| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Estimate from summary statistics |
| Difficulty | Moderate -0.8 This is a straightforward statistics question testing basic calculations (mean, standard deviation) and routine normal distribution applications. Part (a) involves simple formula substitution, part (b) requires standard justification for normal approximation, and part (c) is a direct normal probability calculation. The inverse normal calculation in part (b) is slightly more involved but still a standard technique. All components are textbook exercises with no novel problem-solving required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| 13(a)(i) | Finds mean | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| 100 | AO1.1b | B1 |
| (ii) | States max range for a | |
| n o r m al distribution | AO1.2 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| acceptable | AO2.4 | E1 |
| (iii) | Finds probability | |
| awfw (0.095 – 0.097) | AO1.1b | B1 |
| (b) | Finds z from calculator | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| equation | AO3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| this mark Accept 3.5 or 3.7 | AO1.1b | A1 |
| Total | 8 | |
| Q | Marking Instructions | AO |
Question 13:
--- 13(a)(i) ---
13(a)(i) | Finds mean | AO1.1b | B1 | 3046.14
Mean = X = =30.46 ( 14 )
100
Standard deviation
1746.29
s = = 4.20
99
Finds standard deviation
awfw (4.17 – 4.20)
1746.29
Allow s = = 4.179
100 | AO1.1b | B1
(ii) | States max range for a
n o r m al distribution | AO1.2 | M1 | X +3s ≈ 43
X −3s ≈ 18
Normal dist model acceptable because
data can be regarded as continuous and
the range of amounts is within X ±3s
Explains why normal model
acceptable | AO2.4 | E1
(iii) | Finds probability
awfw (0.095 – 0.097) | AO1.1b | B1 | P( X < 25) = 0.0967
(b) | Finds z from calculator | AO1.1b | B1 | ( )
30 − 29.55
P Z < = 0.55
σ
30 − 29.55
= 0.1257
σ
30 − 29.55
σ= = 3.6
0.1257
Uses Normal model to form
equation | AO3.4 | M1
Finds correct sd
Correct accuracy required for
this mark Accept 3.5 or 3.7 | AO1.1b | A1
Total | 8
Q | Marking Instructions | AO | Marks | Typical SolutionIn the South West region of England, 100 households were randomly selected and, for each household, the weekly expenditure, $£X$, per person on food and drink was recorded.
The maximum amount recorded was £40.48 and the minimum amount recorded was £22.00
The results are summarised below, where $\bar{x}$ denotes the sample mean.
$$\sum x = 3046.14 \quad\quad \sum (x - \bar{x})^2 = 1746.29$$
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item Find the mean of $X$
Find the standard deviation of $X$ [2 marks]
\item Using your results from part (a)(i) and other information given, explain why the normal distribution can be used to model $X$. [2 marks]
\item Find the probability that a household in the South West spends less than £25.00 on food and drink per person per week. [1 mark]
\end{enumerate}
\item For households in the North West of England, the weekly expenditure, $£Y$, per person on food and drink can be modelled by a normal distribution with mean £29.55
It is known that $P(Y < 30) = 0.55$
Find the standard deviation of $Y$, giving your answer to one decimal place. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 Q13 [8]}}