| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF of transformed variable |
| Difficulty | Standard +0.8 This is a Further Maths question requiring transformation of random variables using the CDF method, then finding median and expectation. While the transformation Y=X³ is straightforward and the integration is manageable, students must understand the relationship between CDFs under transformation (G(y) = F(X≤y^(1/3))), perform careful algebraic manipulation, and compute E(Y) either by integration or using E(X³). The multi-step nature and requirement to work with transformed distributions places this above average difficulty, though it follows a standard template for this topic. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(x) = \frac{x^2}{15} + c = \frac{(x^2-1)}{15}\) | M1 A1 | Integrate \(f(x)\) to find \(F(x)\) for \(1 \leq x \leq 4\) |
| \(G(y) = P(Y < y) = P(X^3 < y) = P(X < y^{1/3}) = F(y^{1/3})\) | Relate dist. fn. \(G(y)\) of \(Y\) to \(X\) | |
| \(= \frac{(y^{2/3}-1)}{15}\) | A.G. M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(G(m) = \frac{1}{2},\quad m^{2/3} = \frac{17}{2}\) | M1 A1 | Find relation for median \(m\) of \(Y\) |
| \(m = 24.8\) | A1 | Evaluate \(m\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER \(g(y) = \frac{2y^{-1/3}}{45}\); \(E(Y) = \int yg(y)\,dy = \int \frac{2y^{2/3}}{45}\,dy\) | M1 A1 | Find \(g(y)\) and formulate \(E(Y)\) |
| OR \(E(Y) = E(X^3) = \int_1^4 \frac{2x^4}{15}\,dx\) | (M1 A1) | Formulate \(E(Y)\) in terms of \(X\) |
| \(E(Y) = \left[\frac{2y^{5/3}}{3 \times 75}\right]_1^{64}\) or \(\left[\frac{2x^5}{75}\right]_1^4\) | M1 A1 | Integrate and apply limits |
| \(= \frac{2(1024-1)}{75} = \frac{682}{25}\) or \(27.3\) | ||
| Subtotal: 4 | Total: [11] |
## Question 7:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(x) = \frac{x^2}{15} + c = \frac{(x^2-1)}{15}$ | M1 A1 | Integrate $f(x)$ to find $F(x)$ for $1 \leq x \leq 4$ |
| $G(y) = P(Y < y) = P(X^3 < y) = P(X < y^{1/3}) = F(y^{1/3})$ | | Relate dist. fn. $G(y)$ of $Y$ to $X$ |
| $= \frac{(y^{2/3}-1)}{15}$ | **A.G.** M1 A1 | |
**Subtotal: 4**
### Question 7(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $G(m) = \frac{1}{2},\quad m^{2/3} = \frac{17}{2}$ | M1 A1 | Find relation for median $m$ of $Y$ |
| $m = 24.8$ | A1 | Evaluate $m$ |
**Subtotal: 3**
### Question 7(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **EITHER** $g(y) = \frac{2y^{-1/3}}{45}$; $E(Y) = \int yg(y)\,dy = \int \frac{2y^{2/3}}{45}\,dy$ | M1 A1 | Find $g(y)$ and formulate $E(Y)$ |
| **OR** $E(Y) = E(X^3) = \int_1^4 \frac{2x^4}{15}\,dx$ | (M1 A1) | Formulate $E(Y)$ in terms of $X$ |
| $E(Y) = \left[\frac{2y^{5/3}}{3 \times 75}\right]_1^{64}$ or $\left[\frac{2x^5}{75}\right]_1^4$ | M1 A1 | Integrate and apply limits |
| $= \frac{2(1024-1)}{75} = \frac{682}{25}$ or $27.3$ | | |
**Subtotal: 4 | Total: [11]**
---7 The continuous random variable $X$ has probability density function f given by
$$f ( x ) = \begin{cases} \frac { 2 } { 15 } x & 1 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
The random variable $Y$ is defined by $Y = X ^ { 3 }$. Show that the distribution function G of $Y$ is given by
$$\mathrm { G } ( y ) = \begin{cases} 0 & y < 1 \\ \frac { 1 } { 15 } \left( y ^ { \frac { 2 } { 3 } } - 1 \right) & 1 \leqslant y \leqslant 64 \\ 1 & y > 64 \end{cases}$$
Find\\
(i) the median value of $Y$,\\
(ii) $\mathrm { E } ( Y )$.
\hfill \mbox{\textit{CAIE FP2 2012 Q7 [11]}}