AQA Paper 3 2019 June — Question 17 12 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2019
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeFind p then binomial probability
DifficultyStandard +0.3 This is a standard normal distribution application question requiring inverse normal calculations to find parameters, then straightforward probability calculations and a binomial probability. The multi-step nature and 12 total marks suggest moderate length, but all techniques are routine A-level statistics procedures with no novel problem-solving required.
Spec2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation
Elizabeth's Bakery makes brownies. It is known that the mass, \(X\) grams, of a brownie may be modelled by a normal distribution. 10\% of the brownies have a mass less than 30 grams. 80\% of the brownies have a mass greater than 32.5 grams.
  1. Find the mean and standard deviation of \(X\). [7 marks]
    1. Find P\((X \neq 35)\) [1 mark]
    2. Find P\((X < 35)\) [2 marks]
  3. Brownies are baked in batches of 13. Calculate the probability that, in a batch of brownies, no more than 3 brownies are less than 35 grams. You may assume that the masses of brownies are independent of each other. [2 marks]
Question 17:
AnswerMarks
17(a)Obtains either z-value from inverse
normal distribution
Condone sign error
AWFW [βˆ’1.29, βˆ’1.28] or
AnswerMarks Guidance
[βˆ’0.85, βˆ’0.84]3.1b B1
𝑃𝑃�𝑍𝑍 < οΏ½ = 0.1
𝜎𝜎
32.5βˆ’πœ‡πœ‡
𝑃𝑃�𝑍𝑍 > οΏ½ =0.8
z= βˆ’1.2816𝜎𝜎 z = βˆ’0.8416
3 0βˆ’πœ‡πœ‡
= βˆ’1.2816
𝜎𝜎
3 2.5βˆ’πœ‡πœ‡
= βˆ’0.8416
𝜎𝜎
2.5 = 0.4 4𝜎𝜎
𝜎𝜎 = 5.68
πœ‡πœ‡ = 37.3
Forms one equation with unknown
ΞΌ and Οƒ using standardised result
and z-value (for 0.1)
Accept z = (βˆ’4, 4) except Β±0.1,
Β±0.2, Β±0.8, Β±0.9
Condone
AnswerMarks Guidance
Must use 301.1a M1
Forms nexπœ‡πœ‡t βˆ’eq3u0ation with unknown
ΞΌ and Οƒ using standardised result
and z-value (for 0.8)
Accept z = (βˆ’4, 4) except Β±0.1,
Β±0.2, Β±0.8, Β±0.9
Condone
AnswerMarks Guidance
Must use 32.51.1a M1
Obtains boπœ‡πœ‡thβˆ’ e3q2u.5ations correctly1.1b A1
Solves their two simultaneous
AnswerMarks Guidance
equations in the form of ΞΌ and Οƒ1.1a M1
Obtains correct value of Οƒ
AWFW (5.2, 5.9)
AnswerMarks Guidance
ISW1.1b A1
Obtains correct value of ΞΌ
AWFW (37.1, 37.5)
AnswerMarks Guidance
ISW1.1b A1
(b)(i) ---
17
AnswerMarks Guidance
(b)(i)States correct probability 1.2
(b)(ii) ---
17
AnswerMarks
(b)(ii)Uses their ΞΌ and their Οƒ to find
PI by correct value of probability
𝑃𝑃us(𝑋𝑋ing< th3e5i)r ΞΌ and their Οƒ or
correctly calculated z-value using
AnswerMarks Guidance
their ΞΌ and their Οƒ1.1a M1
Obtains correct probability to 2
decimal places or better
FT their ΞΌ and their Οƒ
If ΞΌ = (37.1, 37.5) and Οƒ = (5.2, 5.9)
AnswerMarks Guidance
used, answer will be (0.31, 0.37)1.1b A1F
AnswerMarks
17(c)Identifies the Binomial distribution
model with , p
PI by correct value of probability
AnswerMarks Guidance
using their p𝑛𝑛 = 13 = their 0.3443.1b M1
35g in a batch of 13
π‘Œπ‘Œ ∼ 𝐡𝐡(13,0.344)
𝑃𝑃(Y ≀ 3)= 0.294
Obtains correct probability to 2
decimal places or better
FT their p
If p = (0.31, 0.37) answer will be
AnswerMarks Guidance
[0.23, 0.39]1.1b A1F
Total12 
Question 17:
--- 17(a) ---
17(a) | Obtains either z-value from inverse
normal distribution
Condone sign error
AWFW [βˆ’1.29, βˆ’1.28] or
[βˆ’0.85, βˆ’0.84] | 3.1b | B1 | 30βˆ’πœ‡πœ‡
𝑃𝑃�𝑍𝑍 < οΏ½ = 0.1
𝜎𝜎
32.5βˆ’πœ‡πœ‡
𝑃𝑃�𝑍𝑍 > οΏ½ =0.8
z= βˆ’1.2816𝜎𝜎 z = βˆ’0.8416
3 0βˆ’πœ‡πœ‡
= βˆ’1.2816
𝜎𝜎
3 2.5βˆ’πœ‡πœ‡
= βˆ’0.8416
𝜎𝜎
2.5 = 0.4 4𝜎𝜎
𝜎𝜎 = 5.68
πœ‡πœ‡ = 37.3
Forms one equation with unknown
ΞΌ and Οƒ using standardised result
and z-value (for 0.1)
Accept z = (βˆ’4, 4) except Β±0.1,
Β±0.2, Β±0.8, Β±0.9
Condone
Must use 30 | 1.1a | M1
Forms nexπœ‡πœ‡t βˆ’eq3u0ation with unknown
ΞΌ and Οƒ using standardised result
and z-value (for 0.8)
Accept z = (βˆ’4, 4) except Β±0.1,
Β±0.2, Β±0.8, Β±0.9
Condone
Must use 32.5 | 1.1a | M1
Obtains boπœ‡πœ‡thβˆ’ e3q2u.5ations correctly | 1.1b | A1
Solves their two simultaneous
equations in the form of ΞΌ and Οƒ | 1.1a | M1
Obtains correct value of Οƒ
AWFW (5.2, 5.9)
ISW | 1.1b | A1
Obtains correct value of ΞΌ
AWFW (37.1, 37.5)
ISW | 1.1b | A1
--- 17
(b)(i) ---
17
(b)(i) | States correct probability | 1.2 | B1 | 1
--- 17
(b)(ii) ---
17
(b)(ii) | Uses their ΞΌ and their Οƒ to find
PI by correct value of probability
𝑃𝑃us(𝑋𝑋ing< th3e5i)r ΞΌ and their Οƒ or
correctly calculated z-value using
their ΞΌ and their Οƒ | 1.1a | M1 | 𝑃𝑃(𝑋𝑋 < 35)= 0.344
Obtains correct probability to 2
decimal places or better
FT their ΞΌ and their Οƒ
If ΞΌ = (37.1, 37.5) and Οƒ = (5.2, 5.9)
used, answer will be (0.31, 0.37) | 1.1b | A1F
--- 17(c) ---
17(c) | Identifies the Binomial distribution
model with , p
PI by correct value of probability
using their p𝑛𝑛 = 13 = their 0.344 | 3.1b | M1 | Y= no. of brownies less than
35g in a batch of 13
π‘Œπ‘Œ ∼ 𝐡𝐡(13,0.344)
𝑃𝑃(Y ≀ 3)= 0.294
Obtains correct probability to 2
decimal places or better
FT their p
If p = (0.31, 0.37) answer will be
[0.23, 0.39] | 1.1b | A1F
Total | 12
Elizabeth's Bakery makes brownies.

It is known that the mass, $X$ grams, of a brownie may be modelled by a normal distribution.

10\% of the brownies have a mass less than 30 grams.

80\% of the brownies have a mass greater than 32.5 grams.

\begin{enumerate}[label=(\alph*)]
\item Find the mean and standard deviation of $X$. [7 marks]

\item \begin{enumerate}[label=(\roman*)]
\item Find P$(X \neq 35)$ [1 mark]

\item Find P$(X < 35)$ [2 marks]
\end{enumerate}

\item Brownies are baked in batches of 13.

Calculate the probability that, in a batch of brownies, no more than 3 brownies are less than 35 grams.

You may assume that the masses of brownies are independent of each other.
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2019 Q17 [12]}}
This paper (17 questions)
View full paper
Q1 1 Q2 1 Q3 1 Q4 3 Q5 5 Q6 4 Q7 8 Q8 12 Q9 15 Q10 1 Q11 1 Q12 6 Q13 10 Q14 7 Q15 3 Q16 10 Q17 12