| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Expectation of reciprocals and nonlinear functions |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard expectation and variance calculations with a nonlinear transformation. Parts (a) and (b) involve routine application of E(X) and Var(1/X) formulas with simple arithmetic. Part (c) requires basic probability manipulation and conditional probability, but the transformations are direct. The calculations are slightly tedious but require no novel insight—slightly easier than average A-level difficulty. |
| Spec | 2.03d Calculate conditional probability: from first principles5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
4 A discrete random variable $X$ has the probability distribution
$$\mathrm { P } ( X = x ) = \left\{ \begin{array} { c l }
\frac { 3 x } { 40 } & x = 1,2,3,4 \\
\frac { x } { 20 } & x = 5 \\
0 & \text { otherwise }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Calculate $\mathrm { E } ( X )$.
\item Show that:
\begin{enumerate}[label=(\roman*)]
\item $\quad \mathrm { E } \left( \frac { 1 } { X } \right) = \frac { 7 } { 20 }$;\\
(2 marks)
\item $\operatorname { Var } \left( \frac { 1 } { X } \right) = \frac { 7 } { 160 }$.
\end{enumerate}\item The discrete random variable $Y$ is such that $Y = \frac { 40 } { X }$.
Calculate:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( Y < 20 )$;
\item $\mathrm { P } ( X < 4 \mid Y < 20 )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2011 Q4 [14]}}