AS June 2023 Q1
1. The discrete random variable \(X\) has the following distribution
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | \(r\) | \(k\) | \(\dfrac{k}{2}\) | \(\dfrac{k}{3}\) | \(\dfrac{k}{4}\) |
where \(r\) and \(k\) are positive constants.
The standard deviation of \(X\) equals the mean of \(X\)
Find the exact value of \(r\) (6)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{E}(X) = 0 \times r + k + 2 \times \dfrac{k}{2} + 3 \times \dfrac{k}{3} + 4 \times \dfrac{k}{4}\) or \(4k\) | B1 | 2.1 |
| \(\mathrm{E}(X^2) = k + 2^2 \times \dfrac{k}{2} + 3^2 \times \dfrac{k}{3} + 4^2 \times \dfrac{k}{4}\) or \(k + 2k + 3k + 4k\) or \(10k\) | M1 | 1.1b |
| \(\sqrt{\mathrm{Var}(X)} = \mathrm{E}(X) \Rightarrow \mathrm{E}(X^2) = 2[\mathrm{E}(X)]^2\) or \(10k - (4k)^2 = (4k)^2\) | M1 | 1.1b |
| \(10k = 32k^2 \Rightarrow k = \dfrac{5}{16}\) | A1 | 1.1b |
| \(\left[\sum \text{probs} = 1 \Rightarrow\right]\ r + \dfrac{1}{12}(12k + 6k + 4k + 3k) = 1\) \(\left[\Rightarrow r + \dfrac{25}{12}k = 1\right]\) | M1 | 3.1a |
| \(r = 1 - \dfrac{25}{12} \times \dfrac{5}{16} = \dfrac{67}{192}\) | A1 | 1.1b |
| (6) |
Notes
B1 for a correct expression for \(\mathrm{E}(X)\)
1st M1 for attempting \(\mathrm{E}(X^2)\) – at least 3 correct non-zero terms
2nd M1 Use of \(\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2\) to form an expression in \(k\) with \(\sqrt{\mathrm{Var}(X)} = \mathrm{E}(X)\)
ft their \(10k\) and their \(4k\)
Must use a consistent expression for \(\mathrm{E}(X)\) in both side of their equation
1st A1 for a correct equation for \(k\) (e.g. 2TQ) or \(k = \dfrac{5}{16}\) or 0.3125.
They may ignore or reject solution of \(k = 0\)
Allow for \(r = 0.349\) or better
3rd M1 for attempt at an equation in \(r\) and \(k\) using sum of probs.
At least 4 terms correct in terms of \(k\) or numerically using their value of \(k\)
2nd A1 for \(\dfrac{67}{192}\) or exact equivalents e.g. \(0.348958\dot{3}\)
Correct exact answer implies full marks