S4 June 2014 (R) Q5
5. A large company has designed an aptitude test for new recruits. The score, \(S\), for an individual taking the test, has a normal distribution with mean \(\mu\) and standard deviation \(\sigma\).
In order to estimate \(\mu\) and \(\sigma\), a random sample of 15 new recruits were given the test and their scores, \(x\), are summarised as
\[\sum x = 880 \qquad \sum x^2 = 54\,892\]The company wants to ensure that no more than 80% of new recruits pass the test.
| Scheme | Marks |
|---|---|
| (i) \(\bar{x} = \left(\dfrac{880}{15} =\right) 58.\dot{6}\) or awrt 58.7 | B1 |
| \({s_x}^2 = \left(\dfrac{54892 - 15 \times 58.\dot{6}^2}{14} =\right) 233.238\ldots\) | B1 |
| \(t_{14}(0.025)\) cv = 2.145 | B1 |
| 95% CI for \(\mu\) is \(58.\dot{6} \pm 2.145 \times \sqrt{\dfrac{233.238\ldots}{15}}\) | M1 |
| \(= (50.209\ldots, 67.124\ldots)\) = awrt (50.2, 67.1) | A1, A1 |
| (ii) \({\chi_{14}}^2(0.025) = 5.629, \quad {\chi_{14}}^2(0.975) = 26.119\) | B1, B1 |
| 95% CI for \(\sigma^2\) is given by: \(5.629 \lt \dfrac{14{s_x}^2}{\sigma^2} \lt 26.119\) | M1 |
| \(= (125.017\ldots, 580.0911\ldots)\) | A1 |
| So 95% CI for \(\sigma\) is \(= (11.1811\ldots, 24.0850\ldots)\) = awrt (11.2, 24.1) | A1 |
| (11) |
Notes
1st M1 ‘their \(\bar{x}\)’ \(\pm\ t\ \textit{value} \times \dfrac{\text{‘their } s\text{’}}{\sqrt{15}}\)
1st A1 for awrt 50.2
2nd A1 for awrt 67.1
2nd M1 for use of their values in \(\chi^2 \lt \dfrac{14s^2}{\sigma^2} \lt \chi^2\)
3rd A1 for awrt 125 or 580
4th A1 for awrt 11.2 and 24.1
| Scheme | Marks |
|---|---|
| Require \(\mathrm{P}(S \gt d) \leqslant 0.80\) i.e. \(\mathrm{P}\left(Z \gt \dfrac{d - \mu}{\sigma}\right) \leqslant 0.80\) | |
| From tables \(\pm 0.8416\) | B1 |
| So require: \(\dfrac{d - \mu}{\sigma} \gt -0.8416\) | M1 |
| i.e. \(d \gt \mu - 0.8416\sigma\) | A1 |
| Worst case is when \(\mu = \mu_{\max}\) and \(\sigma = \sigma_{\min}\) | M1 |
| So \(d \gt 67.1 - 0.8416 \times 11.2\) \((= 57.674\ldots)\) so they should set a pass mark of 58 | A1 |
| (5) | |
| (16 marks) |
Notes
1st M1 for forming a correct expression in \(d\), \(\mu\), \(\sigma\) and their \(z\) value
2nd M1 for using their top value from CI for \(\mu\) and lowest value for CI for \(\sigma\)