| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | State or write down basic properties |
| Difficulty | Easy -1.2 This is a straightforward S2 question testing basic knowledge of continuous uniform distributions. Parts (a) and (b) require only recall of the standard uniform PDF formula and its rectangular graph. Parts (c) and (d) involve routine calculations: E(X²) uses the standard integral formula, and the probability calculation is a simple rectangle area. No problem-solving insight or multi-step reasoning required—purely procedural application of memorized formulas. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = \begin{cases} \frac{1}{9} & -2 \leq x \leq 7 \\ 0 & \text{otherwise} \end{cases}\) | B1, B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Correct rectangular distribution curve with height \(\frac{1}{9}\) and width from \(-2\) to \(7\) | B1, B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 2.5\), \(\text{Var}(X) = \frac{1}{12}(7+2)^2 = 6.75\) with both | B1 | |
| \(E(X^2) = \text{Var}(X) + E(X)^2 = 6.75 + 2.5^2 = 13\) | M1, A1 | (3 marks) |
| Alternative: \(\int_{-2}^{7} x^2 f(x)dx = \left[\frac{x^3}{27}\right]_{-2}^{7} = 13\) with attempt to integrate and use limits of -2 and 7 | B1, M1, A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(-0.2 < X < 0.6) = \frac{1}{9} \times 0.8 = \frac{4}{45}\) or 0.0889 or equiv; awrt 0.089 | M1, A1 | (2 marks) |
## Part (a)
$f(x) = \begin{cases} \frac{1}{9} & -2 \leq x \leq 7 \\ 0 & \text{otherwise} \end{cases}$ | B1, B1 | (2 marks)
## Part (b)
Correct rectangular distribution curve with height $\frac{1}{9}$ and width from $-2$ to $7$ | B1, B1 | (2 marks)
## Part (c)
$E(X) = 2.5$, $\text{Var}(X) = \frac{1}{12}(7+2)^2 = 6.75$ with both | B1 |
$E(X^2) = \text{Var}(X) + E(X)^2 = 6.75 + 2.5^2 = 13$ | M1, A1 | (3 marks)
**Alternative:** $\int_{-2}^{7} x^2 f(x)dx = \left[\frac{x^3}{27}\right]_{-2}^{7} = 13$ with attempt to integrate and use limits of -2 and 7 | B1, M1, A1 | (3 marks)
## Part (d)
$P(-0.2 < X < 0.6) = \frac{1}{9} \times 0.8 = \frac{4}{45}$ or 0.0889 or equiv; awrt 0.089 | M1, A1 | (2 marks)
---The continuous random variable $X$ is uniformly distributed over the interval $[-2, 7]$.
\begin{enumerate}[label=(\alph*)]
\item Write down fully the probability density function f(x) of $X$. [2]
\item Sketch the probability density function f(x) of $X$. [2]
\end{enumerate}
Find
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item E($X^2$), [3]
\item P($-0.2 < X < 0.6$). [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2009 Q2 [9]}}