S2 June 2012 Q5
5. The queueing time, \(X\) minutes, of a customer at a till of a supermarket has probability density function
\[\mathrm{f}(x) = \begin{cases} \dfrac{3}{32}x(k - x) & 0 \leqslant x \leqslant k \\ 0 & \text{otherwise} \end{cases}\]| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^k \frac{3}{32}x(k - x) = 1\) | M1 |
| \(\dfrac{3}{32}\left[\dfrac{kx^2}{2} - \dfrac{x^3}{3}\right]_0^k = 1\) | A1 |
| \(\dfrac{3k^3}{64} - \dfrac{3k^3}{96} = 1\) | M1 dep |
| \(3k^3 - 2k^3 = 64\) | |
| \(k^3 = 64\) | |
| \(k = 4\) | A1cso |
| (4) |
Notes
1st M1 for an attempt to multiply out bracket and for attempting to integrate \(\mathrm{f}(x)\). Both \(x^n \to x^{n+1}\)
1st A1 for correct integration. Ignore limits for these two marks. Need \(\dfrac{3}{32}\left(\dfrac{kx^2}{2} - \dfrac{x^3}{3}\right)\) oe
2nd M1 Dependent on the previous M mark being awarded. For correct use of correct limits and set equal to 1. No need to see 0 substituted in. For verifying they must have \(\dfrac{3}{32}\left(\dfrac{4^3}{2} - \dfrac{4^3}{3}\right)\)
2nd A1 cso or for verifying \(\dfrac{3}{32}\left(\dfrac{4^3}{2} - \dfrac{4^3}{3}\right) = 1\) oe eg \(3(4)^3 - 2(4)^3 = 64\) and a correct comment “so \(k = 4\)”
| Scheme | Marks |
|---|---|
| \([\mathrm{E}(X) =]\ 2\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X^2) = \displaystyle\int_0^4 \frac{3}{32}x^3(4 - x)\) | M1 |
| \(= \left[\dfrac{3x^4}{32} - \dfrac{3x^5}{160}\right]_0^4\) | |
| \(= \left[\dfrac{3 \times 4^4}{32} - \dfrac{3 \times 4^5}{160}\right]\) | |
| \(= 4.8\) | A1 |
| \(\mathrm{Var}(X) = 4.8 - 4\) | M1 |
| \(= 0.8\) | A1 |
| (4) |
Notes
1st M1 attempt to multiply out bracket and attempting \(\displaystyle\int x^2\mathrm{f}(x)\) Limits not needed. Both \(x^n \to x^{n+1}\)
2nd M1 for their \(\mathrm{E}(X^2)\) – (their mean)\(^2\)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_{1.5}^{2.5} \frac{3}{32}x(4 - x) = \left[\frac{3x^2}{16} - \frac{x^3}{32}\right]_{1.5}^{2.5}\) or \(\displaystyle\int_0^{1.5} \frac{3}{32}x(4 - x) = \left[\frac{3x^2}{16} - \frac{x^3}{32}\right]_0^{1.5}\) | M1 |
| \(= \dfrac{47}{128} = 0.3671875\) \(= \dfrac{81}{256} = 0.31640625\) | |
| \(1 - \dfrac{47}{128} = \dfrac{81}{128}\) awrt 0.633 \(2 \times \dfrac{81}{256} = \dfrac{81}{128}\) awrt 0.633 | M1depA1 |
| (3) | |
| (12 marks) |
Notes
1st M1 Multiply out brackets, attempting to integrate (both \(x^n \to x^{n+1}\)), with either limits (their(b) \(\pm\) 0.5) or (their (b) – 0.5 and 0) Accept 2 sf for their limits.
2nd M1dep on gaining 1st M1. 1 – (using limits (their(b) \(\pm\) 0.5)) or 2 \(\times\) (using limits (their(b) – 0.5 and 0)