A2 June 2023 Q1
1. The discrete random variable \(X\) has probability distribution
| \(x\) | \(-2\) | \(-1\) | 0 | 1 | 3 |
|---|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | 0.25 | \(a\) | \(b\) | \(a\) | 0.30 |
where \(a\) and \(b\) are probabilities.
Given that \(\mathrm{Var}(X) = 3.9\)
The independent random variables \(X_1\) and \(X_2\) each have the same distribution as \(X\)
| Scheme | Marks | AO |
|---|---|---|
| [\(\mathrm{E}(X) =\)] \(-2 \times 0.25 + -1 \times a + 0 \times b + 1 \times a + 3 \times 0.3\) | M1 | 1.1b |
| \(= \underline{0.4}\) | A1 | 1.1b |
| (2) |
Notes
M1 for a correct attempt (at least 3 correct non-zero terms or products and addition)
division by \(k\) \((k \neq 1)\) is M0
A1 for 0.4 o.e. (correct answer only scores 2 out of 2)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{E}(X^2) = (-2)^2 \times 0.25 + (-1)^2 \times a + 0 + 1^2 \times a + 3^2 \times 0.3\ (= 2a + 3.7)\) | M1 | 2.1 |
| [\(\mathrm{Var}(X) =\)] \(3.9 = 2a + 3.7 - \text{“}0.4^2\text{”}\) | dM1 | 1.1b |
| \(a = \underline{\mathbf{0.18}}\) | A1 | 1.1b |
| [Use of sum of probs = 1 implies \(2a + b = 0.45\)] \(b = \underline{\mathbf{0.09}}\) | A1ft | 1.1b |
| (4) |
Notes
1st M1 for a correct attempt at \(\mathrm{E}(X^2)\) (at least 3 correct non-zero products and addition)
Missing brackets around –2 and –1 is M0 unless recovered
2nd dM1 (dep on 1st M1) for use of 3.9 = their \(\mathrm{E}(X^2) - [\mathrm{E}(X)]^2\) ft their \(\mathrm{E}(X) = 0.4\)
1st A1 for \(a = 0.18\) o.e.
2nd A1 (dep on 1st M1 only) for \(b = 0.09\) o.e. or their \(b = 0.45 - 2 \times \text{“}a\text{”}\) (provided both \(a\) and \(b\) are probabilities)
| Scheme | Marks | AO |
|---|---|---|
| \(X_1 + X_2 \gt 3\) when \(\quad X_1 = 3, X_2 = 1 \quad X_1 = 1, X_2 = 3 \quad X_1 = 3, X_2 = 3\) | M1 | 3.4 |
| [\(\mathrm{P}(X_1 + X_2 \gt 3) =\)] \(\text{“}0.18\text{”} \times 0.3 + 0.3 \times \text{“}0.18\text{”} + 0.3 \times 0.3\) or \(2 \times 0.3 \times (0.3 + \text{“}0.18\text{”}) - 0.3^2\) | M1 | 1.1b |
| \(= \underline{\mathbf{0.198}}\) | A1 | 1.1b |
| (3) | ||
| (9 marks) |
Notes
1st M1 for identifying at least 2 cases e.g. \(X_1 = 3, X_2 \geqslant 1\) counts as 2 cases
(ignore extras including any incorrect pairs identified)
implied by at least two correct products of probs. or correct ft products of probs.
2nd M1 for a correct numerical expression for the probability ft their “0.18”
A1 for 0.198 o.e.