| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Moderate -0.3 This is a standard S2 probability density function question requiring routine integration to find c, basic sketching, and straightforward probability calculations. The symmetry insight for the median and the 'hence' part make it slightly easier than average, but it still requires multiple techniques across four parts. |
| 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 |
| \(\frac{3}{4}\int_0^c(cx-x^2)\,dx = 1\) | M1 | Attempt integ \(f(x)\) and \(= 1\); ignore limits |
| \(\frac{3}{4}\left[\frac{cx^2}{2}-\frac{x^3}{3}\right]_0^c = 1\) | A1 | Correct integration and limits (condone \(c=2\)) |
| \(\frac{3}{4}\left(\frac{c^3}{2}-\frac{c^3}{3}\right)=1\) or \(\frac{3}{4}\times\frac{c^3}{6}=1\) or \(\frac{c^3}{8}=1\) | A1 [3] | No errors seen |
| \((c = 2\ \mathbf{AG})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Inverted parabola | B1 | |
| Through \((0,0)\) and \((2,0)\) and zero elsewhere | B1 | Must not extend beyond \([0,2]\) |
| Median \(= 1\) | B1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3}{4}\int_0^{1.5}(2x-x^2)\,dx\) | M1 | Attempt integ \(f(x)\); ignore limits |
| \(= \frac{3}{4}\left[x^2-\frac{x^3}{3}\right]_0^{1.5}\) | A1 | Correct integration; ignore limits |
| \(\frac{3}{4}\left(1.5^2-\frac{1.5^3}{3}\right)\) | B1 | Use of correct limits \([0, 1.5]\) or \(1-[1.5, 2]\) |
| \(= \frac{27}{32}\) or \(0.844\) (3 sf) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \[= \frac{11}{32} \text{ or } 0.344 \text{ (3 sf)}\] | B1f [1] | ft their (iii). For use of symmetry. Note: If do not use "hence" and start again B1 for cwo |
## Question 7:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{4}\int_0^c(cx-x^2)\,dx = 1$ | M1 | Attempt integ $f(x)$ and $= 1$; ignore limits |
| $\frac{3}{4}\left[\frac{cx^2}{2}-\frac{x^3}{3}\right]_0^c = 1$ | A1 | Correct integration and limits (condone $c=2$) |
| $\frac{3}{4}\left(\frac{c^3}{2}-\frac{c^3}{3}\right)=1$ or $\frac{3}{4}\times\frac{c^3}{6}=1$ or $\frac{c^3}{8}=1$ | A1 [3] | No errors seen |
| $(c = 2\ \mathbf{AG})$ | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Inverted parabola | B1 | |
| Through $(0,0)$ and $(2,0)$ and zero elsewhere | B1 | Must not extend beyond $[0,2]$ |
| Median $= 1$ | B1 [3] | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{4}\int_0^{1.5}(2x-x^2)\,dx$ | M1 | Attempt integ $f(x)$; ignore limits |
| $= \frac{3}{4}\left[x^2-\frac{x^3}{3}\right]_0^{1.5}$ | A1 | Correct integration; ignore limits |
| $\frac{3}{4}\left(1.5^2-\frac{1.5^3}{3}\right)$ | B1 | Use of correct limits $[0, 1.5]$ or $1-[1.5, 2]$ |
| $= \frac{27}{32}$ or $0.844$ (3 sf) | A1 [4] | |
## Question (iv):
$$\left(\frac{27}{32} - \frac{1}{2} \text{ or } 0.844 - 0.5\right)$$
$$= \frac{11}{32} \text{ or } 0.344 \text{ (3 sf)}$$ | B1f [1] | ft their (iii). For use of symmetry. Note: If do not use "hence" and start again B1 for cwo |
**Total: 11 marks**
**Total for paper: 50**7 The probability density function of the random variable $X$ is given by
$$f ( x ) = \begin{cases} \frac { 3 } { 4 } x ( c - x ) & 0 \leqslant x \leqslant c \\ 0 & \text { otherwise } \end{cases}$$
where $c$ is a constant.\\
(i) Show that $c = 2$.\\
(ii) Sketch the graph of $y = \mathrm { f } ( x )$ and state the median of $X$.\\
(iii) Find $\mathrm { P } ( X < 1.5 )$.\\
(iv) Hence write down the value of $\mathrm { P } ( 0.5 < X < 1 )$.
\hfill \mbox{\textit{CAIE S2 2015 Q7 [11]}}