| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| 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 techniques: finding k by integrating to 1, calculating E(X), and using P(X<1) for expected frequency. All steps are textbook exercises with no novel problem-solving required, making it slightly easier than average A-level difficulty. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\) m | B1 [1] | allow without units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k\int_0^2 x^2(2-x)\,dx = 1\) | M1 | attempt integrate \(f(x)\) and \(`= 1`\). Ignore limits |
| \(k\left[\frac{2x^3}{3} - \frac{x^4}{4}\right]_0^2\) | A1 | correct integration and limits |
| \(k \times \left[\frac{16}{3} - 4\right] = 1\) or \(k \times \frac{4}{3} = 1\) oe | ||
| \(k = \frac{3}{4}\) AG | A1 [3] | No errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{3}{4}\int_0^2 x^3(2-x)\,dx\) | M1 | attempt integrate \(xf(x)\), condone missing \(k\) |
| \(= \frac{3}{4} \times \left[\frac{2x^4}{4} - \frac{x^5}{5}\right]_0^2\) | A1 | correct integration and limits, condone missing \(k\) |
| \(1.2\) m oe | A1 [3] | allow without units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{3}{4}\int_0^1 x^2(2-x)\,dx\) | M1 | attempt integrate \(f(x)\), 0 to 1, condone missing \(k\) |
| \(\left(= \frac{3}{4} \times \left(\frac{2}{3} - \frac{1}{4}\right)\right)\) | ||
| \(= \frac{5}{16}\) or 0.3125 oe | A1 | |
| \(400 \times \frac{5}{16} = 125\) | A1ft [3] | ft their \(\frac{5}{16}\) |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2$ m | B1 [1] | allow without units |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k\int_0^2 x^2(2-x)\,dx = 1$ | M1 | attempt integrate $f(x)$ and $`= 1`$. Ignore limits |
| $k\left[\frac{2x^3}{3} - \frac{x^4}{4}\right]_0^2$ | A1 | correct integration and limits |
| $k \times \left[\frac{16}{3} - 4\right] = 1$ or $k \times \frac{4}{3} = 1$ oe | | |
| $k = \frac{3}{4}$ **AG** | A1 [3] | No errors seen |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3}{4}\int_0^2 x^3(2-x)\,dx$ | M1 | attempt integrate $xf(x)$, condone missing $k$ |
| $= \frac{3}{4} \times \left[\frac{2x^4}{4} - \frac{x^5}{5}\right]_0^2$ | A1 | correct integration and limits, condone missing $k$ |
| $1.2$ m oe | A1 [3] | allow without units |
### Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3}{4}\int_0^1 x^2(2-x)\,dx$ | M1 | attempt integrate $f(x)$, 0 to 1, condone missing $k$ |
| $\left(= \frac{3}{4} \times \left(\frac{2}{3} - \frac{1}{4}\right)\right)$ | | |
| $= \frac{5}{16}$ or 0.3125 oe | A1 | |
| $400 \times \frac{5}{16} = 125$ | A1ft [3] | ft their $\frac{5}{16}$ |
---6 In each turn of a game, a coin is pushed and slides across a table. The distance, $X$ metres, travelled by the coin has probability density function given by
$$f ( x ) = \begin{cases} k x ^ { 2 } ( 2 - x ) & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.\\
(i) State the greatest possible distance travelled by the coin in one turn.\\
(ii) Show that $k = \frac { 3 } { 4 }$.\\
(iii) Find the mean distance travelled by the coin in one turn.\\
(iv) Out of 400 turns, find the expected number of turns in which the distance travelled by the coin is less than 1 metre.
\hfill \mbox{\textit{CAIE S2 2016 Q6 [10]}}