S1 June 2012 Q1
1. A discrete random variable \(X\) has the probability function
\[\mathrm{P}(X = x) = \begin{cases} k(1 - x)^2 & x = -1, 0, 1 \text{ and } 2 \\ 0 & \text{otherwise} \end{cases}\]| Scheme | Marks | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 | ||||||||||
| \(4k + k + (0) + k = 1\) (Allow verify approach) | A1 | ||||||||||
| \(6k = 1 \;\Rightarrow\; k = \dfrac{1}{6}\) (*) | A1cso | ||||||||||
| (3) |
Notes
M1 for attempt at \(\mathrm{P}(X = x)\) with at least 2 correct. Do not give for 4, 1, etc but \(\frac{4}{6}, \frac{1}{6}\) are OK
1st A1 for at least \(4k + k + k = 1\) seen. Allow \(\frac{4}{6} + \frac{1}{6} + \frac{1}{6} = 1\) [Must see = 1]
2nd A1cso provided previous 2 marks are scored and no incorrect working seen. It’s not essential to see \(\mathrm{P}(X = -1) = 4k\) etc but if wrongly assigned probabilities such as \(\mathrm{P}(X = 2) = 4k\) and \(\mathrm{P}(X = -1) = k\) are seen then the final A1 is lost.
Verify To score final A1cso there must be a comment such as “therefore \(k = \frac{1}{6}\)”
Division by 4 (or any other \(n\)) in (b), (c) or (d) is M0. Do not apply ISW
| Scheme | Marks |
|---|---|
| \([\mathrm{E}(X)] = -4k\ (+ 0 + 0) + 2k\) or \(-2k\) or \(-1\times\dfrac{4}{6} + 2\times\dfrac{1}{6}\) | M1 |
| \(= -\dfrac{1}{3}\) (or \(-0.\dot{3}\)) | A1 |
| (2) |
Notes
M1 for a full correct expression for \(\mathrm{E}(X)\), ft their probabilities. Allow in terms of \(k\).
A1 for \(-\dfrac{1}{3}\) or exact equivalent only. Just \(-\dfrac{1}{3}\) scores M1A1
| Scheme | Marks |
|---|---|
| \(\left[\mathrm{E}\left(X^2\right)\right] = (-1)^2\times 4k + (0 + 0) + 2^2k\) or \(4k + 4k\) or \((-1)^2\times\dfrac{4}{6} + 2^2\times\dfrac{1}{6}\) (o.e.) | M1 |
| \(= \dfrac{4}{3}\) (*) | A1cso |
| (2) |
Notes
M1 for evidence of both non-zero terms seen. May be simplified but 2 terms needed.
A1cso for M1 seen leading to \(\dfrac{4}{3}\) or any exact equivalent. Condone \(-1^2\times 4k\) but not \(-4k\)
| Scheme | Marks |
|---|---|
| \([\mathrm{Var}(X)] = \dfrac{4}{3} - \left(-\dfrac{1}{3}\right)^2\) or \(8k - 4k^2 = \left[\dfrac{11}{9}\right]\) | M1 |
| \(\mathrm{Var}(1 - 3X) = (-3)^2\,\mathrm{Var}(X)\) or \(9\mathrm{Var}(X)\) | M1 |
| \(= 11\) | A1 cao |
| (3) | |
| (10 marks) |
Notes
1st M1 for correct attempt at \(\mathrm{Var}(X)\) - follow through their \(\mathrm{E}(X)\) and allow in terms of \(k\). Award if a correct formula is seen and some correct substitution made.
2nd M1 for correct use of \(\mathrm{Var}(aX + b)\). Condone \(-3^2\,\mathrm{Var}(X)\) if it eventually yields \(9\mathrm{Var}(X)\)
A1cao for 11 only
Alternative
| Scheme | Marks |
|---|---|
| \(Y = 1 - 3X\): 4 1 \(-2\) \(-5\) Prob: \(4k\) \(k\) 0 \(k\) And \(\mathrm{E}(Y) = 12k\) | M1 |
| \(\mathrm{E}(Y^2) = 90k\) and \(\mathrm{Var}(Y) = 90k - 144k^2\) | M1 |
| \(= 11\) | A1 cao |