| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Probability distribution table |
| Difficulty | Moderate -0.8 This is a straightforward application of basic probability and expectation. Part (a) requires simple probability calculation (1/5 correct, 4/5 incorrect) and basic expectation formula. Part (b) extends this with adjusted probabilities (1/3, 2/3) and scaling by 24. All steps are routine with no problem-solving insight required, making it easier than average A-level questions. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(\boldsymbol { x }\) | 4 | - 1 |
| \(\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Table: \(x = 4, P(X=x) = \frac{1}{5}\) and \(x = -1, P(X=x) = \frac{4}{5}\) | B1 | 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = \left(4 \times \frac{1}{5}\right) + \left(-1 \times \frac{4}{5}\right) = 0\) | M1 | \((p > 0, \sum p = 1)\) |
| Correct answer \(= 0\) | A1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Table: \(x = 4, P(X=x) = \frac{1}{3}\) and \(x = -1, P(X=x) = \frac{2}{3}\) | B1 | |
| \(E(X) = \left(4 \times \frac{1}{3}\right) + \left(-1 \times \frac{2}{3}\right) = \frac{2}{3}\) | B1 | \((p > 0, \sum p = 1)\) |
| \(E(24X) = 24 \times E(X)\) | ||
| \(= 24 \times \frac{2}{3}\) | M1 | |
| \(= 16\) | A1 | 4 marks total |
## Question 7:
### Part (a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Table: $x = 4, P(X=x) = \frac{1}{5}$ and $x = -1, P(X=x) = \frac{4}{5}$ | B1 | 1 mark total |
### Part (a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = \left(4 \times \frac{1}{5}\right) + \left(-1 \times \frac{4}{5}\right) = 0$ | M1 | $(p > 0, \sum p = 1)$ |
| Correct answer $= 0$ | A1 | 2 marks total |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Table: $x = 4, P(X=x) = \frac{1}{3}$ and $x = -1, P(X=x) = \frac{2}{3}$ | B1 | |
| $E(X) = \left(4 \times \frac{1}{3}\right) + \left(-1 \times \frac{2}{3}\right) = \frac{2}{3}$ | B1 | $(p > 0, \sum p = 1)$ |
| $E(24X) = 24 \times E(X)$ | | |
| $= 24 \times \frac{2}{3}$ | M1 | |
| $= 16$ | A1 | 4 marks total |
---7 On a multiple choice examination paper, each question has five alternative answers given, only one of which is correct. For each question, candidates gain 4 marks for a correct answer but lose 1 mark for an incorrect answer.
\begin{enumerate}[label=(\alph*)]
\item James guesses the answer to each question.
\begin{enumerate}[label=(\roman*)]
\item Copy and complete the following table for the probability distribution of $X$, the number of marks obtained by James for each question.
\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
$\boldsymbol { x }$ & 4 & - 1 \\
\hline
$\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )$ & & \\
\hline
\end{tabular}
\end{center}
\item Hence find $\mathrm { E } ( X )$.
\end{enumerate}\item Karen is able to eliminate two of the incorrect answers from the five alternative answers given for each question before guessing the answer from those remaining.
Given that the examination paper contains 24 questions, calculate Karen's expected total mark.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2007 Q7 [7]}}