AQA Paper 3 2021 June — Question 18 10 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2021
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeStandard two probabilities given
DifficultyModerate -0.3 This is a standard A-level statistics question on normal distribution. Parts (a)(i)-(ii) are routine probability calculations requiring basic z-score work or calculator use. Parts (b)(i)-(ii) involve setting up simultaneous equations from inverse normal calculationsβ€”a common textbook exercise type. While part (b)(ii) requires solving two equations, this is straightforward algebra once set up. The question tests core technique rather than problem-solving insight, making it slightly easier than average overall.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation
A factory produces jars of jam and jars of marmalade.
  1. The weight, \(X\) grams, of jam in a jar can be modelled as a normal variable with mean 372 and a standard deviation of 3.5
    1. Find the probability that the weight of jam in a jar is equal to 372 grams. [1 mark]
    2. Find the probability that the weight of jam in a jar is greater than 368 grams. [2 marks]
  2. The weight, \(Y\) grams, of marmalade in a jar can be modelled as a normal variable with mean \(\mu\) and standard deviation \(\sigma\)
    1. Given that \(P(Y < 346) = 0.975\), show that $$346 - \mu = 1.96\sigma$$ Fully justify your answer. [3 marks]
    2. Given further that $$P(Y < 336) = 0.14$$ find \(\mu\) and \(\sigma\) [4 marks]
Question 18:
AnswerMarks Guidance
18(a)(i)States 0 1.2
Subtotal1
QMarking Instructions AO
AnswerMarks
18(a)(ii)Uses the normal distribution
model to calculate P(X < 368)
or shows
PI by correct answer
AnswerMarks Guidance
𝑃𝑃(𝑋𝑋 > 368) = 1 βˆ’ 𝑃𝑃(𝑋𝑋 < 368)3.1b M1
Obtains correct probability
AnswerMarks Guidance
AWRT 0.871.1b A1
Subtotal2
QMarking Instructions AO
AnswerMarks
18(b)(i)Explains that the 1.96 is
obtained through inverse
normal distribution function or
shows 1.96 on a diagram
AnswerMarks Guidance
PI if 1.959(…) seen2.4 E1
1.95996398 for the area of 0.975
346βˆ’πœ‡πœ‡
𝑃𝑃�𝑍𝑍 < οΏ½ = 0.975
𝜎𝜎
3 46βˆ’πœ‡πœ‡
= 1.96
Hen𝜎𝜎ce
346βˆ’πœ‡πœ‡ = 1.96 𝜎𝜎
Forms an equation with
unknown ΞΌ and Οƒ using
standardised result and their z-
value
Accept z = (βˆ’4, 4) except
Β±0.975
AnswerMarks Guidance
Condone3.1b M1
πœ‡πœ‡βˆ’346
Completes rigorous argument
by forming a correct equation
using 1.96 and rearranging the
AnswerMarks Guidance
equation2.1 R1
Subtotal3
QMarking Instructions AO
AnswerMarks
18(b)(ii)Obtains either z-value from
inverse normal distribution
Condone sign error
AnswerMarks Guidance
AWFW [βˆ’1.1, βˆ’1.08]1.1b B1
z = 1.08
336βˆ’Β΅
=βˆ’1.08
Οƒ
336βˆ’Β΅=βˆ’1.08Οƒ
Οƒ=3.29
Β΅=340
Forms second equation with
unknown ΞΌ and Οƒ using
standardised result and their z-
value
Accept z = (βˆ’4, 4) except Β±0.14
AnswerMarks Guidance
Condone1.1a M1
Obtains coπœ‡πœ‡rrβˆ’ec3t3 v6alue of Οƒ
AWRT 3.3
AnswerMarks Guidance
ISW1.1b A1
Obtains correct value of ΞΌ
AWRT 340
AnswerMarks Guidance
ISW1.1b A1
Subtotal4
Question Total10 
Question 18:
--- 18(a)(i) ---
18(a)(i) | States 0 | 1.2 | B1 | 0
Subtotal | 1
Q | Marking Instructions | AO | Marks | Typical Solution
--- 18(a)(ii) ---
18(a)(ii) | Uses the normal distribution
model to calculate P(X < 368)
or shows
PI by correct answer
𝑃𝑃(𝑋𝑋 > 368) = 1 βˆ’ 𝑃𝑃(𝑋𝑋 < 368) | 3.1b | M1 | 𝑃𝑃(𝑋𝑋 > 368) = 0.87345
Obtains correct probability
AWRT 0.87 | 1.1b | A1
Subtotal | 2
Q | Marking Instructions | AO | Marks | Typical Solution
--- 18(b)(i) ---
18(b)(i) | Explains that the 1.96 is
obtained through inverse
normal distribution function or
shows 1.96 on a diagram
PI if 1.959(…) seen | 2.4 | E1 | Using inverse normal, the z- value is
1.95996398 for the area of 0.975
346βˆ’πœ‡πœ‡
𝑃𝑃�𝑍𝑍 < οΏ½ = 0.975
𝜎𝜎
3 46βˆ’πœ‡πœ‡
= 1.96
Hen𝜎𝜎ce
346βˆ’πœ‡πœ‡ = 1.96 𝜎𝜎
Forms an equation with
unknown ΞΌ and Οƒ using
standardised result and their z-
value
Accept z = (βˆ’4, 4) except
Β±0.975
Condone | 3.1b | M1
πœ‡πœ‡βˆ’346
Completes rigorous argument
by forming a correct equation
using 1.96 and rearranging the
equation | 2.1 | R1
Subtotal | 3
Q | Marking Instructions | AO | Marks | Typical Solution
--- 18(b)(ii) ---
18(b)(ii) | Obtains either z-value from
inverse normal distribution
Condone sign error
AWFW [βˆ’1.1, βˆ’1.08] | 1.1b | B1 | βˆ’
z = 1.08
336βˆ’Β΅
=βˆ’1.08
Οƒ
336βˆ’Β΅=βˆ’1.08Οƒ
Οƒ=3.29
Β΅=340
Forms second equation with
unknown ΞΌ and Οƒ using
standardised result and their z-
value
Accept z = (βˆ’4, 4) except Β±0.14
Condone | 1.1a | M1
Obtains coπœ‡πœ‡rrβˆ’ec3t3 v6alue of Οƒ
AWRT 3.3
ISW | 1.1b | A1
Obtains correct value of ΞΌ
AWRT 340
ISW | 1.1b | A1
Subtotal | 4
Question Total | 10
A factory produces jars of jam and jars of marmalade.

\begin{enumerate}[label=(\alph*)]
\item The weight, $X$ grams, of jam in a jar can be modelled as a normal variable with mean 372 and a standard deviation of 3.5
\begin{enumerate}[label=(\roman*)]
\item Find the probability that the weight of jam in a jar is equal to 372 grams.
[1 mark]

\item Find the probability that the weight of jam in a jar is greater than 368 grams.
[2 marks]
\end{enumerate}

\item The weight, $Y$ grams, of marmalade in a jar can be modelled as a normal variable with mean $\mu$ and standard deviation $\sigma$
\begin{enumerate}[label=(\roman*)]
\item Given that $P(Y < 346) = 0.975$, show that
$$346 - \mu = 1.96\sigma$$

Fully justify your answer.
[3 marks]

\item Given further that
$$P(Y < 336) = 0.14$$

find $\mu$ and $\sigma$
[4 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2021 Q18 [10]}}
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