| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Find constant k in PDF |
| Difficulty | Moderate -0.8 This is a straightforward application of rectangular/uniform distribution properties. Part (a) requires knowing that the pdf integrates to 1 (k=10), part (b) is a simple probability calculation, and part (c) involves standard expectation formulas with basic integration of polynomials. All techniques are routine for S2 with no problem-solving or novel insight required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | 10 | B1 |
| Total: 1 | ||
| (b) | \(0.07 \times 10 = 0.7\) | B1 |
| Total: 1 | ||
| (c)(i) | \(E(X) = \frac{1}{20}\) | B1 |
| Total: 1 | ||
| (ii) | \(E(X^2) = \int_0^{0.1} 10x^2 dx\) | M1 |
| \(= \left[\frac{10}{3}x^3\right]_0^{0.1} = \frac{1}{300}\) | A1 | Or at least 3sf equivalent |
| Total: 2 | ||
| (iii) | \(Var(X) = E(X^2) - E(X)^2 = \frac{1}{300} - (\frac{1}{20})^2\) | M1 |
| \(= \frac{1}{1200}\) So sd = \(\sqrt{\frac{1}{1200}} = \frac{\sqrt{3}}{60} = 0.0289\) | A1 | Equivalent surd or AWFW 0.0288 to 0.0289; Correct answer but not derived from (ii) scores M0 B1. |
| Total: 2 | ||
| TOTAL FOR Q4: 7 |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | 10 | B1 | CAO |
| | **Total: 1** | | |
| (b) | $0.07 \times 10 = 0.7$ | B1 | OE |
| | **Total: 1** | | |
| (c)(i) | $E(X) = \frac{1}{20}$ | B1 | Decimal or fraction |
| | **Total: 1** | | |
| (ii) | $E(X^2) = \int_0^{0.1} 10x^2 dx$ | M1 | Integration, correct limits, their k |
| | $= \left[\frac{10}{3}x^3\right]_0^{0.1} = \frac{1}{300}$ | A1 | Or at least 3sf equivalent |
| | **Total: 2** | | |
| (iii) | $Var(X) = E(X^2) - E(X)^2 = \frac{1}{300} - (\frac{1}{20})^2$ | M1 | Correct expression – allow use of 0.003 or 0.0033 or better for M1 |
| | $= \frac{1}{1200}$ So sd = $\sqrt{\frac{1}{1200}} = \frac{\sqrt{3}}{60} = 0.0289$ | A1 | Equivalent surd or AWFW 0.0288 to 0.0289; Correct answer but not derived from (ii) scores M0 B1. |
| | **Total: 2** | | |
| | **TOTAL FOR Q4: 7** | | |
---A digital thermometer measures temperatures in degrees Celsius. The thermometer rounds down the actual temperature to one decimal place, so that, for example, 36.23 and 36.28 are both shown as 36.2. The error, $X$ °C, resulting from this rounding down can be modelled by a rectangular distribution with the following probability density function.
$$f(x) = \begin{cases}
k & 0 \leqslant x \leqslant 0.1 \\
0 & \text{otherwise}
\end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item State the value of $k$. [1 mark]
\item Find the probability that the error resulting from this rounding down is greater than 0.03 °C. [1 mark]
\item
\begin{enumerate}[label=(\roman*)]
\item State the value for E($X$).
\item Use integration to find the value for E($X^2$).
\item Hence find the value for the standard deviation of $X$.
\end{enumerate}
[5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2016 Q4 [7]}}