| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Validity of normal model |
| Difficulty | Moderate -0.8 This is a straightforward applied statistics question testing basic normal distribution concepts (checking suitability, using tables/calculator for probabilities, independence) and summary statistics calculations. All parts are routine A-level procedures with no problem-solving insight required—significantly easier than average. |
| Spec | 2.02g Calculate mean and standard deviation2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks |
|---|---|
| 18(a) | Calculates 1.78 ± 2 × 0.23 or |
| Answer | Marks | Guidance |
|---|---|---|
| 1.33 – 1.78 2.22 – 1.78 | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | 2.4 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2.22 | 2.2b | R1 |
| Subtotal | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 18(b)(i) | States 0 | 1.2 |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 18(b)(ii) | Calculates the correct |
| Answer | Marks | Guidance |
|---|---|---|
| AWFW [0.335, 0.34] | 1.1b | B1 |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 18(b)(iii) | Finds the value of their answer |
| Answer | Marks | Guidance |
|---|---|---|
| at least 2sf | 3.1b | B1F |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 18(c) | Obtains 1.73 |
| Answer | Marks | Guidance |
|---|---|---|
| Ignore missing or incorrect units | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| � 40 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Ignore missing or incorrect units | 1.1b | A1 |
| Subtotal | 3 |
| Answer | Marks |
|---|---|
| s = | 2.81 |
| Answer | Marks | Guidance |
|---|---|---|
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 18(d) | Uses their mean and 1.78 to |
| Answer | Marks | Guidance |
|---|---|---|
| about the same on average | 2.2b | E1F |
| Answer | Marks | Guidance |
|---|---|---|
| includes ‘range’ or ‘variety’ | 2.2b | E1F |
| Subtotal | 2 | |
| Question 18 Total | 11 | |
| Q | Marking instructions | AO |
Question 18:
--- 18(a) ---
18(a) | Calculates 1.78 ± 2 × 0.23 or
1.78 ± 1.96 × 0.23
or
calculates P(1.33 < x < 2.22)
PI by 0.9469 or 0.947
or
calculates and
1.33 – 1.78 2.22 – 1.78 | 3.1b | M1 | 1.78 2×0.23 = 1.32
1.78 + 2×0.23 = 2.24
–
1.32 ≈ 1.33
2.24 ≈ 2.22
Height is continuous data and 95%
of heights lies within two standard
deviations of the mean so normal
may be a suitable model.
0.23 0.23
Obtains 1.32 and 2.24 and
states they are approximately
1.33 and 2.22
or
obtains 1.33 and 2.23 and
states they are approximately
1.33 and 2.22
or
obtains 0.9469 or 0.947 and
states is approximately 0.95
or
obtains -1.96 and 1.91 and
states are approximately -2 and
2 | 2.4 | A1
Infers that the normal
distribution may be suitable
because height is continuous
data and 95% of heights lies
within two standard deviations of
the mean
or
Infers that the normal
distribution may be suitable
because height is continuous
data and 94.69% or 94.7% of
heights lies between 1.33 and
2.22 | 2.2b | R1
Subtotal | 3
≈
≈
Q | Marking instructions | AO | Marks | Typical solution
--- 18(b)(i) ---
18(b)(i) | States 0 | 1.2 | B1 | 0
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 18(b)(ii) ---
18(b)(ii) | Calculates the correct
probability
AWFW [0.335, 0.34] | 1.1b | B1 | 0.335
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 18(b)(iii) ---
18(b)(iii) | Finds the value of their answer
to (b)(ii) squared
Their answer must be correct to
at least 2sf | 3.1b | B1F | 2
0.335 = 0.112
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 18(c) ---
18(c) | Obtains 1.73
CAO
Ignore missing or incorrect units | 1.1b | B1 | Mean = 1.73
Standard deviation =
2.81
� 40
= 0.265
Uses the correct formula for
standard deviation
eg s =
2.81
Do not � al 3 lo 9 w variance =
2.81
� 40 | 1.1a | M1
Obtains the correct standard
deviation
AWFW [0.265, 0.27]
Allow if not labelled but if
labelled, must be correct
Ignore missing or incorrect units | 1.1b | A1
Subtotal | 3
Standard deviation =
s = | 2.81
Do not � al 3 lo 9 w variance =
Q | Marking instructions | AO | Marks | Typical solution
--- 18(d) ---
18(d) | Uses their mean and 1.78 to
compare heights
Comparison must include on
average.
Follow through their answer to
part (c)
Do not allow ‘general’
Allow statement that they’re
about the same on average | 2.2b | E1F | Summer athletes are taller on
average than Winter athletes.
Summer athletes’ heights are less
varied than the heights of Winter
athletes.
Uses their standard deviation
and 0.23 to compare heights
Comparison must include
‘varies’, ‘spread’ ‘disperse’ ‘more
variation’
or ‘consistent’
Follow through their answer to
part (c)
Allow statement that they are
about the same
Do not allow comparison that
includes ‘range’ or ‘variety’ | 2.2b | E1F
Subtotal | 2
Question 18 Total | 11
Q | Marking instructions | AO | Marks | Typical solutionIn a particular year, the height of a male athlete at the Summer Olympics has a mean 1.78 metres and standard deviation 0.23 metres.
The heights of 95% of male athletes are between 1.33 metres and 2.22 metres.
\begin{enumerate}[label=(\alph*)]
\item Comment on whether a normal distribution may be suitable to model the height of a male athlete at the Summer Olympics in this particular year.
[3 marks]
\item You may assume that the height of a male athlete at the Summer Olympics may be modelled by a normal distribution with mean 1.78 metres and standard deviation 0.23 metres.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that the height of a randomly selected male athlete is 1.82 metres.
[1 mark]
\item Find the probability that the height of a randomly selected male athlete is between 1.70 metres and 1.90 metres.
[1 mark]
\item Two male athletes are chosen at random.
Calculate the probability that both of their heights are between 1.70 metres and 1.90 metres.
[1 mark]
\end{enumerate}
\item The summarised data for the heights, $h$ metres, of a random sample of 40 male athletes at the Winter Olympics is given below.
$$\sum h = 69.2 \quad\quad \sum (h - \bar{h})^2 = 2.81$$
Use this data to calculate estimates of the mean and standard deviation of the heights of male athletes at the Winter Olympics.
[3 marks]
\item Using your answers from part (c), compare the heights of male athletes at the Summer Olympics and male athletes at the Winter Olympics.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2022 Q18 [11]}}