S2 June 2015 Q1
1. In a survey it is found that barn owls occur randomly at a rate of 9 per 1000 km2.
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(N \geqslant 10) = 1 - \mathrm{P}(N \leqslant 9)\) \(= 0.4126\) | M1 A1 |
Notes
M1: using or writing \(1 - \mathrm{P}(N \leqslant 9)\) or \(1 - \mathrm{P}(N \lt 10)\)
A1: awrt 0.413
| Scheme | Marks |
|---|---|
| \(Y\) represents number of owls per 200 km2 \(\Rightarrow Y \sim \mathrm{Po}(1.8)\) | B1 |
| \(\mathrm{P}(Y = 2) = \dfrac{\mathrm{e}^{-1.8}1.8^2}{2!}\) \(= 0.2678\) | M1 A1 |
Notes
B1: using or writing Po(1.8)
M1: for a single term of the form \(\dfrac{\mathrm{e}^{-\lambda}\lambda^2}{2!}\) with any value for \(\lambda\) or \(\mathrm{P}(X \leqslant 2) - \mathrm{P}(X \leqslant 1)\)
A1: awrt 0.268
| Scheme | Marks |
|---|---|
| Normal approximation | M1 |
| \(\mu = 50 \times 9 = 450 \quad \sigma^2 = 450\) | M1 |
| \(\mathrm{P}(X \geqslant 470) \approx 1 - \mathrm{P}\left(Z \lt \dfrac{469.5 - 450}{\sqrt{450}}\right)\) | M1 dM1 A1 |
| \(= 0.1788\) | A1 |
| (6) |
Notes
M1: Using or writing, normal approximation with mean = 450
M1: Using or writing the mean = variance. Does not need to be 450. May be seen in the standardisation calculation.
M1: \(\pm\left(\dfrac{(470 \text{ or } 469.5 \text{ or } 470.5) - \text{their mean}}{\text{their sd}}\right)\) May be implied by a correct answer or \(z\) = awrt 0.92
M1: dep on previous method mark being awarded. Using a continuity correction \(470 \pm 0.5\) May be implied by a correct answer or \(z\) = awrt 0.92
A1: correct standardisation no need to subtract from 1. Award for \(\dfrac{469.5 - 450}{\sqrt{450}}\) or awrt 0.92 or a correct answer
A1: awrt 0.179