S1 June 2014 Q5
5. The discrete random variable \(X\) has the probability function
\[\mathrm{P}(X = x) = \begin{cases} kx & x = 2, 4, 6 \\ k(x - 2) & x = 8 \\ 0 & \text{otherwise} \end{cases}\]where \(k\) is a constant.
| Scheme | Marks |
|---|---|
| \(2k + 4k + 6k + k(8 - 2) = 1\) (commas instead of + or a table OK if \(18k = 1\) seen later) | M1 |
| \(k = \frac{1}{18}\) (*) | A1cso |
| (2) |
Notes
M1 for \(2k + 4k + 6k + k(8 - 2) = 1\) A1 for \(k = \dfrac{1}{18}\) NB cso so no incorrect working seen
or M1 for \(2\times\frac{1}{18} + 4\times\frac{1}{18} + 6\times\frac{1}{18} + \frac{1}{18}(8 - 2)\) A1 for =1 and “therefore \(k = \frac{1}{18}\)”
| Scheme | Marks |
|---|---|
| \([2k + 4k] = \dfrac{6}{18} = \dfrac{1}{3}\) (\(\frac{1}{3}\) or any exact numerical equivalent) | B1 |
| (1) |
Notes
If in parts (c), (d) and (e) there is a correct expression worthy of M1 but later they incorrectly go on and multiply or divide by some number \(n\), then allow the M1 but mark their final answer (A0)
Answers only in (b), (c), (d) and (e) score all the marks.
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = \left(2\times\dfrac{1}{9}\right) + \left(4\times\dfrac{2}{9}\right) + \left(6\times\dfrac{1}{3}\right) + \left(8\times\dfrac{1}{3}\right)\) or \((2\times 2k) + (4\times 4k) + (6\times 6k) + (8\times 6k)\) | M1 |
| \(= 5\dfrac{7}{9}\) (or exact equivalent e.g. \(\dfrac{52}{9}\)) | A1 |
| (2) |
Notes
M1 for an expression for \(\mathrm{E}(X)\) with at least 3 correct terms (products) allow use of \(k\) e.g. \(104k\)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X^2) = \left(4\times\dfrac{1}{9}\right) + \left(16\times\dfrac{2}{9}\right) + \left(36\times\dfrac{1}{3}\right) + \left(64\times\dfrac{1}{3}\right)\) or \((4\times 2k) + (16\times 4k) + (36\times 6k) + (64\times 6k)\) | M1 |
| \(= 37\dfrac{1}{3}\) (or exact equivalent e.g. \(\dfrac{112}{3}\)) | A1 |
| (2) |
Notes
M1 for an expression for \(\mathrm{E}(X^2)\) with at least 3 correct terms (products) allow use of \(k\) e.g. \(672k\)
A1 for any exact equivalent only. E.g. 37.3 is A0 but, of course, \(37.\dot{3}\) is OK
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(X) = 37\dfrac{1}{3} - \left(5\dfrac{7}{9}\right)^2 \quad \left[= 3.95\ldots \text{ or } \dfrac{320}{81}\right]\) | M1 |
| \(\mathrm{Var}(3 - 4X) = 16\times 3.95\ldots\) | M1 |
| \(=\) awrt 63.2 (allow \(\frac{5120}{81}\)) | A1 |
| (3) | |
| (10 marks) |
Notes
1st M1 for \(\mathrm{E}(X^2) - [\mathrm{E}(X)]^2\) ft their answers to (c) and (d). Must see values used correctly.
2nd M1 for statement “\(4^2\times\mathrm{Var}(X)\)”seen or for \(4^2\times\) their \(\mathrm{Var}(X)\) provided their \(\mathrm{Var}(X) \gt 0\). Do not allow for \(16\times\mathrm{E}(X^2)\) but can score M0M1. NB condone \(-4^2\times\mathrm{Var}(X)\) if the answer later becomes positive.
A1 for exact fraction (\(\frac{5120}{81}\) o.e.) or decimal approximation that is awrt 63.2
Beware: rounding to 3sf in (c) (5.78) and (d) (37.3) gives 62.3 which could be misread as 63.2