(Y\). [5]