S2 June 2016 Q7
7. The weight, \(X\) kg, of staples in a bin full of paper has probability density function
\[\mathrm{f}(x) = \begin{cases} \dfrac{9x - 3x^2}{10} & 0 \leqslant x \lt 2 \\ 0 & \text{otherwise} \end{cases}\]Use integration to find
Peter raises money by collecting paper and selling it for recycling. A bin full of paper is sold for £50 but if the weight of the staples exceeds 1.5 kg it sells for £25
Peter could remove all the staples before the paper is sold but the time taken to remove the staples means that Peter will have 20% fewer bins full of paper to sell.
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^2 \frac{9x^2}{10} - \frac{3x^3}{10}\,\mathrm{d}x = \left[\frac{3x^3}{10} - \frac{3x^4}{40}\right]_0^2\) | M1A1 |
| \(= \left(\dfrac{3\times2^3}{10} - \dfrac{3\times2^4}{40}\right)\) | M1d |
| \(= 1.2\) | A1 |
| (4) |
Notes
M1: using \(\displaystyle\int x\mathrm{f}(x)\) and attempting to integrate. At least one \(x^n \to x^{n+1}\). Ignore limits
A1: Correct integration - Ignore limits
M1d: substituting correct limits -dependent on previous Method mark being awarded
A1: 1.2 oe. Allow 1.20
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X^2) = \displaystyle\int_0^2 \frac{9x^3}{10} - \frac{3x^4}{10}\,\mathrm{d}x\) \(= \left[\dfrac{9x^4}{40} - \dfrac{3x^5}{50}\right]_0^2\) | M1 |
| \(= \dfrac{42}{25} = 1.68\) | A1 |
| \(\mathrm{Var}(X) = 1.68 - 1.2^2\) | M1d |
| \(= 0.24\) | A1 |
| (4) |
Notes
M1 using \(\displaystyle\int x^2\mathrm{f}(x)\) and attempting to integrate. At least one \(x^n \to x^{n+1}\). Ignore limits
A1: Allow equivalent fractions. May be implied by a correct answer. Condone Var(\(X\)) = 1.68
M1d: use of \(\mathrm{E}(X^2) - \mathrm{E}(X)^2\)
A1: cao allow 0.240 or 6/25oe
| Scheme | Marks |
|---|---|
| \([\mathrm{P}(X \gt 1.5) =]\) \(\displaystyle\int_{1.5}^2 \frac{9x}{10} - \frac{3x^2}{10}\,\mathrm{d}x\) or \(\displaystyle 1 - \int_0^{1.5} \frac{9x}{10} - \frac{3x^2}{10}\,\mathrm{d}x\) | M1 |
| \(= \left[\dfrac{9x^2}{20} - \dfrac{3x^3}{30}\right]_{1.5}^2\) or \(1 - \left[\dfrac{9x^2}{20} - \dfrac{3x^3}{30}\right]_0^{1.5}\) | A1 |
| \(= \dfrac{13}{40} = 0.325\) | A1cso |
| (3) |
Notes
M1: writing or using \(\displaystyle\int_{1.5}^2 \frac{9x}{10} - \frac{3x^2}{10}\,\mathrm{d}x\) or \(\displaystyle 1 - \int_0^{1.5} \frac{9x}{10} - \frac{3x^2}{10}\,\mathrm{d}x\) Must have correct limits or using 1 – F(1.5) for this distribution
A1 Correct Integration. Condone missing 1-
A1cso: 0.325 or 13/40 oe
NB Watch out for using 1 – f(1.5) or \(1 - \dfrac{9(1.5) - 3(1.5)^2}{10}\). This gets M0A0A0
| Scheme | Marks |
|---|---|
| \((0.325) \times 25 + (1 - 0.325) \times 50 =\) £41.875 | M1A1 |
| (2) |
Notes
M1 \((\textit{their}(c)) \times 25 + (1 - \textit{their}(c)) \times 50\) Allow use of their part (c) or 0.325 ie they may restart. Allow 50 – (part(c))×25
A1: awrt 41.9
| Scheme | Marks |
|---|---|
| £\(50 \times 0.8\) or £40 or 0.4 or awrt 0.038 or awrt 0.163 Peter should not remove the staples as the expected amount earned per bin will be less. | M1 A1ft |
| (2) | |
| (15 marks) |
Notes
M1: Allow \((50 \times 0.8)n\) or £\(40n\) \((n \ne 0)\)
NB Allow 20% off (of) 50 = £40
A1ft: Correct statement containing the word staples and one of the 4 comparisons (ft on (c) or (d)) or the difference in these values must be seen.
£\(40n\) < part(d)×\(n\)
or 0.4 < their part (c) or 0.6 < 1-their part(c)
or awrt 0.838 > 0.8 or 0.162 < 0.2