S3 June 2010 Q2
2. Philip and James are racing car drivers. Philip’s lap times, in seconds, are normally distributed with mean 90 and variance 9. James’ lap times, in seconds, are normally distributed with mean 91 and variance 12. The lap times of Philip and James are independent. Before a race, they each take a qualifying lap.
The race is made up of 60 laps. Assuming that they both start from the same starting line and lap times are independent,
| Scheme | Marks |
|---|---|
| [\(P \sim \mathrm{N}(90, 9)\) and \(J \sim \mathrm{N}(91, 12)\)] \((J - P) \sim \mathrm{N}(1, 21)\) | M1, A1 |
| \(\mathrm{P}(J \lt P) \quad = \mathrm{P}(J - P \lt 0)\) \(= \mathrm{P}\left(Z \lt \dfrac{0 - 1}{\sqrt{21}}\right)\) | dM1 |
| \(= \mathrm{P}(Z \lt -0.2182\ldots)\) \(= 1 - 0.5871 = 0.4129\) awrt (0.413 ~ 0.414) calculator (0.4136….) | A1 |
| (4) |
Notes
1st M1 for attempting \(J - P\) and \(\mathrm{E}(J - P)\) or \(P - J\) and \(\mathrm{E}(P - J)\)
1st A1 for variance of 21 (Accept 9 + 12). Ignore any slip in \(\mu\) here.
2nd dM1 for attempting the correct probability and standardising with their mean and sd.
This mark is dependent on previous M so if \(J - P\) ( or \(P - J\)) is not being used score M0
If their method is not crystal clear then they must be attempting P(\(Z\)< -ve value) or P(\(Z\) > +ve value) i.e. their probability after standardisation should lead to a prob. < 0.5 so e.g. \(\mathrm{P}(J - P \lt 0)\) leading to 0.5871 is M0A0 unless the M1 is clearly earned.
2nd A1 for awrt 0.413 or 0.414
The first 3 marks may be implied by a correct answer
| Scheme | Marks |
|---|---|
| \(X = (J_1 + J_2 + \ldots + J_{60}) - (P_1 + P_2 + \ldots + P_{60})\) | M1 |
| \(\mathrm{E}(X) = 60 \times 91 - 60 \times 90 = 60\) [stated as \(\mathrm{E}(X) = 60\) or \(X \sim \mathrm{N}(60, \ldots)\)] | B1 |
| \(\mathrm{Var}(X) = 60 \times 9 + 60 \times 12 = 1260\) | A1 |
| \(\mathrm{P}(X \gt 120) \quad = \mathrm{P}\left(Z \gt \dfrac{120 - 60}{\sqrt{1260}}\right)\) | M1 |
| \(= \mathrm{P}(Z \gt 1.69030\ldots)\) \(= 1 - 0.9545 = 0.0455\) awrt (0.0455) | A1 |
| (5) | |
| (9 marks) |
Notes
1st M1 for a clear attempt to identify a correct form for \(X\). This may be implied by correct variance of 1260
B1 for \(\mathrm{E}(X) = 60\). Can be awarded even if they are using \(X = 60J - 60P\). Allow \(P - J\) and -60
1st A1 for a correct variance. If 1260 is given the M1 is scored by implication.
2nd M1 for attempting a correct probability and standardising with 120 and their 60 and 1260
If the answer is incorrect a full expression must be seen following through their values for M1 e.g. \(\mathrm{P}\left(Z \gt \dfrac{120 - \text{their } 60}{\sqrt{\text{their variance}}}\right)\). If using -60, should get \(\mathrm{P}\left(Z \lt \dfrac{-120 - -60}{\sqrt{\text{their variance}}}\right)\)
Use of means Attempt to use \(\overline{J} - \overline{P}\) for 1st M1, \(\mathrm{E}(\overline{J} - \overline{P}) = 1\) for B1 and \(\mathrm{Var}(\overline{J} - \overline{P}) = 0.35\) for A1
Then 2nd M1 for standardisation with 2, and their 1 and 0.35