| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 15 |
| 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 straightforward S2 question testing standard PDF operations: sketching, calculating mean via integration, finding probabilities by integration, and applying conditional probability. All techniques are routine with no novel insight required, making it slightly easier than average. |
| Spec | 2.03d Calculate conditional probability: from first principles5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Straight line from \((1, 0.5)\) to \((3, \frac{1}{6})\). Horizontal straight line from \((3, \frac{1}{6})\) to \((5, \frac{1}{6})\). | B2,1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \frac{1}{6}\int_{1}^{3}x(4-x)dx + \frac{1}{6}\int_{3}^{5}1 dx\) | M1 | ignore limits (both parts attempted) |
| \(= \frac{1}{6}\left[2x^2 - \frac{x^3}{3}\right] + \frac{1}{6}\left[\frac{x^2}{2}\right]_{1}^{5}\) | A1 | ignore limits (both correct) |
| \(= \frac{1}{6}[(18-9) - (2-\frac{1}{3})] + \frac{1}{6}[\frac{25}{2} - \frac{9}{2}]\) | m1 | use of correct limits. dep on M1A1 |
| \(= \frac{1}{6}[7\frac{1}{3} + 8]\) | ||
| \(= 2\frac{5}{9}\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbb{P}(X > 2.5) = \frac{1}{3} + \frac{1}{2} \times (0.25 + \frac{1}{6}) \times \frac{1}{2}\) | M1 | Or \(1 - \int_{1}^{2.5}(4-x)dx = 1 - \left[\frac{1}{6}(4x - \frac{x^2}{2})\right]_1\) |
| \(= \frac{7}{16}\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbb{P}(1.5 < X < 4.5) = \frac{1}{3} \times (\frac{5}{12} + \frac{1}{6}) \times 1.5\) | M1 | Or \(\int_{1}^{4.5}\frac{1}{6}(4-x)dx + \int_{3}^{4.5}\frac{1}{6}dx\) |
| \(+ (4.5 - 3) \times \frac{1}{6}\) | ||
| \(= \frac{7}{16} + \frac{1}{4}\) | A1 | |
| \(= \frac{11}{16}\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbb{P}(X > 2.5 \text{ and } 1.5 < X < 4.5) = \mathbb{P}(2.5 < X < 4.5)\) | ||
| \(= \frac{1}{2} \times (0.25 + \frac{1}{6}) \times 0.5 + \frac{1}{4}\) | M1 | |
| \(= \frac{5}{48} + \frac{1}{4} = \frac{17}{48}\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbb{P}(X > 2.5 \mid 1.5 < X < 4.5) = \frac{17/48}{11/16} = \frac{17}{33}\) | M1 | their \(\frac{\text{(iii)}}{\text{(ii)}}\) iff \(0 < p'S < 1\) |
| A1 | 2 | cao (allow 0.51) |
| 15 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(x) = \begin{cases} 0 & x < 1 \\ \frac{1}{12}(x-1)(7-x) & 1 \leq x < 3 \\ \frac{1}{6}(x+1) & 3 \leq x < 5 \\ 1 & x \geq 5 \end{cases}\) | ||
| (i) \(\mathbb{P}(X > 2.5) = 1 - F(2.5) = 1 - \frac{1}{12}(2.5-1)(7-2.5) = 1 - \frac{1}{12} \times 1.5 \times 4.5 = 1 - 0.5625 = 0.4375\) or \(\frac{7}{16}\) | (M1) | |
| (A1) | cao | |
| (ii) \(\mathbb{P}(1.5 < X < 4.5) = F(4.5) - F(1.5) = \frac{1}{6}(4.5+1) - \frac{1}{12}(1.5-1)(7-1.5) = \frac{11}{12} - \frac{11}{48} = \frac{11}{16}\) or 0.6875 | (M1) | |
| (A1) | cao | |
| (iii) \(\mathbb{P}(X > 2.5 \text{ and } 1.5 < X < 4.5) = \mathbb{P}(2.5 < X < 4.5) = F(4.5) - F(2.5) = \frac{11}{12} - \frac{9}{16} = \frac{17}{48}\) | (M1) | |
| (A1) | cao | |
| (iv) \(\mathbb{P}(X > 2.5 \mid 1.5 < X < 4.5) = \frac{F(4.5) - F(2.5)}{F(4.5) - F(1.5)}\) or \(\frac{\text{their (iii)}}{\text{their (ii)}} = \frac{17/48}{11/16} = \frac{17}{33}\) or (allow 0.51) | (M1) | cao |
| (A1) | ||
| 15 | ||
| TOTAL 75 |
### 7(a)
| Straight line from $(1, 0.5)$ to $(3, \frac{1}{6})$. Horizontal straight line from $(3, \frac{1}{6})$ to $(5, \frac{1}{6})$. | B2,1 | 2 | |
### 7(b)
| $E(X) = \frac{1}{6}\int_{1}^{3}x(4-x)dx + \frac{1}{6}\int_{3}^{5}1 dx$ | M1 | ignore limits (both parts attempted) |
| --- | --- | --- |
| $= \frac{1}{6}\left[2x^2 - \frac{x^3}{3}\right] + \frac{1}{6}\left[\frac{x^2}{2}\right]_{1}^{5}$ | A1 | ignore limits (both correct) |
| $= \frac{1}{6}[(18-9) - (2-\frac{1}{3})] + \frac{1}{6}[\frac{25}{2} - \frac{9}{2}]$ | m1 | use of correct limits. dep on M1A1 |
| $= \frac{1}{6}[7\frac{1}{3} + 8]$ | | |
| $= 2\frac{5}{9}$ | A1 | 4 | (AG) |
### 7(c)(i)
| $\mathbb{P}(X > 2.5) = \frac{1}{3} + \frac{1}{2} \times (0.25 + \frac{1}{6}) \times \frac{1}{2}$ | M1 | Or $1 - \int_{1}^{2.5}(4-x)dx = 1 - \left[\frac{1}{6}(4x - \frac{x^2}{2})\right]_1$ |
| --- | --- | --- |
| $= \frac{7}{16}$ | A1 | 2 | cao (0.4375) |
### 7(c)(ii)
| $\mathbb{P}(1.5 < X < 4.5) = \frac{1}{3} \times (\frac{5}{12} + \frac{1}{6}) \times 1.5$ | M1 | Or $\int_{1}^{4.5}\frac{1}{6}(4-x)dx + \int_{3}^{4.5}\frac{1}{6}dx$ |
| --- | --- | --- |
| $+ (4.5 - 3) \times \frac{1}{6}$ | | |
| $= \frac{7}{16} + \frac{1}{4}$ | A1 | |
| $= \frac{11}{16}$ | A1 | 3 | cao ($= \frac{11}{16}$ or 0.6875) |
### 7(c)(iii)
| $\mathbb{P}(X > 2.5 \text{ and } 1.5 < X < 4.5) = \mathbb{P}(2.5 < X < 4.5)$ | | |
| --- | --- | --- |
| $= \frac{1}{2} \times (0.25 + \frac{1}{6}) \times 0.5 + \frac{1}{4}$ | M1 | |
| $= \frac{5}{48} + \frac{1}{4} = \frac{17}{48}$ | A1 | 2 | cao (0.35416) |
### 7(c)(iv)
| $\mathbb{P}(X > 2.5 \mid 1.5 < X < 4.5) = \frac{17/48}{11/16} = \frac{17}{33}$ | M1 | their $\frac{\text{(iii)}}{\text{(ii)}}$ iff $0 < p'S < 1$ |
| --- | --- | --- |
| | A1 | 2 | cao (allow 0.51) |
| | | **15** |
## Question 7(c) - Alternative Solution
| $F(x) = \begin{cases} 0 & x < 1 \\ \frac{1}{12}(x-1)(7-x) & 1 \leq x < 3 \\ \frac{1}{6}(x+1) & 3 \leq x < 5 \\ 1 & x \geq 5 \end{cases}$ | | |
| --- | --- | --- |
| **(i)** $\mathbb{P}(X > 2.5) = 1 - F(2.5) = 1 - \frac{1}{12}(2.5-1)(7-2.5) = 1 - \frac{1}{12} \times 1.5 \times 4.5 = 1 - 0.5625 = 0.4375$ or $\frac{7}{16}$ | (M1) | |
| | (A1) | cao |
| **(ii)** $\mathbb{P}(1.5 < X < 4.5) = F(4.5) - F(1.5) = \frac{1}{6}(4.5+1) - \frac{1}{12}(1.5-1)(7-1.5) = \frac{11}{12} - \frac{11}{48} = \frac{11}{16}$ or 0.6875 | (M1) | |
| | (A1) | cao |
| **(iii)** $\mathbb{P}(X > 2.5 \text{ and } 1.5 < X < 4.5) = \mathbb{P}(2.5 < X < 4.5) = F(4.5) - F(2.5) = \frac{11}{12} - \frac{9}{16} = \frac{17}{48}$ | (M1) | |
| | (A1) | cao |
| **(iv)** $\mathbb{P}(X > 2.5 \mid 1.5 < X < 4.5) = \frac{F(4.5) - F(2.5)}{F(4.5) - F(1.5)}$ or $\frac{\text{their (iii)}}{\text{their (ii)}} = \frac{17/48}{11/16} = \frac{17}{33}$ or (allow 0.51) | (M1) | cao |
| | (A1) | |
| | **15** | |
| | **TOTAL 75** | |7 A continuous random variable $X$ has probability density function defined by
$$f ( x ) = \begin{cases} \frac { 1 } { 6 } ( 4 - x ) & 1 \leqslant x \leqslant 3 \\ \frac { 1 } { 6 } & 3 \leqslant x \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Draw the graph of f on the grid on page 6 .
\item Prove that the mean of $X$ is $2 \frac { 5 } { 9 }$.
\item Calculate the exact value of:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X > 2.5 )$;
\item $\mathrm { P } ( 1.5 < X < 4.5 )$;
\item $\mathrm { P } ( X > 2.5$ and $1.5 < X < 4.5 )$;
\item $\mathrm { P } ( X > 2.5 \mid 1.5 < X < 4.5 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{bc21c177-6cd8-4c79-8782-d17f0238ce17-6_1340_1363_317_383}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2012 Q7 [15]}}