S2 June 2012 Q1
1. A manufacturer produces sweets of length \(L\) mm where \(L\) has a continuous uniform distribution with range [15, 30].
These sweets are randomly packed in bags of 20 sweets.
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(L \gt 24) = \dfrac{1}{15} \times 6\) | M1 |
| \(= \dfrac{2}{5}\) or 0.4 oe | A1 |
| (2) |
Notes
M1 \(\dfrac{1}{15} \times (6 \text{ or } 5.5 \text{ or } 6.5 \text{ or } (30 - 24))\) or \(1 - \dfrac{1}{15}\big((24 - 15) \text{ or } (23.5 - 15) \text{ or } (24.5 - 15)\big)\)
| Scheme | Marks |
|---|---|
| Let \(X\) represent the number of sweets with \(L \gt 24\) | |
| \(X \sim \mathrm{B}(20, 0.4)\) | M1 |
| \(\mathrm{P}(X \geqslant 8) = 1 - \mathrm{P}(X \leqslant 7)\) | M1dep |
| \(= 1 - 0.4159\) | |
| \(= 0.5841\) awrt 0.584 | A1 |
| (3) |
Notes
M1 using B(20, “their (a)”)
M1 dependent on 1st M1. Writing or use of \(1 - \mathrm{P}(X \leqslant 7)\)
NB Use of normal/normal approximation/ Poisson/uniform gets M0 M0 A0
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(\text{both } X \geqslant 8) = (0.5841)^2\) | M1 |
| \(= 0.341\ldots\) | A1 ft |
| (2) | |
| (7 marks) |
Notes
M1 (their(b))\(^2\) or \((0.58)^2\) or \((0.5841)^2\) or \((0.584)^2\)
A1ft – either awrt 0.34 or follow through their answer to part (b) must be to 2sf or better.
Note you will have to check this.