S1 January 2010 Q5
5. The probability function of a discrete random variable \(X\) is given by
\[\mathrm{p}(x) = kx^2 \qquad x = 1, 2, 3\]where \(k\) is a positive constant.
Find
| Scheme | Marks |
|---|---|
| \(k + 4k + 9k = 1\) | M1 |
| \(14k = 1\) \(k = \dfrac{1}{14}\) **given** cso | A1 |
| (2) |
Notes
M1 for clear attempt to use \(\sum \mathrm{p}(x) = 1\), full expression needed and the “1” must be clearly seen. This may be seen in a table.
A1cso for no incorrect working seen. The sum and “= 1” must be explicitly seen somewhere.
A verification approach to (a) must show addition for M1 and have a suitable comment e.g. “therefore \(k = \tfrac{1}{14}\)” for A1 cso
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \geqslant 2) = 1 - \mathrm{P}(X = 1)\) or \(\mathrm{P}(X = 2) + \mathrm{P}(X = 3)\) | M1 |
| \(= 1 - k = \dfrac{13}{14}\) or 0.92857... awrt 0.929 | A1 |
| (2) |
Notes
M1 for 1- \(\mathrm{P}(X \leqslant 1)\) or \(\mathrm{P}(X = 2) + \mathrm{P}(X = 3)\)
A1 for awrt 0.929. Answer only scores 2/2
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = 1 \times k + 2 \times k \times 4 + 3 \times k \times 9\) or \(36k\) | M1 |
| \(= \dfrac{36}{14} = \dfrac{18}{7}\) or \(2\dfrac{4}{7}\) (or exact equivalent) | A1 |
| (2) |
Notes
M1 for a full expression for \(\mathrm{E}(X)\) with at least two terms correct.
NB If there is evidence of division (usually by 3) then score M0
A1 for any exact equivalent - answer only scores 2/2
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(X) = 1 \times k + 4 \times k \times 4 + 9 \times k \times 9, \ -\left(\dfrac{18}{7}\right)^2\) | M1 M1 |
| \(\mathrm{Var}(1 - X) = \mathrm{Var}(X)\) | M1 |
| \(= \dfrac{19}{49}\) or 0.387755... awrt 0.388 | A1 |
| (4) | |
| (10 marks) |
Notes
1st M1 for clear attempt at \(\mathrm{E}(X^2)\), need at least 2 terms correct in \(1 \times k + 4 \times 4k + 9 \times 9k\) or \(\mathrm{E}(X^2) = 7\)
2nd M1 for their \(\mathrm{E}(X^2) - (\text{their } \mu)^2\)
3rd M1 for clearly stating that Var(1 - \(X\)) = Var(\(X\)), wherever seen
A1 accept awrt 0.388. All 3 M marks are required.
Allow 4/4 for correct answer only but must be for Var(1 – \(X\)).