S1 June 2013 (R) Q4
4. The time, in minutes, taken to fly from London to Malaga has a normal distribution with mean 150 minutes and standard deviation 10 minutes.
The time taken to fly from London to Berlin has a normal distribution with mean 100 minutes and standard deviation \(d\) minutes.
Given that 15% of the flights from London to Berlin take longer than 115 minutes,
The time, \(X\) minutes, taken to fly from London to another city has a normal distribution with mean \(\mu\) minutes.
Given that \(\mathrm{P}(X \lt \mu - 15) = 0.35\)
| Scheme | Marks |
|---|---|
| \([\mathrm{P}(M \lt 145) =]\ \mathrm{P}\left(Z \lt \dfrac{145 - 150}{10}\right)\) | M1 |
| \(= \mathrm{P}(Z \lt -0.5)\) or \(\mathrm{P}(Z \gt 0.5)\) | A1 |
| = awrt 0.309 | A1 |
| (3) |
Notes
Condone poor use of notation if a correct line appears later.
M1 for standardising with 145, 150 and 10. Allow \(\pm\) and use of symmetry so 155 instead of 145
1st A1 for \(\mathrm{P}(Z \lt -0.5)\) or \(\mathrm{P}(Z \gt 0.5)\) i.e. a \(z\) value of \(\pm 0.5\) and a correct region indicated
2nd A1 for awrt 0.309 Answer only is 3/3
| Scheme | Marks |
|---|---|
| \([\mathrm{P}(B \gt 115) = 0.15 \Rightarrow]\ \dfrac{115 - 100}{d} = 1.0364\) (Calc gives 1.036433...) | M1B1A1 |
| \(\boldsymbol{d}\) = 14.5 (Calc gives 14.4727...) | A1 |
| (4) |
Notes
M1 for \(\pm\dfrac{115 - 100}{d} = z\) where \(|z| \gt 1\). Condone MR of \(\mu = 150\) instead of 100 for M1B1only
B1 for a standardised expression \(= \pm 1.0364\) (do not allow for use of 1 – 1.0364)
1st A1 for \(z\) = awrt 1.04 and compatible signs i.e. a correct equation with \(z\) = awrt 1.04
2nd A1 for awrt 14.5 (allow awrt 14.4 if \(z\) = awrt 1.04 is seen)
Calc Answer only of awrt 14.473 scores M1B1A1A1
Answer only of awrt 14.48 scores M1B0A1A1
| Scheme | Marks |
|---|---|
| \([\mathrm{P}(X \gt \mu + 15 \mid X \gt \mu - 15) =]\ \dfrac{\mathrm{P}(X \gt \mu + 15)}{\mathrm{P}(X \gt \mu - 15)}\) | M1 |
| \(= \dfrac{0.35}{1 - 0.35}\) | A1 |
| \(= \underline{\dfrac{7}{13}}\) or awrt 0.538 | A1 |
| (3) | |
| (10 marks) |
Notes
M1 for a correct ratio expression need \(\mathrm{P}(X \gt \mu + 15)\) on numerator. Allow use of a value for \(\mu\). May be implied by next line.
NB \(\dfrac{0.35\times 0.65}{0.65} = \dfrac{0.2275}{0.65}\) is M0
1st A1 for a correct ratio of probabilities
2nd A1 for awrt 0.538 or \(\frac{7}{13}\) (o.e.). Allow 0.5385 provided 2nd A1 is scored.