| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find parameter from expectation |
| Difficulty | Standard +0.3 This is a standard S2 question requiring routine application of pdf properties (integrating to find k), expectation formula to find a, and variance formula. The algebra is straightforward with no conceptual surprises—slightly easier than average since it's methodical application of learned techniques with clear signposting. |
| 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 |
| Positive parabola (only), through 0, nothing below \(x\)-axis | M1 | \(k < 0\): M0 even if \(k > 0\) as well. |
| Clear truncation at ends | A1 | Don't need any scales, vertical line at \(a\) etc. Can be vertical at \(A\), needn't be horizontal at \(O\). |
| Withhold if concept misunderstood. Need to have probability of *values* (not of *occurring*); not just shape. Allow for U-shape but nothing else | B1 | E.g.: "More likely to *occur* for \(x\) close to \(a\)": B0. Ignore extra comments like "exponential" |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^a kx^2\,dx = 1 \Rightarrow k = \dfrac{3}{a^3}\) | M1 | Attempt to integrate \(kx^2\), ignore limits. Must attempt integration |
| A1 | Correct limits and equate to 1 | |
| \(\int_0^a \dfrac{3}{a^3}x^3\,dx = \dfrac{9}{2} \Rightarrow a = 6\) | M1 | Attempt to integrate \(kx^3\), ignore limits. Must attempt integration. Don't need \(k\) in terms of \(a\) here |
| A1 | Correct limits and equate to 4.5. \(ka^3 = 3\) or \(ka^4 = 18\), a.e. simplified form | |
| A1 | One correct equation connecting \(k\) and \(a\), can be implied | |
| A1 | Correctly obtain \(a = 6\) only. No marks explicitly for \(k\ [= 1/72\) or \(0.01388\ldots]\) | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^6 \frac{1}{72} x^3 \, dx \quad \left[= \frac{108}{5}\right]\) | M1 | Attempt to integrate \(kx^4\), their \(a\), \(k\) can be algebraic |
| Subtract \(4.5^2\) (given in question) | A1 | Somewhere |
| \(21.6 - 4.5^2 = 1.35\) | A1 | \(\left[=\frac{27}{20}\right]\) |
| [3] | Limits \(0\), \(a\sqrt{}\); 1.35 or exact equivalent only |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Positive parabola (only), through 0, nothing below $x$-axis | M1 | $k < 0$: M0 even if $k > 0$ as well. |
| Clear truncation at ends | A1 | Don't need any scales, vertical line at $a$ etc. Can be vertical at $A$, needn't be horizontal at $O$. |
| Withhold if concept misunderstood. Need to have probability of *values* (not of *occurring*); not just shape. Allow for U-shape but nothing else | B1 | E.g.: "More likely to *occur* for $x$ close to $a$": B0. Ignore extra comments like "exponential" |
| | **[3]** | |
---
## Question 7(ii)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^a kx^2\,dx = 1 \Rightarrow k = \dfrac{3}{a^3}$ | M1 | Attempt to integrate $kx^2$, ignore limits. Must attempt integration |
| | A1 | Correct limits and equate to 1 |
| $\int_0^a \dfrac{3}{a^3}x^3\,dx = \dfrac{9}{2} \Rightarrow a = 6$ | M1 | Attempt to integrate $kx^3$, ignore limits. Must attempt integration. Don't need $k$ in terms of $a$ here |
| | A1 | Correct limits and equate to 4.5. $ka^3 = 3$ or $ka^4 = 18$, a.e. simplified form |
| | A1 | One correct equation connecting $k$ and $a$, can be implied |
| | A1 | Correctly obtain $a = 6$ only. No marks explicitly for $k\ [= 1/72$ or $0.01388\ldots]$ |
| | **[6]** | |
## Question 7:
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^6 \frac{1}{72} x^3 \, dx \quad \left[= \frac{108}{5}\right]$ | M1 | Attempt to integrate $kx^4$, their $a$, $k$ can be algebraic |
| Subtract $4.5^2$ (given in question) | A1 | Somewhere |
| $21.6 - 4.5^2 = 1.35$ | A1 | $\left[=\frac{27}{20}\right]$ |
| | **[3]** | Limits $0$, $a\sqrt{}$; 1.35 or exact equivalent only |
---7 The continuous random variable $X$ has probability density function
$$f ( x ) = \begin{cases} k x ^ { 2 } & 0 \leqslant x \leqslant a \\ 0 & \text { otherwise } \end{cases}$$
where $a$ and $k$ are constants.\\
(i) Sketch the graph of $y = \mathrm { f } ( x )$ and explain in non-technical language what this tells you about $X$.\\
(ii) Given that $\mathrm { E } ( X ) = 4.5$, find
\begin{enumerate}[label=(\alph*)]
\item the value of $a$,
\item $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2012 Q7 [12]}}