S3 June 2005 Q7
7. A manufacturer produces two flavours of soft drink, cola and lemonade. The weights, \(C\) and \(L\), in grams, of randomly selected cola and lemonade cans are such that \(C \sim \mathrm{N}(350, 8)\) and \(L \sim \mathrm{N}(345, 17)\).
(a) Find the probability that the weights of two randomly selected cans of cola will differ by more than 6 g. (6)
One can of each flavour is selected at random.
(b) Find the probability that the can of cola weighs more than the can of lemonade. (6)
Cans are delivered to shops in boxes of 24 cans. The weights of empty boxes are normally distributed with mean 100 g and standard deviation 2 g.
(c) Find the probability that a full box of cola cans weighs between 8.51 kg and 8.52 kg. (6)
(d) State an assumption you made in your calculation in part (c). (1)
| Scheme | Marks |
|---|---|
| Let \(W = C_1 - C_2\) \(\therefore W \sim \mathrm{N}(0, 16)\) Normal; 0; 16 | M1 A1; A1 |
| \(\therefore \mathrm{P}(|W| \gt 6) = 2\,\mathrm{P}(W \gt 6)\) | M1 |
| \(= 2 \times \mathrm{P}\left(Z \gt \dfrac{6 - 0}{\sqrt{16}}\right)\) Standardising, their \(\sigma\) | M1 |
| \(= 2 \times \mathrm{P}(Z \gt 1.5)\) \(= 2 \times (1 - 0.9332) = \underline{0.1336}\) | A1 |
| (6) |
Notes
NB \(W = C_1 + C_2 \Rightarrow\) M1 A0 M1 only
NB \(W = C - L\) treat as MR. Prob = 0.4346
| Scheme | Marks |
|---|---|
| Let \(W = C - L\) \(\therefore W \sim \mathrm{N}(5, 25)\) 5; 25 | B1; B1 |
| \(\mathrm{P}(W \gt 0) = \mathrm{P}\left(Z \gt \dfrac{\pm 5}{\sqrt{25}}\right)\) | M1 A1 |
| \(= \mathrm{P}(Z \lt 1)\) | M1 (\(p \gt 0.5\)) |
| \(= \underline{0.8413}\) | A1 |
| (6) |
| Scheme | Marks |
|---|---|
| Let \(W = C_1 + \cdots + C_{24} + B\) \(\therefore \mathrm{E}(W) = 24 \times 350 + 100 = \underline{8500}\) | B1 |
| \(\mathrm{Var}(W) = 24 \times 8 + 2^2 = \underline{196}\) | B1 |
| \(\mathrm{P}(8510 \leqslant W \leqslant 8520) = \mathrm{P}\left(\dfrac{8510 - 8500}{\sqrt{196}} \leqslant Z \leqslant \dfrac{8520 - 8500}{\sqrt{196}}\right)\) | M1 |
| \(= \mathrm{P}(0.71\ldots \leqslant Z \leqslant 1.43\ldots)\) AWRT | A1ft A1ft |
| \(= 0.9236 - 0.7611\) \(= \underline{0.1625}\) 0.161 – 0.163 | A1 |
| (6) |
| Scheme | Marks |
|---|---|
| All random variables are independent. | B1 |
| (1) | |
| (19 marks) |