| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | PDF to CDF derivation |
| Difficulty | Standard +0.3 This is a standard S2 question on piecewise probability density functions requiring routine techniques: sketching a triangular distribution, identifying the mode by inspection, integrating to find the CDF in three regions, and solving F(x)=0.5 for the median. While multi-part with 18 marks total, each component follows textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks |
|---|---|
| (a) Triangular graph with vertices at \((0, 0)\), \((2, \frac{1}{3})\), \((6, 0)\) | B4 |
| (b) \(2\) | A1 |
| (c) 0 to 2: \(F(t) = \int_0^t \frac{1}{6}x \, dx = [\frac{1}{12}x^2]_0^t = \frac{1}{12}t^2\); 2 to 6: \(F(t) = \frac{1}{3} + \frac{1}{12} \times 2 \times \frac{1}{3} + \int_2^t \frac{1}{6}(6-x) \, dx = \frac{1}{3} + \frac{1}{12}[6x - \frac{1}{2}x^2]_2^t = \frac{1}{3} + \frac{1}{12}[(6t - \frac{1}{2}t^2) - (12 - 2)] = \frac{1}{24}(12t - t^2 - 12)\) | M1 M1 A1 M1 M1 A1 |
| \[F(x) = \begin{cases} 0, & x < 0, \\ \frac{1}{12}x^2 & 0 \leq x \leq 2, \\ \frac{1}{24}(12x - x^2 - 12) & 2 \leq x \leq 6, \\ 1, & x > 6. \end{cases}\] | A1 |
| (d) \(\frac{1}{24}(12m - m^2 - 12) = \frac{1}{2}\); \(m^2 - 12m + 24 = 0\); \(m = 2.536\) or \(9.464\); \(2 \leq m \leq 6 \therefore\) median \(= 2.536\) (4sf) | M1 A1 M1 A1 |
(a) Triangular graph with vertices at $(0, 0)$, $(2, \frac{1}{3})$, $(6, 0)$ | B4 |
(b) $2$ | A1 |
(c) 0 to 2: $F(t) = \int_0^t \frac{1}{6}x \, dx = [\frac{1}{12}x^2]_0^t = \frac{1}{12}t^2$; 2 to 6: $F(t) = \frac{1}{3} + \frac{1}{12} \times 2 \times \frac{1}{3} + \int_2^t \frac{1}{6}(6-x) \, dx = \frac{1}{3} + \frac{1}{12}[6x - \frac{1}{2}x^2]_2^t = \frac{1}{3} + \frac{1}{12}[(6t - \frac{1}{2}t^2) - (12 - 2)] = \frac{1}{24}(12t - t^2 - 12)$ | M1 M1 A1 M1 M1 A1 |
$$F(x) = \begin{cases} 0, & x < 0, \\ \frac{1}{12}x^2 & 0 \leq x \leq 2, \\ \frac{1}{24}(12x - x^2 - 12) & 2 \leq x \leq 6, \\ 1, & x > 6. \end{cases}$$ | A1 |
(d) $\frac{1}{24}(12m - m^2 - 12) = \frac{1}{2}$; $m^2 - 12m + 24 = 0$; $m = 2.536$ or $9.464$; $2 \leq m \leq 6 \therefore$ median $= 2.536$ (4sf) | M1 A1 M1 A1 |The continuous random variable $X$ has the following probability density function:
$$f(x) = \begin{cases}
\frac{1}{8}x, & 0 \leq x \leq 2, \\
\frac{1}{12}(6-x), & 2 \leq x \leq 6, \\
0, & \text{otherwise}.
\end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Sketch $f(x)$ for all values of $x$. [4 marks]
\item State the mode of $X$. [1 mark]
\item Define fully the cumulative distribution function $F(x)$ of $X$. [9 marks]
\item Show that the median of $X$ is 2.536, correct to 4 significant figures. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q6 [18]}}