S2 June 2017 Q3
3. The lifetime, \(X\), in tens of hours, of a battery is modelled by the probability density function
\[\mathrm{f}(x) = \begin{cases} \dfrac{1}{9}x(4 - x) & 1 \leqslant x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}\]Use algebraic integration to find
A radio runs using 2 of these batteries, both of which must be working. Two fully-charged batteries are put into the radio.
Given that the radio is working after 16 hours of use,
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = \dfrac{1}{9}\displaystyle\int_1^4 \left(4x^2 - x^3\right)\mathrm{d}x\) | M1 |
| \(= \dfrac{1}{9}\left[\dfrac{4x^3}{3} - \dfrac{x^4}{4}\right]_1^4\) | A1 |
| \(= \dfrac{1}{9}\left[\dfrac{4\times4^3}{3} - \dfrac{4^4}{4}\right] - \dfrac{1}{9}\left[\dfrac{4}{3} - \dfrac{1}{4}\right]\) | M1d |
| \(= \dfrac{9}{4}\) or 2.25 | A1 |
| (4) |
Notes
1st M1 Using \(\displaystyle\int x\mathrm{f}(x)\,\mathrm{d}x\), multiplying out and at least one of \(x^2 \to x^3\) or \(x^3 \to x^4\) ignore limits
1st A1 correct integration, ignore limits
2nd M1d subst in correct limits (allow 1 sign error)
2nd A1 cao allow equivalent fractions
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \gt 2.5) = \dfrac{1}{9}\displaystyle\int_{2.5}^4 x(4 - x)\,\mathrm{d}x\) | M1 |
| \(= \dfrac{1}{9}\left[2x^2 - \dfrac{x^3}{3}\right]_{2.5}^4\) | A1 |
| \(= \dfrac{3}{8}\) oe or 0.375 | A1 |
| (3) |
Notes
M1 for using \(\dfrac{1}{9}\displaystyle\int_{2.5}^4 x(4 - x)\,\mathrm{d}x\) or \(1 - \dfrac{1}{9}\displaystyle\int_1^{2.5} x(4 - x)\,\mathrm{d}x\) correct limits needed at some point Or \(1 - \left(\dfrac{2}{9}x^2 - \dfrac{1}{27}x^3 - \dfrac{5}{27}\right)\) and attempt to subst 2.5
1st A1 correct integration with correct limits at some point
2nd A1 allow equivalent fractions
| Scheme | Marks |
|---|---|
| P(both batteries working after 25 hours) \(= (0.375)^2\) | M1 |
| \(= 0.140625\) or \(\dfrac{9}{64}\) | A1 |
| (2) |
Notes
M1 (their part(b))2 or writing (P(\(X\) >2.5))2
A1 awrt 0.141
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \gt 1.6) = \dfrac{1}{9}\displaystyle\int_{1.6}^4 x(4 - x)\,\mathrm{d}x\) \(= \dfrac{96}{125}\) or 0.768 | B1 |
| P(works for 25 hours|worked for 16 hours) \(= \dfrac{0.140625}{(0.768)^2}\) | M1 |
| \(= 0.2384\ldots\) | A1 |
| (3) | |
| (12 marks) |
Notes
B1 0.768 or awrt 0.77 or 0.5898…or awrt 0.59. These may be seen in the conditional probability or implied by a correct final answer
M1 \(\dfrac{\text{their part(c)}}{\textit{prob}}\) or \(\dfrac{(\text{their}(b))^2}{\textit{prob}}\) and numerator < denominator
A1 awrt 0.238
NB if use one battery rather than 2 they could get B1 M0 A0