| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Geometric/graphical PDF with k |
| Difficulty | Moderate -0.3 This is a straightforward S2 PDF question requiring standard techniques: using the area-under-curve property to find c, calculating a probability from the geometric shape, and finding E(X) by integration. The geometric shapes (likely triangles/rectangles) make calculations simpler than general integration. Slightly easier than average due to the visual/geometric nature reducing algebraic complexity. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2} \times 3 \times c = 1\) | B1 | Must see this line, oe, and result. Alternative method involving equation of line \(y = \left(\frac{-c}{3}\right)x + c\) must have all relevant working shown |
| \(c = \frac{2}{3}\) AG | ||
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(\frac{1}{3}\right)^2\) | M1 | Allow M1 for \(\frac{1}{3}\) seen as a linear scale factor; or attempt to find equation of line (of form \(y = mx + c\) where \(c \neq 0\)) and interval from 2 to 3 OE; or attempt to find the point \(\left(2, \frac{2}{9}\right)\) using the equation of the line (of form \(y = mx + c\) where \(c \neq 0\)) and then use area of triangle |
| \(= \frac{1}{9}\) or \(0.111\) (3sf) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equation of line is \(y = \frac{2}{3} - \left(\frac{2}{3} \div 3\right)x\) | *M1 | OE Must be of form \(y = mx + c\) (\(c \neq 0\)) |
| \(E(X) = \int_0^3 \left(\frac{2}{3}x - \frac{2}{9}x^2\right)dx\) | DM1 | Attempt integration \(x \times\) *their* \(f(x)\), ignore limits |
| \(= \left[\frac{x^2}{3} - \frac{2x^3}{27}\right]_0^3\) | A1 FT | Correct integration and limits. FT *their* equation of line |
| \(= 1\) | A1 | |
| Total: 4 |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2} \times 3 \times c = 1$ | B1 | Must see this line, oe, and result. Alternative method involving equation of line $y = \left(\frac{-c}{3}\right)x + c$ must have all relevant working shown |
| $c = \frac{2}{3}$ **AG** | | |
| Total: 1 | | |
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(\frac{1}{3}\right)^2$ | M1 | Allow M1 for $\frac{1}{3}$ seen as a linear scale factor; or attempt to find equation of line (of form $y = mx + c$ where $c \neq 0$) and interval from 2 to 3 OE; or attempt to find the point $\left(2, \frac{2}{9}\right)$ using the equation of the line (of form $y = mx + c$ where $c \neq 0$) and then use area of triangle |
| $= \frac{1}{9}$ or $0.111$ (3sf) | A1 | |
| Total: 2 | | |
## Question 3(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equation of line is $y = \frac{2}{3} - \left(\frac{2}{3} \div 3\right)x$ | *M1 | OE Must be of form $y = mx + c$ ($c \neq 0$) |
| $E(X) = \int_0^3 \left(\frac{2}{3}x - \frac{2}{9}x^2\right)dx$ | DM1 | Attempt integration $x \times$ *their* $f(x)$, ignore limits |
| $= \left[\frac{x^2}{3} - \frac{2x^3}{27}\right]_0^3$ | A1 FT | Correct integration and limits. FT *their* equation of line |
| $= 1$ | A1 | |
| Total: 4 | | |3\\
\includegraphics[max width=\textwidth, alt={}, center]{ec7cab36-683b-4022-9cac-fb3b4e64778a-04_332_1100_260_520}
A random variable $X$ takes values between 0 and 3 only and has probability density function as shown in the diagram, where $c$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $c = \frac { 2 } { 3 }$.
\item Find $\mathrm { P } ( X > 2 )$.
\item Calculate $\mathrm { E } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q3 [7]}}