| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Piecewise PDF with multiple regions |
| Difficulty | Standard +0.3 This is a standard S2 piecewise PDF question requiring routine integration and probability calculations. While it involves multiple regions and several parts, the techniques are straightforward: sketching a linear piecewise function, integrating polynomials to find probabilities, and computing expectation using the standard formula. No novel insight or complex problem-solving is required—just careful application of learned methods. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Line segment on \(0-3\) | B1 | |
| Line segment on \(3-5\) | B1 | |
| Scales (\(0-0.4\) vertical; \(0-5\) horizontal) | B1 | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T \leq 2) = \frac{1}{2} \times 2 \times \frac{4}{15}\) | M1 | |
| \(= \frac{4}{15}\) | A1 | 2 marks; (0.267) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(2 < T < 4) = 1 - (P(T < 2) + P(T > 4))\) | M1 | |
| \(= 1 - \left(\frac{4}{15} + \frac{1}{2} \times \frac{1}{5}\right)\) | A1 | For \(P(T>4) = \frac{1}{10}\); \(\frac{1}{2}d[(f_2+f_4)+2f_3]\); \(f_2=\frac{4}{15}\); \(f_4=\frac{1}{5}\); \(f_3=\frac{2}{5}\); \(d=1\) |
| \(= 1 - \frac{4}{15} - \frac{1}{10} = \frac{19}{30}\) | A1 | 3 marks; (0.633) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(T) = \int_0^3 \frac{2}{15}t^2\, dt + \int_3^5 t\left(1 - \frac{1}{5}t\right)dt\) | M1 | Both integrals seen |
| \(= \left[\frac{2}{45}t^3\right]_0^3 + \left[\frac{1}{2}t^2 - \frac{1}{15}t^3\right]_3^5\) | B1B1 | |
| \(= \frac{6}{5} + \frac{25}{6} - \frac{27}{10} = 2\frac{2}{3}\) | A1 | 4 marks; OE |
## Question 4:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Line segment on $0-3$ | B1 | |
| Line segment on $3-5$ | B1 | |
| Scales ($0-0.4$ vertical; $0-5$ horizontal) | B1 | 3 marks total |
### Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T \leq 2) = \frac{1}{2} \times 2 \times \frac{4}{15}$ | M1 | |
| $= \frac{4}{15}$ | A1 | 2 marks; (0.267) |
### Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(2 < T < 4) = 1 - (P(T < 2) + P(T > 4))$ | M1 | |
| $= 1 - \left(\frac{4}{15} + \frac{1}{2} \times \frac{1}{5}\right)$ | A1 | For $P(T>4) = \frac{1}{10}$; $\frac{1}{2}d[(f_2+f_4)+2f_3]$; $f_2=\frac{4}{15}$; $f_4=\frac{1}{5}$; $f_3=\frac{2}{5}$; $d=1$ |
| $= 1 - \frac{4}{15} - \frac{1}{10} = \frac{19}{30}$ | A1 | 3 marks; (0.633) |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(T) = \int_0^3 \frac{2}{15}t^2\, dt + \int_3^5 t\left(1 - \frac{1}{5}t\right)dt$ | M1 | Both integrals seen |
| $= \left[\frac{2}{45}t^3\right]_0^3 + \left[\frac{1}{2}t^2 - \frac{1}{15}t^3\right]_3^5$ | B1B1 | |
| $= \frac{6}{5} + \frac{25}{6} - \frac{27}{10} = 2\frac{2}{3}$ | A1 | 4 marks; OE |
---4 The delay, in hours, of certain flights from Australia may be modelled by the continuous random variable $T$, with probability density function
$$\mathrm { f } ( t ) = \left\{ \begin{array} { c c }
\frac { 2 } { 15 } t & 0 \leqslant t \leqslant 3 \\
1 - \frac { 1 } { 5 } t & 3 \leqslant t \leqslant 5 \\
0 & \text { otherwise }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of f.
\item Calculate:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( T \leqslant 2 )$;
\item $\mathrm { P } ( 2 < T < 4 )$.
\end{enumerate}\item Determine $\mathrm { E } ( T )$.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2008 Q4 [12]}}