AQA Paper 3 2023 June — Question 16 9 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeMixed calculations with boundaries
DifficultyStandard +0.3 Part (a) involves three routine normal distribution calculations using standardization (z-scores) and tables - completely standard A-level fare worth 3 marks total. Part (b) is slightly more challenging, requiring students to set up and solve simultaneous equations from two inverse normal conditions, but this is a well-practiced technique in A-level statistics. The 6-mark allocation reflects the working required rather than conceptual difficulty. Overall, this is a straightforward application question slightly above average due to part (b).
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation
A farm supplies apples to a supermarket. The diameters of the apples, \(D\) centimetres, are normally distributed with mean 6.5 and standard deviation 0.73
    1. Find \(P(D < 5.2)\) [1 mark]
    2. Find \(P(D > 7)\) [1 mark]
    3. The supermarket only accepts apples with diameters between 5 cm and 8 cm. Find the proportion of apples that the supermarket accepts. [1 mark]
  2. The farm also supplies plums to the supermarket. These plums have diameters that are normally distributed. It is found that 60% of these plums have a diameter less than 5.9 cm. It is found that 20% of these plums have a diameter greater than 6.1 cm. Find the mean and standard deviation of the diameter, in centimetres, of the plums supplied by the farm. [6 marks]
Question 16:
AnswerMarks
16(a)(i)Obtains correct probability
AWFW [0.037, 0.038]
Ignore incorrect rounding after
AnswerMarks Guidance
correct probability seen1.1b B1
Subtotal1
QMarking instructions AO
AnswerMarks
16(a)(ii)Obtains correct probability
AWFW [0.246, 0.25]
Ignore incorrect rounding after
AnswerMarks Guidance
correct probability seen1.1b B1
Subtotal1
QMarking instructions AO
AnswerMarks
16(a)(iii)Obtains correct probability
AWFW [0.96, 0.9602]
Ignore incorrect rounding after
AnswerMarks Guidance
correct probability seen3.3 B1
Subtotal1
QMarking instructions AO
AnswerMarks
16(b)Obtains either z-value from
inverse normal distribution
AWFW [0.25, 0.26] or
AWFW [0.84, 0.85]
AnswerMarks Guidance
Ignore signs3.1b B1
= 0.6
−𝜇𝜇
PīŋŊ𝑧𝑧 < 𝜎𝜎 īŋŊ
6.1
= 0.2
−𝜇𝜇
PīŋŊ𝑧𝑧 > 𝜎𝜎 īŋŊ
z = 0.2533 and z = 0.8416
= 0.2533
5.9 − 𝜇𝜇
𝜎𝜎
= 0.8416
6.1 − 𝜇𝜇
=𝜎𝜎 5.81 and = 0.34
𝜇𝜇 𝜎𝜎
Forms an equation with
unknown and using
standardised result and z-value
𝜇𝜇 𝜎𝜎
for 0.6
Accept z = AWFW [−4, 4] but do
not allow 0, Âą0.2, Âą0.4, Âą0.6 or
Âą0.8
Condone
Must use 5.9
AnswerMarks Guidance
𝜇𝜇−5.93.3 M1
Forms an equation with
unknown and using
standardised result and z-value
𝜇𝜇 𝜎𝜎
for 0.2
Accept z = AWFW [−4, 4] but do
not allow 0, Âą0.2, Âą0.4, Âą0.6 or
Âą0.8
Condone
Must use 6.1
AnswerMarks Guidance
𝜇𝜇−6.13.3 M1
Obtains both equations correctly1.1b A1
Obtains correct value of
AWFW [5.8, 5.82]
𝜇𝜇
AnswerMarks Guidance
ISW1.1b A1
Obtains correct value of
AWFW [0.33, 0.35]
𝜎𝜎
AnswerMarks Guidance
ISW1.1b A1
Subtotal6
Question 16 Total9
QMarking instructions AO 
Question 16:
--- 16(a)(i) ---
16(a)(i) | Obtains correct probability
AWFW [0.037, 0.038]
Ignore incorrect rounding after
correct probability seen | 1.1b | B1 | 0.0375
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 16(a)(ii) ---
16(a)(ii) | Obtains correct probability
AWFW [0.246, 0.25]
Ignore incorrect rounding after
correct probability seen | 1.1b | B1 | 0.2467
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 16(a)(iii) ---
16(a)(iii) | Obtains correct probability
AWFW [0.96, 0.9602]
Ignore incorrect rounding after
correct probability seen | 3.3 | B1 | 0.9601
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 16(b) ---
16(b) | Obtains either z-value from
inverse normal distribution
AWFW [0.25, 0.26] or
AWFW [0.84, 0.85]
Ignore signs | 3.1b | B1 | 5.9
= 0.6
−𝜇𝜇
PīŋŊ𝑧𝑧 < 𝜎𝜎 īŋŊ
6.1
= 0.2
−𝜇𝜇
PīŋŊ𝑧𝑧 > 𝜎𝜎 īŋŊ
z = 0.2533 and z = 0.8416
= 0.2533
5.9 − 𝜇𝜇
𝜎𝜎
= 0.8416
6.1 − 𝜇𝜇
=𝜎𝜎 5.81 and = 0.34
𝜇𝜇 𝜎𝜎
Forms an equation with
unknown and using
standardised result and z-value
𝜇𝜇 𝜎𝜎
for 0.6
Accept z = AWFW [−4, 4] but do
not allow 0, Âą0.2, Âą0.4, Âą0.6 or
Âą0.8
Condone
Must use 5.9
𝜇𝜇−5.9 | 3.3 | M1
Forms an equation with
unknown and using
standardised result and z-value
𝜇𝜇 𝜎𝜎
for 0.2
Accept z = AWFW [−4, 4] but do
not allow 0, Âą0.2, Âą0.4, Âą0.6 or
Âą0.8
Condone
Must use 6.1
𝜇𝜇−6.1 | 3.3 | M1
Obtains both equations correctly | 1.1b | A1
Obtains correct value of
AWFW [5.8, 5.82]
𝜇𝜇
ISW | 1.1b | A1
Obtains correct value of
AWFW [0.33, 0.35]
𝜎𝜎
ISW | 1.1b | A1
Subtotal | 6
Question 16 Total | 9
Q | Marking instructions | AO | Marks | Typical solution
A farm supplies apples to a supermarket.

The diameters of the apples, $D$ centimetres, are normally distributed with mean 6.5 and standard deviation 0.73

\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find $P(D < 5.2)$

[1 mark]

\item Find $P(D > 7)$

[1 mark]

\item The supermarket only accepts apples with diameters between 5 cm and 8 cm.

Find the proportion of apples that the supermarket accepts.

[1 mark]
\end{enumerate}

\item The farm also supplies plums to the supermarket.

These plums have diameters that are normally distributed.

It is found that 60% of these plums have a diameter less than 5.9 cm.

It is found that 20% of these plums have a diameter greater than 6.1 cm.

Find the mean and standard deviation of the diameter, in centimetres, of the plums supplied by the farm.

[6 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2023 Q16 [9]}}
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