S1 January 2005 Q4
4. The random variable \(X\) has probability function
\[\mathrm{P}(X = x) = kx, \qquad x = 1, 2, \ldots, 5.\](a) Show that \(k = \dfrac{1}{15}\). (2)
Find
(b) \(\mathrm{P}(X \lt 4)\), (2)
(c) \(\mathrm{E}(X)\), (2)
(d) \(\mathrm{E}(3X - 4)\). (2)
| Scheme | Marks |
|---|---|
| \(k + 2k + 3k + 4k + 5k = 1\) \(15k = 1\) | M1 |
| \(k = \dfrac{1}{15}\) | A1 |
| (2) |
Notes
M1 \(\sum \mathrm{P}(X = x) = 1\)
A1 cso
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \lt 4) = \mathrm{P}(1) + \mathrm{P}(2) + \mathrm{P}(3) = \dfrac{1}{15} + \dfrac{2}{15} + \dfrac{3}{15}\) | M1 |
| \(= \dfrac{2}{5}\) | A1 |
| (2) |
Notes
M1 sum of 3 probabilities
A1 \(\dfrac{6}{15}\) or \(\dfrac{2}{5}\)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = 1 \times \dfrac{1}{15} + 2 \times \dfrac{2}{15} + 3 \times \dfrac{3}{15} + 4 \times \dfrac{4}{15} + 5 \times \dfrac{5}{15}\) | M1 |
| \(= \dfrac{11}{3}\) | A1 |
| (2) |
Notes
M1 use of \(\sum x\mathrm{P}(X = x)\)
A1 \(\dfrac{55}{15}\) or \(\dfrac{11}{3}\) or \(3\tfrac{2}{3}\) or \(3.\dot{6}\) or 3.67
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(3X - 4) = 3\mathrm{E}(X) - 4 = 11 - 4\) | M1 |
| \(= 7\) | A1 |
| (2) | |
| (8 marks) |
Notes
M1 3 × theirs − 4
Alternative
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(3X - 4) = -1 \times \dfrac{1}{15} + 2 \times \dfrac{2}{15} + 5 \times \dfrac{3}{15} + 8 \times \dfrac{4}{15} + 11 \times \dfrac{5}{15}\) | M1 |
| \(= 7\) | A1 |
M1 \(\sum (3x - 4)kx\); A1 cao