| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Calculate probabilities and expectations |
| Difficulty | Easy -1.2 This is a straightforward application of standard uniform distribution formulas with no problem-solving required. Students need only recall and apply the formulas for E(X) = (a+b)/2, Var(X) = (b-a)²/12, and P(c<X<d) = (d-c)/(b-a), plus sketch a simple rectangular PDF. All parts are direct substitution into memorized formulas. |
| Spec | 5.02e Discrete uniform distribution5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Horizontal line at \(f(x) = \frac{1}{6}\) from \(x=-1\) to \(x=5\) | B1 | value \(-1, 5\) shown |
| \(\frac{1}{6}\) marked on \(y\)-axis | B1 | |
| Correct axes with \(-1\) and \(5\) labelled | B1 |
| Answer | Marks |
|---|---|
| \(E(X) = 2\) by symmetry | B1 |
| Answer | Marks |
|---|---|
| \(\text{Var}(X) = \frac{1}{12}(5+1)^2\) or \(\int \frac{x^2}{6}\,dx - 4 = \left[\frac{x^3}{18}\right]_{-1}^{5} - 4\) | M1 |
| \(= 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(-0.3 < X < 3.3) = \frac{3.6}{6}\) or \(\int_{-0.3}^{3.3} \frac{1}{6}\,dx = \left[\frac{x}{6}\right]_{-0.3}^{3.3}\) | M1 | full correct method for the correct area |
| \(= 0.6\) | A1 |
# Question 3:
## Part (a)
| Horizontal line at $f(x) = \frac{1}{6}$ from $x=-1$ to $x=5$ | B1 | value $-1, 5$ shown |
| $\frac{1}{6}$ marked on $y$-axis | B1 | |
| Correct axes with $-1$ and $5$ labelled | B1 | |
## Part (b)
| $E(X) = 2$ by symmetry | B1 | |
## Part (c)
| $\text{Var}(X) = \frac{1}{12}(5+1)^2$ or $\int \frac{x^2}{6}\,dx - 4 = \left[\frac{x^3}{18}\right]_{-1}^{5} - 4$ | M1 | |
| $= 3$ | A1 | |
## Part (d)
| $P(-0.3 < X < 3.3) = \frac{3.6}{6}$ or $\int_{-0.3}^{3.3} \frac{1}{6}\,dx = \left[\frac{x}{6}\right]_{-0.3}^{3.3}$ | M1 | full correct method for the correct area |
| $= 0.6$ | A1 | |
---3. The random variable $X$ is uniformly distributed over the interval $[ - 1,5 ]$.
\begin{enumerate}[label=(\alph*)]
\item Sketch the probability density function $\mathrm { f } ( x )$ of $X$.
Find
\item $\mathrm { E } ( X )$,
\item $\operatorname { Var } ( \mathrm { X } )$,
\item $\mathrm { P } ( - 0.3 < X < 3.3 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2006 Q3 [8]}}