| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Geometric applications |
| Difficulty | Moderate -0.8 This question tests basic understanding of continuous uniform (rectangular) distributions with straightforward probability calculations. Parts (a) and (b) are direct applications of standard formulas for uniform distributions. Parts (c) and (d) require visualizing rectangular regions and calculating areas, but involve only simple arithmetic with no conceptual challenges beyond recognizing independence of X and Y. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks |
|---|---|
| (a) \(1.6 \times \frac{1}{20} = 0.08\) | M1 A1 |
| (b) mean \(= 10\); variance \(= \frac{1}{12}(20 - 0)^2 = \frac{100}{3}\) | A1 M1 A1 |
| (c) \(= P(X \text{ in middle 4 cm}) \times P(Y \text{ in middle 4 cm}) = (4 \times \frac{1}{20}) \times (4 \times \frac{1}{16}) = \frac{1}{5} \times \frac{1}{4} = \frac{1}{20}\) | M1 M1 A1 |
| (d) \(= 1 - [P(X \text{ in middle 16 cm}) \times P(Y \text{ in middle 12 cm})] = 1 - [(16 \times \frac{1}{20}) \times (12 \times \frac{1}{16})] = 1 - (\frac{4}{5} \times \frac{3}{4}) = 1 - \frac{3}{5} = \frac{2}{5}\) | M1 M1 A1 |
(a) $1.6 \times \frac{1}{20} = 0.08$ | M1 A1 |
(b) mean $= 10$; variance $= \frac{1}{12}(20 - 0)^2 = \frac{100}{3}$ | A1 M1 A1 |
(c) $= P(X \text{ in middle 4 cm}) \times P(Y \text{ in middle 4 cm}) = (4 \times \frac{1}{20}) \times (4 \times \frac{1}{16}) = \frac{1}{5} \times \frac{1}{4} = \frac{1}{20}$ | M1 M1 A1 |
(d) $= 1 - [P(X \text{ in middle 16 cm}) \times P(Y \text{ in middle 12 cm})] = 1 - [(16 \times \frac{1}{20}) \times (12 \times \frac{1}{16})] = 1 - (\frac{4}{5} \times \frac{3}{4}) = 1 - \frac{3}{5} = \frac{2}{5}$ | M1 M1 A1 |
---In a test studying reaction times, white dots appear at random on a black rectangular screen. The continuous random variable $X$ represents the distance, in centimetres, of the dot from the left-hand edge of the screen. The distribution of $X$ is rectangular over the interval $[0, 20]$.
\begin{enumerate}[label=(\alph*)]
\item Find $P(2 < X < 3.6)$. [2 marks]
\item Find the mean and variance of $X$. [3 marks]
\end{enumerate}
The continuous random variable $Y$ represents the distance, in centimetres, of the dot from the bottom edge of the screen. The distribution of $Y$ is rectangular over the interval $[0, 16]$.
Find the probability that a dot appears
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item in a square of side 4 cm at the centre of the screen, [4 marks]
\item within 2 cm of the edge of the screen. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q3 [13]}}