| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Verify geometric PDF from graph |
| Difficulty | Moderate -0.8 This is a straightforward S2 question testing basic PDF properties: verifying area=1 by integration (parts a,b require simple geometric area calculations or basic integration), and computing mean from a given PDF formula. Part (c)(i) requires only qualitative reasoning about median from a graph. All techniques are standard recall with minimal problem-solving. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2} \times 2 \times 1\) or \(\int_0^2 \frac{1}{2}x\, dx = 1\), which is the correct area under a pdf. | B1 | Calculation and result. |
| \(f(x) \geqslant 0\) | B1 | Condone \(f(x) > 0\) or 'Line is above \(x\)-axis' OE. |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}\pi r^2 = 1\) | M1 | Area of semi-circle equated to 1 OE. Missing factor of \(\frac{1}{2}\) gets M1A0. |
| \(r = \sqrt{\frac{2}{\pi}}\) or 0.798 (3sf) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area to the left of 15 is greater than 0.5 | B1 | OE, e.g. 'The distribution of X is skewed to the right / positively skewed, suggesting the median will be less than the mid-point of the interval.' or 'The distribution of X is skewed to the right / positively skewed' or 'It is a decreasing function suggesting the median will be less than the mid-point of the interval'. |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_{10}^{20}\left(\frac{40}{x} - \frac{x}{10}\right)dx\) | M1 | Integration of \(xh(x)\) attempted. Ignore limits. |
| \(\left[40\ln x - \frac{x^2}{20}\right]_{10}^{20}\) | A1 | Correct integration and limits (can be implied by final answer). |
| \(= 40\ln 2 - 15\) or 12.7 (3sf) | A1 | |
| Total: 3 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times 2 \times 1$ or $\int_0^2 \frac{1}{2}x\, dx = 1$, which is the correct area under a pdf. | B1 | Calculation and result. |
| $f(x) \geqslant 0$ | B1 | Condone $f(x) > 0$ or 'Line is above $x$-axis' OE. |
| | **2** | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\pi r^2 = 1$ | M1 | Area of semi-circle equated to 1 OE. Missing factor of $\frac{1}{2}$ gets M1A0. |
| $r = \sqrt{\frac{2}{\pi}}$ or 0.798 (3sf) | A1 | |
| **Total: 2** | | |
## Question 2(c)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area to the left of 15 is greater than 0.5 | B1 | OE, e.g. 'The distribution of X is skewed to the right / positively skewed, suggesting the median will be less than the mid-point of the interval.' or 'The distribution of X is skewed to the right / positively skewed' or 'It is a decreasing function suggesting the median will be less than the mid-point of the interval'. |
| **Total: 1** | | |
## Question 2(c)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_{10}^{20}\left(\frac{40}{x} - \frac{x}{10}\right)dx$ | M1 | Integration of $xh(x)$ attempted. Ignore limits. |
| $\left[40\ln x - \frac{x^2}{20}\right]_{10}^{20}$ | A1 | Correct integration and limits (can be implied by final answer). |
| $= 40\ln 2 - 15$ or 12.7 (3sf) | A1 | |
| **Total: 3** | | |2
\begin{enumerate}[label=(\alph*)]
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{b0960fa7-ddbe-47b7-929e-f62f72f9dc93-04_324_574_264_813}
The graph of the function f is a straight line segment from $( 0,0 )$ to $( 2,1 )$.\\
Show that $f$ could be a probability density function.
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{b0960fa7-ddbe-47b7-929e-f62f72f9dc93-04_364_592_1466_804}
The graph of the function g is a semicircle, centre $( 0,0 )$, entirely above the $x$-axis.\\
Given that g is a probability density function, find the radius of the semicircle.
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{b0960fa7-ddbe-47b7-929e-f62f72f9dc93-05_369_826_264_689}
The time, $X$ minutes, taken by a large number of students to complete a test has probability density function h , as shown in the diagram.
\begin{enumerate}[label=(\roman*)]
\item Without calculation, use the diagram to explain how you can tell that the median time is less than 15 minutes.\\
It is now given that
$$h ( x ) = \begin{cases} \frac { 40 } { x ^ { 2 } } - \frac { 1 } { 10 } & 10 \leqslant x \leqslant 20 \\ 0 & \text { otherwise. } \end{cases}$$
\item Find the mean time.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q2 [8]}}