| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find parameter from median |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring standard techniques: verifying a pdf by integration (routine), finding a parameter using the median condition (solving a simple cubic equation), and calculating E(X) by integration. All steps are mechanical applications of definitions with no novel insight 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 integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3}{a^3} \int_0^a x^2 \, dx = \frac{3}{a^3} \left[\frac{x^3}{3}\right]_0^a\) | M1 | Attempt to integrate \(f(x)\) with limits 0 and \(a\) (condone missing \(\frac{3}{a^3}\)) |
| \(= \frac{3a^3}{3a^3}\) | A1 | \(\frac{3a^3}{3a^3} - 0\) or better seen |
| \(= 1\) Hence f is pdf for all \(a\) | A1 | Answer \(= 1\) and comment |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3}{a^3} \int_0^2 x^2 \, dx = 0.5\); \(\frac{3}{a^3} \left[\frac{x^3}{3}\right]_0^2 = 0.5\) | M1 | Attempt to integrate \(f(x) = 0.5\), limits 0 and 2 oe, condone missing \(\frac{3}{a^3}\) |
| \(\frac{3}{a^3} \times \frac{8}{3} = 0.5\) oe | A1 | \(\frac{2^3}{3} - 0\) or better, condone missing \(\frac{3}{a^3}\) |
| \(a^3 = 16\) or \(a = \sqrt[3]{16}\) \((= 2.52\ \mathbf{AG})\) | A1 | Convincingly obtained; Note: Attempt to verify 2.52, M1 as stated except not equated to 0.5. A1 as stated, A1 for evaluation to 0.499..approx 0.5 |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3}{16} \int_0^{2.52} x^3 \, dx\) or \(\frac{3}{16} \int_0^a x^3 \, dx\); \(= \frac{3}{16} \left[\frac{x^4}{4}\right]_0^{2.52}\) or \(\frac{3}{16} \left[\frac{x^4}{4}\right]_0^a\) | M1 | Attempt to integrate \(xf(x)\), correct limits, condone missing \(\frac{3}{a^3}\) |
| \(= \frac{3}{16} \times \frac{40.317}{4}\) | A1 | \(\frac{2.52^4}{4} - 0\) or better, condone missing \(\frac{3}{a^3}\) |
| \(= 1.89\) (3 sf) | A1 | |
| Total: 3 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{a^3} \int_0^a x^2 \, dx = \frac{3}{a^3} \left[\frac{x^3}{3}\right]_0^a$ | M1 | Attempt to integrate $f(x)$ with limits 0 and $a$ (condone missing $\frac{3}{a^3}$) |
| $= \frac{3a^3}{3a^3}$ | A1 | $\frac{3a^3}{3a^3} - 0$ or better seen |
| $= 1$ Hence f is pdf for all $a$ | A1 | Answer $= 1$ and comment |
| **Total: 3** | | |
---
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{a^3} \int_0^2 x^2 \, dx = 0.5$; $\frac{3}{a^3} \left[\frac{x^3}{3}\right]_0^2 = 0.5$ | M1 | Attempt to integrate $f(x) = 0.5$, limits 0 and 2 oe, condone missing $\frac{3}{a^3}$ |
| $\frac{3}{a^3} \times \frac{8}{3} = 0.5$ oe | A1 | $\frac{2^3}{3} - 0$ or better, condone missing $\frac{3}{a^3}$ |
| $a^3 = 16$ or $a = \sqrt[3]{16}$ $(= 2.52\ \mathbf{AG})$ | A1 | Convincingly obtained; Note: Attempt to verify 2.52, M1 as stated except not equated to 0.5. A1 as stated, A1 for evaluation to 0.499..approx 0.5 |
| **Total: 3** | | |
---
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{16} \int_0^{2.52} x^3 \, dx$ or $\frac{3}{16} \int_0^a x^3 \, dx$; $= \frac{3}{16} \left[\frac{x^4}{4}\right]_0^{2.52}$ or $\frac{3}{16} \left[\frac{x^4}{4}\right]_0^a$ | M1 | Attempt to integrate $xf(x)$, correct limits, condone missing $\frac{3}{a^3}$ |
| $= \frac{3}{16} \times \frac{40.317}{4}$ | A1 | $\frac{2.52^4}{4} - 0$ or better, condone missing $\frac{3}{a^3}$ |
| $= 1.89$ (3 sf) | A1 | |
| **Total: 3** | | |
---6 A function f is defined by
$$f ( x ) = \begin{cases} \frac { 3 x ^ { 2 } } { a ^ { 3 } } & 0 \leqslant x \leqslant a \\ 0 & \text { otherwise } \end{cases}$$
where $a$ is a constant.\\
(i) Show that f is a probability density function for all positive values of $a$.\\
The random variable $X$ has probability density function f and the median of $X$ is 2 .\\
(ii) Show that $a = 2.52$, correct to 3 significant figures.\\
(iii) Find $\mathrm { E } ( X )$.\\
\hfill \mbox{\textit{CAIE S2 2019 Q6 [9]}}