S4 June 2014 (R) Q2
2. The cloth produced by a certain manufacturer has defects that occur randomly at a constant rate of \(\lambda\) per square metre. If \(\lambda\) is thought to be greater than 1.5 then action has to be taken.
Using \(\mathrm{H}_0 : \lambda = 1.5\) and \(\mathrm{H}_1 : \lambda \gt 1.5\) a quality control officer takes a 4 m\(^2\) sample of cloth and rejects \(\mathrm{H}_0\) if there are 11 or more defects. If there are 8 or fewer defects she accepts \(\mathrm{H}_0\). If there are 9 or 10 defects a second sample of 4 m\(^2\) is taken and \(\mathrm{H}_0\) is rejected if there are 11 or more defects in this second sample, otherwise it is accepted.
| Scheme | Marks |
|---|---|
| [\(X\) = no. of defects in 4 square metres.] \(X \sim \mathrm{Po}(6)\) | |
| [Size =] \(\mathrm{P}(X \gt 10) + \mathrm{P}(X = 9 \text{ or } 10)\mathrm{P}(X \gt 10)\) | M1 |
| \(= (1 - 0.9574) + (0.9574 - 0.8472)(1 - 0.9574)\) | M1A1 |
| \(= 0.04729\ldots\) = awrt 0.0473 | A1 |
| (4) |
Notes
1st M1 for a correct expression/selection of probabilities
2nd M1 for use of Po(6) and at least one correct prob. seen
May see \(\mathrm{P}(X = 9) = \dfrac{\mathrm{e}^{-6}6^9}{9!} = 0.06883\ldots\) or \(\mathrm{P}(X = 10) = \dfrac{\mathrm{e}^{-6}6^{10}}{10!} = 0.04130\ldots\)
1st A1 for a fully correct expression
2nd A1 for awrt 0.0473
| Scheme | Marks |
|---|---|
| \(Y \sim \mathrm{Po}(8)\) | B1 |
| Power \(= 1 - \left(\mathrm{P}(X \leqslant 8) + [\mathrm{P}(X = 9) + \mathrm{P}(X = 10)] \times \mathrm{P}(X \leqslant 10)\right)\) Or \(\left(1 - \mathrm{P}(X \leqslant 10)\right) + [\mathrm{P}(X = 9) + \mathrm{P}(X = 10)] \times \left(1 - \mathrm{P}(X \leqslant 10)\right)\) | M1 |
| \(= (1 - 0.8159) + (0.8159 - 0.5925)(1 - 0.8159)\) \(= 0.22522\ldots\) = awrt 0.225 | A1 |
| (3) | |
| (7 marks) |
Notes
B1 for evidence of use of Po(8)
M1 for an expression of the correct form with at least one correct prob.
A1 for awrt 0.225