| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(X) from table |
| Difficulty | Moderate -0.5 This is a straightforward S2 question requiring standard variance calculation from a probability distribution table using the formula Var(R) = E(R²) - [E(R)]². The calculations involve simple fractions and basic arithmetic with no conceptual challenges or problem-solving required beyond routine application of memorized formulas. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(\boldsymbol { r }\) | 1 | 2 | 4 |
| \(\mathbf { P } ( \boldsymbol { R } = \boldsymbol { r } )\) | \(\frac { 1 } { 4 }\) | \(\frac { 1 } { 2 }\) | \(\frac { 1 } { 4 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(R) = \left(1 \times \frac{1}{4}\right) + \left(2 \times \frac{1}{2}\right) + \left(4 \times \frac{1}{4}\right) = 2.25\) | M1A1 | \(2\frac{1}{4}\) |
| \(E(R^2) = \left(1 \times \frac{1}{4}\right) + \left(4 \times \frac{1}{2}\right) + \left(16 \times \frac{1}{4}\right) = 6.25\) | \(6\frac{1}{4}\) | |
| \(\text{Var}(R) = 6.25 - (2.25)^2 = 1.1875\) | M1, A1\(\checkmark\) (4 marks) | \(1\frac{3}{16}\) on their \(E(R)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\) | \(1\) | \(\frac{1}{4}\) |
| \(P(X=x)\) | \(\frac{1}{4}\) | \(\frac{1}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{1}{4} + \frac{1}{8} + \frac{1}{64} = \frac{16+8+1}{64} = \frac{25}{64}\) | M1, A1 (3 marks) | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = \left(R + \frac{8}{R}\right) \times \frac{8}{R} = 8 + \frac{64}{R^2}\) | M1 | Attempt at area |
| \(E(A) = E\!\left(8 + \frac{64}{R^2}\right) = 8 + E\!\left(\frac{64}{R^2}\right) = 8 + 64 \times E(X)\) | M1 | |
| \(= 8 + 64 \times \frac{25}{64} = 33\) | A1 (3 marks) | CAO |
## Question 5:
### Part (a)
$E(R) = \left(1 \times \frac{1}{4}\right) + \left(2 \times \frac{1}{2}\right) + \left(4 \times \frac{1}{4}\right) = 2.25$ | M1A1 | $2\frac{1}{4}$
$E(R^2) = \left(1 \times \frac{1}{4}\right) + \left(4 \times \frac{1}{2}\right) + \left(16 \times \frac{1}{4}\right) = 6.25$ | | $6\frac{1}{4}$
$\text{Var}(R) = 6.25 - (2.25)^2 = 1.1875$ | M1, A1$\checkmark$ (4 marks) | $1\frac{3}{16}$ on their $E(R)$
### Part (b)(i)
| $x$ | $1$ | $\frac{1}{4}$ | $\frac{1}{16}$ |
|------|-----|----------------|-----------------|
| $P(X=x)$ | $\frac{1}{4}$ | $\frac{1}{2}$ | $\frac{1}{4}$ |
B1
$E(X) = \left(1 \times \frac{1}{4}\right) + \left(\frac{1}{4} \times \frac{1}{2}\right) + \left(\frac{1}{16} \times \frac{1}{4}\right)$
$= \frac{1}{4} + \frac{1}{8} + \frac{1}{64} = \frac{16+8+1}{64} = \frac{25}{64}$ | M1, A1 (3 marks) | AG
### Part (b)(ii)
$A = \left(R + \frac{8}{R}\right) \times \frac{8}{R} = 8 + \frac{64}{R^2}$ | M1 | Attempt at area
$E(A) = E\!\left(8 + \frac{64}{R^2}\right) = 8 + E\!\left(\frac{64}{R^2}\right) = 8 + 64 \times E(X)$ | M1
$= 8 + 64 \times \frac{25}{64} = 33$ | A1 (3 marks) | CAO
---5 The discrete random variable $R$ has the following probability distribution.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$\boldsymbol { r }$ & 1 & 2 & 4 \\
\hline
$\mathbf { P } ( \boldsymbol { R } = \boldsymbol { r } )$ & $\frac { 1 } { 4 }$ & $\frac { 1 } { 2 }$ & $\frac { 1 } { 4 }$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Calculate exact values for $\mathrm { E } ( R )$ and $\operatorname { Var } ( R )$.
\item \begin{enumerate}[label=(\roman*)]
\item By tabulating the probability distribution for $X = \frac { 1 } { R ^ { 2 } }$, show that $\mathrm { E } ( X ) = \frac { 25 } { 64 }$.
\item Hence find the value of the mean of the area of a rectangle which has sides of length $\frac { 8 } { R }$ and $\left( R + \frac { 8 } { R } \right)$.\\
(3 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2005 Q5 [10]}}