S1 June 2005 Q5
5. The random variable \(X\) has probability function
\[\mathrm{P}(X = x) = \begin{cases} kx, & x = 1, 2, 3, \\ k(x + 1), & x = 4, 5, \end{cases}\]where \(k\) is a constant.
(a) Find the value of \(k\). (2)
(b) Find the exact value of \(\mathrm{E}(X)\). (2)
(c) Show that, to 3 significant figures, \(\mathrm{Var}(X) = 1.47\). (4)
(d) Find, to 1 decimal place, \(\mathrm{Var}(4 - 3X)\). (2)
| Scheme | Marks |
|---|---|
| \(k + 2k + 3k + 5k + 6k = 1\) \(17k = 1\) | M1 |
| \(k = \dfrac{1}{17} = 0.0588\) | A1 |
| (2) |
Notes
M1 use of \(\sum \mathrm{P}(X = x) = 1\)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = 1 \times \dfrac{1}{17} + 2 \times \dfrac{2}{17} + \ldots + 5 \times \dfrac{6}{17} = \dfrac{64}{17}\) | M1 |
| \(= 3\tfrac{13}{17}\) | A1 |
| (2) |
Notes
M1 use of \(\sum x\mathrm{P}(X = x)\) and at least 2 prob correct
A1 Do not ignore subsequent working
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X^2) = 1^2 \times \dfrac{1}{17} + 2^2 \times \dfrac{2}{17} + \ldots + 5^2 \times \dfrac{6}{17} = \left(\dfrac{266}{17} = 15.6\right)\) | M1 A1 |
| \(\mathrm{Var}(X) = \dfrac{266}{17} - \left(\dfrac{64}{17}\right)^2\) | M1 |
| \(= 1.4740\ldots\) | A1 |
| (4) |
Notes
M1 A1 use of \(\sum x^2\mathrm{P}(X = x)\) and at least 2 prob correct
M1 use of \(\sum x^2\mathrm{P}(X = x) - (\mathrm{E}(X))^2\)
A1 awrt 1.47
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(4 - 3X) = 9\,\mathrm{Var}(X) = 9 \times 1.47 = 13.23 \Rightarrow 13.2\) or \(9 \times 1.4740\ldots = 13.266 \Rightarrow 13.3\) | M1 A1 |
| (2) | |
| (10 marks) |
Notes
M1 \(9\,\mathrm{Var}\,X\)
A1 cao