| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Probability |
| Type | Deriving CDF from area proportionality |
| Difficulty | Standard +0.3 This is a straightforward S2 continuous distributions question requiring standard techniques: deriving a geometric CDF, differentiating to find PDF, computing E(R) by integration, and interpreting a decreasing PDF. The geometric reasoning in part (a) is intuitive (area of circle), and all calculus is routine. Slightly easier than average due to the guided structure and standard methods. |
| 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) Must land on board, so \(F(r) = 0\) (\(r < 0\)), \(F(r) = 1\) (\(r > a\)) | B1 | |
| By definition, \(F(r) = P(X < r) = \frac{\pi r^2}{\pi a^2} = \frac{r^2}{a^2}\) \((0 \leq r \leq a)\) | M1 A1 A1 | |
| (b) \(f(r) = F'(r) = \frac{2r}{a^2}\) \((0 \leq r \leq a)\); \(f(r) = 0\) otherwise | M1 A1 M1 B1 | |
| \(E(R) = \int_0^a \frac{2r^2}{a^2} \, dr = \frac{2}{a}\left[\frac{r^3}{3}\right]_0^a = \frac{2a}{3}\) | M1 A1 A1 | |
| (c) \(f(x) = F'(x) = \frac{2}{a} - \frac{2x}{a^2}\) \((0 \leq x \leq a)\); \(f(x) = 0\) otherwise | M1 A1 B1 | |
| \(f(x)\) decreases from \(x = 0\) to \(x = a\), so more likely to land near \(O\) | M1 A1 | Total: 16 |
(a) Must land on board, so $F(r) = 0$ ($r < 0$), $F(r) = 1$ ($r > a$) | B1 |
By definition, $F(r) = P(X < r) = \frac{\pi r^2}{\pi a^2} = \frac{r^2}{a^2}$ $(0 \leq r \leq a)$ | M1 A1 A1 |
(b) $f(r) = F'(r) = \frac{2r}{a^2}$ $(0 \leq r \leq a)$; $f(r) = 0$ otherwise | M1 A1 M1 B1 |
$E(R) = \int_0^a \frac{2r^2}{a^2} \, dr = \frac{2}{a}\left[\frac{r^3}{3}\right]_0^a = \frac{2a}{3}$ | M1 A1 A1 |
(c) $f(x) = F'(x) = \frac{2}{a} - \frac{2x}{a^2}$ $(0 \leq x \leq a)$; $f(x) = 0$ otherwise | M1 A1 B1 |
$f(x)$ decreases from $x = 0$ to $x = a$, so more likely to land near $O$ | M1 A1 | **Total: 16**
---Two people are playing darts. Peg hits points randomly on the circular board, whose radius is $a$. If the distance from the centre $O$ of the point that she hits is modelled by the variable $R$,
\begin{enumerate}[label=(\alph*)]
\item explain why the cumulative distribution function $F(r)$ is given by
$$F(r) = 0 \quad r < 0,$$
$$F(r) = \frac{r^2}{a^2} \quad 0 \leq r \leq a,$$
$$F(r) = 1 \quad r > a.$$ [4 marks]
\item By first finding the probability density function of $R$, show that the mean distance from $O$ of the points that Peg hits is $\frac{2a}{3}$. [7 marks]
Bob, a more experienced player, aims for $O$, and his points have a distance $X$ from $O$ whose cumulative distribution function is
$$F(x) = 0, \quad x < 0; \quad F(x) = \frac{x}{a}\left(2 - \frac{x}{a}\right), \quad 0 \leq x \leq a; \quad F(x) = 1, \quad x > a.$$
\item Find the probability density function of $X$, and explain why it shows that Bob is aiming for $O$. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q6 [16]}}