| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Standard +0.3 This is a standard S2 probability density function question requiring routine integration techniques: finding the constant using ∫f(x)dx=1, solving ∫f(x)dx=0.5 for the median, and computing E(X)=∫xf(x)dx. While it involves multiple parts and some algebraic manipulation, all steps follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| 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 percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^{\sqrt{2}}(ax - x^3)\,dx = 1\) | M1 | Attempted integration of \(f(x)\) and equated to 1 |
| \(\left[a\frac{x^2}{2} - \frac{x^4}{4}\right]_0^{\sqrt{2}} = 1\) | A1 | Correct integration and substitute correct limits |
| \(a - \frac{4}{4} = 1\) | ||
| \(a = 2\) | A1 | AG. Convincingly obtained and no errors seen. |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^m(2x - x^3)\,dx = \frac{1}{2}\) | M1 | Attempt integrate \(f(x)\) with limits 0 to \(m\) (or \(m\) to \(\sqrt{2}\)) and equate to \(\frac{1}{2}\) |
| \(m^2 - \frac{m^4}{4} = \frac{1}{2}\) | A1 | For correct quartic in any form |
| \(m^4 - 4m^2 + 2 = 0 \Rightarrow m^2 = \frac{4\pm\sqrt{16-8}}{2}\) \([= 2\pm\sqrt{2}]\) | M1 | For solving their three term quartic to find \(m^2\) |
| \(m = \sqrt{2-\sqrt{2}}\) or \(0.765\) (3sf) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^{\sqrt{2}}(2x^2 - x^4)\,dx\) | M1 | Attempt to integrate \(xf(x)\). Ignore limits. |
| \(\left[\frac{2x^3}{3} - \frac{x^5}{5}\right]_0^{\sqrt{2}}\) | A1 | Correct integration and correct limits |
| \(\left[= \frac{4\sqrt{2}}{3} - \frac{4\sqrt{2}}{5}\right] = \frac{8}{15}\sqrt{2}\) | A1 | OE. For single exact term. |
| Total: 3 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{\sqrt{2}}(ax - x^3)\,dx = 1$ | M1 | Attempted integration of $f(x)$ and equated to 1 |
| $\left[a\frac{x^2}{2} - \frac{x^4}{4}\right]_0^{\sqrt{2}} = 1$ | A1 | Correct integration and substitute correct limits |
| $a - \frac{4}{4} = 1$ | | |
| $a = 2$ | A1 | AG. Convincingly obtained and no errors seen. |
| **Total: 3** | | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^m(2x - x^3)\,dx = \frac{1}{2}$ | M1 | Attempt integrate $f(x)$ with limits 0 to $m$ (or $m$ to $\sqrt{2}$) and equate to $\frac{1}{2}$ |
| $m^2 - \frac{m^4}{4} = \frac{1}{2}$ | A1 | For correct quartic in any form |
| $m^4 - 4m^2 + 2 = 0 \Rightarrow m^2 = \frac{4\pm\sqrt{16-8}}{2}$ $[= 2\pm\sqrt{2}]$ | M1 | For solving their three term quartic to find $m^2$ |
| $m = \sqrt{2-\sqrt{2}}$ or $0.765$ (3sf) | A1 | |
| **Total: 4** | | |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{\sqrt{2}}(2x^2 - x^4)\,dx$ | M1 | Attempt to integrate $xf(x)$. Ignore limits. |
| $\left[\frac{2x^3}{3} - \frac{x^5}{5}\right]_0^{\sqrt{2}}$ | A1 | Correct integration and correct limits |
| $\left[= \frac{4\sqrt{2}}{3} - \frac{4\sqrt{2}}{5}\right] = \frac{8}{15}\sqrt{2}$ | A1 | OE. For single exact term. |
| **Total: 3** | | |
---5 A random variable $X$ has probability density function f given by
$$f ( x ) = \begin{cases} a x - x ^ { 3 } & 0 \leqslant x \leqslant \sqrt { 2 } \\ 0 & \text { otherwise } \end{cases}$$
where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $a = 2$ .
\item Find the median of $X$ .
\item Find the exact value of $\mathrm { E } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q5 [10]}}