| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Measurement error modeling |
| Difficulty | Easy -1.2 This is a straightforward application of the rectangular (continuous uniform) distribution to rounding errors. Part (a) requires basic understanding of rounding (1 mark), part (b) involves direct recall of formulas for mean and standard deviation of uniform distribution, and part (c) is a simple probability calculation. All parts are routine with no problem-solving or novel insight required—easier than average A-level questions. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| 21.05 and 21.15 | B1 | both (allow 21.049 and 21.149) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 0\) (symmetry) | B1 | For \(R[-a,a]: E(x) = 0\) iff \(a = 0.05, 0.1, 0.5\) then: \(\text{Var}(X) = \frac{1}{12}(a - a)^2\) or their \(a = 0.049\) to \(0.05\) used for M1 |
| \(\text{Var}(X) = \frac{1}{12}(0.05 - (-0.05))^2 = \frac{1}{12} \times \frac{1}{100}\) | M1 | |
| \(\Rightarrow sd(X) = \sqrt{\frac{1}{12} \times \frac{1}{100}} = \frac{1}{20\sqrt{3}}\) | A1 | or \(\frac{\sqrt{3}}{60}\) or \(\sqrt{\frac{1}{1200}}\) or 0.0289 (3sf) A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(-0.01 \leq X \leq 0.03) = 0.04 \times 10 = 0.4\) | B1 | cao from correct value used \(\frac{0.03}{∫_{-0.01}^{0.03}10dx = [10x]_{-0.01}^{0.03} = 0.4}\) or OE |
## (a)
| 21.05 and 21.15 | B1 | both (allow 21.049 and 21.149) |
## (b)
| $E(X) = 0$ (symmetry) | B1 | For $R[-a,a]: E(x) = 0$ iff $a = 0.05, 0.1, 0.5$ then: $\text{Var}(X) = \frac{1}{12}(a - a)^2$ or their $a = 0.049$ to $0.05$ used for M1 |
| $\text{Var}(X) = \frac{1}{12}(0.05 - (-0.05))^2 = \frac{1}{12} \times \frac{1}{100}$ | M1 | |
| $\Rightarrow sd(X) = \sqrt{\frac{1}{12} \times \frac{1}{100}} = \frac{1}{20\sqrt{3}}$ | A1 | or $\frac{\sqrt{3}}{60}$ or $\sqrt{\frac{1}{1200}}$ or 0.0289 (3sf) A0 |
## (c)
| $P(-0.01 \leq X \leq 0.03) = 0.04 \times 10 = 0.4$ | B1 | cao from correct value used $\frac{0.03}{∫_{-0.01}^{0.03}10dx = [10x]_{-0.01}^{0.03} = 0.4}$ or OE |
**Total: 5 marks**
---1 Josephine accurately measures the widths of A4 sheets of paper and then rounds the widths to the nearest 0.1 cm . The rounding error, $X$ centimetres, follows a rectangular distribution.
A randomly selected A4 sheet of paper is measured to be 21.1 cm in width.
\begin{enumerate}[label=(\alph*)]
\item Write down the limits between which the true width of this A4 sheet of paper lies.\\
(1 mark)
\item Write down the value of $\mathrm { E } ( X )$ and determine the exact value of the standard deviation of $X$.
\item Calculate $\mathrm { P } ( - 0.01 \leqslant X \leqslant 0.03 )$.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2012 Q1 [5]}}