| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Moderate -0.8 This is a straightforward continuous probability distribution question requiring only standard techniques: integrating the pdf to find k (routine application of ∫f(x)dx=1), then solving ∫f(x)dx=0.5 for the median. Both parts involve simple polynomial integration with no conceptual challenges or problem-solving insight needed. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(k\int_1^2(3-x)\,dx = 1\) | M1 | Attempt \(\int f(x)=1\), ignore limits or \(\frac{k}{2}(h_1+h_2)=1\) |
| \(k\left[3x - \frac{x^2}{2}\right]_1^2 = 1\) | A1 | Correct integration & limits or \(\frac{k}{2}(2+1)=1\) |
| \(k \times 1.5 = 1\) or \(k \times \frac{3}{2} = 1\) or \(k = \frac{1}{1.5}\) oe | ||
| \(k = \frac{2}{3}\) AG | A1 [3] | No errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{2}{3}\int_1^m(3-x)\,dx = 0.5\) oe \(\int\) from \(m\) to 2 | M1* | Attempt Int \(f(x)=0.5\), ignore limits oe. Or use of area of trapezium |
| \(\left(\frac{2}{3}\left[3x - \frac{x^2}{2}\right]_1^m = 0.5\right)\) | ||
| \(\frac{2}{3}\left[3m - \frac{m^2}{2} - 2.5\right] = 0.5\) | dep M1* | Sub of correct limits into their integral. Or trapezium using 1 and \(m/m\) and 2. Any correct 3-term QE \(= 0\) or \((m-3)^2 = 2.5\) |
| \(m^2 - 6m + 6.5 = 0\) oe | A1 | |
| \(m = \frac{6 \pm \sqrt{36 - 4 \times 6.5}}{2} = 1.42\) or \(4.58\) | ||
| \(m = 1.42\) (3 sf) | A1 [4] | or \(\frac{6-\sqrt{10}}{2}\) oe; single correct answer |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $k\int_1^2(3-x)\,dx = 1$ | M1 | Attempt $\int f(x)=1$, ignore limits or $\frac{k}{2}(h_1+h_2)=1$ |
| $k\left[3x - \frac{x^2}{2}\right]_1^2 = 1$ | A1 | Correct integration & limits or $\frac{k}{2}(2+1)=1$ |
| $k \times 1.5 = 1$ or $k \times \frac{3}{2} = 1$ or $k = \frac{1}{1.5}$ oe | | |
| $k = \frac{2}{3}$ **AG** | A1 [3] | No errors seen |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2}{3}\int_1^m(3-x)\,dx = 0.5$ oe $\int$ from $m$ to 2 | M1* | Attempt Int $f(x)=0.5$, ignore limits oe. Or use of area of trapezium |
| $\left(\frac{2}{3}\left[3x - \frac{x^2}{2}\right]_1^m = 0.5\right)$ | | |
| $\frac{2}{3}\left[3m - \frac{m^2}{2} - 2.5\right] = 0.5$ | dep M1* | Sub of correct limits into their integral. Or trapezium using 1 and $m/m$ and 2. Any correct 3-term QE $= 0$ or $(m-3)^2 = 2.5$ |
| $m^2 - 6m + 6.5 = 0$ oe | A1 | |
| $m = \frac{6 \pm \sqrt{36 - 4 \times 6.5}}{2} = 1.42$ or $4.58$ | | |
| $m = 1.42$ (3 sf) | A1 [4] | or $\frac{6-\sqrt{10}}{2}$ oe; single correct answer |
---4 A random variable $X$ has probability density function given by
$$\mathrm { f } ( x ) = \begin{cases} k ( 3 - x ) & 1 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.\\
(i) Show that $k = \frac { 2 } { 3 }$.\\
(ii) Find the median of $X$.
\hfill \mbox{\textit{CAIE S2 2015 Q4 [7]}}