| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(X) from given distribution |
| Difficulty | Easy -1.2 This is a straightforward application of standard formulas for expectation and variance from a given discrete probability distribution, followed by basic probability calculations with given percentages. Part (a) requires only E(X) = Σxp(x) and Var(X) = E(X²) - [E(X)]² with simple fractions. Part (b) involves multiplying probabilities by 32 and computing a weighted sum. No problem-solving insight or novel techniques required—purely mechanical calculation that any S2 student should handle routinely. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(\boldsymbol { r }\) | 1 | 2 | 3 | 4 |
| \(\mathbf { P } ( \boldsymbol { R } = \boldsymbol { r } )\) | \(\frac { 7 } { 16 }\) | \(\frac { 5 } { 16 }\) | \(\frac { 3 } { 16 }\) | \(\frac { 1 } { 16 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(R) = \sum_{\text{all } r} r \cdot P(R = r) = \left(1 \times \frac{7}{16}\right) + \left(2 \times \frac{5}{16}\right) + \left(3 \times \frac{3}{16}\right) + \left(4 \times \frac{1}{16}\right) = \frac{30}{16} = 1\frac{7}{8}\) | B1 | (1.875) |
| \(E(R^2) = \sum_{\text{all } r} r^2 \cdot P(R = r) = \frac{70}{16} \text{ or } 4\frac{3}{8}\) | B1 | (4.375) |
| \(\text{Var}(R) = 4\frac{3}{8} - \left(1\frac{7}{8}\right)^2 = \frac{220}{256} \text{ or } \frac{55}{64}\) | M1, A1 | (0.859375) |
| Answer | Marks |
|---|---|
| \(32 \times \frac{1}{4} = 8\) | B1 |
| Answer | Marks |
|---|---|
| \(= \left(32 \times \frac{7}{16} \times \frac{1}{5}\right) + \left(32 \times \frac{5}{16} \times \frac{1}{2}\right) + 8 \times \frac{9}{10}\) | M1 |
| \(= 2.8 + 5 + 7.2 = 15\) | A1 |
## 3(a)
$E(R) = \sum_{\text{all } r} r \cdot P(R = r) = \left(1 \times \frac{7}{16}\right) + \left(2 \times \frac{5}{16}\right) + \left(3 \times \frac{3}{16}\right) + \left(4 \times \frac{1}{16}\right) = \frac{30}{16} = 1\frac{7}{8}$ | B1 | (1.875)
$E(R^2) = \sum_{\text{all } r} r^2 \cdot P(R = r) = \frac{70}{16} \text{ or } 4\frac{3}{8}$ | B1 | (4.375)
$\text{Var}(R) = 4\frac{3}{8} - \left(1\frac{7}{8}\right)^2 = \frac{220}{256} \text{ or } \frac{55}{64}$ | M1, A1 | (0.859375)
## 3(b)(i)
$32 \times \frac{1}{4} = 8$ | B1 |
## 3(b)(ii)
$= \left(32 \times \frac{7}{16} \times \frac{1}{5}\right) + \left(32 \times \frac{5}{16} \times \frac{1}{2}\right) + 8 \times \frac{9}{10}$ | M1 |
$= 2.8 + 5 + 7.2 = 15$ | A1 |
---3 Morecrest football team always scores at least one goal but never scores more than four goals in each game. The number of goals, $R$, scored in each game by the team can be modelled by the following probability distribution.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$\boldsymbol { r }$ & 1 & 2 & 3 & 4 \\
\hline
$\mathbf { P } ( \boldsymbol { R } = \boldsymbol { r } )$ & $\frac { 7 } { 16 }$ & $\frac { 5 } { 16 }$ & $\frac { 3 } { 16 }$ & $\frac { 1 } { 16 }$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Calculate exact values for the mean and variance of $R$.
\item Next season the team will play 32 games. They expect to win $90 \%$ of the games in which they score at least three goals, half of the games in which they score exactly two goals and $20 \%$ of the games in which they score exactly one goal.
Find, for next season:
\begin{enumerate}[label=(\roman*)]
\item the number of games in which they expect to score at least three goals;
\item the number of games that they expect to win.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2006 Q3 [8]}}