| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | March |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Geometric/graphical PDF with k |
| Difficulty | Standard +0.3 This is a straightforward PDF question requiring integration of a linear function to find k (using area = 1), then computing E(X) with standard formula, and solving a probability equation. All steps are routine S2 techniques with no conceptual challenges beyond applying standard methods to a simple geometric shape. |
| 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 |
| \(\frac{1}{2} \times \frac{1}{2}k \times k = 1\) | M1 | Or use of \(\int_0^k\left(-\frac{1}{2}x+\frac{1}{2}k\right)dx=1\) and attempt at integral |
| \(k = 2\) | A1 | Unsupported answers M0 A0. Do not accept \(\pm 2\) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x) = -\frac{1}{2}x + 1\) | B1 FT | FT *their* \(k\) from \(y = -\frac{1}{2}x + \frac{1}{2}k\) |
| \(\int_0^2\left(-\frac{1}{2}x^2 + x\right)dx = \left[-\frac{x^3}{6}+\frac{x^2}{2}\right]_0^2\) | M1 | Attempt integration of \(xf(x)\) limits 0 to \(k\). FT *their* \(f(x)\). Could be in terms of \(k\) |
| \(\frac{2}{3}\) or \(0.667\) (3 sf) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_p^1\left(-\frac{1}{2}x+1\right)dx\ [=0.25]\) | M1 | FT their equation of line; correct integral and limits (could be reversed) stated or \(\frac{1}{2}(1-p)(1-\frac{1}{2}p+\frac{1}{2})[=0.25]\) |
| \(\left[-\frac{x^2}{4}+x\right]_p^1 = 0.25\) | M1 | Attempt substitution of correct limits (not reversed) into their integral or attempt expand must equal 0.25. OE |
| \(-\frac{1}{4}+1+\frac{p^2}{4}-p=0.25\) | ||
| \(p^2 - 4p + 2 = 0\) | M1 | Obtain 3-term quadratic set equal to 0, obtain at least 1 solution |
| \(p = 2 - \sqrt{2}\) or \(0.586\) | A1 | CAO |
| 4 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times \frac{1}{2}k \times k = 1$ | M1 | Or use of $\int_0^k\left(-\frac{1}{2}x+\frac{1}{2}k\right)dx=1$ and attempt at integral |
| $k = 2$ | A1 | Unsupported answers M0 A0. Do not accept $\pm 2$ |
| | **2** | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = -\frac{1}{2}x + 1$ | B1 FT | FT *their* $k$ from $y = -\frac{1}{2}x + \frac{1}{2}k$ |
| $\int_0^2\left(-\frac{1}{2}x^2 + x\right)dx = \left[-\frac{x^3}{6}+\frac{x^2}{2}\right]_0^2$ | M1 | Attempt integration of $xf(x)$ limits 0 to $k$. FT *their* $f(x)$. Could be in terms of $k$ |
| $\frac{2}{3}$ or $0.667$ (3 sf) | A1 | |
| | **3** | |
## Question 2(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_p^1\left(-\frac{1}{2}x+1\right)dx\ [=0.25]$ | M1 | FT their equation of **line**; correct integral and limits (could be reversed) stated or $\frac{1}{2}(1-p)(1-\frac{1}{2}p+\frac{1}{2})[=0.25]$ |
| $\left[-\frac{x^2}{4}+x\right]_p^1 = 0.25$ | M1 | Attempt substitution of correct limits (not reversed) into their integral or attempt expand must equal 0.25. OE |
| $-\frac{1}{4}+1+\frac{p^2}{4}-p=0.25$ | | |
| $p^2 - 4p + 2 = 0$ | M1 | Obtain 3-term quadratic set equal to 0, obtain at least 1 solution |
| $p = 2 - \sqrt{2}$ or $0.586$ | A1 | CAO |
| | **4** | |2\\
\includegraphics[max width=\textwidth, alt={}, center]{image-not-found}
The diagram shows the graph of the probability density function, f , of a random variable $X$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of the constant $k$.
\item Using this value of $k$, find $\mathrm { f } ( x )$ for $0 \leqslant x \leqslant k$ and hence find $\mathrm { E } ( X )$.
\item Find the value of $p$ such that $\mathrm { P } ( p < X < 1 ) = 0.25$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q2 [9]}}