| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Interquartile range calculation |
| Difficulty | Standard +0.3 This is a standard continuous probability distribution question requiring integration to find constants and then calculating quartiles. Part (a) involves routine application of probability conditions (integrating pdf to 1, using given probability), while part (b) requires solving for quartiles by integration—straightforward Further Maths content with multiple steps but no novel insight required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X \leqslant 2) = \frac{1}{3}\), so \(\left[\frac{mx^2}{2}\right]_0^2 = \frac{1}{3}\); \(m\left(\frac{4}{2}\right) = \frac{1}{3}\), \(m = \frac{1}{6}\) | B1 | AG, reasoning required |
| \(\int_2^6 \left(\frac{k}{x^2} + c\right)dx = \frac{2}{3}\), so \(\left[-\frac{k}{x} + cx\right]_2^6 = 4c + \frac{1}{3}k = \frac{2}{3}\) or \(k + 12c = 2\) | M1 | Attempt to integrate and correct use of correct limits to form linear equation in \(k\) and \(c\) |
| Also, \(2m = \frac{k}{4} + c\) or \(\frac{1}{3} = \frac{k}{4} + c\) or \(3k + 12c = 4\) | M1 | Matching at \(x = 2\) |
| Solve: \(k = 1\), \(c = \frac{1}{12}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| LQ: \(\int_0^L \frac{1}{6}x\,dx = \frac{1}{4}\), \(L = \sqrt{3}\) | B1 | |
| UQ: \(\frac{1}{3} + \int_2^U \left(\frac{k}{x^2} + c\right)dx = \frac{3}{4}\) | M1 | Or find and use CDF. May be in terms of \(m\) |
| \(-\frac{k}{U} + Uc + \frac{k}{2} - 2c = \frac{5}{12}\); \(U^2 - U - 12 = 0\) | M1 | Attempt at integral and correct use of correct limits. Simplify to quadratic equation, may be in terms of \(k\) and \(c\) |
| \(U = 4\) | A1 | |
| \(\text{IQR} = 4 - \sqrt{3}\) | A1 FT | FT their UQ \(-\) their LQ, exact values only |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X \leqslant 2) = \frac{1}{3}$, so $\left[\frac{mx^2}{2}\right]_0^2 = \frac{1}{3}$; $m\left(\frac{4}{2}\right) = \frac{1}{3}$, $m = \frac{1}{6}$ | B1 | AG, reasoning required |
| $\int_2^6 \left(\frac{k}{x^2} + c\right)dx = \frac{2}{3}$, so $\left[-\frac{k}{x} + cx\right]_2^6 = 4c + \frac{1}{3}k = \frac{2}{3}$ or $k + 12c = 2$ | M1 | Attempt to integrate and correct use of correct limits to form linear equation in $k$ and $c$ |
| Also, $2m = \frac{k}{4} + c$ or $\frac{1}{3} = \frac{k}{4} + c$ or $3k + 12c = 4$ | M1 | Matching at $x = 2$ |
| Solve: $k = 1$, $c = \frac{1}{12}$ | A1 | |
---
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| LQ: $\int_0^L \frac{1}{6}x\,dx = \frac{1}{4}$, $L = \sqrt{3}$ | B1 | |
| UQ: $\frac{1}{3} + \int_2^U \left(\frac{k}{x^2} + c\right)dx = \frac{3}{4}$ | M1 | Or find and use CDF. May be in terms of $m$ |
| $-\frac{k}{U} + Uc + \frac{k}{2} - 2c = \frac{5}{12}$; $U^2 - U - 12 = 0$ | M1 | Attempt at integral and correct use of correct limits. Simplify to quadratic equation, may be in terms of $k$ and $c$ |
| $U = 4$ | A1 | |
| $\text{IQR} = 4 - \sqrt{3}$ | A1 FT | FT their UQ $-$ their LQ, exact values only |
---\begin{enumerate}[label=(\alph*)]
\item Given that $\mathrm { P } ( X \leqslant 2 ) = \frac { 1 } { 3 }$, show that $m = \frac { 1 } { 6 }$ and find the values of $k$ and $c$.
\item Find the exact numerical value of the interquartile range of $X$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2023 Q4 [9]}}