| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Standard +0.3 This is a standard S2 probability density function question requiring routine integration to find k, calculate E(X), set up and solve a cumulative probability equation. The quartic equation reduces to a quadratic via substitution (explicitly given). All techniques are textbook exercises 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 |
|---|---|---|
| (i) \(1 = k\int_0^1 (9x - x^3) \, dx = k\left[\frac{9}{2}x^2 - \frac{1}{4}x^4\right]_0^1 = \frac{81}{4}k\) | M1 | For equating to 1 and integrating |
| A1 | For showing given answer correctly |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(\text{E}(X) = \frac{4}{81}\int_0^1 x^2(9-x^2) \, dx = \frac{4}{81}\left[3x^3 - \frac{1}{5}x^5\right]_0^1 = 1.6\) | M1 | For attempt at \(\int_0^1 xf(x) \, dx\) |
| A1 | For correct indefinite integral, in any form | |
| A1 | For correct answer 1.6 | |
| (iii)(a) \(\frac{3}{5} = \frac{4}{81}\int_0^y x(9-x^2) \, dx = \frac{4}{81}\left[\frac{9}{2}x^2 - \frac{1}{4}x^4\right]_0^y\) | M1 | For attempt at \(\int_0^y f(x) \, dx = \frac{3}{5}\) |
| B1 | For correct indefinite integral, in any form | |
| Hence \(\frac{3}{5} = \frac{4}{81}\left(\frac{9}{2}y^2 - \frac{1}{4}y^4\right)\) | M1 | Use limits to produce relevant equation in \(y\) |
| i.e. \(5y^4 - 90y^2 + 243 = 0\) | A1 | For showing given answer correctly |
| (iii)(b) \(w = \frac{90 \pm \sqrt{(90)^2 - 4 \times 5 \times 243}}{10} = 3.31 \text{ or } 14.7\) | M1 | For use of quadratic formula to find \(w\) |
| A1 | For either value found correctly | |
| Hence \(y = \sqrt{3.31} = 1.82\) | A1 | For correct (unique) answer 1.82 |
(i) $1 = k\int_0^1 (9x - x^3) \, dx = k\left[\frac{9}{2}x^2 - \frac{1}{4}x^4\right]_0^1 = \frac{81}{4}k$ | M1 | For equating to 1 and integrating
| A1 | For showing given answer correctly
Hence $k = \frac{4}{81}$
(ii) $\text{E}(X) = \frac{4}{81}\int_0^1 x^2(9-x^2) \, dx = \frac{4}{81}\left[3x^3 - \frac{1}{5}x^5\right]_0^1 = 1.6$ | M1 | For attempt at $\int_0^1 xf(x) \, dx$
| A1 | For correct indefinite integral, in any form
| A1 | For correct answer 1.6
(iii)(a) $\frac{3}{5} = \frac{4}{81}\int_0^y x(9-x^2) \, dx = \frac{4}{81}\left[\frac{9}{2}x^2 - \frac{1}{4}x^4\right]_0^y$ | M1 | For attempt at $\int_0^y f(x) \, dx = \frac{3}{5}$
| B1 | For correct indefinite integral, in any form
Hence $\frac{3}{5} = \frac{4}{81}\left(\frac{9}{2}y^2 - \frac{1}{4}y^4\right)$ | M1 | Use limits to produce relevant equation in $y$
i.e. $5y^4 - 90y^2 + 243 = 0$ | A1 | For showing given answer correctly
(iii)(b) $w = \frac{90 \pm \sqrt{(90)^2 - 4 \times 5 \times 243}}{10} = 3.31 \text{ or } 14.7$ | M1 | For use of quadratic formula to find $w$
| A1 | For either value found correctly
Hence $y = \sqrt{3.31} = 1.82$ | A1 | For correct (unique) answer 1.82
**Total: 12 marks**
---7 The time, in minutes, for which a customer is prepared to wait on a telephone complaints line is modelled by the random variable $X$. The probability density function of $X$ is given by
$$\mathrm { f } ( x ) = \begin{cases} k x \left( 9 - x ^ { 2 } \right) & 0 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Show that $k = \frac { 4 } { 81 }$.
\item Find $\mathrm { E } ( X )$.
\item (a) Show that the value $y$ which satisfies $\mathrm { P } ( X < y ) = \frac { 3 } { 5 }$ satisfies
$$5 y ^ { 4 } - 90 y ^ { 2 } + 243 = 0 .$$
(b) Using the substitution $w = y ^ { 2 }$, or otherwise, solve the equation in part (a) to find the value of $y$.
\end{enumerate}
\hfill \mbox{\textit{OCR S2 Q7 [12]}}