CAIE S2 2022 November — Question 6 10 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2022
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeMixed sum threshold probability
DifficultyStandard +0.3 This is a standard application of linear combinations of normal distributions requiring knowledge that sums of independent normals are normal, calculation of means/variances using linearity properties, and straightforward standardization. While it involves multiple bags and two parts, the techniques are routine for S2 level with no conceptual surprises—slightly easier than average since it's purely procedural application of well-practiced formulas.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions
The masses, in grams, of small and large bags of flour have the distributions N(510, 100) and N(1015, 324) respectively. André selects 4 small bags of flour and 2 large bags of flour at random.
  1. Find the probability that the total mass of these 6 bags of flour is less than 4130 g. [5]
  2. Find the probability that the total mass of the 4 small bags is more than the total mass of the 2 large bags. [5]
Question 6:
AnswerMarks
6Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9709/62 Cambridge International AS & A Level – Mark Scheme October/November 2022
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2022 Page 5 of 12
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(a)E(T) = 4  510 + 2  1015 [= 4070] B1
Var(T) = 4 × 100 + 2 × 324 [= 1048]B1 or (4100+2324)  =32.4(3sf)  .
4130−4070
[= 1.853]
AnswerMarks Guidance
'1048'M1 Standardising with their values.
Variance must be from a combination attempt.
M1 can be implied by correct final answer.
AnswerMarks Guidance
Φ(‘1.853’)M1 For area consistent with their values.
M1 can be implied by correct final answer.
AnswerMarks Guidance
= 0.968 (3 sf)A1 As final answer.
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(b)E(D) = 4  510 – 2  1015 [= 10] B1
Var(D) = 4  100 + 2  324 [= 1048]B1 Or (4100+2324) =32.4(3sf) .
 
0−'10'
[= –0.309]
AnswerMarks Guidance
'1048'M1 Standardising with their values.
Variance must be from a combination attempt.
M1 can be implied by correct final answer.
AnswerMarks Guidance
1 – Φ(‘–0.309’) = Φ(‘0.309’)M1 For area consistent with their values.
M1 can be implied by correct final answer.
AnswerMarks Guidance
= 0.621A1 As final answer.
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9709/62 Cambridge International AS & A Level – Mark Scheme October/November 2022
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2022 Page 5 of 12
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | E(T) = 4  510 + 2  1015 [= 4070] | B1
Var(T) = 4 × 100 + 2 × 324 [= 1048] | B1 | or (4100+2324)  =32.4(3sf)  .
4130−4070
[= 1.853]
'1048' | M1 | Standardising with their values.
Variance must be from a combination attempt.
M1 can be implied by correct final answer.
Φ(‘1.853’) | M1 | For area consistent with their values.
M1 can be implied by correct final answer.
= 0.968 (3 sf) | A1 | As final answer.
5
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | E(D) = 4  510 – 2  1015 [= 10] | B1
Var(D) = 4  100 + 2  324 [= 1048] | B1 | Or (4100+2324) =32.4(3sf) .
 
0−'10'
[= –0.309]
'1048' | M1 | Standardising with their values.
Variance must be from a combination attempt.
M1 can be implied by correct final answer.
1 – Φ(‘–0.309’) = Φ(‘0.309’) | M1 | For area consistent with their values.
M1 can be implied by correct final answer.
= 0.621 | A1 | As final answer.
5
Question | Answer | Marks | Guidance
The masses, in grams, of small and large bags of flour have the distributions N(510, 100) and N(1015, 324) respectively. André selects 4 small bags of flour and 2 large bags of flour at random.

\begin{enumerate}[label=(\alph*)]
\item Find the probability that the total mass of these 6 bags of flour is less than 4130 g. [5]

\item Find the probability that the total mass of the 4 small bags is more than the total mass of the 2 large bags. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2022 Q6 [10]}}
This paper (7 questions)
View full paper
Q1 7 Q2 5 Q3 6 Q4 8 Q5 6 Q6 10 Q7 8