| 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 algebraic PDF formula |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring verification that a PDF integrates to 1 (using the area formula for a semicircle) and basic geometry/trigonometry to find a probability. The semicircle setup makes the integration trivial, and finding the angle uses standard inverse trig. Slightly easier than average due to the geometric shortcut avoiding actual integration. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2}\pi\left(\sqrt{\frac{2}{\pi}}\right)^2\) | M1 | |
| \(= 1\), which is the area under a PDF \([\text{and } f(x) \geq 0]\) | A1 | Result and statement are both needed. |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\cos^{-1}\left(\frac{\sqrt{\frac{1}{\pi}}}{\sqrt{\frac{2}{\pi}}}\right) = \frac{\pi}{4}\) | B1 | AG. Accept alternative approaches, e.g. using Pythagoras, tangent, or isosceles right-angle triangles. Answer should be convincingly obtained and all correct. |
| Area of sector \(= \frac{1}{4}\) | B1 | |
| Area of triangle \(AOB = \frac{1}{2}OA \times OB = \frac{1}{2} \times \sqrt{\frac{1}{\pi}} \times \sqrt{\frac{2}{\pi} - \frac{1}{\pi}}\) or Area of triangle \(AOB = \frac{1}{2}OA \times OB \times \sin(AOB) = \frac{1}{2} \times \sqrt{\frac{1}{\pi}} \times \sqrt{\frac{2}{\pi}} \sin\frac{\pi}{4}\) | M1 | Accept alternative approaches. Note: \(AB = \sqrt{0.7979^2 - 0.5642^2}\ [= 0.5642]\). Allow values to 3sf. |
| \(\frac{1}{2\pi}\) or \(0.1592\) | A1 | |
| \(\frac{1}{4} - \frac{1}{2\pi}\) or \(0.25 - 0.1592\) | M1 | Attempt area of sector \(-\) area of triangle \(AOB\). |
| \(= \frac{1}{4} - \frac{1}{2\pi}\) or \(0.0908\) (3sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Find equation of curve \(x^2 + y^2 = \frac{2}{\pi}\) | M1 | |
| \(y = \sqrt{\frac{2}{\pi} - x^2}\) | A1 | |
| Attempt to integrate (any limits) | M1 | |
| Use of correct limits \(\sqrt{\frac{1}{\pi}}\) to \(\sqrt{\frac{2}{\pi}}\) | B1 | |
| Correct integration with correct limits | A1 | |
| \(= \frac{1}{4} - \frac{1}{2\pi}\) or \(0.0908\) (3sf) | A1 | Correct final answer. |
| Total | 6 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}\pi\left(\sqrt{\frac{2}{\pi}}\right)^2$ | M1 | |
| $= 1$, which is the area under a PDF $[\text{and } f(x) \geq 0]$ | A1 | Result and statement are both needed. |
| | **2** | |
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cos^{-1}\left(\frac{\sqrt{\frac{1}{\pi}}}{\sqrt{\frac{2}{\pi}}}\right) = \frac{\pi}{4}$ | B1 | AG. Accept alternative approaches, e.g. using Pythagoras, tangent, or isosceles right-angle triangles. Answer should be convincingly obtained and all correct. |
| Area of sector $= \frac{1}{4}$ | B1 | |
| Area of triangle $AOB = \frac{1}{2}OA \times OB = \frac{1}{2} \times \sqrt{\frac{1}{\pi}} \times \sqrt{\frac{2}{\pi} - \frac{1}{\pi}}$ or Area of triangle $AOB = \frac{1}{2}OA \times OB \times \sin(AOB) = \frac{1}{2} \times \sqrt{\frac{1}{\pi}} \times \sqrt{\frac{2}{\pi}} \sin\frac{\pi}{4}$ | M1 | Accept alternative approaches. Note: $AB = \sqrt{0.7979^2 - 0.5642^2}\ [= 0.5642]$. Allow values to 3sf. |
| $\frac{1}{2\pi}$ or $0.1592$ | A1 | |
| $\frac{1}{4} - \frac{1}{2\pi}$ or $0.25 - 0.1592$ | M1 | Attempt area of sector $-$ area of triangle $AOB$. |
| $= \frac{1}{4} - \frac{1}{2\pi}$ or $0.0908$ (3sf) | A1 | |
## Question 7(b) Alternative Method (Using Integration):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Find equation of curve $x^2 + y^2 = \frac{2}{\pi}$ | M1 | |
| $y = \sqrt{\frac{2}{\pi} - x^2}$ | A1 | |
| Attempt to integrate (any limits) | M1 | |
| Use of correct limits $\sqrt{\frac{1}{\pi}}$ to $\sqrt{\frac{2}{\pi}}$ | B1 | |
| Correct integration with correct limits | A1 | |
| $= \frac{1}{4} - \frac{1}{2\pi}$ or $0.0908$ (3sf) | A1 | Correct final answer. |
| **Total** | **6** | |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{10cf346f-dee2-4223-8caa-2a49f1eaa99f-10_547_880_260_621}
A random variable $X$ has probability density function f , where the graph of $y = \mathrm { f } ( x )$ is a semicircle with centre $( 0,0 )$ and radius $\sqrt { \frac { 2 } { \pi } }$, entirely above the $x$-axis. Elsewhere $\mathrm { f } ( x ) = 0$ (see diagram).
\begin{enumerate}[label=(\alph*)]
\item Verify that f can be a probability density function.\\
$A$ and $B$ are the points where the line $x = \sqrt { \frac { 1 } { \pi } }$ meets the $x$-axis and the semicircle respectively.
\item Show that angle $A O B$ is $\frac { 1 } { 4 } \pi$ radians and hence find $\mathrm { P } \left( X > \sqrt { \frac { 1 } { \pi } } \right)$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q7 [8]}}