| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Finding binomial parameters from properties |
| Difficulty | Moderate -0.8 This is a straightforward binomial distribution question testing standard recall and basic calculations. Part (a) requires stating textbook assumptions, parts (b)(i-iii) involve routine binomial probability calculations using either formula or tables, and part (c) uses the variance formula for binomial distribution to solve a quadratic equation. All techniques are standard AS-level statistics with no problem-solving insight required, making it easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks |
|---|---|
| 14(a) | States one of the following |
| Answer | Marks | Guidance |
|---|---|---|
| calls | 3.5b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| context | 3.5b | B1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 14(b)(i) | Calculates the correct |
| Answer | Marks | Guidance |
|---|---|---|
| AWFW [0.0068, 0.007] | 1.1b | B1 |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 14(b)(ii) | FindsP(X ≤4 )orP(X ≤3 ) |
| Answer | Marks | Guidance |
|---|---|---|
| Allow figures to 2sf | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AWFW [0.107, 0.11] | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 14(b)(iii) | Finds their or |
| Answer | Marks | Guidance |
|---|---|---|
| Allow figures to 2sf | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AWFW [0.0479, 0.048] | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 14(c) | Uses np(1 p) | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| ACF | 3.1a | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 10±√10 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 3 | |
| Question 14 Total | 10 | |
| Q | Marking instructions | AO |
Question 14:
--- 14(a) ---
14(a) | States one of the following
assumptions in context
• probability of complaint call
is constant
• calls or complaint calls occur
independently of each other
or have no effect on each
other
• only two outcomes of
complaint or non-complaint
calls
Condone ‘complaint’ for
‘complaint call’
Do not allow probability being
independent
Do not allow fixed number of
calls | 3.5b | B1 | The probability of getting a
complaint call is fixed
Calls occur independently of each
other
States a second assumption in
context | 3.5b | B1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 14(b)(i) ---
14(b)(i) | Calculates the correct
probability ACF
AWFW [0.0068, 0.007] | 1.1b | B1 | 0.00684
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 14(b)(ii) ---
14(b)(ii) | FindsP(X ≤4 )orP(X ≤3 )
PI by 0.2375 or 0.107
Allow figures to 2sf | 1.1a | M1 | P(X ≤3 )=0.107
Obtains correct probability ACF
AWFW [0.107, 0.11] | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 14(b)(iii) ---
14(b)(iii) | Finds their or
or or
𝑃𝑃(𝑋𝑋 ≤ 9)
𝑃𝑃P(I 𝑋𝑋by≤ 01.905)2 or𝑃𝑃 0(𝑋𝑋.98≥3 1o0r) 0.0172
𝑃𝑃or( 𝑋𝑋co>rre1c0t) answer
Allow figures to 2sf | 1.1a | M1 | 𝑃𝑃(𝑋𝑋 ≥ 10)= 1−𝑃𝑃 (𝑋𝑋 ≤ 9)
= 1−0.952
= 0.048
Obtains correct probability ACF
AWFW [0.0479, 0.048] | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 14(c) ---
14(c) | Uses np(1 p) | 1.1a | M1 | 10p(1 p) = 1.52
10p2− + 10p 2.25 = 0
−p = 0.34 and – 0.66
−
Forms a correct equation in p
using 10 and 1.5
ACF | 3.1a | A1
Obtains values for p
ACF
ISW
Allow AWFW [0.34, 0.342] and
[0.658, 0.66] for p
or
10±√10 | 1.1b | A1
𝑝𝑝 = 20
Subtotal | 3
Question 14 Total | 10
Q | Marking instructions | AO | Marks | Typical solutionA customer service centre records every call they receive.
It is found that 30% of all calls made to this centre are complaints.
A sample of 20 calls is selected.
The number of calls in the sample which are complaints is denoted by the random variable $X$.
\begin{enumerate}[label=(\alph*)]
\item State two assumptions necessary for $X$ to be modelled by a binomial distribution.
[2 marks]
\item Assume that $X$ can be modelled by a binomial distribution.
\begin{enumerate}[label=(\roman*)]
\item Find P($X = 1$)
[1 mark]
\item Find P($X < 4$)
[2 marks]
\item Find P($X \geq 10$)
[2 marks]
\end{enumerate}
\item In a random sample of 10 calls to a school, the number of calls which are complaints, $Y$, may be modelled by a binomial distribution
$$Y \sim \text{B}(10, p)$$
The standard deviation of $Y$ is 1.5
Calculate the possible values of $p$.
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2022 Q14 [10]}}