| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Moderate -0.5 This is a straightforward S2 question testing standard pdf properties: recognizing x as a value of the random variable, using the integral condition to find k, and applying variance formulas. All steps are routine applications of well-practiced techniques with no problem-solving insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Values taken by \(X\) | B1 | This answer only. Not "values taken by f" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^a kx\,dx = 1 \Rightarrow k = \dfrac{2}{a^2}\) | M1, A1 | Use definite integral and equate to 1. Correctly obtain \(2/a^2\). *Or* clear argument from triangle area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^a kx^2\,dx = \left[k\dfrac{x^3}{3}\right]_0^a = \tfrac{2}{3}a\) | M1, B1, A1\(\sqrt{}\) | Attempt to integrate \(xf(x)\), limits 0 and \(a\). Correct indefinite integral seen. Correct mean *or* correct \(E(X^2)\ [=a^2/2]\), \(\sqrt{}\) on \(k\). Either here or for \(x^2f(x)\); can be in terms of \(k\) |
| \(\int_0^a kx^3\,dx = \left[k\dfrac{x^4}{4}\right]_0^a = \dfrac{a^2}{2}\) | M1* | Attempt to integrate \(x^2f(x)\), limits 0, \(a\) |
| \(\dfrac{a^2}{2} - \left(\tfrac{2}{3}a\right)^2 = \tfrac{1}{18}a^2\) | depM1, A1 | Subtract their \(\mu^2\). Correct final answer, ae exact f, no \(k\) now. Or decimal, \(0.056a^2\) or better |
## Question 4:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Values taken by $X$ | B1 | This answer only. Not "values taken by f" |
**[1 mark]**
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^a kx\,dx = 1 \Rightarrow k = \dfrac{2}{a^2}$ | M1, A1 | Use definite integral and equate to 1. Correctly obtain $2/a^2$. *Or* clear argument from triangle area |
**[2 marks]**
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^a kx^2\,dx = \left[k\dfrac{x^3}{3}\right]_0^a = \tfrac{2}{3}a$ | M1, B1, A1$\sqrt{}$ | Attempt to integrate $xf(x)$, limits 0 and $a$. Correct indefinite integral seen. Correct mean *or* correct $E(X^2)\ [=a^2/2]$, $\sqrt{}$ on $k$. Either here or for $x^2f(x)$; can be in terms of $k$ |
| $\int_0^a kx^3\,dx = \left[k\dfrac{x^4}{4}\right]_0^a = \dfrac{a^2}{2}$ | M1* | Attempt to integrate $x^2f(x)$, limits 0, $a$ |
| $\dfrac{a^2}{2} - \left(\tfrac{2}{3}a\right)^2 = \tfrac{1}{18}a^2$ | depM1, A1 | Subtract their $\mu^2$. Correct final answer, ae exact f, no $k$ now. Or decimal, $0.056a^2$ or better |
**[6 marks]**
---4 A continuous random variable $X$ has probability density function
$$\mathrm { f } ( x ) = \left\{ \begin{array} { c l }
k x & 0 \leqslant x \leqslant a \\
0 & \text { otherwise }
\end{array} \right.$$
where $k$ and $a$ are constants.\\
(i) State what the letter $x$ represents.\\
(ii) Find $k$ in terms of $a$.\\
(iii) Find $\operatorname { Var } ( X )$ in terms of $a$.
\hfill \mbox{\textit{OCR S2 2013 Q4 [9]}}