| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 18 |
| 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 techniques: finding k by integrating to 1, sketching a simple quadratic, and computing means using E(X) = ∫xf(x)dx. All steps are textbook procedures with straightforward polynomial integration, making it slightly easier than average for A-level. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\int_0^3 k(t-3)^2 \, dt = 1\); \(k\int_0^3 t^2 - 6t + 9 \, dt = 1\); \(\therefore k[\frac{1}{3}t^3 - 3t^2 + 9t]_0^3 = 1\); \(\therefore k(9 - 27 + 27) - (0) = 1; 9k = 1; k = \frac{1}{9}\) | M1 M1 A1 M1 A1 | |
| (b) Graph showing decreasing curve from f(0) = 1 on y-axis, passing through point 3 on x-axis, with appropriate shape | B3 | |
| (c) \(E(T) = \int_0^3 t \times \frac{1}{9}(t-3)^2 \, dt = \frac{1}{9}\int_0^3 t^3 - 6t^2 + 9t \, dt = \frac{1}{9}[\frac{1}{4}t^4 - 2t^3 + \frac{9}{2}t^2]_0^3 = \frac{1}{9}[(\frac{81}{4} - 54 + \frac{81}{2}) - (0)] = \frac{3}{4}\); \(\therefore\) mean time \(= \frac{3}{4} \times 10 = 7.5\) s | M1 A1 M1 A1 A1 | |
| (d) \(E(S) = \int_0^3 s \times \frac{1}{12}(8 - s^3) \, ds = \frac{1}{12}\int_0^3 8s - s^4 \, ds = \frac{1}{12}[4s^2 + \frac{1}{5}s^5]_0^3 = \frac{1}{12}[(16 - \frac{32}{5}) - (0)] = \frac{4}{5}\); \(\therefore\) new mean \(= \frac{4}{5} \times 10 = 8\) s \(\therefore\) increased by 0.5 s | M1 A1 M1 A1 A1 | (18 marks total) |
(a) $\int_0^3 k(t-3)^2 \, dt = 1$; $k\int_0^3 t^2 - 6t + 9 \, dt = 1$; $\therefore k[\frac{1}{3}t^3 - 3t^2 + 9t]_0^3 = 1$; $\therefore k(9 - 27 + 27) - (0) = 1; 9k = 1; k = \frac{1}{9}$ | M1 M1 A1 M1 A1 |
(b) Graph showing decreasing curve from f(0) = 1 on y-axis, passing through point 3 on x-axis, with appropriate shape | B3 |
(c) $E(T) = \int_0^3 t \times \frac{1}{9}(t-3)^2 \, dt = \frac{1}{9}\int_0^3 t^3 - 6t^2 + 9t \, dt = \frac{1}{9}[\frac{1}{4}t^4 - 2t^3 + \frac{9}{2}t^2]_0^3 = \frac{1}{9}[(\frac{81}{4} - 54 + \frac{81}{2}) - (0)] = \frac{3}{4}$; $\therefore$ mean time $= \frac{3}{4} \times 10 = 7.5$ s | M1 A1 M1 A1 A1 |
(d) $E(S) = \int_0^3 s \times \frac{1}{12}(8 - s^3) \, ds = \frac{1}{12}\int_0^3 8s - s^4 \, ds = \frac{1}{12}[4s^2 + \frac{1}{5}s^5]_0^3 = \frac{1}{12}[(16 - \frac{32}{5}) - (0)] = \frac{4}{5}$; $\therefore$ new mean $= \frac{4}{5} \times 10 = 8$ s $\therefore$ increased by 0.5 s | M1 A1 M1 A1 A1 | (18 marks total)
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**Total: 75 marks**7. In a competition at a funfair, participants have to stay on a log being rotated in a pool of water for as long as possible. The length of time, in tens of seconds, that the competitors stay on the log is modelled by the random variable $T$ with the following probability density function:
$$\mathrm { f } ( t ) = \begin{cases} k ( t - 3 ) ^ { 2 } , & 0 \leq t \leq 3 \\ 0 , & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 1 } { 9 }$.
\item Sketch f $( t )$ for all values of $t$.
\item Show that the mean time that competitors stay on the $\log$ is 7.5 seconds.
When the competition is next run the organisers decide to make it easier at first by spinning the log more slowly and then increasing the speed of rotation. The length of time, in tens of seconds, that the competitors now stay on the log is modelled by the random variable $S$ with the following probability density function:
$$f ( s ) = \begin{cases} \frac { 1 } { 12 } \left( 8 - s ^ { 3 } \right) , & 0 \leq s \leq 2 \\ 0 , & \text { otherwise } \end{cases}$$
\item Find the change in the mean time that competitors stay on the log.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q7 [18]}}