| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| 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 to find k, expected value, and variance. Part (i) uses the fundamental property that total probability equals 1, part (ii) applies standard E(X) and Var(X) formulas with straightforward polynomial integration, and part (iii) is a simple probability calculation. The symmetry of the distribution makes E(X) = 0.5 immediately obvious. While it requires multiple techniques, all are textbook applications with no novel problem-solving required, making it slightly easier than average. |
| 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 |
| \(k\int_0^1(x-x^2)\,dx = 1\) | M1 | Attempt integ \(f(x)\) and "= 1", ignore limits |
| \(= k\left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = 1\) | A1 | correct integration, limits 0 and 1 |
| \(= k\left[\frac{1}{2} - \frac{1}{3}\right] = 1\) or \(\frac{k}{6} = 1\) | A1 | correctly obtained, no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = 0.5\) | B1 | |
| \(6\int_0^1(x^3 - x^4)\,dx\) | M1 | Attempt integ \(x^2f(x)\), limits 0 to 1 |
| \(\left(= 6\left[\frac{1}{4} - \frac{1}{5}\right] = 0.3\right)\) "0.3" \(-\) "0.5"\(^2\) | M1 | their int \(x^2f(x)\) – their \((E(X))^2\) dep +ve result |
| \(= 0.05\,(= 1/20)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(6\int_{0.4}^1(x - x^2)\,dx\) | M1 | ignore limits, eg M1 for \(6\int_{0.4}^2(x-x^2)\,dx\) |
| \(= 6\left\{\frac{1}{2} - \frac{1}{3} - \left(\frac{0.4^2}{2} - \frac{0.4^3}{3}\right)\right\}\) | A1FT | subst correct limits into correct integration |
| \(= 0.648\,(= 81/125)\) | A1 | condone incorrect "k" for A1 |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k\int_0^1(x-x^2)\,dx = 1$ | M1 | Attempt integ $f(x)$ and "= 1", ignore limits |
| $= k\left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = 1$ | A1 | correct integration, limits 0 and 1 |
| $= k\left[\frac{1}{2} - \frac{1}{3}\right] = 1$ or $\frac{k}{6} = 1$ | A1 | correctly obtained, no errors seen |
**Total: 3**
---
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = 0.5$ | B1 | |
| $6\int_0^1(x^3 - x^4)\,dx$ | M1 | Attempt integ $x^2f(x)$, limits 0 to 1 |
| $\left(= 6\left[\frac{1}{4} - \frac{1}{5}\right] = 0.3\right)$ "0.3" $-$ "0.5"$^2$ | M1 | their int $x^2f(x)$ – their $(E(X))^2$ dep +ve result |
| $= 0.05\,(= 1/20)$ | A1 | |
**Total: 4**
---
## Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $6\int_{0.4}^1(x - x^2)\,dx$ | M1 | ignore limits, eg M1 for $6\int_{0.4}^2(x-x^2)\,dx$ |
| $= 6\left\{\frac{1}{2} - \frac{1}{3} - \left(\frac{0.4^2}{2} - \frac{0.4^3}{3}\right)\right\}$ | A1FT | subst correct limits into correct integration |
| $= 0.648\,(= 81/125)$ | A1 | condone incorrect "k" for A1 |
**Total: 3**
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{395f7f2c-42db-4fb6-9b22-3b0f46ad16d3-08_355_670_260_735}
The diagram shows the graph of the probability density function, f , of a continuous random variable $X$, where f is defined by
$$\mathrm { f } ( x ) = \begin{cases} k \left( x - x ^ { 2 } \right) & 0 \leqslant x \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$
(i) Show that the value of the constant $k$ is 6 .\\
(ii) State the value of $\mathrm { E } ( X )$ and find $\operatorname { Var } ( X )$.\\
(iii) Find $\mathrm { P } ( 0.4 < X < 2 )$.\\
\hfill \mbox{\textit{CAIE S2 2017 Q6 [10]}}