| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Find constant k in PDF |
| Difficulty | Easy -1.2 This is a straightforward textbook exercise on continuous uniform distributions requiring only standard formulas and basic integration. Part (a) involves proving well-known results using simple integration, while part (b) applies these formulas to a numerical example with routine calculations. No problem-solving insight or novel techniques are needed. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03d E(g(X)): general expectation formula |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(= k(b-a) = 1\) ⇒ \(k = \frac{1}{b-a}\) | E1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \[E(X) = \int_a^b kx\,dx\] | M1 | |
| \[= \left[\frac{kx^2}{2}\right]_a^b\] | A1 | |
| \[= \frac{1}{2}k(b^2-a^2)\] | ||
| \[= \frac{1}{2} \times \frac{1}{b-a} \times (b-a)(a+b)\] | M1A1 | |
| \[= \frac{1}{2}(a+b)\] | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu = 1\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \[\sigma^2 = \text{Var}(X) = \frac{1}{12}(b-a)^2 = \frac{1}{12} \times 6^2 = 3\] | M1 | |
| \[\therefore \sigma = \sqrt{3}\] | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \[P\left(X < \frac{2-\mu}{\sigma}\right) = P\left(X < \frac{1}{\sqrt{3}}\right)\] | M1√ | |
| \[= \frac{1}{6} \times 2.577\] | M1√ | |
| \[= 0.430\] | A1 | 3 marks |
**4(a)(i)**
Area $= k(b-a) = 1$ ⇒ $k = \frac{1}{b-a}$ | E1 | 1 mark | AG
**4(a)(ii)**
$$E(X) = \int_a^b kx\,dx$$ | M1 | |
$$= \left[\frac{kx^2}{2}\right]_a^b$$ | A1 | |
$$= \frac{1}{2}k(b^2-a^2)$$ | |
$$= \frac{1}{2} \times \frac{1}{b-a} \times (b-a)(a+b)$$ | M1A1 | | (factors shown)
$$= \frac{1}{2}(a+b)$$ | | 4 marks | AG
**4(b)(i)**
$\mu = 1$ | B1 | 1 mark
**4(b)(ii)**
$$\sigma^2 = \text{Var}(X) = \frac{1}{12}(b-a)^2 = \frac{1}{12} \times 6^2 = 3$$ | M1 | |
$$\therefore \sigma = \sqrt{3}$$ | A1 | 2 marks | 1.7321
**4(b)(iii)**
$$P\left(X < \frac{2-\mu}{\sigma}\right) = P\left(X < \frac{1}{\sqrt{3}}\right)$$ | M1√ | | (on their μ and σ)
$$= \frac{1}{6} \times 2.577$$ | M1√ | |
$$= 0.430$$ | A1 | 3 marks | cao
**Question 4 Total: 11 marks**
---4
\begin{enumerate}[label=(\alph*)]
\item A random variable $X$ has probability density function defined by
$$\mathrm { f } ( x ) = \begin{cases} k & a < x < b \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\roman*)]
\item Show that $k = \frac { 1 } { b - a }$.
\item Prove, using integration, that $\mathrm { E } ( X ) = \frac { 1 } { 2 } ( a + b )$.
\end{enumerate}\item The error, $X$ grams, made when a shopkeeper weighs out loose sweets can be modelled by a rectangular distribution with the following probability density function:
$$f ( x ) = \begin{cases} k & - 2 < x < 4 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\roman*)]
\item Write down the value of the mean, $\mu$, of $X$.
\item Evaluate the standard deviation, $\sigma$, of $X$.
\item Hence find $\mathrm { P } \left( X < \frac { 2 - \mu } { \sigma } \right)$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2006 Q4 [11]}}