S3 June 2013 (R) Q6
6. The continuous random variable \(X\) is uniformly distributed over the interval
\[[a - 1,\ a + 5]\]where \(a\) is a constant.
Fifty observations of \(X\) are taken, giving a sample mean of 17.2
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(X) = \dfrac{(a + 5 - a + 1)^2}{12}\) [=3] | M1 |
| \(\bar{X} \sim \mathrm{N}\left(a + 2, \dfrac{3}{50}\right)\) | A1, A1ft |
| (3) |
Notes
M1 for a correct expression for \(\mathrm{Var}(X)\) in terms of \(a\) or \(\mathrm{Var}(X) = 3\)
1st A1 for normal and correct mean must be \(a + 2\)
NB \(\mathrm{N}(17.2, \ldots)\) is A0 and \(\mathrm{N}\left(17.2, \tfrac{3}{50}\right)\) is M1A0A1
2nd A1ft for correct \(\mathrm{Var}(\bar{X})\), i.e. (their “3”)/50
| Scheme | Marks |
|---|---|
| \(17.2 - 1.96 \times \sqrt{\dfrac{3}{50}} \lt \mu \lt 17.2 + 1.96 \times \sqrt{\dfrac{3}{50}}\) | B1 M1 |
| \(17.2 - 1.96 \times \sqrt{\dfrac{3}{50}} \lt a + 2 \lt 17.2 + 1.96 \times \sqrt{\dfrac{3}{50}}\) | B1 |
| \(14.7 \lt a \lt 15.7\) | A1 |
| (4) | |
| (7 marks) |
Notes
1st B1 for correct use of \(z = 1.96\) in an attempt e.g. \(\bar{x} \pm z\sigma\) or \(\bar{x} \pm z\sigma^2\)
M1 for \(17.2 \pm z \times \sqrt{\dfrac{\text{"3"}}{50}}\) where \(|z| \gt 1.5\) accept just + or just −
Answer of (16.7, 17.7) scores B1M1B0A0
2nd B1 for either of the inequalities with \(a + 2\) and any \(z\) (\(|z| \gt 1.5\)) or \(a = 15.2 \pm z \times \sqrt{\dfrac{\text{"3"}}{50}}\)
A1 for awrt 14.7 and 15.7