| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Calculate and compare mean, median, mode |
| Difficulty | Standard +0.3 This is a straightforward continuous probability distribution question requiring standard techniques: sketching a pdf, finding mean by integration, finding mode by differentiation, and calculating a probability using given variance. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(P( | X - \mu | < \sigma) = \frac{4}{27} \int_{1.2}^{2.4} (3x^2 - x^3) dx\) (Limits required.) M1 |
**Question 6(i)**
Above $x$-axis between $(0, 0)$ to $(3, 0)$ **B1**
Correct concavity. (Do not condone parabolas.) **B1** [2]
**Question 6(ii)**
$\mu = \frac{4}{27} \int_0^3 (3x^3 - x^4) dx$ (Limits required.) **M1**
$= \frac{4}{27} \left[ \frac{3x^4}{4} - \frac{x^5}{5} \right]_0^3 = 1.8$ **A1A1**
$f'(x) = \frac{4}{27}(6x - 3x^2) = 0$ **M1A1**
$\Rightarrow x = 0, 2 \quad \therefore \text{Mode} = 2$ **A1** [6]
**Question 6(iii)**
Mean less than mode in **(ii)** matches negative skew in sketch. **B1** [1]
**Question 6(iv)**
$P(|X - \mu| < \sigma) = \frac{4}{27} \int_{1.2}^{2.4} (3x^2 - x^3) dx$ (Limits required.) **M1**
$= \frac{4}{27} \left[ x^3 - \frac{x^4}{4} \right]_{1.2}^{2.4} = 0.64$ **A1A1** [3] **[12]**6 The lengths of time, in years, that sales representatives for a certain company keep their company cars may be modelled by the distribution with probability density function $\mathrm { f } ( x )$, where
$$f ( x ) = \begin{cases} \frac { 4 } { 27 } x ^ { 2 } ( 3 - x ) & 0 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
(i) Draw a sketch of this probability density function.\\
(ii) Calculate the mean and the mode of $X$.\\
(iii) Comment briefly on the values obtained in part (ii) in relation to the sketch in part (i).\\
(iv) Given that $\sigma ^ { 2 } = 0.36$, find $\mathrm { P } ( | X - \mu | < \sigma )$, where $\mu$ and $\sigma ^ { 2 }$ denote the mean and the variance of $X$ respectively.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2012 Q6 [12]}}