| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | One unknown from sum constraint only |
| Difficulty | Easy -1.3 This is a straightforward S2 question requiring only basic probability distribution properties: finding a missing probability from the sum-to-1 constraint, calculating E(X) and Var(X) using standard formulas, and applying linear transformation rules. All steps are routine recall with no problem-solving or insight required. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(\boldsymbol { x }\) | 1 | 2 | 3 | 4 | 5 | 6 | \(\geqslant 7\) |
| \(\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )\) | 0.01 | 0.05 | 0.14 | 0.30 | \(k\) | 0.12 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum p = 1\); \(k = 1-(0.01+0.05+0.14+0.30+0.12) = 0.38\) | B1 | Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = \sum_{\text{all }x} x\,P(X=x) = 4.35\) | B1 | \(\dfrac{87}{20}\), Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(X) = \sum_{\text{all }x} x^2\,P(X=x) - \mu^2\) | M1 | \(E(X^2)\) attempted |
| \(= 20.09 - 18.9225\) | M1 | \(\sum x^2 P(X=x) - \mu^2\) |
| \(= 1.1675\) | A1 | \(\dfrac{467}{400}\) (AWFW \(1.16\)–\(1.17\)), Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(Y) = 5E(X) + 2 = 5 \times 4.35 + 2 = 23.75\) | M1 | Their (b)(i) \(\times 5 + 2\), Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(Y) = 25\,\text{Var}(X) = 29.1875\) | M1 | Their (b)(ii) \(\times 25\) |
| Standard deviation \(= 5.40\) | m1 A1 | \(\sqrt{\phantom{x}}\); \((5.40\)–\(5.41)\), Total: 3 |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum p = 1$; $k = 1-(0.01+0.05+0.14+0.30+0.12) = 0.38$ | B1 | **Total: 1** |
### Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = \sum_{\text{all }x} x\,P(X=x) = 4.35$ | B1 | $\dfrac{87}{20}$, **Total: 1** |
### Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(X) = \sum_{\text{all }x} x^2\,P(X=x) - \mu^2$ | M1 | $E(X^2)$ attempted |
| $= 20.09 - 18.9225$ | M1 | $\sum x^2 P(X=x) - \mu^2$ |
| $= 1.1675$ | A1 | $\dfrac{467}{400}$ (AWFW $1.16$–$1.17$), **Total: 3** |
### Part (c)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(Y) = 5E(X) + 2 = 5 \times 4.35 + 2 = 23.75$ | M1 | Their (b)(i) $\times 5 + 2$, **Total: 1** |
### Part (c)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(Y) = 25\,\text{Var}(X) = 29.1875$ | M1 | Their (b)(ii) $\times 25$ |
| Standard deviation $= 5.40$ | m1 A1 | $\sqrt{\phantom{x}}$; $(5.40$–$5.41)$, **Total: 3** |
---4 The number of fish, $X$, caught by Pearl when she goes fishing can be modelled by the following discrete probability distribution:
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
$\boldsymbol { x }$ & 1 & 2 & 3 & 4 & 5 & 6 & $\geqslant 7$ \\
\hline
$\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )$ & 0.01 & 0.05 & 0.14 & 0.30 & $k$ & 0.12 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
\item Find:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { E } ( X )$;
\item $\operatorname { Var } ( X )$.
\end{enumerate}\item When Pearl sells her fish, she earns a profit, in pounds, given by
$$Y = 5 X + 2$$
Find:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { E } ( Y )$;
\item the standard deviation of $Y$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2007 Q4 [9]}}