S2 January 2013 Q2
2. In a village, power cuts occur randomly at a rate of 3 per year.
| Scheme | Marks |
|---|---|
| Let \(X\) be the random variable the number power cuts. | |
| \(X \sim \mathrm{Po}(3)\) | B1 |
| (i) \(\mathrm{P}(X = 7) = \mathrm{P}(X \leqslant 7) - \mathrm{P}(X \leqslant 6)\) or \(\dfrac{\mathrm{e}^{-3}3^7}{7!}\) | M1 |
| \(= 0.9881 - 0.9665\) | |
| \(= 0.0216\) awrt 0.0216 | A1 |
| (ii) \(\mathrm{P}(X \geqslant 4) = 1 - \mathrm{P}(X \leqslant 3)\) | M1 |
| \(= 1 - 0.6472\) | |
| \(= 0.3528\) awrt 0.353 | A1 |
| (5) |
Notes
B1 Writing or using Po(3) in either (i) or (ii)
(i) M1 writing or using \(\mathrm{P}(X \leqslant 7) - \mathrm{P}(X \leqslant 6)\) or \(\dfrac{\mathrm{e}^{-\lambda}\lambda^7}{7!}\)
(ii) M1 writing or using \(1 - \mathrm{P}(X \leqslant 3)\). (Do not accept writing \(1 - \mathrm{P}(X \lt 4)\) unless they have used \(1 - \mathrm{P}(X \leqslant 3)\)).
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{Po}(30)\) | |
| N(30,30) | M1A1 |
| \(\mathrm{P}(X \lt 20) = \mathrm{P}\left(Z \lt \dfrac{19.5 - 30}{\sqrt{30}}\right)\) | M1M1 A1 |
| \(= \mathrm{P}(Z \lt -1.92)\) | |
| \(= 1 - 0.9726\) | |
| \(= 0.0274 - 0.0276\) | A1 |
| (6) | |
| (11 marks) |
Notes
1st M1 for writing or using a normal approximation
1st A1 for correct mean and sd (may be given if correct in standardisation formula)
2nd M1 Standardising using their mean and their sd and using [18.5, 19, 19.5, 20 or 20.5] and for finding correct area by doing 1 – P(\(Z \leqslant\) “their 1.92”) If they have not written down a mean and sd then these need to be correct here to award the mark
3rd M1 for attempting a continuity correction (\(19 \pm 0.5\)) i.e. 18.5 or 19.5 only.
2nd A1 for \(\pm\dfrac{19.5 - 30}{\sqrt{30}}\) or \(\pm\) awrt 1.9 or better.
3rd A1 awrt 0.0274, 0.0275 or 0.0276
SC using \(\mathrm{P}(X \lt 20.5/19.5) - \mathrm{P}(X \lt 19.5/18.5)\) can get M1A1 M0M1A0A0