S2 January 2008 Q3
3.
The number of cars passing an observation point in a 10 minute interval is modelled by a Poisson distribution with mean 1.
The number of other vehicles, other than cars, passing the observation point in a 60 minute interval is modelled by a Poisson distribution with mean 12.
| Scheme | Marks |
|---|---|
| Events occur at a constant rate. Events occur independently or randomly. Events occur singly. | B1 B1 |
| (2) |
Notes
any two of the 3
B1 B1 Need the word events at least once.
Independently and randomly are the same reason.
Award the first B1 if they only gain 1 mark
Special case. If they have 2 of the 3 lines without the word events they get B0 B1
| Scheme | Marks |
|---|---|
| Let \(X\) be the random variable the number of cars passing the observation point. \(\mathrm{Po}(6)\) | B1 |
| (i) \(\mathrm{P}(X \leqslant 4) - \mathrm{P}(X \leqslant 3) = 0.2851 - 0.1512\) or \(\dfrac{\mathrm{e}^{-6}6^4}{4!}\) | M1 |
| \(= 0.1339\) | A1 |
| (ii) \(1 - \mathrm{P}(X \leqslant 4) = 1 - 0.2851\) or \(1 - \mathrm{e}^{-6}\left(\dfrac{6^4}{4!} + \dfrac{6^3}{3!} + \dfrac{6^2}{2!} + \dfrac{6}{1!} + 1\right)\) | M1 |
| \(= 0.7149\) | A1 |
| (5) |
Notes
B1 Using Po(6) in (i) or (ii)
(i) M1 Attempting to find \(\mathrm{P}(X \leqslant 4) - \mathrm{P}(X \leqslant 3)\) or \(\dfrac{e^{-\lambda}\lambda^4}{4!}\)
A1 awrt 0.134
(ii) M1 Attempting to find \(1 - \mathrm{P}(X \leqslant 4)\)
A1 awrt 0.715
| Scheme | Marks |
|---|---|
| P (0 car and 1 others) + P (1 cars and 0 other) | B1 |
| \(= \mathrm{e}^{-1} \times 2\mathrm{e}^{-2} + 1\mathrm{e}^{-1} \times \mathrm{e}^{-2}\) \(= 0.3679 \times 0.2707 + 0.3679 \times 0.1353\) | M1 A1 |
| \(= 0.0996 + 0.0498\) \(= 0.149\) | A1 |
| (4) | |
| (11 marks) |
Notes
B1 Attempting to find both possibilities. May be implied by doing \(\mathrm{e}^{-\lambda_1} \times \lambda_2\mathrm{e}^{-\lambda_2} + \mathrm{e}^{-\lambda_2} \times \lambda_1\mathrm{e}^{-\lambda_1}\) any values of \(\lambda_1\) and \(\lambda_2\)
M1 finding one pair of form \(\mathrm{e}^{-\lambda_1} \times \lambda_2\mathrm{e}^{-\lambda_2}\) any values of \(\lambda_1\) and \(\lambda_2\)
A1 one pair correct
A1 awrt 0.149
(corrected from the printed mark scheme: the second product is printed as \(0.3674 \times 0.1353\); \(\mathrm{e}^{-1} = 0.3679\))
alternative
| Scheme | Marks |
|---|---|
| \(\mathrm{P_o}(1 + 2) = \mathrm{P_o}(3)\) | B1 |
| \(\mathrm{P}(X = 1) = 3\mathrm{e}^{-3}\) | M1 A1 |
| \(= 0.149\) | A1 |
B1 for Po(3)
M1 for attempting to find \(\mathrm{P}(X = 1)\) with Po(3)
A1 \(3\mathrm{e}^{-3}\)
A1 awrt 0.149