A2 October 2021 Q7
7. The weights of a particular type of apple, \(A\) grams, and a particular type of orange, \(R\) grams, each follow independent normal distributions.
\[A \sim \mathrm{N}(160,\ 12^2) \qquad\qquad R \sim \mathrm{N}(140,\ 10^2)\]A box contains 4 apples and 1 orange only. Jesse selects 2 pieces of fruit at random from the box.
From a large number of apples and oranges, Celeste selects \(m\) apples and 1 orange at random. The random variable \(W\) is given by
\[W = \left(\sum_{i=1}^{m} A_i\right) - n \times R\]where \(n\) is a positive integer.
Given that the middle 95% of the distribution of \(W\) lies between 1100.08 and 1499.92 grams,
| Scheme | Marks | AO |
|---|---|---|
| \((A + R) \sim \mathrm{N}(300,\ 12^2 + 10^2)\) \((A_1 + A_2) \sim \mathrm{N}(320,\ 2 \times 12^2)\) | M1 A1 A1 | 3.3 1.1b 1.1b |
| (3) |
Notes
M1: Setting up either model for the weights of the two fruit
A1: Correct distribution for 1 apple 1 orange
A1: Correct distribution for 2 apples
| Scheme | Marks | AO |
|---|---|---|
| P(both are apples) \(\left[= \tfrac{4}{5} \times \tfrac{3}{4}\right] = \tfrac{3}{5}\) P(one apple and one orange) \(= \tfrac{2}{5}\) | M1 | 2.1 |
| \(\text{‘}\tfrac{3}{5}\text{’}\mathrm{P}(A_1 + A_2 \gt 310) + \text{‘}\tfrac{2}{5}\text{’}\mathrm{P}(A + R \gt 310)\) | M1 | 2.1 |
| \(= 0.5377\ldots\) awrt 0.538 | A1 | 1.1b |
| (3) |
Notes
M1: Finding probability for each possible outcome
M1: Fully correct method for finding the required probability
A1: awrt 0.538
| Scheme | Marks | AO |
|---|---|---|
| \(\left[W = \displaystyle\sum_{1}^{m} A - n \times R\right]\) \(W \sim \mathrm{N}(160m - 140n,\ m \times 12^2 + n^2 \times 10^2)\) | M1 A1 | 3.3 1.1b |
| \(160m - 140n = (1100.08 + 1499.92) \div 2\ [= 1300]\) | M1 | 2.1 |
| \(2 \times 1.96 \times \sqrt{m \times 12^2 + n^2 \times 10^2} = (1499.92 - 1100.08)\) \(\left[\sqrt{m \times 12^2 + n^2 \times 10^2} = 102\right]\) | B1 M1 | 1.1b 1.1b |
| \(m = \dfrac{1300 + 140n}{160} \rightarrow \sqrt{\left(\tfrac{1300 + 140n}{160}\right) \times 12^2 + n^2 \times 10^2} = 102\) \(100n^2 + 126n - 9234 = 0\) | dM1 | 2.1 |
| \(n = 9\) (\(n = -10.26\) reject) | A1 | 1.1b |
| \(m = 16\) | A1 | 1.1b |
| (8) | ||
| (14 marks) |
Notes
M1: Setting up model for \(W\)
A1: correct distribution
M1: Using given interval to set up equation for mean
B1: 1.96
M1: Using given interval to set up equation for variance
dM1: Solving simultaneously leading to a 3TQ (dep on previous M mark)
A1: \(n = 9\) (only)
A1: \(m = 16\) (only)