| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find expectation E(X) |
| Difficulty | Moderate -0.3 This is a straightforward S2 question requiring standard techniques: sketching given PDFs, computing E(X) using a simple polynomial integral, and making a qualitative comparison of variances from graphs. The integration is routine (polynomial of degree 3), and the variance comparison requires only visual reasoning about spread, not calculation. Slightly easier than average due to the direct application of formulas without problem-solving insight. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Upwards parabola, not below \(x\)-axis | M1 | *[scales/annotations not needed]* |
| Correct place, not extending beyond limits, ignore pointed at \(a\) | A1 | Touching axes (not asymptotic) |
| Horizontal straight line, not beyond limits, \(y\)-intercept below curve (unless curve makes this meaningless) | B1 | Don't need vertical lines. i.e., 3/3 only if wholly right |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\displaystyle\int_0^a \frac{3}{a^3} x(x-a)^2\,dx\) | M1 | Attempt this integral, correct limits seen somewhere |
| \(= \displaystyle\int_0^a \frac{3}{a^3}(x^3 - 2ax^2 + a^2x)\,dx\) | M1 | Method for \(\int xf(x)\), e.g. multiply out or parts, independent of first M1. Multiplication: needs 3 terms |
| Correct form for integration, e.g. multiplied out correctly, or correct first stage of parts | A1 | E.g. \(\dfrac{3}{a^3}x\dfrac{(x-a)^3}{3} - \displaystyle\int \dfrac{3}{a^3}\dfrac{(x-a)^3}{3}\,dx\) |
| \(= \left[\dfrac{3}{a^3}\!\left(\dfrac{x^4}{4} - \dfrac{2ax^3}{3} + \dfrac{a^2x^2}{2}\right)\right]_0^a\) | B1 | Correct indefinite integral. E.g. \(\dfrac{3}{a^3}x\dfrac{(x-a)^3}{3} - \dfrac{3}{a^3}\dfrac{(x-a)^4}{12}\) |
| \(= \dfrac{a}{4}\) | A1 | \(\dfrac{a}{4}\) or exact equivalent (e.g. \(0.25a\)) only. Limits not seen anywhere: can get M0M1A0B1A0 |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S\) is concentrated more towards 0, therefore \(T\) has bigger variance | M1 A1 | M1: Reason that shows understanding of PDF. A1: Correct conclusion. *Not*, e.g., "\(T\) is constant" |
| Total: 2 |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Upwards parabola, not below $x$-axis | M1 | *[scales/annotations not needed]* |
| Correct place, not extending beyond limits, ignore pointed at $a$ | A1 | Touching axes (not asymptotic) |
| Horizontal straight line, not beyond limits, $y$-intercept below curve (unless curve makes this meaningless) | B1 | Don't need vertical lines. i.e., 3/3 only if wholly right |
| | **Total: 3** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\displaystyle\int_0^a \frac{3}{a^3} x(x-a)^2\,dx$ | M1 | Attempt this integral, correct limits seen somewhere |
| $= \displaystyle\int_0^a \frac{3}{a^3}(x^3 - 2ax^2 + a^2x)\,dx$ | M1 | Method for $\int xf(x)$, e.g. multiply out or parts, independent of first M1. Multiplication: needs 3 terms |
| Correct form for integration, e.g. multiplied out correctly, or correct first stage of parts | A1 | E.g. $\dfrac{3}{a^3}x\dfrac{(x-a)^3}{3} - \displaystyle\int \dfrac{3}{a^3}\dfrac{(x-a)^3}{3}\,dx$ |
| $= \left[\dfrac{3}{a^3}\!\left(\dfrac{x^4}{4} - \dfrac{2ax^3}{3} + \dfrac{a^2x^2}{2}\right)\right]_0^a$ | B1 | Correct indefinite integral. E.g. $\dfrac{3}{a^3}x\dfrac{(x-a)^3}{3} - \dfrac{3}{a^3}\dfrac{(x-a)^4}{12}$ |
| $= \dfrac{a}{4}$ | A1 | $\dfrac{a}{4}$ or exact equivalent (e.g. $0.25a$) only. Limits not seen anywhere: can get M0M1A0B1A0 |
| | **Total: 5** | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S$ is concentrated more towards 0, therefore $T$ has bigger variance | M1 A1 | M1: Reason that shows understanding of PDF. A1: Correct conclusion. *Not*, e.g., "$T$ is constant" |
| | **Total: 2** | |
---5 Two random variables $S$ and $T$ have probability density functions given by
$$\begin{aligned}
& f _ { S } ( x ) = \begin{cases} \frac { 3 } { a ^ { 3 } } ( x - a ) ^ { 2 } & 0 \leqslant x \leqslant a \\
0 & \text { otherwise } \end{cases} \\
& f _ { T } ( x ) = \begin{cases} c & 0 \leqslant x \leqslant a \\
0 & \text { otherwise } \end{cases}
\end{aligned}$$
where $a$ and $c$ are constants.\\
(i) On a single diagram sketch both probability density functions.\\
(ii) Calculate the mean of $S$, in terms of $a$.\\
(iii) Use your diagram to explain which of $S$ or $T$ has the bigger variance. (Answers obtained by calculation will score no marks.)
\hfill \mbox{\textit{OCR S2 2013 Q5 [10]}}