S3 June 2010 Q3
3. A woodwork teacher measures the width, \(w\) mm, of a board. The measured width, \(X\) mm, is normally distributed with mean \(w\) mm and standard deviation 0.5 mm.
The same board is measured 16 times and the results are recorded.
Given that the mean of these 16 measurements is 35.6 mm,
| Scheme | Marks |
|---|---|
| \(E \sim \mathrm{N}(0, 0.5^2)\) or \(X \sim \mathrm{N}\left(w, 0.5^2\right)\) \(\mathrm{P}(|E| \lt 0.6) \quad = \mathrm{P}\left(|Z| \lt \dfrac{0.6}{0.5}\right)\) or \(\mathrm{P}\left(|X - w| \lt 0.6\right) = \mathrm{P}\left(|Z| \lt \dfrac{0.6}{0.5}\right)\) | M1 |
| \(= \mathrm{P}(|Z| \lt 1.2)\) \(= 2 \times 0.8849 - 1 = 0.7698\) awrt 0.770 | A1 |
| (2) |
Notes
1st M1 for identifying a correct probability (they must have the 0.6) and attempting to standardise. Need \(|\ |\). This mark can be given for 0.8849 - 0.1151 seen as final answer.
1st A1 for awrt 0.770. NB an answer of 0.3849 or 0.8849 scores M0A0 (since it implies no \(|\ |\))
M1 may be implied by a correct answer
| Scheme | Marks |
|---|---|
| \(\overline{E} \sim \mathrm{N}\left(0, \dfrac{1}{64}\right)\) or \(\overline{X} \sim \mathrm{N}\left(w, \dfrac{0.5^2}{16}\right)\) | M1 |
| \(\mathrm{P}\left(\left|\overline{E}\right| \lt 0.3\right) \quad = \mathrm{P}\left(|Z| \lt \dfrac{0.3}{\frac{1}{8}}\right)\) or \(\mathrm{P}\left(\left|\overline{X} - w\right| \lt 0.3\right) = \mathrm{P}\left(|Z| \lt \dfrac{0.3}{\frac{1}{8}}\right)\) | M1, A1 |
| \(= \mathrm{P}(|Z| \lt 2.4)\) \(= 2 \times 0.9918 - 1 = 0.9836\) awrt 0.984 | A1 |
| (4) |
Notes
1st M1 for a correct attempt to define \(\overline{E}\) or \(\overline{X}\) but must attempt \(\dfrac{\sigma^2}{n}\). Condone labelling as \(E\) or \(X\)
This mark may be implied by standardisation in the next line.
2nd M1 for identifying a correct probability statement using \(\overline{E}\) or \(\overline{X}\). Must have 0.3 and \(|\ |\)
1st A1 for correct standardisation as printed or better
2nd A1 for awrt 0.984
The M marks may be implied by a correct answer.
Sum of 16, not means
1st M1 for correct attempt at suitable sum distribution with correct variance (\(= 16 \times \tfrac{1}{4}\))
2nd M1 for identifying a correct probability. Must have 4.8 and \(|\ |\)
1st A1 for correct standardisation i.e. need to see \(\dfrac{4.8}{\sqrt{4}}\) or better
| Scheme | Marks |
|---|---|
| \(35.6 \pm 2.3263 \times \dfrac{1}{8}\) | M1 B1 |
| (35.3, 35.9) | A1,A1 |
| (4) | |
| (10 marks) |
Notes
M1 for \(35.6 \pm z \times \dfrac{0.5}{\sqrt{16}}\)
B1 for 2.3263 or better. Use of 2.33 will lose this mark but can still score ¾
1st A1 for awrt 35.3
2nd A1 for awrt 35.9