S2 June 2011 Q5
5. Defects occur at random in planks of wood with a constant rate of 0.5 per 10 cm length. Jim buys a plank of length 100 cm.
Shivani buys 6 planks each of length 100 cm.
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{Po}(5)\); \(\mathrm{P}(X \leqslant 3) = 0.2650\) | M1 A1 |
| (2) |
Notes
M1 for identifying Po(5) - it should be clearly seen somewhere or implied
A1 for correct probability. Allow 0.265
| Scheme | Marks |
|---|---|
| Let \(Y\) = the no.of planks with at most 3 defects, \(Y \sim\) Binomial \(Y \sim \mathrm{B}(6, 0.265)\) | M1 A1ft |
| \(\mathrm{P}(Y \lt 2) = \mathrm{P}(Y \leqslant 1)\) | M1 |
| \(= \left[0.735^6 + 6 \times 0.265 \times 0.735^5\right]\) | A1 |
| \(= 0.4987\ldots\) awrt 0.499 or 0.498 | A1 |
| (5) |
Notes
1st M1 for writing or using the binomial - may be implied by use of \(nq^x(1 - q)^{6-x}\) with \(n \geqslant 1\)
1st A1ft for \(n = 6\) and \(p\) = their (a) may be implied by \(6p(1 - p)^5\) or \((1 - p)^6\)
NB if they write B(6,(a)) they get M1 A1
2nd M1 for writing \(\mathrm{P}(Y \leqslant 1)\) or \(\mathrm{P}(Y = 0) + \mathrm{P}(Y = 1)\) or \((1 - q)^6 + nq(1 - q)^5\) with \(n \geqslant 1\)
2nd A1 \((1 - p)^6 + 6p(1 - p)^5\) where \(p\) = their (a)
3rd A1 for awrt 0.499
SC use of a probability in the tables – lose last two marks – could get M1A1M1 M0 A0
| Scheme | Marks |
|---|---|
| Let \(T\) = total number of defects on 6 planks, \(T \sim \mathrm{Po}(30)\) so \(T \approx S \sim\) Normal \(S \sim \mathrm{N}(30, 30)\) | M1 A1 |
| \(\mathrm{P}(T \lt 18) = \mathrm{P}(S \lt 17.5)\) | M1 |
| \(= P\left(z \lt \dfrac{17.5 - 30}{\sqrt{30}}\right)\) | M1 |
| \(= \mathrm{P}(Z \lt -2.28\ldots)\) | A1 |
| \(= 0.01123\ldots\) awrt 0.0112 or 0.0113 | A1 |
| (6) | |
| (13 marks) |
Notes
1st M1 for a normal approx
1st A1 for correct mean and sd
2nd M1 for use of continuity correction, either 17.5 or 18.5 or 42.5 or 41.5 seen
3rd M1 Standardising with their mean and their sd and 17.5 or 18 or 18.5 or 41.5 or 42 or 42.5
NB if they have not written down a mean and sd then they need to be correct in the standardisation to gain this mark.
2nd A1 for \(z = \pm 2.28\) or better. May be awarded for \(\pm\dfrac{17.5 - 30}{\sqrt{30}}\) [NB no continuity correction \(z\) = 2.19]
3rd A1 for awrt 0.0112 or 0.0113 [NB no approximation gives 0.00727…]
SC using P(\(X\)<18.5) – P(\(X\)<17.5) can get M1 A1 M1 M0A0A0