| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Calculate and compare mean, median, mode |
| Difficulty | Standard +0.3 This is a standard S2 continuous distribution question requiring routine calculus techniques: differentiation for the mode, integration for mean and CDF, and solving F(m)=0.5 for the median. All parts follow predictable methods with no novel problem-solving required, though the algebra is moderately involved. Slightly easier than average due to its formulaic nature. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f'(x) = \frac{2}{27}(6x - 3x^2) = 0\) when \(x = 2\), so mode \(= 2\) | M1 A1 A1 | |
| (b) \(E(x) = \frac{2}{27}\int_0^3 3x^2 - x^3 \, dx = \frac{2}{27}\left[2x^3 - \frac{x^4}{5}\right]_0^3 = 1.8\) | M1 A1 A1 | |
| (c) \(F(x) = 0\) \((x \leq 0)\), \(F(x) = \frac{2}{27}\left(x^3 - \frac{x^4}{4}\right)\) \((0 \leq x \leq 3)\), | B1 M1 A1 | |
| \(F(x) = 1\) \((x > 3)\) | B1 | |
| (d) For median \(m\), \(F(m) = 0.5\). Now \(F(1.84) = 0.498 < 0.5\) and | M1 A1 | |
| \(F(1.85) = 0.504 > 0.5\), so \(1.84 < m < 1.85\) | A1 A1 | |
| (e) Mean \(<\) median \(<\) mode | B1 | Negative skew |
(a) $f'(x) = \frac{2}{27}(6x - 3x^2) = 0$ when $x = 2$, so mode $= 2$ | M1 A1 A1 |
(b) $E(x) = \frac{2}{27}\int_0^3 3x^2 - x^3 \, dx = \frac{2}{27}\left[2x^3 - \frac{x^4}{5}\right]_0^3 = 1.8$ | M1 A1 A1 |
(c) $F(x) = 0$ $(x \leq 0)$, $F(x) = \frac{2}{27}\left(x^3 - \frac{x^4}{4}\right)$ $(0 \leq x \leq 3)$, | B1 M1 A1 |
$F(x) = 1$ $(x > 3)$ | B1 |
(d) For median $m$, $F(m) = 0.5$. Now $F(1.84) = 0.498 < 0.5$ and | M1 A1 |
$F(1.85) = 0.504 > 0.5$, so $1.84 < m < 1.85$ | A1 A1 |
(e) Mean $<$ median $<$ mode | B1 | Negative skew
**Total: 15 marks**
---A random variable $X$ has a probability density function given by
$$f(x) = \frac{4x^2(3-x)}{27} \quad 0 \leq x \leq 3,$$
$$f(x) = 0 \quad \text{otherwise}.$$
\begin{enumerate}[label=(\alph*)]
\item Find the mode of $X$. [3 marks]
\item Find the mean of $X$. [3 marks]
\item Specify completely the cumulative distribution function of $X$. [4 marks]
\item Deduce that the median, $m$, of $X$ satisfies the equation $m^4 - 4m^3 + 13·5 = 0$, and hence show that $1·84 < m < 1·85$. [4 marks]
\item What do these results suggest about the skewness of the distribution? [1 mark]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q6 [15]}}