AQA Paper 3 2024 June — Question 17 14 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2024
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeProbability calculation plus find unknown boundary
DifficultyModerate -0.8 This is a straightforward normal distribution question covering standard techniques: labeling a diagram, using symmetry properties, calculating probabilities with z-scores, finding inverse normal values, and conducting a basic one-tailed hypothesis test. All parts are routine A-level statistics procedures with no novel problem-solving required. The hypothesis test is formulaic with clearly stated parameters. Easier than average due to its entirely procedural nature.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05e Hypothesis test for normal mean: known variance
In 2019, the lengths of new-born babies at a clinic can be modelled by a normal distribution with mean 50 cm and standard deviation 4 cm.
  1. This normal distribution is represented in the diagram below. Label the values 50 and 54 on the horizontal axis. [2 marks] \includegraphics{figure_17a}
  2. State the probability that the length of a new-born baby is less than 50 cm. [1 mark]
  3. Find the probability that the length of a new-born baby is more than 56 cm. [1 mark]
  4. Find the probability that the length of a new-born baby is more than 40 cm but less than 60 cm. [1 mark]
  5. Determine the length exceeded by 95% of all new-born babies at the clinic. [2 marks]
  6. In 2020, the lengths of 40 new-born babies at the clinic were selected at random. The total length of the 40 new-born babies was 2060 cm. Carry out a hypothesis test at the 10% significance level to investigate whether the mean length of a new-born baby at the clinic in 2020 has **increased** compared to 2019. You may assume that the length of a new-born baby is still normally distributed with standard deviation 4 cm. [7 marks]
Question 17:
AnswerMarks
17(a)Labels 50 on the horizontal axis
below the vertex
AnswerMarks Guidance
Condone label at the vertex3.3 B1
Labels 54 on the horizontal axis
below the right-hand point of
inflection
Condone label at the right-hand
AnswerMarks Guidance
point of inflection3.3 B1
Subtotal2
QMarking instructions AO
AnswerMarks Guidance
17(b)States 0.5 1.2
Subtotal1
QMarking instructions AO
AnswerMarks Guidance
17(c)Obtains AWFW [0.0668, 0.067] 1.1b
Subtotal1
QMarking instructions AO
AnswerMarks Guidance
17(d)Obtains AWFW [0.987, 0.99] 1.1b
Subtotal1
QMarking instructions AO
AnswerMarks
17(e)x – 5 0
Forms = –1.6449
4
or
forms 50 + 4×(–1.6449)
PI by correct answer or
AWFW [56.56, 56.6]cm or 57cm
Allow [–4, 4] except ±0.95 or
AnswerMarks Guidance
±0.05 or 0 for –1.64493.1b M1
= –1.6449
4
Minimum length is 43.4 cm
Obtains AWFW [43.4, 43.44] cm
or 43 cm
AnswerMarks Guidance
Condone missing units1.1b A1
Subtotal2
QMarking instructions AO
States
H : μ = 50
0
H : μ > 50
AnswerMarks Guidance
12.5 B1
H : μ = 50
0
H : μ > 50
1
𝑥 = 51.5
42
x̄ ~ 𝑁(50, )
40
P(x̄ > 51.5) = 0.0089
0.0089 < 0.1
Reject H
0
There is sufficient evidence to
suggest that the mean length of a
new-born baby at the clinic in 2020
has increased compared to 2019.
AnswerMarks Guidance
Obtains 51.5 OE1.1b B1
States or uses correct model
PI by normal with mean 50 and
42
variance or 0.4
40
or standard deviation 0 .4 or
0.63 or better OE
or by correct probability
AWFW [0.0086, 0.009]
51.5−50
or test statistic (±)
4
40
FT their 51.5 for test statistic
or test statistic value
AWFW (±)[2.37, 2.4]
or critical value
AnswerMarks Guidance
AWFW [50.8, 51]1.1a M1
Obtains AWFW [0.0086, 0.009]
or the correct value of the test
statistic AWFW [2.37, 2.4]
or acceptance region
 AWFW [50.8, 51]
allow strict inequality
or critical region
≥ AWFW [50.8, 51]
allow strict inequality
or critical value
AnswerMarks Guidance
AWFW [50.8, 51]1.1b A1
Correctly compares their value
of P(> or ≥ their sample mean)
with 0.1
or correctly compares their
positive test statistic with AWFW
[1.28, 1.282]
or correctly compares 51.5 with
their acceptance region or
critical region or critical value
FT their sample mean
AnswerMarks Guidance
May be seen on a diagram3.5a M1
Infers H rejected
0
FT their comparison
Condone accept H
AnswerMarks Guidance
12.2b A1F
Concludes, from a fully correct
comparison, in context by
referring to an increase in the
mean length of new-born baby
at the clinic.
Conclusion must not be definite,
eg use of ‘suggest’, ‘support’ etc
To be awarded R1, marks
B0B1M1A1M1A1 must be
AnswerMarks Guidance
scored as the minimum3.2a R1
Subtotal7
Question 17 Total14
QMarking instructions AO 
Question 17:
--- 17(a) ---
17(a) | Labels 50 on the horizontal axis
below the vertex
Condone label at the vertex | 3.3 | B1
Labels 54 on the horizontal axis
below the right-hand point of
inflection
Condone label at the right-hand
point of inflection | 3.3 | B1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 17(b) ---
17(b) | States 0.5 | 1.2 | B1 | 0.5
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 17(c) ---
17(c) | Obtains AWFW [0.0668, 0.067] | 1.1b | B1 | 0.0668
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 17(d) ---
17(d) | Obtains AWFW [0.987, 0.99] | 1.1b | B1 | 0.9876
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 17(e) ---
17(e) | x – 5 0
Forms = –1.6449
4
or
forms 50 + 4×(–1.6449)
PI by correct answer or
AWFW [56.56, 56.6]cm or 57cm
Allow [–4, 4] except ±0.95 or
±0.05 or 0 for –1.6449 | 3.1b | M1 | x– 50
= –1.6449
4
Minimum length is 43.4 cm
Obtains AWFW [43.4, 43.44] cm
or 43 cm
Condone missing units | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
States
H : μ = 50
0
H : μ > 50
1 | 2.5 | B1 | X = length of new-born baby
H : μ = 50
0
H : μ > 50
1
𝑥 = 51.5
42
x̄ ~ 𝑁(50, )
40
P(x̄ > 51.5) = 0.0089
0.0089 < 0.1
Reject H
0
There is sufficient evidence to
suggest that the mean length of a
new-born baby at the clinic in 2020
has increased compared to 2019.
Obtains 51.5 OE | 1.1b | B1
States or uses correct model
PI by normal with mean 50 and
42
variance or 0.4
40
or standard deviation 0 .4 or
0.63 or better OE
or by correct probability
AWFW [0.0086, 0.009]
51.5−50
or test statistic (±)
4
40
FT their 51.5 for test statistic
or test statistic value
AWFW (±)[2.37, 2.4]
or critical value
AWFW [50.8, 51] | 1.1a | M1
Obtains AWFW [0.0086, 0.009]
or the correct value of the test
statistic AWFW [2.37, 2.4]
or acceptance region
 AWFW [50.8, 51]
allow strict inequality
or critical region
≥ AWFW [50.8, 51]
allow strict inequality
or critical value
AWFW [50.8, 51] | 1.1b | A1
Correctly compares their value
of P(> or ≥ their sample mean)
with 0.1
or correctly compares their
positive test statistic with AWFW
[1.28, 1.282]
or correctly compares 51.5 with
their acceptance region or
critical region or critical value
FT their sample mean
May be seen on a diagram | 3.5a | M1
Infers H rejected
0
FT their comparison
Condone accept H
1 | 2.2b | A1F
Concludes, from a fully correct
comparison, in context by
referring to an increase in the
mean length of new-born baby
at the clinic.
Conclusion must not be definite,
eg use of ‘suggest’, ‘support’ etc
To be awarded R1, marks
B0B1M1A1M1A1 must be
scored as the minimum | 3.2a | R1
Subtotal | 7
Question 17 Total | 14
Q | Marking instructions | AO | Marks | Typical solution
In 2019, the lengths of new-born babies at a clinic can be modelled by a normal distribution with mean 50 cm and standard deviation 4 cm.

\begin{enumerate}[label=(\alph*)]
\item This normal distribution is represented in the diagram below.

Label the values 50 and 54 on the horizontal axis.
[2 marks]

\includegraphics{figure_17a}

\item State the probability that the length of a new-born baby is less than 50 cm.
[1 mark]

\item Find the probability that the length of a new-born baby is more than 56 cm.
[1 mark]

\item Find the probability that the length of a new-born baby is more than 40 cm but less than 60 cm.
[1 mark]

\item Determine the length exceeded by 95% of all new-born babies at the clinic.
[2 marks]

\item In 2020, the lengths of 40 new-born babies at the clinic were selected at random.

The total length of the 40 new-born babies was 2060 cm.

Carry out a hypothesis test at the 10% significance level to investigate whether the mean length of a new-born baby at the clinic in 2020 has **increased** compared to 2019.

You may assume that the length of a new-born baby is still normally distributed with standard deviation 4 cm.
[7 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2024 Q17 [14]}}
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