| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Probability of range of values |
| Difficulty | Moderate -0.8 This is a straightforward binomial probability question requiring direct application of the formula for part (a)(i) and summing probabilities for part (a)(ii). Part (b) tests basic understanding of binomial assumptions. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial since it requires calculator work with multiple probability calculations. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| 16(a)(i) | Finds P(X = 8) correctly | |
| AWRT 0.207 | 3.4 | B1 |
| Subtotal | 1 |
| Answer | Marks |
|---|---|
| 16(a)(ii) | Finds P(X ≤ 11) – P(X ≤ 6). |
| Answer | Marks | Guidance |
|---|---|---|
| added, at most one error | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| (AWRT 0.81) | 1.1b | A1 |
| Subtotal | 2 |
| Answer | Marks |
|---|---|
| 16(b) | States one clear assumption in |
| Answer | Marks | Guidance |
|---|---|---|
| fail or the probability is 0.6. | 3.5b | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Question Total | 4 | |
| Q | Marking Instructions | AO |
Question 16:
--- 16(a)(i) ---
16(a)(i) | Finds P(X = 8) correctly
AWRT 0.207 | 3.4 | B1 | 0.207
Subtotal | 1
--- 16(a)(ii) ---
16(a)(ii) | Finds P(X ≤ 11) – P(X ≤ 6).
Must be two terms subtracted with
at least one term correct
Or
Terms for P(X = 7,8,9,10,11)
added, at most one error | 1.1a | M1 | P(X ≤ 11) – P(X ≤ 6)
= 0.96020841 – 0.1501401
= 0.810
Obtains correct probability
(AWRT 0.81) | 1.1b | A1
Subtotal | 2
--- 16(b) ---
16(b) | States one clear assumption in
context linked to independence or
the constant probability of success.
Do not accept a fixed number of
trials or she either fails or does not
fail or the probability is 0.6. | 3.5b | E1 | Being able to solve one puzzle is
independent of being able to solve
any other puzzle
Subtotal | 1
Question Total | 4
Q | Marking Instructions | AO | Marks | Typical SolutionA mathematical puzzle is published every day in a newspaper.
Over a long period of time Paula is able to solve the puzzle correctly 60% of the time.
\begin{enumerate}[label=(\alph*)]
\item For a randomly chosen 14-day period find the probability that:
\begin{enumerate}[label=(\roman*)]
\item Paula correctly solves exactly 8 puzzles
[1 mark]
\item Paula correctly solves at least 7 but not more than 11 puzzles.
[2 marks]
\end{enumerate}
\item State one assumption that is necessary for the distribution used in part (a) to be valid.
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2020 Q16 [4]}}