AS June 2023 Q4
4. The random variable \(X\) has a continuous uniform distribution over the interval \([-3, k]\)
Given that \(\mathrm{P}(-4 \lt X \lt 2) = \dfrac{1}{3}\)
A computer generates a random number, \(Y\), where
- \(Y\) has a continuous uniform distribution over the interval \([a, b]\)
- \(\mathrm{E}(Y) = 6\)
- \(\mathrm{Var}(Y) = 192\)
The computer generates 5 random numbers.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(-4 \lt X \lt 2) = \mathrm{P}(-3 \lt X \lt 2)\) | M1 | 1.1b |
| \(\dfrac{2+3}{k+3} = \dfrac{1}{3}\) or \(k = 2 + 2 \times (2 - -3)\) | M1 | 1.1b |
| \(k = 12\) | A1 | 1.1b |
| (3) |
Notes
M1: for realising they need to consider \(\mathrm{P}(-3 \lt X \lt 2)\)
M1: for a correct equation for \(k\) or allow \(\dfrac{2+4}{k+3} = \dfrac{1}{3}\ [\rightarrow k = 15]\) or \(k = 2 + 2 \times (2 - -4)\,[= 14]\)
A1: 12 cao [NB M0M1A0 is possible here, often implied by \(k = 15\)]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{a+b}{2} = 6\) and \(\dfrac{1}{12}(b-a)^2 = 192\) | M1 | 3.1a |
| \(\dfrac{1}{12}\big(a - (12 - a)\big)^2 = 192\) or \(\dfrac{1}{12}\big((12 - b) - b\big)^2 = 192\) (oe) | M1 | 1.1b |
| \(a = -18\) or \(b = 30\) | A1 | 1.1b |
| \(\mathrm{P}(Y \gt 7.5) = \dfrac{\text{“}30\text{”} - 7.5}{\text{“}30\text{”} - (\text{“}-18\text{”})}\) [\(= \dfrac{15}{32}\) or 0.46875 (accept 0.469 or better)] | M1 | 3.4 |
| \(R \sim \mathrm{B}(5, \text{“}0.46875\text{”})\) | M1 | 3.3 |
| \(\mathrm{P}(R \geqslant 2) = 0.7710\ldots\) | A1 | 1.1b |
| (6) | ||
| (9 marks) |
Notes
M1: for translating the problem into 2 correct equations (using 6 and 192)
M1: for a correct method used to eliminate \(a\) or \(b\) (e.g. an equation in \(a\) or \(b\))
also allow for \(a + b = 12\) and \(b - a = 48\)
A1: for a correct single value for \(a\) or \(b\) (\(a = 30\) and \(b = -18\) is A0)
M1: for using their model to find \(\mathrm{P}(Y \gt 7.5) = p\) [\(\mathrm{P}(Y \lt 7.5) = \frac{17}{32}\) is M0 unless it leads to 0.771]
M1: for stating or using the correct model i.e. \(\mathrm{B}(5, p)\) where \(p\) is a probability based on \(Y\) and 7.5
NB use of \(\mathrm{B}(5, \frac{17}{32})\) is OK here and typically scores M0M1A0
A1: for awrt 0.771