AS June 2022 Q5
5. The random variable \(X\) has the continuous uniform distribution over the interval \([0.5, 2.5]\)
Talia selects a number, \(T\), at random from the distribution of \(X\)
Malik takes Talia’s number, \(T\), and calculates his number, \(M\), where \(M = \dfrac{1}{T^2}\)
Raja and Greta play a game many times.
Each time they play they use a number, \(R\), randomly selected from the distribution of \(X\)
Raja’s score is \(R\)
Greta’s score is \(G\), where \(G = \dfrac{2}{R^2}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\left[\mathrm{P}(T \lt 1) = \dfrac{1 - 0.5}{2.5 - 0.5}\right] = \dfrac{1}{4}\) | B1 | 3.4 |
| (1) |
Notes
B1: 0.25 oe
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}\left(\{T \lt 2.25\} \cap \left\{\dfrac{1}{T^2} \lt 2.25\right\}\right) = \mathrm{P}\left(\{T \lt 2.25\} \cap \left\{T^2 \gt \tfrac{4}{9}\right\}\right)\) | M1 | 2.1 |
| \(\mathrm{P}\left(\tfrac{2}{3} \lt T \lt 2.25\right) = \dfrac{2.25 - \frac{2}{3}}{2.5 - 0.5}\) | M1 | 1.1b |
| \(= \dfrac{19}{24}\) | A1 | 1.1b |
| (3) |
Notes
M1: Determining the conditions for both numbers to be smaller than 2.25
M1: Use of uniform distribution for their region for \(T\)
A1: allow awrt 0.792
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{E}(R) = 1.5\) | B1 | 1.1b |
| \(\mathrm{E}\left(\dfrac{2}{R^2}\right) = \displaystyle\int_{0.5}^{2.5} \left(\frac{1}{2.5 - 0.5}\right)\frac{2}{r^2}\,\mathrm{d}r\) | M1 | 3.1b |
| \(\left[-\dfrac{1}{r}\right]_{0.5}^{2.5}\) | dM1 | 1.1b |
| \(= 1.6\) | A1 | 1.1b |
| Greta is the expected winner since she has the higher expected value \((1.6 \gt 1.5)\) | A1 | 2.2b |
| (5) | ||
| (9 marks) |
Notes
B1: 1.5
M1: Attempt to set up an integral for Greta’s expectation
dM1: (dep on previous M1) for integration of expectation
A1: 1.6
A1: Greta with correct supporting reason and all previous marks scored in (c)
SC: Use of \(R = \dfrac{2}{R^2} \rightarrow R = \sqrt[3]{2} \rightarrow 1.5 \gt \sqrt[3]{2}\ (= 1.25\ldots)\) therefore Raja is more likely to win a single game, scores B1M0M0A0A1.