S2 June 2014 (R) Q7
7. A piece of string \(AB\) has length 9 cm. The string is cut at random at a point \(P\) and the random variable \(X\) represents the length of the piece of string \(AP\).
The two pieces of string \(AP\) and \(PB\) are used to form two sides of a rectangle.
The random variable \(R\) represents the area of the rectangle.
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{U}[0, 9]\) | B1 |
| (1) |
Notes
B1 for \(X \sim \mathrm{U}[0, 9]\) or “continuous uniform”/“rectangular” distribution with correct range
Or allow the pdf \(\mathrm{f}(x) = \begin{cases} \dfrac{1}{9} & 0 \leqslant x \leqslant 9 \\ 0 & \text{otherwise} \end{cases}\)
| Scheme | Marks |
|---|---|
| \(\left[\mathrm{P}(X \gt 6) =\right] \tfrac{1}{3}\) oe allow awrt 0.333 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(R = X(9 - X),\ = 9X - X^2\) | M1, A1 |
| (2) |
Notes
M1 for \(X(9 - X)\) or \(9X - X^2\) may be implied by a correct answer
A1 for \(9X - X^2\) or \(a = -1\) and \(b = 9\)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = 4.5\) | B1 |
| \(\mathrm{Var}(X) = \dfrac{81}{12} = \dfrac{27}{4}\) or \(\mathrm{E}(X^2) = \displaystyle\int_0^9 \dfrac{x^2}{9}\,\mathrm{d}x\) | B1 |
| \(\mathrm{E}(X^2) = \mathrm{Var}(X) + [\mathrm{E}(X)]^2\) or \(= \left[\dfrac{x^3}{27}\right]_0^9\) | M1 |
| \(\mathrm{E}(X^2) = 27\) | A1 |
| So \(\mathrm{E}(R) = 9 \times 4.5 - 27 = 13.5\) | dM1A1 |
| (6) |
Notes
1st B1 for 4.5 or may be implied
2nd B1 for \(\dfrac{81}{12}\) or \(\dfrac{27}{4}\) or \(\displaystyle\int_0^9 \dfrac{x^2}{9}\) ignore limits
1st M1 for full method for \(\mathrm{E}(X^2)\) using their Var(\(X\)) and E(\(X\)) or attempt to integrate \(x^n \to x^{n+1}\) leading to a value for \(\mathrm{E}(X^2)\). Need to be using \(\displaystyle\int_0^9 \dfrac{x^2}{9}\) ignore limits.
1st A1 for \(\mathrm{E}(X^2) = 27\), may be implied.
d2nd M1 for using \(9\mathrm{E}(X) - \mathrm{E}(X^2)\). With their E(\(X\)) and \(\mathrm{E}(X^2)\). This may be implied by a correct answer. Dep on first M
Alternative method
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^9 \dfrac{\left(9x - x^2\right)}{9}\,\mathrm{d}x = \left[\dfrac{9x^2}{18} - \dfrac{x^3}{27}\right]_0^9\) | B1 B1 M1A1 |
| \(= \dfrac{81}{2} - \dfrac{81}{3}\) | dM1 |
| \(= 13.5\) | A1 |
B1 \(\displaystyle\int_0^9 \dfrac{\left(9x - x^2\right)}{9}\,\mathrm{d}x\) ignore limits, ft their (c) whch must be of the form \(aX^2 + b\)
B1 \(\displaystyle\int_0^9 \dfrac{\left(9x - x^2\right)}{9}\,\mathrm{d}x\) with correct limits, ft their (c)
M1 attempt to integrate at least one \(x^n \to x^{n+1}\). Need to be using their \(\displaystyle\int_0^9 \dfrac{\left(9x - x^2\right)}{9}\,\mathrm{d}x\) condone limits missing
A1 Correct Integration
dM1 subst in limits, need to see 9 substituted. Condone missing 0
| Scheme | Marks |
|---|---|
| \(R \gt 2X^2\) or \(9X - X^2 \gt 2X^2\) | M1 |
| \(9X \gt 3X^2\) | A1 |
| So \(\mathrm{P}(X \lt 3)\) | M1 |
| \(= \dfrac{1}{3}\) | A1 |
| (4) | |
| (14 marks) |
Notes
Allow \(\leqslant\) instead of < and \(\geqslant\) instead of > in this part
1st M1 for forming a suitable inequality in \(R\) and \(X\) or just \(X\). May be implied by a correct probability in \(X\).
1st A1 for simplifying to \(9X \gt 3X^2\) or \(3 \gt X\). May be implied by a correct probability in \(X\)
2nd M1 for forming a correct probability in \(X\)
2nd A1 for \(\frac{1}{3}\) or exact equivalent