| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Compare mean and median using probability |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring standard techniques: sketching a piecewise pdf, finding median by area consideration, computing mean by integration, and evaluating a probability. Part (b) requires minimal insight (recognizing equal areas), part (c) is routine integration shown to a given answer, and part (d) is simple substitution and integration. Slightly above average due to the piecewise function and multiple parts, but all techniques are standard textbook exercises. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sketch: straight line \(0 \leq x \leq 1\) from \((0, 0.5)\) to \((1, 0.5)\); straight line \(1 \leq x \leq 3\) from \((1, 0.5)\) to \((3, 0)\); axes correct with \((0,0.5)\), \((1,0)\) and \((3,0)\) labelled | B3 | 1 mark each line + 1 for axes [must have at least \((0,0.5)\), \((1,0)\) and \((3,0)\) labelled] |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X \leq \eta) = F(\eta) = 0.5\) | M1 | |
| \(\Rightarrow \eta = 1\) (from graph) | A1 | AG |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mu = E(X) = \int_0^1 \left(\dfrac{x}{2}\right)dx + \int_1^3 x\left(\dfrac{3-x}{4}\right)dx\) | M1 | Both integrals stated |
| \(= \left[\dfrac{x^2}{4}\right]_0^1 + \dfrac{1}{4}\left[\dfrac{3x^2}{2} - \dfrac{x^3}{3}\right]_1^3\) | A1 | Either |
| \(= \dfrac{1}{4} + \dfrac{1}{4}\left[\left(\dfrac{27}{2}-9\right)-\left(\dfrac{3}{2}-\dfrac{1}{3}\right)\right]\) | ml | Correct limits used on both integrals + combined dep M1 |
| \(= \dfrac{1}{4} + \dfrac{5}{6} \quad (0.25 + 0.8\overline{3})\) | ||
| \(= 1\dfrac{1}{12}\) | A1 | CAO |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of \(\triangle\): \(P\!\left(X > 2\tfrac{1}{4}\right) = \tfrac{1}{2} \times \tfrac{3}{4} \times \dfrac{3 - 2\tfrac{1}{4}}{4}\) | M1ft | |
| \(= \dfrac{3}{32} \times \dfrac{3}{4} = \dfrac{9}{128}\) | ||
| \(\therefore P\!\left(X < 2\tfrac{1}{4}\right) = 1 - \dfrac{9}{128}\) | M1ft | Alternative: For \(1 \leq x \leq 3\): \(F(x) = 1 - \tfrac{1}{8}(3-x)^2\); \(F\!\left(2\tfrac{1}{4}\right) = 1 - \tfrac{1}{8}\times\tfrac{9}{16}\) (M1ft) \(= \dfrac{119}{128}\) (M1ft) |
| \(= \dfrac{119}{128}\ (0.9296875)\) | A1 | CAO |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_{2\frac{1}{4}}^{3} \frac{3-x}{4} dx \left(= \frac{9}{128}\right)\) | M1 ft | |
| \(= 1 - \int_{2\frac{1}{4}}^{3} \frac{3-x}{4} dx\) | M1 ft | |
| \(= 1 - \frac{1}{4}\left[3x - \frac{x^2}{2}\right]_{2\frac{1}{4}}^{3}\) | ||
| \(= 1 - \frac{1}{4}\left[9 - \frac{9}{2} - \frac{27}{4} + \frac{81}{32}\right]\) | ||
| \(= 1 - \frac{1}{4} \times \frac{9}{32} = \frac{119}{128}\) | A1 | |
| Alternative: \(f\left(2\frac{1}{4}\right) = \frac{3}{16} = 0.1875\) | ||
| \(P\left(X < 3\mu - \eta\right) = P\left(X < 2\frac{1}{4}\right)\) | ||
| \(= \frac{1}{2} + \left[\frac{1}{2}\left(\frac{3}{16} + \frac{1}{2}\right) \times 1\frac{1}{4}\right]\) | M1 ft | |
| \(= \frac{1}{2} + \frac{55}{128}(0.4296875)\) | M1 ft | |
| \(= \frac{119}{128}\ (0.930)\) | A1 | |
| Total | 12 |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sketch: straight line $0 \leq x \leq 1$ from $(0, 0.5)$ to $(1, 0.5)$; straight line $1 \leq x \leq 3$ from $(1, 0.5)$ to $(3, 0)$; axes correct with $(0,0.5)$, $(1,0)$ and $(3,0)$ labelled | B3 | 1 mark each line + 1 for axes [must have at least $(0,0.5)$, $(1,0)$ and $(3,0)$ labelled] |
| **Total** | **3** | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X \leq \eta) = F(\eta) = 0.5$ | M1 | |
| $\Rightarrow \eta = 1$ (from graph) | A1 | AG |
| **Total** | **2** | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mu = E(X) = \int_0^1 \left(\dfrac{x}{2}\right)dx + \int_1^3 x\left(\dfrac{3-x}{4}\right)dx$ | M1 | Both integrals stated |
| $= \left[\dfrac{x^2}{4}\right]_0^1 + \dfrac{1}{4}\left[\dfrac{3x^2}{2} - \dfrac{x^3}{3}\right]_1^3$ | A1 | Either |
| $= \dfrac{1}{4} + \dfrac{1}{4}\left[\left(\dfrac{27}{2}-9\right)-\left(\dfrac{3}{2}-\dfrac{1}{3}\right)\right]$ | ml | Correct limits used on both integrals + combined dep M1 |
| $= \dfrac{1}{4} + \dfrac{5}{6} \quad (0.25 + 0.8\overline{3})$ | | |
| $= 1\dfrac{1}{12}$ | A1 | CAO |
| **Total** | **4** | |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of $\triangle$: $P\!\left(X > 2\tfrac{1}{4}\right) = \tfrac{1}{2} \times \tfrac{3}{4} \times \dfrac{3 - 2\tfrac{1}{4}}{4}$ | M1ft | |
| $= \dfrac{3}{32} \times \dfrac{3}{4} = \dfrac{9}{128}$ | | |
| $\therefore P\!\left(X < 2\tfrac{1}{4}\right) = 1 - \dfrac{9}{128}$ | M1ft | **Alternative:** For $1 \leq x \leq 3$: $F(x) = 1 - \tfrac{1}{8}(3-x)^2$; $F\!\left(2\tfrac{1}{4}\right) = 1 - \tfrac{1}{8}\times\tfrac{9}{16}$ (M1ft) $= \dfrac{119}{128}$ (M1ft) |
| $= \dfrac{119}{128}\ (0.9296875)$ | A1 | CAO |
| **Total** | **3** | |
# Question 4(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_{2\frac{1}{4}}^{3} \frac{3-x}{4} dx \left(= \frac{9}{128}\right)$ | M1 ft | |
| $= 1 - \int_{2\frac{1}{4}}^{3} \frac{3-x}{4} dx$ | M1 ft | |
| $= 1 - \frac{1}{4}\left[3x - \frac{x^2}{2}\right]_{2\frac{1}{4}}^{3}$ | | |
| $= 1 - \frac{1}{4}\left[9 - \frac{9}{2} - \frac{27}{4} + \frac{81}{32}\right]$ | | |
| $= 1 - \frac{1}{4} \times \frac{9}{32} = \frac{119}{128}$ | A1 | |
| **Alternative:** $f\left(2\frac{1}{4}\right) = \frac{3}{16} = 0.1875$ | | |
| $P\left(X < 3\mu - \eta\right) = P\left(X < 2\frac{1}{4}\right)$ | | |
| $= \frac{1}{2} + \left[\frac{1}{2}\left(\frac{3}{16} + \frac{1}{2}\right) \times 1\frac{1}{4}\right]$ | M1 ft | |
| $= \frac{1}{2} + \frac{55}{128}(0.4296875)$ | M1 ft | |
| $= \frac{119}{128}\ (0.930)$ | A1 | |
| **Total** | **12** | |
---4 The continuous random variable $X$ has probability density function given by
$$f ( x ) = \left\{ \begin{array} { c c }
\frac { 1 } { 2 } & 0 \leqslant x \leqslant 1 \\
\frac { 3 - x } { 4 } & 1 \leqslant x \leqslant 3 \\
0 & \text { otherwise }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of f.
\item Explain why the value of $\eta$, the median of $X$, is 1 .
\item Show that the value of $\mu$, the mean of $X$, is $\frac { 13 } { 12 }$.
\item Find $\mathrm { P } ( X < 3 \mu - \eta )$.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2009 Q4 [12]}}