| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 21 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Piecewise PDF with multiple regions |
| Difficulty | Standard +0.3 This is a standard S2 continuous distribution question requiring routine integration techniques across piecewise intervals. While it involves multiple parts (sketch, mean, variance, CDF) and careful handling of the piecewise function, all methods are textbook procedures with no novel problem-solving required. The 21 total marks reflect length rather than conceptual difficulty, making it slightly easier than the average A-level question. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Graph sketched: straight lines joining \((0, 0)\), \((1, \frac{2}{3})\) and \((3, 0)\) | B3 | |
| (b) \(E(X) = \int_0^1 \frac{2}{3}x \, dx + \int_1^3 x - \frac{1}{3}x^2 \, dx = \left[\frac{2x^2}{3}\right]_0^1 + \left[\frac{x^2}{2} - \frac{x^3}{9}\right]_1^3\) | M1 A1 M1 A1 | |
| \(= \frac{2}{3} + \frac{9}{2} - 3 - \frac{1}{2} + \frac{1}{9} = 1\frac{1}{3}\) | A1 | |
| (c) \(E(X^2) = \int_0^1 \frac{2}{3}x^2 \, dx + \int_1^3 x^2 - \frac{1}{3}x^3 \, dx = \left[\frac{x^4}{6}\right]_0^1 + \left[\frac{x^3}{3} - \frac{x^4}{12}\right]_1^3\) | M1 A1 M1 A1 | |
| \(= \frac{1}{6} + 9 - \frac{81}{12} - \frac{1}{3} + \frac{1}{12} = 2\frac{1}{8}\) s.d. \(= \sqrt{0.389} = 0.624\) | A1 M1 A1 | |
| (d) \(F(x) = \int_0^x \frac{2}{3}u \, du = \frac{2x^2}{3}\) \((0 \leq x < 1)\) | M1 | |
| \(F(x) = \frac{1}{3} + \int_1^x 1 - \frac{1}{3}u \, du = \left[u - \frac{1}{6}u^2\right]_1^x + \frac{1}{3} = x - \frac{1}{6}x^2 - \frac{1}{2}\) \((1 \leq x \leq 3)\) | M1 A1 M1 A1 | Total: 21 marks |
(a) Graph sketched: straight lines joining $(0, 0)$, $(1, \frac{2}{3})$ and $(3, 0)$ | B3 |
(b) $E(X) = \int_0^1 \frac{2}{3}x \, dx + \int_1^3 x - \frac{1}{3}x^2 \, dx = \left[\frac{2x^2}{3}\right]_0^1 + \left[\frac{x^2}{2} - \frac{x^3}{9}\right]_1^3$ | M1 A1 M1 A1 |
$= \frac{2}{3} + \frac{9}{2} - 3 - \frac{1}{2} + \frac{1}{9} = 1\frac{1}{3}$ | A1 |
(c) $E(X^2) = \int_0^1 \frac{2}{3}x^2 \, dx + \int_1^3 x^2 - \frac{1}{3}x^3 \, dx = \left[\frac{x^4}{6}\right]_0^1 + \left[\frac{x^3}{3} - \frac{x^4}{12}\right]_1^3$ | M1 A1 M1 A1 |
$= \frac{1}{6} + 9 - \frac{81}{12} - \frac{1}{3} + \frac{1}{12} = 2\frac{1}{8}$ s.d. $= \sqrt{0.389} = 0.624$ | A1 M1 A1 |
(d) $F(x) = \int_0^x \frac{2}{3}u \, du = \frac{2x^2}{3}$ $(0 \leq x < 1)$ | M1 |
$F(x) = \frac{1}{3} + \int_1^x 1 - \frac{1}{3}u \, du = \left[u - \frac{1}{6}u^2\right]_1^x + \frac{1}{3} = x - \frac{1}{6}x^2 - \frac{1}{2}$ $(1 \leq x \leq 3)$ | M1 A1 M1 A1 | **Total: 21 marks**A continuous random variable $X$ has probability density function f(x) given by
$$\text{f(x)} = \frac{2x}{3} \quad 0 \leq x < 1,$$
$$\text{f(x)} = 1 - \frac{x}{3} \quad 1 \leq x \leq 3,$$
$$\text{f(x)} = 0 \quad \text{otherwise}.$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of f(x) for all $x$. [3 marks]
\item Find the mean of $X$. [5 marks]
\item Find the standard deviation of $X$. [7 marks]
\item Show that the cumulative distribution function of $X$ is given by
$$\text{F(x)} = \frac{x^2}{3} \quad 0 \leq x < 1,$$
and find F(x) for $1 \leq x \leq 3$. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q7 [21]}}