A-LEVEL

Further Mathematics
Paper 3 – Statistics
Mark scheme

Practice paper – Set 2

Version 1.0
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Mark scheme instructions to examiners
General
The mark scheme for each question shows:
• the marks available for each part of the question
• the total marks available for the question
• marking instructions that indicate when marks should be awarded or withheld including the
principle on which each mark is awarded. Information is included to help the examiner make his or
her judgement and to delineate what is creditworthy from that not worthy of credit
• a typical solution. This response is one we expect to see frequently. However credit must be given
on the basis of the marking instructions.

If a student uses a method which is not explicitly covered by the marking instructions the same
principles of marking should be applied. Credit should be given to any valid methods. Examiners
should seek advice from their senior examiner if in any doubt.

Key to mark types

M mark is for method

R mark is for reasoning
A mark is dependent on M marks and is for accuracy
B mark is independent of M marks and is for method and accuracy
E mark is for explanation
F follow through from previous incorrect result

Key to mark scheme abbreviations

CAO correct answer only

CSO correct solution only
ft follow through from previous incorrect result
‘their’ indicates that credit can be given from previous incorrect result
AWFW anything which falls within
AWRT anything which rounds to
ACF any correct form
AG answer given
SC special case
OE or equivalent
NMS no method shown
PI possibly implied
SCA substantially correct approach
sf significant figure(s)
dp decimal place(s)
Examiners should consistently apply the following general marking principles

No method shown

Where the question specifically requires a particular method to be used, we must usually see
evidence of use of this method for any marks to be awarded.

Where the answer can be reasonably obtained without showing working and it is very unlikely that the
correct answer can be obtained by using an incorrect method, we must award full marks. However,
the obvious penalty to candidates showing no working is that incorrect answers, however close, earn
no marks.

Where a question asks the candidate to state or write down a result, no method need be shown for
full marks.

Where the permitted calculator has functions which reasonably allow the solution of the question
directly, the correct answer without working earns full marks, unless it is given to less than the
degree of accuracy accepted in the mark scheme, when it gains no marks.

Otherwise we require evidence of a correct method for any marks to be awarded.

Diagrams

Diagrams that have working on them should be treated like normal responses. If a diagram has been
written on but the correct response is within the answer space, the work within the answer space
should be marked. Working on diagrams that contradicts work within the answer space is not to be
considered as choice but as working, and is not, therefore, penalised.

Work erased or crossed out

Erased or crossed out work that is still legible and has not been replaced should be marked. Erased
or crossed out work that has been replaced can be ignored.

Choice

When a choice of answers and/or methods is given and the student has not clearly indicated which
answer they want to be marked, mark positively, awarding marks for all the student’s best attempts.
Withhold marks for final accuracy and conclusions if there are conflicting complete answers or when
an incorrect solution (or part thereof) is referred to in the final answer.
 MARK SCHEME – A-LEVEL FURTHER MATHEMATICS – PAPER 1 – PRACTICE PAPER 2

Q Marking instructions AO Marks Typical solution

1 Circles correct answer. 1
AO1.1b B1 6

Total 1

Q Marking instructions AO Marks Typical solution

2 Circles correct answer. AO1.2 B1 64
Total 1

Q Marking instructions AO Marks Typical solution

3(a) Uses correct probability AO3.1a M1 −λ 5 −λ 2
e λ 9 e λ
expression for P(X = 2) or =
P(X = 5) 5! 20 2!

λ =3
Forms an equation in λ AO1.1a M1

Solves equation to give λ = 3 AO1.1b A1

Total 3
3(b) Correctly obtains (or clearly AO3.1b M1 X + Y ~ Po(4.5)
uses) the distribution of X + Y
using Po(their λ + 1.5) PI
P(X + Y = 6) = 0.128
Obtains correct answer AO1.1b A1F
FT their λ
Total 2

5 of 10
 MARK SCHEME – A-LEVEL FURTHER MATHEMATICS – PAPER 3 – PRACTICE PAPER 2

Q Marking instructions AO Marks Typical solution

4(a) States either that sample size is AO2.4 E1 The population variance is
less than 30 or that population unknown and the sample size is
variance is unknown. less than 30 so we need to use the
t distribution.
States second reason and that t AO2.4 E1
Critical value for 7 degrees of
distribution must be used.
freedom for 90% confidence
interval is 1.895
Uses t7 value 1.895 AO3.3 B1
1.895×15.824
Constructs confidence interval AO1.1a M1 448.125 ±
8
value×sd
of form x ± 90% confidence interval is
8
Condone use of normal (437.4, 458.6)
distribution and of either
standard deviation.

Uses 15.824 AO1.1b B1

Calculates interval correctly AO1.1b A1

AWRT (437, 459)
Total 6
4(b) Deduces that the confidence AO2.2a M1 The confidence interval does not
interval does not contain 425 contain 425 so the claim is unlikely
to be justified.
Explains that the claim is AO3.5a A1
unlikely to be justified.
Total 2

6 of 10
 MARK SCHEME – A-LEVEL FURTHER MATHEMATICS – PAPER 3 – PRACTICE PAPER 2

Q Marking instructions AO Marks Typical solution

H0: No association between
5 Clearly states null and
AO2.5 B1 favourite flavour and age
alternative hypotheses.
H1: Association between favourite
Calculates expected AO1.1a M1 flavour and age
frequencies. 120×130
= 31.2
500
Calculates at least six expected AO1.1b A1
frequencies correctly. (35 − 33.8)2
33.8
2 AO1.1a M1
Calculates χ statistic.
No of d.f. = (4 – 1)(3 – 1) = 6
2
Finds correct value AWRT 6.21 AO1.1b A1 Critical χ (6) value = 12.592

Finds correct critical value. AO1.1b B1 As 6.21 < 12.592 we do not reject
H0
2 AO2.2b B1
Compares their χ statistic to
There is insufficient evidence to
their critical value and infers that
suggest that there is an association
H0 is not rejected.
between favourite flavour of crisp
and age.
Completes test and concludes AO3.2a E1
correctly in context.
Total 8

7 of 10
 MARK SCHEME – A-LEVEL FURTHER MATHEMATICS – PAPER 3 – PRACTICE PAPER 2

Marking instructions AO Marks Typical solution

6(a) Models probabilities of all AO3.1a M1 Pay 0 1 5 10 50
outcomes. out

Sets up a model using at least AO3.3 M1 Prob 0.77 31b 2a 5a b

one of the pay-out ratios.
0.77 + 31b + 2 a + 5 a + b =
1
Uses both ratios to create AO1.1b A1
correct probability models. 31b + 10 a + 50 a + 50 b =
0.945
Uses Σp = 1 to obtain an AO1.1a M1
equation. 7 a + 32 b =
0.23
Uses E(X) = 0.945 to obtain an AO1.1a M1
equation. 60 a + 81b =
0.945

Obtains two correct equations in AO1.1b A1

two unknowns. Hence a = 0.00858, b = 0.00531
Probability of winning £5 is 0.01716
Solves equations. AO1.1a M1

Finds correct probability. AO1.1b A1

AWRT 0.017
Total 8
Expected payout in euros =
6(b) Converts expected gain in £ to AO1.1a M1
euros. 0.945 × 1.25 =
1.18125
Pays to play
Finds winnings over 18 games. AO1.1a A1 1.50 × 18 =
27
Earns
Finds correct total amount left. AO3.2a A1 1.18125 × 18 =
21.26
Amount left =
35 − 27 + 21.26 =
29.26 euros
Total 3

8 of 10
 MARK SCHEME – A-LEVEL FURTHER MATHEMATICS – PAPER 1 – PRACTICE PAPER 2

Q Marking instructions AO Marks Typical solution

7(a) States that the manufacturing Hazelnuts are distributed randomly
process must distribute AO3.5b E1 throughout the mixture used to
hazelnuts randomly. make the bars.
Total 1
7(b) States both hypotheses using AO2.5 B1 H0: λ = 40
correct language.
H1: λ < 40
(Condone λ = 5)
Let N = number of hazelnuts in bar
Selects and uses Po(40) AO3.3 M1 Assuming H0: N ~ Po(40)
distribution either P( ≤ 33)
P(N ≤ 33) = 0.1514
or P( < 33)
As 0.151 > 0.1 we do not reject H0
Obtains 0.151 (AWRT) AO1.1b A1 There is insufficient evidence to
say that the average number of
Evaluates Poisson model by AO2.2b M1 hazelnuts is less than 5 per bar.
comparing ‘their’ p value with
0.1 and infers H0 is not rejected.

Concludes in context (not too AO3.2a E1

definitive.)
Total 5
7(c) Identifies need to find AO3.4 M1 Conclusion is incorrect if do not
acceptance region for test in (b). reject H0 given that λ = 4.8
P(N ≤ 32) = 0.1153
Identifies acceptance region for AO1.1b A1
test in (b). P(N ≤ 31) = 0.0855

Calculates P(N ≥ 32) for λ = 4.8 AO1.1a M1 Acceptance region is N ≥ 32

P(N ≥ 32) when λ = 4.8
States correct answer. AO1.1b A1
Prob = 0.869
Total 4

Q Marking instructions AO Marks Typical solution

8(a) Recalls the pdf or cdf of an −3λ 13
exponential distribution.
AO1.2 B1 1− e =
1−
50
Forms a correct equation using AO3.1b M1 λ = 0.449
pdf or cdf to find λ
1
Hence =
mean = 2.2
Obtains the correct value of λ AO1.1b A1 0.449

Find the correct mean time. AO1.1b B1F

FT their λ
Total 4

9 of 10
 MARK SCHEME – A-LEVEL FURTHER MATHEMATICS – PAPER 3 – PRACTICE PAPER 2

8(b) Uses no memory property of AO3.1a M1 Exponential distribution has no

exponential distribution to memory so need P(X ≤ 1)
consider P(X ≤ 1)
−0.449
=1− e
Obtains correct answer. AO1.1b A1F
= 0.3617
FT their λ
Total 2

10 of 10