| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Variance of transformed variable |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths statistics question requiring standard techniques: computing variance of a transformed variable using E(X) and E(X²) integrals, finding PDF of Y=X² via Jacobian transformation, and solving for the median. While it involves several steps and careful algebra with the given PDF, all methods are textbook procedures for Further Maths students with no novel insight required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles5.07a Non-parametric tests: when to use |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(x) = \begin{cases} \frac{x}{4}(4-x^2) & 0 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}\) | B1 | |
| \(E(X) = \int_0^2 \frac{1}{4}(4x^2 - x^4)\,dx = \frac{1}{4}\!\left[\frac{4}{3}x^3 - \frac{1}{5}x^5\right]_0^2 = \frac{16}{15}\ (=1.067)\) | ||
| \(E(\sqrt{X}) = \int_0^2 \frac{1}{4}(4x^{1.5} - x^{3.5})\,dx = \frac{1}{4}\!\left[\frac{8}{5}x^{2.5} - \frac{2}{9}x^{4.5}\right]_0^2 \left[= \frac{32}{45}\sqrt{2} = 1.005(7)\right]\) | M1 | Integration, correct powers required, ignore limits. |
| \(\text{Var}(\sqrt{X}) = E(X) - \left(E(\sqrt{X})\right)^2\) used | M1 | Correct formula used following attempt at integration of both terms. |
| \(\left[\frac{16}{15} - \left(\frac{32}{45}\sqrt{2}\right)^2\right] = 0.0553\) | A1 | \(\frac{112}{2025}\) |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(F(x) = \frac{1}{16}(8x^2 - x^4)\quad [0 \leq x \leq 2]\) | B1 | |
| \(G(y) = \frac{1}{16}(8y - y^2)\quad [0 \leq y \leq 4]\) | M1 | Change variable in their \(F(x)\). |
| \(g(y) = \frac{1}{16}(8 - 2y)\quad [0 \leq y \leq 4]\) | M1 | Differentiate their \(G(y)\). |
| \(g(y) = \begin{cases} \frac{1}{16}(8-2y) & 0 \leq y \leq 4 \\ 0 & \text{otherwise} \end{cases}\) | A1 | Complete and correct. |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{16}(8m - m^2) = \frac{1}{2}\) | M1 | Equate their \(G(y)\) to \(\frac{1}{2}\). |
| \([m^2 - 8m + 8 = 0,]\quad m = 4 - 2\sqrt{2}\) | A1 | Single answer in an exact form. |
| Total: 2 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(x) = \begin{cases} \frac{x}{4}(4-x^2) & 0 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}$ | B1 | |
| $E(X) = \int_0^2 \frac{1}{4}(4x^2 - x^4)\,dx = \frac{1}{4}\!\left[\frac{4}{3}x^3 - \frac{1}{5}x^5\right]_0^2 = \frac{16}{15}\ (=1.067)$ | | |
| $E(\sqrt{X}) = \int_0^2 \frac{1}{4}(4x^{1.5} - x^{3.5})\,dx = \frac{1}{4}\!\left[\frac{8}{5}x^{2.5} - \frac{2}{9}x^{4.5}\right]_0^2 \left[= \frac{32}{45}\sqrt{2} = 1.005(7)\right]$ | M1 | Integration, correct powers required, ignore limits. |
| $\text{Var}(\sqrt{X}) = E(X) - \left(E(\sqrt{X})\right)^2$ used | M1 | Correct formula used following attempt at integration of both terms. |
| $\left[\frac{16}{15} - \left(\frac{32}{45}\sqrt{2}\right)^2\right] = 0.0553$ | A1 | $\frac{112}{2025}$ |
| **Total: 4** | | |
---
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $F(x) = \frac{1}{16}(8x^2 - x^4)\quad [0 \leq x \leq 2]$ | B1 | |
| $G(y) = \frac{1}{16}(8y - y^2)\quad [0 \leq y \leq 4]$ | M1 | Change variable in their $F(x)$. |
| $g(y) = \frac{1}{16}(8 - 2y)\quad [0 \leq y \leq 4]$ | M1 | Differentiate their $G(y)$. |
| $g(y) = \begin{cases} \frac{1}{16}(8-2y) & 0 \leq y \leq 4 \\ 0 & \text{otherwise} \end{cases}$ | A1 | Complete and correct. |
| **Total: 4** | | |
---
## Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{16}(8m - m^2) = \frac{1}{2}$ | M1 | Equate their $G(y)$ to $\frac{1}{2}$. |
| $[m^2 - 8m + 8 = 0,]\quad m = 4 - 2\sqrt{2}$ | A1 | Single answer in an exact form. |
| **Total: 2** | | |7 The continuous random variable $X$ has probability density function f given by
$$f ( x ) = \left\{ \begin{array} { c c }
\frac { x } { 4 } \left( 4 - x ^ { 2 } \right) & 0 \leqslant x \leqslant 2 \\
0 & \text { otherwise }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Find $\operatorname { Var } ( \sqrt { X } )$.\\
The continuous random variable $Y$ is defined by $Y = X ^ { 2 }$.
\item Find the probability density function of $Y$.
\item Find the exact value of the median of $Y$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q7 [10]}}