S4 June 2014 (R) Q6
6. Emily is monitoring the level of pollution in a river. Over a period of time she has found that the amount of pollution, \(X\), in a 100 ml sample of river water has a continuous distribution with probability density function \(\mathrm{f}(x)\) given by
\[\mathrm{f}(x) = \begin{cases} \dfrac{2x}{a^2} & 0 \leqslant x \leqslant a \\ 0 & \text{otherwise} \end{cases}\]where \(a\) is a constant.
Emily takes a random sample \(X_1, X_2, X_3, \ldots, X_n\) to try to estimate the value of \(a\).
The random variable \(S = p\bar{X}\), where \(p\) is a constant, is an unbiased estimator of \(a\).
Felix suggests using the statistic \(M = \max\{X_1, X_2, X_3, \ldots, X_n\}\) as an estimator of \(a\).
He calculates \(\mathrm{E}(M) = \dfrac{2n}{2n + 1}a\) and \(\mathrm{Var}(M) = \dfrac{n}{(n + 1)(2n + 1)^2}a^2\)
The random variable \(T = qM\), where \(q\) is a constant, is an unbiased estimator of \(a\).
Emily took a sample of 5 values of \(X\) and obtained the following:
5.3 4.3 5.7 7.8 6.9
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = \displaystyle\int_0^a x\tfrac{2}{a^2}x\,\mathrm{d}x = \left[\dfrac{2}{a^2}\dfrac{x^3}{3}\right]_0^a = \underline{\dfrac{2a}{3}}\) | B1cso |
| \(\mathrm{E}(X^2) = \displaystyle\int_0^a x^2\tfrac{2}{a^2}x\,\mathrm{d}x = \left[\dfrac{2}{a^2}\dfrac{x^4}{4}\right]_0^a = \dfrac{a^2}{2}\) so \(\sigma^2 = \dfrac{a^2}{2} - \dfrac{4a^2}{9} = \underline{\dfrac{a^2}{18}}\) | M1 A1 |
| So \(\mathrm{E}(\bar{X}) = \mu = \dfrac{2a}{3}\) and \(\mathrm{Var}(\bar{X}) = \dfrac{\sigma^2}{n} = \dfrac{a^2}{18n}\) | A1cso |
| (4) |
Notes
1st B1 for some working to establish \(\mu\). Allow median of triangle for example.
1st M1 for correct method for \(\sigma^2\)
| Scheme | Marks |
|---|---|
| \(p = \underline{\dfrac{3}{2}}\) and \(\mathrm{Var}(S) = \dfrac{9}{4}\mathrm{Var}(\bar{X}) = \underline{\dfrac{a^2}{8n}}\) | B1, B1ft |
| (2) |
Notes
2nd B1ft ft their value of \(p\)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(M) \to a\) as \(n \to \infty\), and \(\mathrm{Var}(M) \to 0\) as \(n \to \infty\) So \(M\) is a consistent estimator of \(a\) | B1, B1 dB1 |
| (3) |
Notes
3rd dB1 dependent on both of first 2 Bs in (c) for concluding that \(M\) is consistent
| Scheme | Marks |
|---|---|
| \(q = \underline{\dfrac{2n + 1}{2n}}, \qquad \mathrm{Var}(T) = \dfrac{\cancel{(2n + 1)^2}}{4n^{\cancel{2}}} \times \dfrac{\cancel{n}}{(n + 1)\cancel{(2n + 1)^2}}a^2, = \underline{\dfrac{a^2}{4n(n + 1)}}\) | B1, M1, A1 |
| (3) |
Notes
M1 for correct use of \(\mathrm{Var}(T) = q^2\,\mathrm{Var}(M)\) for their \(q\).
| Scheme | Marks |
|---|---|
| \(\dfrac{a^2}{4n(n + 1)} \lt \dfrac{a^2}{8n} \iff 2 \lt n + 1 \iff 1 \lt n\) So \(\mathrm{Var}(T) \lt \mathrm{Var}(S)\) | M1 A1 |
| So (since both are unbiased) choose \(T\) since it has the lower variance | A1cso. |
| (3) |
Notes
M1 for attempt to compare \(\mathrm{Var}(T)\) and \(\mathrm{Var}(S)\)
1st A1 for clearly establishing that \(\mathrm{Var}(T) \lt \mathrm{Var}(S)\)
2nd A1 for choosing \(T\) and stating variance is smaller
SC M0 A0 B1 for T because it has a smaller variance
| Scheme | Marks |
|---|---|
| \(m = 7.8\) so using \(t\) gives estimate of \(\dfrac{11}{10} \times 7.8, = 8.58\) [NB \(\bar{x} = 6\) and \(s\) gives 9] | M1, A1ft |
| (2) |
Notes
M1 for using their estimator chosen in (e)
| Scheme | Marks |
|---|---|
| Using \(\mathrm{Var}(T) = \frac{a^2}{120}\); so standard error is \(\frac{8.58}{\sqrt{120}}\), = awrt 0.78 [NB \(s\) gives \(\frac{a}{\sqrt{40}} = 1.42\)] | M1;A1 |
| (2) | |
| (19 marks) |
Notes
M1 for using their Variance formula to calculate std. error. subst in \(n = 5\) and their (f)
(Corrected from the printed mark scheme: the note printed “subst in \(n\)=4”; the sample has \(n = 5\), giving \(4n(n + 1) = 120\).)