S2 January 2012 Q4
4. A website receives hits at a rate of 300 per hour.
Find the probability of
The website will go down if there are more than 70 hits in 10 minutes.
| Scheme | Marks |
|---|---|
| Poisson | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| Hits occur singly in time Hits are independent or Hits occur randomly Hits occur at a constant rate | B1B1 |
| (2) |
Notes
1st B1 Any one of the 3 statements - no context required. NB It must be a constant (mean) rate and not a constant probability or a constant mean.
2nd B1 A different statement with context of hits. NB random and independent are the same statement.
If only one mark awarded give the 1st B1. Never award B0 B1
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{Po}(5)\) | B1 |
| \(P(X = 10) = \mathrm{P}(X \leqslant 10) - \mathrm{P}(X \leqslant 9)\) or \(\dfrac{e^{-5}5^{10}}{10!}\) | M1 |
| \(= 0.9863 - 0.9682\) \(= 0.0181\) awrt 0.0181 | A1 |
| (3) |
Notes
B1 writing or using Po(5)
M1 writing or using \(\mathrm{P}(X \leqslant 10) - \mathrm{P}(X \leqslant 9)\) or \(\dfrac{e^{-5}5^{10}}{10!}\)
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{Po}(10)\) | B1 |
| \(\mathrm{P}(X \geqslant 15) = 1 - \mathrm{P}(X \leqslant 14)\) | M1 |
| \(= 1 - 0.9165\) \(= 0.0835\) awrt 0 .0835 | A1 |
| (3) |
Notes
B1 writing or using Po(10)
M1 writing or using \(1 - \mathrm{P}(X \leqslant 14)\)
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{Po}(50)\) Approximated by N(50,50) | B1B1 |
| \(\mathrm{P}(X \gt 70) = \mathrm{P}\left(Z \gt \dfrac{70.5 - 50}{\sqrt{50}}\right)\) | M1M1 |
| \(= \mathrm{P}(Z \gt 2.899...)\) | A1 |
| \(= 1 - 0.9981\) | M1 |
| \(= 0.0019\) awrt 0.0019 | A1 |
| (7) | |
| (16 marks) |
Notes
1st B1 for a normal approximation 2nd B1 for correct mean and sd (may be seen in standardisation formula
1st M1 for attempting a continuity correction (71 \(\pm\) 0.5)
2nd M1 Standardising using their mean and their sd and using [69.5, 70, 70.5, 71 or 71.5] allow \(\pm\) z
NB if they have not written down a mean and sd then they need to be correct in the standardisation to gain this mark.
1st A1 for \(z = \pm\) awrt 2.9 or better. May be awarded for \(\pm\dfrac{70.5 - 50}{\sqrt{50}}\)
3rd M1 for 1 - tables value
SC using P(\(X\)< 70.5/71.5) – P(\(X\)<69.5/70.5) can get B1B1 M0M1A0 M0A0