| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Standard +0.3 This is a straightforward continuous probability distribution question requiring standard techniques: integration to find a probability, calculating E(X) using the standard formula, and setting up the median equation by integrating to 0.5. All steps are routine S2 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 integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{18}\int_{0}^{1.2}(9 - x^2)\,dx\) | M1 | Attempt to integrate \(f(x)\), ignore limits. Must see an increase of power. |
| \(\frac{1}{18}\left[9x - \frac{x^3}{3}\right]_{0}^{1.2}\) | A1 | Correct integration and correct limits. |
| \(\frac{71}{125}\) or \(0.568\) | A1 | SC unsupported answer scores B2 only. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{18}\int_{0}^{3}(9x - x^3)\,dx\) | M1 | Attempt to integrate \(xf(x)\), ignore limits. Must see an increase of power. |
| \(\frac{1}{18}\left[\frac{9x^2}{2} - \frac{x^4}{4}\right]_{0}^{3}\) | A1 | Correct integration and correct limits. |
| \(\frac{9}{8}\) or \(1.125\) | A1 | SC unsupported answer scores B2 only. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{18}\left[9x - \frac{x^3}{3}\right]_{0}^{m} = 0.5\) | M1 | Attempt to integrate \(f(x)\) with correct limits and \(= 0.5\). OE. Accept limits \(m\) to 3. Allow \(x\) instead of \(m\). |
| \(\frac{1}{18}\left[9m - \frac{m^3}{3}\right] - 0.5 = 0\) | A1 | Any correct cubic equation in \(m\) or \(x\). |
| \(m^3 - 27m + 27 = 0\) | A1 | AG. Correctly obtain this equation. No errors seen. |
| 3 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{18}\int_{0}^{1.2}(9 - x^2)\,dx$ | M1 | Attempt to integrate $f(x)$, ignore limits. Must see an increase of power. |
| $\frac{1}{18}\left[9x - \frac{x^3}{3}\right]_{0}^{1.2}$ | A1 | Correct integration and correct limits. |
| $\frac{71}{125}$ or $0.568$ | A1 | SC unsupported answer scores **B2** only. |
| | **3** | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{18}\int_{0}^{3}(9x - x^3)\,dx$ | M1 | Attempt to integrate $xf(x)$, ignore limits. Must see an increase of power. |
| $\frac{1}{18}\left[\frac{9x^2}{2} - \frac{x^4}{4}\right]_{0}^{3}$ | A1 | Correct integration and correct limits. |
| $\frac{9}{8}$ or $1.125$ | A1 | SC unsupported answer scores **B2** only. |
| | **3** | |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{18}\left[9x - \frac{x^3}{3}\right]_{0}^{m} = 0.5$ | M1 | Attempt to integrate $f(x)$ with correct limits and $= 0.5$. OE. Accept limits $m$ to 3. Allow $x$ instead of $m$. |
| $\frac{1}{18}\left[9m - \frac{m^3}{3}\right] - 0.5 = 0$ | A1 | Any correct cubic equation in $m$ or $x$. |
| $m^3 - 27m + 27 = 0$ | A1 | AG. Correctly obtain this equation. No errors seen. |
| | **3** | |
---4 A random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} \frac { 1 } { 18 } \left( 9 - x ^ { 2 } \right) & 0 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X < 1.2 )$.
\item Find $\mathrm { E } ( X )$.\\
The median of $X$ is $m$.
\item Show that $m ^ { 3 } - 27 m + 27 = 0$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q4 [9]}}