5.03g Cdf of transformed variables

6 \includegraphics[max width=\textwidth, alt={}, center]{f45a7c6f-ebb4-43a2-8751-11aab3561d3c-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]{f45a7c6f-ebb4-43a2-8751-11aab3561d3c-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]{f45a7c6f-ebb4-43a2-8751-11aab3561d3c-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 \(0.69 < a < 0.70\).

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 \(0.69 < a < 0.70\).

3 The continuous random variable \(X\) has cumulative distribution function F given by $$F ( x ) = \begin{cases} 0 & x < 0 \\ \frac { 1 } { 81 } x ^ { 2 } & 0 \leqslant x \leqslant 9 \\ 1 & x > 9 \end{cases}$$

1. Find \(\mathrm { E } ( \sqrt { X } )\).
2. Find \(\operatorname { Var } ( \sqrt { X } )\).
3. The random variable \(Y\) is given by \(Y ^ { 3 } = X\). Find the probability density function of \(Y\).

3 The continuous random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 3 } { 16 } ( 2 - \sqrt { x } ) & 0 \leqslant x < 1 \\ \frac { 3 } { 16 \sqrt { x } } & 1 \leqslant x \leqslant 9 \\ 0 & \text { otherwise } \end{cases}$$

1. Find \(\mathrm { E } ( X )\).
   The random variable \(Y\) is such that \(Y = \sqrt { X }\).
2. Find the probability density function of \(Y\).

6 The continuous random variable \(X\) has cumulative distribution function F given by $$F ( x ) = \begin{cases} 0 & x < 0 \\ \frac { 1 } { 60 } \left( 16 x - x ^ { 2 } \right) & 0 \leqslant x \leqslant 6 \\ 1 & x > 6 \end{cases}$$

1. Find the interquartile range of \(X\).
2. Find \(\mathrm { E } \left( X ^ { 3 } \right)\).
   The random variable \(Y\) is such that \(Y = \sqrt { X }\).
3. Find the probability density function of \(Y\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

4 The continuous random variable \(X\) has cumulative distribution function F given by $$F ( x ) = \begin{cases} 0 & x < 2 \\ \frac { 1 } { 60 } x ^ { 2 } - \frac { 1 } { 15 } & 2 \leqslant x \leqslant 8 \\ 1 & x > 8 \end{cases}$$

1. Find \(\mathrm { P } ( 3 \leqslant X \leqslant 6 )\).
2. Find \(\mathrm { E } ( \sqrt { X } )\).
3. Find \(\operatorname { Var } ( \sqrt { X } )\).
4. The random variable \(Y\) is defined by \(Y = X ^ { 3 }\). Find the probability density function of \(Y\).

5 The continuous random variable \(X\) has cumulative distribution function F given by $$F ( x ) = \begin{cases} 0 & x < 0 \\ 1 - \frac { 1 } { 144 } ( 12 - x ) ^ { 2 } & 0 \leqslant x \leqslant 12 \\ 1 & x > 12 \end{cases}$$

1. Find the upper quartile of \(X\).
2. Find \(\operatorname { Var } \left( X ^ { 2 } \right)\).
   The random variable \(Y\) is given by \(Y = \sqrt { X }\).
3. Find the probability density function of \(Y\).

4 The random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 21 } ( x - 1 ) ^ { 2 } & 2 \leqslant x \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$

1. Find the cumulative distribution function of \(X\).
   The random variable \(Y\) is defined by \(Y = ( X - 1 ) ^ { 4 }\).
2. Find the probability density function of \(Y\). \includegraphics[max width=\textwidth, alt={}, center]{b9cbf607-4f40-41bb-8374-6b2c39f945ac-09_2725_35_99_20}
3. Find the median value of \(Y\).
4. Find \(\mathrm { E } ( Y )\).

7 The continuous random variable \(T\) has probability density function given by $$f ( t ) = \begin{cases} 4 t ^ { 3 } & 0 < t \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$

1. Obtain the cumulative distribution function of \(T\).
2. Find the cumulative distribution function of \(H\), where \(H = \frac { 1 } { T ^ { 4 } }\), and hence show that the probability density function of \(H\) is given by \(\mathrm { g } ( h ) = \frac { 1 } { h ^ { 2 } }\) over an interval to be stated.
3. Find \(\mathrm { E } \left( 1 + 2 H ^ { - 1 } \right)\).

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

1. Given that \(Y = \frac { 1 } { X }\), find the (cumulative) distribution function of \(Y\), and deduce that \(Y\) and \(X\) have identical distributions.
2. Find \(\mathrm { E } ( X + 1 )\) and deduce the value of \(\mathrm { E } \left( \frac { 1 } { X } \right)\).

7 The continuous random variable \(X\) has (cumulative) distribution function given by $$\mathrm { F } ( x ) = \begin{cases} 0 & x < 1 \\ 1 - \frac { 1 } { x ^ { 4 } } & x \geqslant 1 \end{cases}$$

1. Find the (cumulative) distribution function, \(\mathrm { G } ( y )\), of the random variable \(Y\), where \(Y = \frac { 1 } { X ^ { 2 } }\).
2. Hence show that the probability density function of \(Y\) is given by $$g ( y ) = \begin{cases} 2 y & 0 < y \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$
3. Find \(\mathrm { E } ( \sqrt [ 3 ] { Y } )\).

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

1. Find the median value of \(X\).
2. Find the (cumulative) distribution function of \(Y\), where \(Y = \frac { 1 } { X ^ { 2 } }\), and hence find the probability density function of \(Y\).
3. Evaluate \(\mathrm { E } \left( 2 - \frac { 2 } { X ^ { 2 } } \right)\).

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 \(\mathrm { E } ( \sqrt { } Y )\).

4 The continuous random variable \(V\) has (cumulative) distribution function given by $$\mathrm { F } ( v ) = \begin{cases} 0 & v < 1 \\ 1 - \frac { 8 } { ( 1 + v ) ^ { 3 } } & v \geqslant 1 \end{cases}$$ The random variable \(Y\) is given by \(Y = \frac { 1 } { 1 + V }\).

1. Show that the (cumulative) distribution function of \(Y\) is \(8 y ^ { 3 }\), over an interval to be stated, and find the probability density function of \(Y\).
2. Find \(\mathrm { E } \left( \frac { 1 } { Y ^ { 2 } } \right)\).

4 The continuous random variable \(X\) has probability density function given by $$f ( x ) = \begin{cases} \frac { 3 } { 2 } \sqrt { x } & 0 < x \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is given by \(Y = \frac { 1 } { \sqrt { X } }\).

1. Find the (cumulative) distribution function of \(Y\), and hence show that its probability density function is given by $$\mathrm { g } ( y ) = \frac { 3 } { y ^ { 4 } }$$ for a set of values of \(y\) to be stated.
2. Find the value of \(\mathrm { E } \left( Y ^ { 2 } \right)\).

6 The function \(\mathrm { F } ( t )\) is defined as follows. $$\mathrm { F } ( t ) = \begin{cases} 0 & t < 0 \\ \sin ^ { 4 } t & 0 \leqslant t \leqslant \frac { 1 } { 2 } \pi \\ 1 & t > \frac { 1 } { 2 } \pi \end{cases}$$

1. Verify that F is a (cumulative) distribution function. The continuous random variable \(T\) has (cumulative) distribution function \(\mathrm { F } ( t )\).
2. Find the lower quartile of \(T\).
3. Find the (cumulative) distribution function of \(Y\), where \(Y = \sin T\), and obtain the probability density function of \(Y\).
4. Find the expected value of \(\frac { 1 } { Y ^ { 3 } + 2 Y ^ { 4 } }\).

8 The continuous random variable \(S\) has probability density function given by $$f ( s ) = \begin{cases} \frac { 8 } { 3 s ^ { 3 } } & 1 \leqslant s \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$ An isosceles triangle has equal sides of length \(S\), and the angle between them is \(30 ^ { \circ }\) (see diagram).

1. Find the (cumulative) distribution function of the area \(X\) of the triangle, and hence show that the probability density function of \(X\) is \(\frac { 1 } { 3 x ^ { 2 } }\) over an interval to be stated.
2. Find the median value of \(X\). www.ocr.org.uk after the live examination series.
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6 \includegraphics[max width=\textwidth, alt={}, center]{054e0081-afce-4a87-93f5-650dad40b313-3_508_611_262_719} The diagram shows the probability density function f of the continuous random variable \(T\), given by $$f ( t ) = \begin{cases} a t & 0 \leqslant t \leqslant 1 \\ a & 1 < t \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$ where \(a\) is a constant.

1. Find the value of \(a\).
2. Obtain the cumulative distribution function of \(T\).
3. Find the cumulative distribution of \(Y\), where \(Y = T ^ { \frac { 1 } { 2 } }\), and hence find the probability density function of \(Y\).

5 The continuous random variable \(Y\) has probability density function given by $$\mathrm { f } ( y ) = \begin{cases} \ln ( y ) & 1 \leqslant y \leqslant \mathrm { e } \\ 0 & \text { otherwise } \end{cases}$$

1. Verify that this is a valid probability density function.
2. Show that the (cumulative) distribution function of \(Y\) is given by $$\mathrm { F } ( y ) = \begin{cases} 0 & y < 1 \\ y \ln y - y + 1 & 1 \leqslant y \leqslant \mathrm { e } \\ 1 & \text { otherwise } \end{cases}$$
3. Verify that the upper quartile of \(Y\) lies in the interval [2.45, 2.46].
4. Find the (cumulative) distribution function of \(X\) where \(X = \ln Y\).

8 The radius, \(R\), of a sphere is a random variable with a continuous uniform distribution between 0 and 10 .

1. Find the cumulative distribution function and probability density function of \(A\), the surface area of the sphere.
2. Find \(\mathrm { P } ( \mathrm { A } \leqslant 200 \pi )\). \section*{END OF QUESTION PAPER}

The continuous random variable \(T\) has probability density function given by $$\mathrm { f } ( t ) = \begin{cases} 0 & t < 2 \\ \frac { 2 } { ( t - 1 ) ^ { 3 } } & t \geqslant 2 \end{cases}$$

1. Find the distribution function of \(T\), and find also \(\mathrm { P } ( T > 5 )\).
2. Consecutive independent observations of \(T\) are made until the first observation that exceeds 5 is obtained. The random variable \(N\) is the total number of observations that have been made up to and including the observation exceeding 5. Find \(\mathrm { P } ( N > \mathrm { E } ( N ) )\).
3. Find the probability density function of \(Y\), where \(Y = \frac { 1 } { T - 1 }\).

The continuous random variable \(X\) takes values in the interval \(0 \leqslant x \leqslant 3\) only. For \(0 \leqslant x \leqslant 3\) the graph of its probability density function f consists of two straight line segments meeting at the point \(( 1 , k )\), as shown in the diagram. Find \(k\) and hence show that the distribution function F is given by $$\mathrm { F } ( x ) = \begin{cases} 0 & x \leqslant 0 , \\ \frac { 1 } { 3 } x ^ { 2 } & 0 < x \leqslant 1 , \\ x - \frac { 1 } { 2 } - \frac { 1 } { 6 } x ^ { 2 } & 1 < x \leqslant 3 , \\ 1 & x > 3 . \end{cases}$$ The random variable \(Y\) is given by \(Y = X ^ { 2 }\). Find

1. the probability density function of \(Y\),
2. the median value of \(Y\).

The continuous random variable \(X\) takes values in the interval \(0 \leqslant x \leqslant 3\) only. For \(0 \leqslant x \leqslant 3\) the graph of its probability density function f consists of two straight line segments meeting at the point \(( 1 , k )\), as shown in the diagram. Find \(k\) and hence show that the distribution function F is given by $$\mathrm { F } ( x ) = \begin{cases} 0 & x \leqslant 0 , \\ \frac { 1 } { 3 } x ^ { 2 } & 0 < x \leqslant 1 , \\ x - \frac { 1 } { 2 } - \frac { 1 } { 6 } x ^ { 2 } & 1 < x \leqslant 3 , \\ 1 & x > 3 . \end{cases}$$ The random variable \(Y\) is given by \(Y = X ^ { 2 }\). Find

1. the probability density function of \(Y\),
2. the median value of \(Y\).

9 The continuous random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 2 a } & - a \leqslant x \leqslant a \\ 0 & \text { otherwise } \end{cases}$$ where \(a\) is a positive constant. Find the distribution function of \(X\). The random variable \(Y\) is defined by \(Y = \mathrm { e } ^ { X }\). Find the distribution function of \(Y\). Given that \(a = 4\), find the value of \(k\) for which \(\mathrm { P } ( Y \geqslant k ) = 0.25\).

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 \(\mathrm { E } ( Y )\).