AS June 2019 Q4
4. The random variable \(X\) has a continuous uniform distribution over the interval \([5, a]\), where \(a\) is a constant.
Given that \(\mathrm{Var}(X) = \dfrac{27}{4}\)
The continuous random variable \(Y\) has probability density function
\[\mathrm{f}(y) = \begin{cases} \dfrac{1}{20}(2y - 3) & 2 \leqslant y \leqslant 6 \\ 0 & \text{otherwise} \end{cases}\]The random variable \(T = 3(X^2 + X) + 2Y\)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{12}(a - 5)^2 = \dfrac{27}{4}\) | M1 | 3.1a |
| \((a - 5)^2 = 81\) | ||
| \(a - 5 = 9\) or \(a - 5 = -9\) | A1 | 1.1b |
| \(\therefore\) since \(a \gt 5\) \(a = 14\)* | A1cso* | 2.2a |
| (3) |
Notes
M1: translating a problem in mathematical contexts into a correct equation. Allow \(\dfrac{a^3 - 125}{3(a - 5)} - \left(\dfrac{a + 5}{2}\right)^2 = \dfrac{27}{4}\)
A1: for \(a - 5 = 9\) or \(a - 5 = -9\) or \(a^2 - 10a - 56 = 0\) or \(a^3 - 15a^2 - 6a + 280 = 0\)
A1cso*: concluding it is 14 giving a reason why – 4 is rejected
| Scheme | Marks | AO |
|---|---|---|
| Correct method for \(\mathrm{E}(Y)\), \(\mathrm{E}(X)\) and \(\mathrm{E}(X^2)\) or \(\mathrm{E}(Y)\) and \(\mathrm{E}(X^2 + X)\) | M1 | 3.1a |
| \(\mathrm{E}(Y) = \displaystyle\int_2^6 \frac{1}{20}y(2y - 3)\,\mathrm{d}y\) | M1 | 1.1b |
| \(= \dfrac{68}{15}\) | A1 | 1.1b |
| \(\mathrm{E}(X) = \dfrac{5 + 14}{2}\) or 9.5 and \(\dfrac{27}{4} = \mathrm{E}(X^2) - 9.5^2\) or \(\displaystyle\int_5^{14} \frac{x^2}{9}\,\mathrm{d}x\) or \(\displaystyle\int_5^{14} \left(\frac{x^2}{9} + \frac{x}{9}\right)\mathrm{d}x\) or \(3\displaystyle\int_5^{14} \left(\frac{x^2}{9} + \frac{x}{9}\right)\mathrm{d}x\) | M1 | 1.1b |
| \(\mathrm{E}(X^2) = 97\) and \(\mathrm{E}(X) = 9.5\) or \(\mathrm{E}(X^2 + X) = 106.5\) or \(3\mathrm{E}(X^2 + X) = 319.5\) | A1 | 1.1b |
| \(\mathrm{E}(T) = 3 \times \text{“}97\text{”} + 3 \times \text{“}9.5\text{”} + 2 \times \dfrac{68}{15}\) oe | M1 | 1.1b |
| \(\mathrm{E}(T) = \dfrac{9857}{30}\) * | A1*cso | 2.1 |
| (7) | ||
| (10 marks) |
Notes
M1: For a complete method to solve the problem
M1: For an attempt at \(\mathrm{E}(Y)\)
A1: \(= \dfrac{68}{15}\) or awrt 4.53
M1: For an attempt at \(\mathrm{E}(X)\) and \(\mathrm{E}(X^2)\) or \(\mathrm{E}(X^2 + X)\) or \(3\mathrm{E}(X^2 + X)\) Some sort of working must be seen for \(\mathrm{E}(X^2)\) eg \(\dfrac{27}{4} = \mathrm{E}(X^2) - \mathrm{E}(X)^2\). Allow \(\mathrm{Var}(X) = \mathrm{E}(X^2) - \mathrm{E}(X)^2\) leading to \(= \mathrm{E}(X^2)\)
A1: 319.5
M1: Method for finding \(\mathrm{E}(T)\) ft their values
A1*cso: Fully correct solution no errors, must have \(\mathrm{E}(T) = \dfrac{9857}{30}\) *
(corrected from the printed mark scheme: the two integrals for \(\mathrm{E}(X^2 + X)\) are printed as \(\displaystyle\int_5^{14} \left(\frac{x^3}{9} + \frac{x^2}{9}\right)\mathrm{d}x\) and \(3\displaystyle\int_5^{14} \left(\frac{x^3}{9} + \frac{x^2}{9}\right)\mathrm{d}x\); the integrand should be \(\dfrac{x^2}{9} + \dfrac{x}{9}\))