| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find or specify CDF |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring standard techniques: sketching a piecewise pdf, calculating a probability by integration, verifying a given CDF formula by integration, and finding the median by solving F(t)=0.5. All steps are routine applications of definitions with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| [Graph: curve, line, axes shown] | B1 | for curve |
| B1 | for line | |
| B1 | for axes, Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T \geq 1) = \dfrac{1}{2} \times \dfrac{7}{8} \times 2 = \dfrac{7}{8}\) | M1A1 | OE, Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int_1^t \frac{1}{16}(t+5)\,dt = \left[\frac{1}{32}t^2 + \frac{5}{16}t\right]_1^t\) | M1A1 | |
| \(F(1) = \frac{1}{8}\) | B1 | |
| \(F(t) = \frac{1}{8} + \frac{1}{32}t^2 + \frac{5}{16}t - \frac{11}{32}\) | M1 | Use of: \(F(t) = F(1) + \int_1^t \frac{1}{16}(t+5)\,dt\) |
| \(F(t) = \frac{1}{32}(t^2 + 10t - 7)\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int \frac{1}{16}(t+5)\,dt = \frac{1}{16}\left(\frac{1}{2}t^2 + 5t + c\right)\) | (M1)(A1) | |
| \(F(1) = \frac{1}{8}\) | (B1) | |
| \(\Rightarrow c = -3.5\) | (M1) | |
| \(F(t) = \frac{1}{32}(t^2 + 10t - 7)\) | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{32}(m^2 + 10m - 7) = 0.5\) | M1 | |
| \(m^2 + 10m - 23 = 0\) | A1 | |
| \(m = \frac{-10 \pm \sqrt{192}}{2} = -5 \pm \sqrt{48} = -5 \pm 4\sqrt{3}\) | m1 | (or any valid method) |
| \((m > 0)\), \(m = 4\sqrt{3} - 5 = 1.93\) | A1 | (1.9282) |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| [Graph: curve, line, axes shown] | B1 | for curve |
| | B1 | for line |
| | B1 | for axes, **Total: 3** |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T \geq 1) = \dfrac{1}{2} \times \dfrac{7}{8} \times 2 = \dfrac{7}{8}$ | M1A1 | OE, **Total: 2** |
# Question 6(c)(i):
For $1 \leq t \leq 3$
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_1^t \frac{1}{16}(t+5)\,dt = \left[\frac{1}{32}t^2 + \frac{5}{16}t\right]_1^t$ | M1A1 | |
| $F(1) = \frac{1}{8}$ | B1 | |
| $F(t) = \frac{1}{8} + \frac{1}{32}t^2 + \frac{5}{16}t - \frac{11}{32}$ | M1 | Use of: $F(t) = F(1) + \int_1^t \frac{1}{16}(t+5)\,dt$ |
| $F(t) = \frac{1}{32}(t^2 + 10t - 7)$ | A1 | **AG** |
**Alternative:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int \frac{1}{16}(t+5)\,dt = \frac{1}{16}\left(\frac{1}{2}t^2 + 5t + c\right)$ | (M1)(A1) | |
| $F(1) = \frac{1}{8}$ | (B1) | |
| $\Rightarrow c = -3.5$ | (M1) | |
| $F(t) = \frac{1}{32}(t^2 + 10t - 7)$ | (A1) | |
**Total: 5 marks**
---
# Question 6(c)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{32}(m^2 + 10m - 7) = 0.5$ | M1 | |
| $m^2 + 10m - 23 = 0$ | A1 | |
| $m = \frac{-10 \pm \sqrt{192}}{2} = -5 \pm \sqrt{48} = -5 \pm 4\sqrt{3}$ | m1 | (or any valid method) |
| $(m > 0)$, $m = 4\sqrt{3} - 5 = 1.93$ | A1 | (1.9282) |
**Total: 4 marks**
---6 The waiting time, $T$ minutes, before being served at a local newsagents can be modelled by a continuous random variable with probability density function
$$\mathrm { f } ( t ) = \begin{cases} \frac { 3 } { 8 } t ^ { 2 } & 0 \leqslant t < 1 \\ \frac { 1 } { 16 } ( t + 5 ) & 1 \leqslant t \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of f.
\item For a customer selected at random, calculate $\mathrm { P } ( T \geqslant 1 )$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the cumulative distribution function for $1 \leqslant t \leqslant 3$ is given by
$$\mathrm { F } ( t ) = \frac { 1 } { 32 } \left( t ^ { 2 } + 10 t - 7 \right)$$
\item Hence find the median waiting time.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2007 Q6 [14]}}