| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | June |
| Marks | 14 |
| 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 continuous probability distribution question requiring routine techniques: sketching a piecewise PDF, recognizing P(T=3)=0 for continuous distributions, integrating to find probabilities, solving for the median, and calculating the mean. While it involves multiple parts and some algebraic manipulation (especially the quadratic integral), all techniques are textbook-standard with no novel insight required. Slightly easier than average due to the straightforward piecewise structure and clear signposting of methods. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Graph with 2 axes with scales | B1 | |
| Horizontal line at \(\frac{1}{5}\) from \(0\) to \(3\) | B1 | |
| Curve from \(3\) to \(6\) | B1 | |
| Total: 3 marks | B3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(T=3)=0\) | B1 | |
| Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(T \geq 3) = 1 - P(T < 3)\) | M1 | \(\int_3^6 \frac{1}{45}t(6-t)\,dt = \frac{2}{5}\) |
| \(= 1 - \frac{3}{5}\) | ||
| \(= \frac{2}{5}\) | A1 | |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^m \frac{1}{5}\,dt = 0.5\) | M1 | \(P(T \leq 3) = 0.6\), \(\therefore 0 \leq \text{median} < 3\) |
| \(\left(\frac{t}{5}\right)_0^m = 0.5\) | ||
| \(\frac{m}{5} - 0 = 0.5\) | \(\frac{1}{5}m = 0.5\) | |
| \(m = 0.5 \times 5\) | \(m = 5 \times 0.5\) | |
| \(m = 2.5\) | A1 | \(m = 2.5\) AG |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(T) = \int_0^3 \frac{1}{5}t\,dt + \int_3^6 \frac{1}{45}t^2(6-t)\,dt\) | M1 | |
| \(= \left[\frac{1}{10}t^2\right]_0^3 + \left[\frac{2}{45}t^3 - \frac{1}{180}t^4\right]_3^6\) | A1A1 | |
| \(= \frac{9}{10} + 1.65\) | ||
| \(= 2.55\) | A1 | |
| \(\therefore P(\text{median} < T < \text{mean}) = P(2.5 < T < 2.55)\) | M1 | |
| \(= 0.05 \times \frac{1}{5}\) | ||
| \(= 0.01\) | A1 | |
| Total: 6 marks |
# Question 7:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Graph with 2 axes with scales | B1 | |
| Horizontal line at $\frac{1}{5}$ from $0$ to $3$ | B1 | |
| Curve from $3$ to $6$ | B1 | |
| **Total: 3 marks** | B3 | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(T=3)=0$ | B1 | |
| **Total: 1 mark** | | |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(T \geq 3) = 1 - P(T < 3)$ | M1 | $\int_3^6 \frac{1}{45}t(6-t)\,dt = \frac{2}{5}$ |
| $= 1 - \frac{3}{5}$ | | |
| $= \frac{2}{5}$ | A1 | |
| **Total: 2 marks** | | |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^m \frac{1}{5}\,dt = 0.5$ | M1 | $P(T \leq 3) = 0.6$, $\therefore 0 \leq \text{median} < 3$ |
| $\left(\frac{t}{5}\right)_0^m = 0.5$ | | |
| $\frac{m}{5} - 0 = 0.5$ | | $\frac{1}{5}m = 0.5$ |
| $m = 0.5 \times 5$ | | $m = 5 \times 0.5$ |
| $m = 2.5$ | A1 | $m = 2.5$ AG |
| **Total: 2 marks** | | |
## Part (e):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(T) = \int_0^3 \frac{1}{5}t\,dt + \int_3^6 \frac{1}{45}t^2(6-t)\,dt$ | M1 | |
| $= \left[\frac{1}{10}t^2\right]_0^3 + \left[\frac{2}{45}t^3 - \frac{1}{180}t^4\right]_3^6$ | A1A1 | |
| $= \frac{9}{10} + 1.65$ | | |
| $= 2.55$ | A1 | |
| $\therefore P(\text{median} < T < \text{mean}) = P(2.5 < T < 2.55)$ | M1 | |
| $= 0.05 \times \frac{1}{5}$ | | |
| $= 0.01$ | A1 | |
| **Total: 6 marks** | | |
---7 The time, $T$ hours, that the supporters of Bracken Football Club have to queue in order to obtain their Cup Final tickets has the following probability density function.
$$\mathrm { f } ( t ) = \begin{cases} \frac { 1 } { 5 } & 0 \leqslant t < 3 \\ \frac { 1 } { 45 } t ( 6 - t ) & 3 \leqslant t \leqslant 6 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of f.
\item Write down the value of $\mathrm { P } ( T = 3 )$.
\item Find the probability that a randomly selected supporter has to queue for at least 3 hours in order to obtain tickets.
\item Show that the median queuing time is 2.5 hours.
\item Calculate P (median $< T <$ mean).
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2005 Q7 [14]}}