| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Geometric/graphical PDF with k |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths statistics question requiring: (1) finding k using the normalization condition (area = 1 under piecewise linear PDF), (2) integrating piecewise to verify the given CDF, (3) transforming a random variable using the Jacobian method, and (4) finding a median. While it involves several techniques and the transformation of random variables is a Further Maths topic, each step follows standard procedures without requiring novel insight. The piecewise nature adds some algebraic complexity but the question structure is methodical and typical for FP2 level. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| MI of one side of square about midpoint: \(\frac{1}{3}\left(\frac{M}{4}\right)\frac{a^2}{2} = \frac{Ma^2}{24}\) | M1 A1 | |
| MI of side of square (or of square) about \(O\): \(\frac{Ma^2}{24} + \left(\frac{M}{4}\right)\frac{a^2}{2} = \frac{Ma^2}{6}\) | M1 A1 | |
| MI of ring about \(O\): \(2Ma^2\) | B1 | |
| Sum to find MI of system about \(O\): \(I_O = 4 \times \frac{Ma^2}{6} + 2Ma^2 = 8\frac{Ma^2}{3}\) | A1 | A.G. |
| MI of system about axis through \(A\): \(I_A = \frac{1}{2}I_O + 3Ma^2 = 13\frac{Ma^2}{3}\) | M1 A1 | |
| State or imply that speed is max when \(AC\) vertical | M1 | |
| Use energy when \(AC\) vertical (or at general point): \(\frac{1}{2}(I_A + 4Ma^2)\omega^2 = [25\frac{Ma^2}{3}]\) or \(\frac{1}{2}I_A\omega^2 + \frac{1}{2}Mv^2\) | ||
| \(= 5Mga(1 + \cos 60°)\) or \(\frac{15Mga}{2}\) | M1 | A1 for each side of eqn, A.E.F. |
| A1 A1 | ||
| Substitute for \(I_A\) and find max speed \(v = 2a\omega\): \(\omega^2 = \frac{15Mga/2}{25Ma^2/6} = \frac{9g}{5a}\) | ||
| \(v = 6\sqrt{\frac{ga}{5}}\) or \(6\sqrt{2a}\) or \(8.49\sqrt{a}\) | M1 A1 | [6+2+6=14] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Find \(k\) by equating area under graph to 1: \(\frac{1}{2}k + k = 1\), \(k = \frac{2}{3}\) | M1 A1 | |
| Find \(f(x)\) for \(0 \leq x \leq 1\) and \(1 < x \leq 3\): \(kx\left[=\frac{2}{3}x\right]\), \(\frac{1}{2}k(3-x)\left[= 1 - \frac{1}{3}x\right]\) | M1 A1 | |
| Integrate to find \(F(x)\) for \(0 < x \leq 1\) and \(1 < x \leq 3\): \(\frac{1}{3}x^2\), \(x - \frac{1}{2} - \frac{x^2}{6}\) | M1 A1 | A.G. [6] |
| (i) Relate dist. fn. \(G(y)\) of \(Y\) to \(X\): \(G(y) = P(Y < y) = P(X^2 < y)\) | working may be omitted | |
| \(= P(X < y^{1/2}) = F(y^{1/2})\) | ||
| \(= \frac{1}{3}y\), \(y^{1/2} - \frac{1}{2} - \frac{y}{6}\) | M1 A1 | |
| Differentiate to find \(g(y)\): \(g(y) = \frac{1}{3}\) \((0 < y \leq 1)\) | ||
| \(\frac{1}{2}y^{-1/2} - \frac{1}{6}\) \((1 < y \leq 9)\) | ||
| \([0 \text{ otherwise}]\) | M1 A1 | [4] |
| (ii) Use \(G(m) = \frac{1}{2}\) to find eqn for median \(m\): \(m^{1/2} - \frac{1}{2} - \frac{m}{6} = \frac{1}{2}\) | M1 | |
| Solve quadratic for \(\sqrt{m}\): \((\sqrt{m})^2 - 6\sqrt{m} + 6 = 0\), \(\sqrt{m} = 3 \pm \sqrt{3}\) | M1 A1 | |
| Select value of \(m\) in interval: \(m = (3 - \sqrt{3})^2 = 12 - 6\sqrt{3}\) or \(1.61\) | B1 | [4+14] |
# Question 11:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| MI of one side of square about midpoint: $\frac{1}{3}\left(\frac{M}{4}\right)\frac{a^2}{2} = \frac{Ma^2}{24}$ | M1 A1 | |
| MI of side of square (or of square) about $O$: $\frac{Ma^2}{24} + \left(\frac{M}{4}\right)\frac{a^2}{2} = \frac{Ma^2}{6}$ | M1 A1 | |
| MI of ring about $O$: $2Ma^2$ | B1 | |
| Sum to find MI of system about $O$: $I_O = 4 \times \frac{Ma^2}{6} + 2Ma^2 = 8\frac{Ma^2}{3}$ | A1 | **A.G.** |
| MI of system about axis through $A$: $I_A = \frac{1}{2}I_O + 3Ma^2 = 13\frac{Ma^2}{3}$ | M1 A1 | |
| State or imply that speed is max when $AC$ vertical | M1 | |
| Use energy when $AC$ vertical (or at general point): $\frac{1}{2}(I_A + 4Ma^2)\omega^2 = [25\frac{Ma^2}{3}]$ or $\frac{1}{2}I_A\omega^2 + \frac{1}{2}Mv^2$ | | |
| $= 5Mga(1 + \cos 60°)$ or $\frac{15Mga}{2}$ | M1 | A1 for each side of eqn, A.E.F. |
| | A1 A1 | |
| Substitute for $I_A$ and find max speed $v = 2a\omega$: $\omega^2 = \frac{15Mga/2}{25Ma^2/6} = \frac{9g}{5a}$ | | |
| $v = 6\sqrt{\frac{ga}{5}}$ or $6\sqrt{2a}$ or $8.49\sqrt{a}$ | M1 A1 | **[6+2+6=14]** |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Find $k$ by equating area under graph to 1: $\frac{1}{2}k + k = 1$, $k = \frac{2}{3}$ | M1 A1 | |
| Find $f(x)$ for $0 \leq x \leq 1$ and $1 < x \leq 3$: $kx\left[=\frac{2}{3}x\right]$, $\frac{1}{2}k(3-x)\left[= 1 - \frac{1}{3}x\right]$ | M1 A1 | |
| Integrate to find $F(x)$ for $0 < x \leq 1$ and $1 < x \leq 3$: $\frac{1}{3}x^2$, $x - \frac{1}{2} - \frac{x^2}{6}$ | M1 A1 | **A.G.** **[6]** |
| **(i)** Relate dist. fn. $G(y)$ of $Y$ to $X$: $G(y) = P(Y < y) = P(X^2 < y)$ | | working may be omitted |
| $= P(X < y^{1/2}) = F(y^{1/2})$ | | |
| $= \frac{1}{3}y$, $y^{1/2} - \frac{1}{2} - \frac{y}{6}$ | M1 A1 | |
| Differentiate to find $g(y)$: $g(y) = \frac{1}{3}$ $(0 < y \leq 1)$ | | |
| $\frac{1}{2}y^{-1/2} - \frac{1}{6}$ $(1 < y \leq 9)$ | | |
| $[0 \text{ otherwise}]$ | M1 A1 | **[4]** |
| **(ii)** Use $G(m) = \frac{1}{2}$ to find eqn for median $m$: $m^{1/2} - \frac{1}{2} - \frac{m}{6} = \frac{1}{2}$ | M1 | |
| Solve quadratic for $\sqrt{m}$: $(\sqrt{m})^2 - 6\sqrt{m} + 6 = 0$, $\sqrt{m} = 3 \pm \sqrt{3}$ | M1 A1 | |
| Select value of $m$ in interval: $m = (3 - \sqrt{3})^2 = 12 - 6\sqrt{3}$ or $1.61$ | B1 | **[4+14]** |\begin{center}
\includegraphics[max width=\textwidth, alt={}]{e8a16ec8-b6b7-4b0c-b0c1-8f5f7a9e4fa6-5_383_839_1635_651}
\end{center}
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\\
(i) the probability density function of $Y$,\\
(ii) the median value of $Y$.
\hfill \mbox{\textit{CAIE FP2 2011 Q11 OR}}