| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2023 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find multiple parameters from system |
| Difficulty | Challenging +1.8 This is a challenging Further Maths statistics question requiring students to use multiple properties of PDFs simultaneously: normalization condition, quartile condition, and continuity at x=4. The system of equations requires algebraic manipulation and the expectation of √X involves non-trivial integration. However, the structure is methodical with clear signposting through parts (a)-(c), making it accessible to well-prepared Further Maths students despite requiring several sophisticated techniques. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(UQ = 4\) so \(\int_{4}^{6} c\, dx = \frac{1}{4}\): \(c[6-4] = \frac{1}{4}\), \(c = \frac{1}{8}\) | B1 | AG. Some indication of reasoning required. |
| \(\int_{0}^{4} \frac{1}{128}\left(4ax - bx^3\right)dx = \frac{3}{4}\); \(\frac{1}{128}\left[2ax^2 - \frac{b}{4}x^4\right] = \frac{3}{4}\), \(32a - 64b = 96\) | M1 | Integrate and use correct limits to form equation in \(a\) and \(b\). |
| Equate at \(x = 4\): \(a - 4b = 1\) | M1 | |
| Solve: \(a = 5, b = 1\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_{0}^{m} \frac{1}{128}\left(20x - x^3\right)dx = \frac{1}{2}\), OR \(F(x) = \frac{1}{128}\left(2ax^2 - \frac{b}{4}x^4\right) = \frac{5}{64}x^2 - \frac{1}{512}x^4\) | M1* | Attempt at integral and use of correct limits. Could be in terms of \(a,b,c\). Or attempt at relevant part of CDF and set \(F(m) = \frac{1}{2}\). Could be in terms of \(a,b,c\). |
| \(m^4 - 40m^2 + 256 = 0\) | M1 | Simplify to quartic. Could be in terms of \(a,b,c\). |
| \(m = 2\sqrt{2}\) | A1 | Positive answer only. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E\left(\sqrt{X}\right) = \frac{1}{128}\int_{0}^{4} 20x^{\frac{3}{2}} - x^{\frac{7}{2}}\, dx + \int_{4}^{6} \frac{1}{8}x^{\frac{1}{2}}\, dx\) | M1 | *Their* PDF multiplied by \(\sqrt{x}\) with correct limits. Could be in terms of \(a,b,c\). |
| \(\frac{1}{128}\left[8x^{\frac{5}{2}} - \frac{2}{9}x^{\frac{9}{2}}\right]_{0}^{4} + \frac{1}{12}\left[x^{\frac{3}{2}}\right]_{4}^{6}\) | M1 | Correct use of correct limits in *their* integral. Could be in terms of \(a,b,c\). May be implied by correct final answer. |
| \(\frac{1}{128}\left(2^8 - \frac{1}{9}\cdot 2^{10}\right) + \frac{1}{12}\left(6\sqrt{6} - 8\right) = \frac{4}{9} + \frac{1}{2}\sqrt{6} \approx 1.67\) | A1 | CAO. 1.67 following (first) M0 SC B2 |
| 3 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $UQ = 4$ so $\int_{4}^{6} c\, dx = \frac{1}{4}$: $c[6-4] = \frac{1}{4}$, $c = \frac{1}{8}$ | B1 | AG. Some indication of reasoning required. |
| $\int_{0}^{4} \frac{1}{128}\left(4ax - bx^3\right)dx = \frac{3}{4}$; $\frac{1}{128}\left[2ax^2 - \frac{b}{4}x^4\right] = \frac{3}{4}$, $32a - 64b = 96$ | M1 | Integrate and use correct limits to form equation in $a$ and $b$. |
| Equate at $x = 4$: $a - 4b = 1$ | M1 | |
| Solve: $a = 5, b = 1$ | A1 | |
| | **4** | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_{0}^{m} \frac{1}{128}\left(20x - x^3\right)dx = \frac{1}{2}$, OR $F(x) = \frac{1}{128}\left(2ax^2 - \frac{b}{4}x^4\right) = \frac{5}{64}x^2 - \frac{1}{512}x^4$ | M1* | Attempt at integral and use of correct limits. Could be in terms of $a,b,c$. Or attempt at relevant part of CDF and set $F(m) = \frac{1}{2}$. Could be in terms of $a,b,c$. |
| $m^4 - 40m^2 + 256 = 0$ | M1 | Simplify to quartic. Could be in terms of $a,b,c$. |
| $m = 2\sqrt{2}$ | A1 | Positive answer only. |
| | **3** | |
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E\left(\sqrt{X}\right) = \frac{1}{128}\int_{0}^{4} 20x^{\frac{3}{2}} - x^{\frac{7}{2}}\, dx + \int_{4}^{6} \frac{1}{8}x^{\frac{1}{2}}\, dx$ | M1 | *Their* PDF multiplied by $\sqrt{x}$ with correct limits. Could be in terms of $a,b,c$. |
| $\frac{1}{128}\left[8x^{\frac{5}{2}} - \frac{2}{9}x^{\frac{9}{2}}\right]_{0}^{4} + \frac{1}{12}\left[x^{\frac{3}{2}}\right]_{4}^{6}$ | M1 | Correct use of correct limits in *their* integral. Could be in terms of $a,b,c$. May be implied by correct final answer. |
| $\frac{1}{128}\left(2^8 - \frac{1}{9}\cdot 2^{10}\right) + \frac{1}{12}\left(6\sqrt{6} - 8\right) = \frac{4}{9} + \frac{1}{2}\sqrt{6} \approx 1.67$ | A1 | CAO. 1.67 following (first) M0 SC B2 |
| | **3** | |4\\
\includegraphics[max width=\textwidth, alt={}, center]{a9f9cf66-0734-4316-99ae-c57090d08135-08_353_1141_255_463}
The diagram shows the continuous random variable $X$ with probability density function f given by
$$f ( x ) = \begin{cases} \frac { 1 } { 128 } \left( 4 a x - b x ^ { 3 } \right) & 0 \leqslant x \leqslant 4 \\ c & 4 \leqslant x \leqslant 6 \\ 0 & \text { otherwise } \end{cases}$$
where $a , b$ and $c$ are constants.\\
The upper quartile of $X$ is equal to 4 .
\begin{enumerate}[label=(\alph*)]
\item Show that $c = \frac { 1 } { 8 }$ and find the values of $a$ and $b$.
\item Find the exact value of the median of $X$.
\item Find $\mathrm { E } ( \sqrt { X } )$, giving your answer correct to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2023 Q4 [10]}}