S2 June 2013 Q2
2. The number of defects per metre in a roll of cloth has a Poisson distribution with mean 0.25
Find the probability that
(a) a randomly chosen metre of cloth has 1 defect, (2)
(b) the total number of defects in a randomly chosen 6 metre length of cloth is more than 2 (3)
A tailor buys 300 metres of cloth.
(c) Using a suitable approximation find the probability that the tailor’s cloth will contain less than 90 defects. (5)
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X = 1) = 0.25\mathrm{e}^{-0.25} = 0.1947\) awrt 0.195 | M1A1 |
| (2) |
Notes
M1 \(0.25\mathrm{e}^{-0.25}\) o.e
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{Po}(1.5)\) | B1 |
| \(\mathrm{P}(X \gt 2) = 1 - \mathrm{P}(X \leqslant 2)\) | M1 |
| \(= 1 - 0.8088\) | |
| \(= 0.1912\) awrt 0.191 | A1 |
| (3) |
Notes
B1 stating or using Po(1.5)
M1 stating or using \(1 - \mathrm{P}(X \leqslant 2)\)
| Scheme | Marks |
|---|---|
| \([\lambda = 300 \times 0.25 = 75]\) | |
| \(X \sim \mathrm{N}(75, 75)\) | B1 B1 |
| \(\mathrm{P}(X \lt 90) = \mathrm{P}\left(X \leqslant \frac{89.5 - 75}{\sqrt{75}}\right)\) | M1M1 |
| \(= \mathrm{P}(Z \leqslant 1.6743..)\) | |
| = awrt 0.953 or 0.952 | A1 |
| (5) | |
| (10 marks) |
Notes
1st B1 for normal approximation and correct mean
2nd B1 Var \((X) = 75\) or sd \(= \sqrt{75}\) or awrt 8.66 (may be given if correct in standardisation formula)
1st M1 using either 89.5 or 88.5
2nd M1 Standardising using their mean and their sd, using [89.5, 88.5 or 89] and for finding correct area
NB use of Poisson gives an answer of 0.9498 and gains no marks