A2 June 2024 Q8
8. A company packs chickpeas into small bags and large bags.
The weight of a small bag of chickpeas is normally distributed with mean 500 g and standard deviation 5 g
A random sample of 3 small bags of chickpeas is taken.
The weight of a large bag of chickpeas is normally distributed with mean 1020 g and standard deviation 20 g
One large bag and one small bag of chickpeas are chosen at random.
Show your working clearly. (6)
| Scheme | Marks | AO |
|---|---|---|
| Let \(T = S_1 + S_2 + S_3\) then \(\mathrm{E}(T) = 1500\) | M1 | 3.3 |
| \(\mathrm{Var}(T) = 75\) | M1 | 2.1 |
| \(\mathrm{P}(1490 \lt T \lt 1530) = 0.8756\ldots\) | A1 | 1.1b |
| (3) |
Notes
M1: Selecting and using the appropriate model and attempting \(3 \times 500\)
M1: For realising the need to use \(\mathrm{Var}(S) + \mathrm{Var}(S) + \mathrm{Var}(S) = 3 \times 5^2\)
A1: awrt 0.876
| Scheme | Marks | AO |
|---|---|---|
| Let \(W = \pm(L - 2S - 30)\) then \(\mathrm{E}(W) = \pm(1020 - 2 \times 500 - 30)\) or Let \(X = \pm(L - 2S)\) then \(\mathrm{E}(X) = \pm(1020 - 2 \times 500)\) | M1 | 3.3 |
| \(\mathrm{E}(W) = -10\) (or 10) or \(\mathrm{E}(X) = 20\) (or \(-20\)) | A1 | 1.1b |
| \(\mathrm{Var}(\ldots) = 20^2 + 4 \times 5^2\) | M1 | 2.1 |
| \(\mathrm{Var}(\ldots) = 500\) | A1 | 1.1b |
| \(\mathrm{P}(W \gt 0)\) or \(\mathrm{P}(X \gt 30)\) (or \(\mathrm{P}(W \lt 0)\) or \(\mathrm{P}(X \lt 30)\)) | M1 | 2.1 |
| \(= 0.3273\ldots\) | A1 | 1.1b |
| (6) | ||
| (9 marks) |
Notes
M1: Selecting and using the appropriate model \(\pm(L - 2S - 30)\) or \(\pm(L - 2S)\) in an attempt to find the expected value
A1: \(-10\) or 20 (or 10 or \(-20\))
M1: For realising they need to use \(\mathrm{Var}(L) + 4\mathrm{Var}(S) = 20^2 + 4 \times 5^2\)
A1: 500 only
M1: dependent on using an appropriate model and realising that \(\mathrm{P}(W \gt 0)\) (or \(\mathrm{P}(W \lt 0)\)) or \(\mathrm{P}(X \gt 30)\) (or \(\mathrm{P}(X \lt 30)\)) is required. May be implied by awrt 0.327
Using standardisation look for e.g.
\(\mathrm{P}\left(Z \gt \dfrac{0 - \text{“}-10\text{”}}{\sqrt{500}}\right)\) or \(\mathrm{P}\left(Z \gt \dfrac{30 - \text{“}20\text{”}}{\sqrt{500}}\right)\) \((= \mathrm{P}(Z \gt 0.4472\ldots))\)
A1: awrt 0.327