S2 June 2009 Q8
8. A cloth manufacturer knows that faults occur randomly in the production process at a rate of 2 every 15 metres.
A retailer buys a large amount of this cloth and sells it in pieces of length \(x\) metres. He chooses \(x\) so that the probability of no faults in a piece is 0.80
The retailer sells 1200 of these pieces of cloth. He makes a profit of 60p on each piece of cloth that does not contain a fault but a loss of £1.50 on any pieces that do contain faults.
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{Po}(2)\) \(\mathrm{P}(X = 4) = \dfrac{\mathrm{e}^{-2} \times 2^4}{4!} = 0.0902\) awrt 0.09 | M1 A1 |
| (2) |
Notes
M1 for use of Po(2) may be implied
A1 awrt 0.09
| Scheme | Marks |
|---|---|
| \(Y \sim \mathrm{Po}(8)\) | B1 |
| \(\mathrm{P}(Y \gt 10) = 1 - \mathrm{P}(Y \leqslant 10) = 1 - 0.8159 = 0.18411\ldots\) awrt 0.184 | M1A1 |
| (3) |
Notes
B1 for Po(8) seen or used
M1 for \(1 - \mathrm{P}(Y \leqslant 10)\) oe
A1 awrt 0.184
| Scheme | Marks |
|---|---|
| \(F\) = no. of faults in a piece of cloth of length \(x\) \(F \sim \mathrm{Po}\left(x \times \tfrac{2}{15}\right)\) | |
| \(\mathrm{e}^{-\frac{2x}{15}} = 0.80\) | M1A1 |
| \(\mathrm{e}^{-\frac{2}{15} \times 1.65} = 0.8025\ldots,\quad \mathrm{e}^{-\frac{2}{15} \times 1.75} = 0.791\ldots\) | M1 |
| These values are either side of 0.80 therefore \(x = 1.7\) to 2 sf | A1 |
| (4) |
Notes
1st M1 for forming a suitable Poisson distribution of the form \(\mathrm{e}^{-\lambda} = 0.8\)
1st A1 for use of lambda as \(\dfrac{2x}{15}\) (this may appear after taking logs)
2nd M1 for attempt to consider a range of values that will prove 1.7 is correct OR for use of logs to show lambda = …
2nd A1 correct solution only. Either get 1.7 from using logs or stating values either side
S.C for \(\mathrm{e}^{-\frac{2}{15} \times 1.7} = 0.797\ldots \approx 0.80 \quad \therefore x = 1.7\) to 2 sf allow 2nd M1A0
| Scheme | Marks |
|---|---|
| Expected number with no faults \(= 1200 \times 0.8 = 960\) Expected number with some faults \(= 1200 \times 0.2 = 240\) | M1 A1 |
| So expected profit \(= 960 \times 0.60 - 240 \times 1.50,\) = £216 | M1, A1 |
| (4) | |
| (13 marks) |
Notes
1st M1 for one of the following 1200 p or 1200 (1 – p) where p = 0.8 or 2/15.
1st A1 for both expected values being correct or two correct expressions.
2nd M1 for an attempt to find expected profit, must consider with and without faults
2nd A1 correct answer only.