| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Measurement error modeling |
| Difficulty | Moderate -0.8 This is a straightforward continuous uniform distribution question requiring only standard formula application: E(T) = (a+b)/2, Var(T) = (b-a)²/12, and basic probability calculation using the uniform pdf. All parts are direct recall with minimal problem-solving, making it easier than average A-level questions. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(T) = \frac{1}{2}(25 + -5) = 10\) | B1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(T) = \frac{1}{12}(25 - -5)^2 = 75\) | B1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(-2 < T < 2) = \frac{2}{15}\) (OE) | B1 | Diagram optional |
| \(P(\text{magnitude at least 2 minutes}) = 1 - P(-2 < T < 2) = 1 - \frac{4}{30}\) | M1 | |
| \(= \frac{13}{15}\) (OE) \(= 0.867\) | A1 | CAO (AWRT) |
| Alternative: \(P(T > 2) = \frac{23}{30}\ (0.766)\) or \(P(T < -2) = \frac{1}{10}\) | B1 | |
| \(P(\text{magnitude at least 2}) = P(T<-2) + P(T>2) = \frac{13}{15}\) | for M1A1 |
## Question 2(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(T) = \frac{1}{2}(25 + -5) = 10$ | B1 | CAO |
## Question 2(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(T) = \frac{1}{12}(25 - -5)^2 = 75$ | B1 | CAO |
## Question 2(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(-2 < T < 2) = \frac{2}{15}$ (OE) | B1 | Diagram optional |
| $P(\text{magnitude at least 2 minutes}) = 1 - P(-2 < T < 2) = 1 - \frac{4}{30}$ | M1 | |
| $= \frac{13}{15}$ (OE) $= 0.867$ | A1 | CAO (AWRT) |
| **Alternative:** $P(T > 2) = \frac{23}{30}\ (0.766)$ or $P(T < -2) = \frac{1}{10}$ | B1 | |
| $P(\text{magnitude at least 2}) = P(T<-2) + P(T>2) = \frac{13}{15}$ | | for M1A1 |
---2 The error, in minutes, made by Paul in estimating the time that he takes to complete a college assignment may be modelled by the random variable $T$ with probability density function
$$f ( t ) = \left\{ \begin{array} { c c }
\frac { 1 } { 30 } & - 5 \leqslant t \leqslant 25 \\
0 & \text { otherwise }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Find:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { E } ( T )$;\\
(1 mark)
\item $\quad \operatorname { Var } ( T )$.
\end{enumerate}\item Calculate the probability that Paul will make an error of magnitude at least 2 minutes when estimating the time that he takes to complete a given assignment.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2010 Q2 [5]}}