S1 June 2014 (R) Q1
1. The discrete random variable \(X\) has probability distribution
| \(x\) | \(-4\) | \(-2\) | 1 | 3 | 5 |
|---|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | 0.4 | \(p\) | 0.05 | 0.15 | \(p\) |
Find
Given that \(\mathrm{Var}(X) = 13.35\)
| Scheme | Marks |
|---|---|
| \(0.4 + p + 0.05 + 0.15 + p = 1\) or verify \(0.4 + 0.2 + 0.05 + 0.15 + 0.2 = 1\) | M1 |
| \(2p = 0.4\) \(p = 0.2\) (verify: conclusion \(p = 0.2\) must be stated) | A1cso |
| (2) |
Notes
M1 for equating sum of all probabilities to 1
The minimum working required is: \(0.6 + 2p = 1\) but \(2p = 1 - 0.6\) or \(2p = 0.4\) is M0
BUT allow \(1 - 0.4 - 0.05 - 0.15 = 0.4\) followed by \(2p = 0.4\) or \(1 - 0.4 - 0.05 - 0.15 = 2p\)
Since all of the probabilities are seen.
A1cso for a correct solution with no incorrect working seen
(For verify method, they must conclude that \(p = 0.2\))
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = 0.4 \times -4 + 0.2 \times -2 + 0.05 \times 1 + 0.15 \times 3 + 0.2 \times 5 = -0.5\) | M1 A1 |
| (2) |
Notes
M1 for a correct expression with at least 3 correct terms
May be: \(-1.6 - 0.4 + 0.05 + 0.45 + 1\)
A1 for \(-0.5\)
| Scheme | Marks |
|---|---|
| \(\left[\mathrm{F}(0) = \mathrm{P}(X = -2) + \mathrm{P}(X = -4) = 0.2 + 0.4\right] = 0.6\) | B1 |
| (1) |
Notes
B1 for 0.6
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(3X + 2 \gt 5) = \mathrm{P}(X \gt 1)\) \(\mathrm{P}(3X + 2 \gt 5) = \mathrm{P}(X = 3) + \mathrm{P}(X = 5)\) | M1 |
| \(\mathrm{P}(3X + 2 \gt 5) = 0.35\) | A1 |
| (2) |
Notes
M1 for identifying \(X = 3\) and \(X = 5\) only (\(X \gt 1\) is not sufficient)
A1 for 0.35
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(aX + 3) = a^2\,\mathrm{Var}(X)\) | M1 |
| \(53.4 = a^2 13.35\) \(a = \pm 2\) | A1 |
| (2) | |
| (9 marks) |
Notes
M1 for \(\mathrm{Var}(aX + 3) = a^2\,\mathrm{Var}(X)\) but this may be implied by seeing \(a = 2\) or \(a = -2\)
A1 for both correct values \(+2\) and \(-2\)