Oxford Cambridge and RSA Level 3 Certificate Quantitative ...€¦ · Average all rule • Find the mean of all five scores. Eliminate extremes rule • Cross out the highest and

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Oxford Cambridge and RSA

Level 3 CertificateQuantitative Reasoning (MEI)H866/02 Critical Maths

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The polygraph

The polygraph is known in everyday language as a lie detector. It measures blood pressure, heart rate, breathing rate and sweating. Someone who is lying is likely to show signs of stress, including changes in these bodily functions.

Of course, taking a lie detector test is itself stressful so interviews include some control questions for comparison purposes.

Someone who is identified as lying by the polygraph has failed the polygraph test. Otherwise, the person has passed the polygraph test.

The British Psychological Society has reported that

• between 42% and 89% of the time, someone who is lying fails the polygraph test (a true positive),

• between 2% and 47% of the time, someone who is telling the truth fails the polygraph test (a false positive).

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INSTRUCTIONS• Use black ink. You may use an HB pencil for graphs and diagrams.• Complete the boxes above with your name, centre number and candidate number.• Answer all the questions. • Write your answer to each question in the space provided. If additional space is

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Level 3 CertificateQuantitative Reasoning (MEI)H866/02 Critical Maths

Wednesday 23 May 2018 – MorningTime allowed: 2 hours

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© OCR 2018

Answer all the questions.

1 The chart below shows the changes in home ownership and renting for households in England and Wales.

1918

77%

50%

50%

23%

31%

69%

36%

64%

Ren

ting

Ow

ners

hip

1939 1953 1961 1971 1981 1991 2001 2011

12

...but from 1953owning started toincrease fasterthan previously...

...and by 1971equal % owningand renting...

...howeverbetween 2001and 2011 firstfall in ownershipover the century

In 1918...majority ofhouseholdsrented...

3

4A century of homeownership and renting

(i) (A) Estimate the percentage for “renting” in 1991.

(B) Find the percentage for “ownership” in 1991. [2]

(ii) Write down the percentage point decrease in “ownership” from 2001 to 2011. [1]

1(i) (A)

(B)

1(ii)

3

Turn over© OCR 2018

(iii) There were 21.7 million households in England and Wales in 2001 and 23.4 million in 2011. Were there fewer households in the “ownership” category in 2011 than there were in 2001? You must explain your reasoning. [4]

1(iii)

(iv) The total number of households has changed over time but the sum of the lengths of the “ownership” and “renting” bars for each year is the same. Explain why. [1]

1(iv)

4

© OCR 2018

2 A competition has five judges. Each judge gives each contestant a whole number score between 0 and 10 (inclusive). The five scores are combined to decide who has won the competition. Two different rules are proposed for combining the scores of each contestant.

Average all rule

• Find the mean of all five scores.

Eliminate extremes rule

• Cross out the highest and lowest score.• If there is more than one highest score, just one

of them is crossed out. Similarly for lowest scores.

• Find the mean of the remaining three scores.

(i) A contestant receives the following five scores from the judges.

7 4 4 2 5 (A) Find the combined score using the average all rule.

(B) Find the combined score using the eliminate extremes rule. [3]

2(i) (A)

(B)

(ii) Find a possible set of five scores where the eliminate extremes rule gives a higher combined score than the average all rule. Give evidence that your answer is correct. [3]

(iii) Find a possible set of five scores where the eliminate extremes rule gives the same combined score as the average all rule. Give evidence that your answer is correct. [2]

5


2(ii)

2(iii)

(iv) (A) Give one advantage of the average all rule.

(B) Give one advantage of the eliminate extremes rule. [2]

2(iv) (A)

(B)

6

© OCR 2018

3 An advert for a product to help people stop smoking contains the following words.

People who used it were 180% more likely to stop smoking.

(i) For every 200 people who try to stop smoking without the product, on average 10 people succeed.

Assume the claim in the advert is true. Work out how many on average would succeed out of 200 people using the product. [2]

3(i)

Number stopping smoking without the product

Number stopping smoking with the product

10

(ii) Would it be correct to say that people who try to stop smoking without using the product are 180% less likely to stop? Explain your reasoning. [1]

3(ii)

7


Stoptober is a national campaign to encourage smokers to stop smoking for 28 days in October. The advertising for Stoptober 2015 included the following words.

Stop smoking for 28 days and you’re 5 times more likely to stop for good.

(iii) In general, 5% of attempts to stop smoking are successful. Fill in the missing percentage in the table below. [1]

3(iii) Percentage stopping smoking for

good out of all who startPercentage stopping smoking for good after stopping for 28 days

5%

(iv) A quarter of a million people started the 28 days of Stoptober 2014. About half of them completed the full 28 days; estimate how many of these people stopped smoking for good. [3]

3(iv)

8

© OCR 2018

The graphs show the percentage of adults who smoke in Great Britain and the percentage of households with internet access.

01998 2002 2004 20082000 2006

Year20122010

102030405060708090

% adults smoking% households with internet access

%

005

1015202530

20 40 60% households with internet access

% a

dults

smok

ing

80 100

(v) Describe the relationship between the percentage of adults smoking and the percentage of households with internet access. [1]

(vi) Stella says that the greater availability of the internet has caused a reduction in smoking. Comment on her statement. [1]

3(v)

3(vi)

9


BLANK PAGE

PLEASE DO NOT WRITE ON THIS PAGE

10

© OCR 2018

4 Ofcom reports regularly on communications technology use by UK adults aged 16+.

People in the UK took an estimated 1.2 billion ‘selfies’ in the past year. Nearly a third (31%) of UK adults admit to taking a selfie, with one in ten (11%) doing so at least once a week.

Ofcom, August 2015

A selfie is a photograph of oneself. A billion is 1000 million.

(i) (A) Write down an estimate of the total population of the UK.

(B) Use this to estimate how many adults admit to taking selfies. Show your working. [6]

(ii) Assume all the 1.2 billion selfies were taken by the adults in part (i). How often, on average, did each of them take a selfie? Show your working. [3]

(iii) Use the Ofcom statement to find the median number of selfies taken by each UK adult. [1]

4(i) (A)

(B)

11


4(ii)

4(iii)

12

© OCR 2018

5 (i) Sports competitors are sometimes given medals on ribbon to hang around their necks.

For a competition, 30 medals like this will be given out. Estimate the total length of ribbon needed. Show your working. [3]

5(i)

13


(ii) Ribbon is sold in rolls A and B. The roll of ribbon is wound around a cardboard cylinder.

12 cm

Roll A Roll B

6 cm

The total diameter of roll A of ribbon is 12 cm; the diameter of the centre cylinder is 7 cm. The total diameter of roll B of ribbon is 6 cm; the diameter of the centre cylinder is 1 cm. The ribbon on both rolls is of the same kind.

Decide which, if either, roll has a longer length of ribbon on it. Explain your reasoning. [3]

5(ii)

14

© OCR 2018

6 This question refers to the article ‘The polygraph’. This was given out as pre-release material and is available as an insert.

(i) Using the information from the British Psychological Society, find upper and lower limits for the probability that someone who is telling the truth will pass the polygraph test. [3]

6(i)

A businessman reads that one third of workers cheat their bosses. He runs a large business and decides to have all 300 of his workers take a polygraph test.

(ii) Assume that one third of the workers will lie and two thirds will tell the truth. Assume that the polygraph will correctly identify 85% of people who are lying and will falsely identify as liars 5% of people who are telling truth.

(A) Find the probability that a worker who fails the test is actually telling the truth.

(B) Find the probability that a worker who passes the test is actually lying. [7]

6(ii)

(A)

(B)

15

© OCR 2018

John assumes that a polygraph correctly identifies liars 50% of the time. In an experiment, 400 people deliberately lie during a polygraph test.

(iii) Assume John is correct.

(A) On average, how many of the 400 people will the polygraph identify as liars? [1]

(B) What is the standard deviation of the number identified as liars? [2]

6(iii) (A)

(B)

(C) The graph below models the likelihood of different numbers of liars being identified by the experiment. Number the horizontal axis appropriately. [2]

Like

lihoo

d

Number identified as liars

(iv) The polygraph in the experiment identifies 257 liars. What can you say about John’s assumption? [2]

6(iv)

END OF QUESTION PAPER

16

© OCR 2018

ADDITIONAL ANSWER SPACE

If additional space is required, you should use the following lined page(s). The question number(s) must be clearly shown in the margin(s).

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Copyright Information

OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download from our public website (www.ocr.org.uk) after the live examination series.

If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.

For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.

OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

Oxford Cambridge and RSA Examinations

Level 3 Certificate

Quantitative Reasoning (MEI)

Unit H866/02 Statistical problem solving

OCR Level 3 Certificate

Mark Schemes for June 2018

OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. © OCR 2018

H866/02 Mark Scheme June 2018

3

Annotations and abbreviations

Annotation in scoris Meaning

and

BOD Benefit of doubt

FT Follow through

ISW Ignore subsequent working

M0, M1 Method mark awarded 0, 1

A0, A1 Accuracy mark awarded 0, 1

B0, B1 Independent mark awarded 0, 1

SC Special case

^ Omission sign

MR Misread

Highlighting

Other abbreviations in mark scheme

Meaning

E1 Mark for explaining

U1 Mark for correct units

G1 Mark for a correct feature on a graph

M1 dep* Method mark dependent on a previous mark, indicated by *

cao Correct answer only

oe Or equivalent

rot Rounded or truncated

soi Seen or implied

www Without wrong working


4

Subject-specific Marking Instructions a Annotations should be used whenever appropriate during your marking.

The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.

b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader.

c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks.


5

E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.

d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.

e The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.

f Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. 2 or 4 significant figures where 3 is expected) should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader.

g Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests.


6

If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.

h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some components. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.

i Anything in the mark scheme which is in square brackets […] is not required for the mark to be earned, but if present it must be correct.


7

Question Answer Marks Guidance

1 (i) (A) Answer in range 31 to 35[%] inclusive B1

(B) Answer in range 65 to 69[%] inclusive B1

[2]

OR FT 100 – their A

(ii) 5[%] B1

[1]

Condone -5 [%] decrease

Accept 0.05 but 0.05% is B0

(iii) 0.69 × 21.7 [million] oe

0.64 × 23.4 [million] oe

14.973 [million]

14.976 [million]

No [there were not fewer in 2011] oe

[3000] more in 2011

M1

A1

A1

B1 FT

[4]

For either percentage calculation (implied by correct answer)

Allow M1 with place value slip in households

14 973 000 14.97m

14 976 000 14.98m

Less in 2001 or [about] the same number

FT conclusion must be consistent with their working.

Qualitative work only: SC1 is possible eg ‘population has increased so lower %

could still be a bigger number’

(iv) [Each bar represents] 100% oe B1

[1]

Allow comments that say it should not be 100% with a reason, e.g. does not

include the homeless.

2 (i) (A) 4.4 B1

5

25447

(B) 2 and 7 eliminated

4.3[3333....]

M1

A1

[3]

SC1 for 3⅓

(ii) Attempt to calculate BOTH combined scores for any

new set of 5 scores

Either correct combined score for that set

Both scores correct for a set of scores 0 – 10 that

demonstrates ‘eliminate extremes’ is higher

M1

A1

A1

[3]

Implied by correct set of scores

Must include both combined scores or other justification eg for 0, 5, 5, 5, 9

average of middle three scores is more than average of highest and lowest

Condone rounding or truncating if conclusion unaffected

A wrong set of scores can earn M1A1A0


8


(iii) Attempt to calculate BOTH combined scores for any

new set of 5 scores

Both scores correct for a set of scores 0 -10 that

demonstrates combined scores are equal

M1

A1

[2]

Implied by correct set of scores

Must include both combined scores or other justification eg for 5, 5, 5, 5, 5 eg for

1, 2, 3, 4, 5 average of middle three scores is equal to average of highest and

lowest

Must work exactly ie both scores correct with no rounding or truncation errors

(iv) (A) Uses all values oe B1

Soi any reference to using all numbers [therefore fairer or true average]

Do not accept ‘easy to calculate’

(iv) (B) Removes outliers / anomalies / biased scoring oe

B1

[2]

Soi reference to fairer because uses middle values or highest/lowest excluded

3 (i) 10 × 2.8 oe

28

M1

A1

[2]

Implied by correct answer or implied by 10% + 8%

If zero scored, SC B1 for 18

(ii) No because can’t reduce by over 100% oe

B1 FT

[1]

Must have reason eg ‘% decrease is not the same as % increase’ or ‘reducing by

180% makes it negative’ Allow good qualitative argument eg ref to proportions

or % of different amounts

Justification could be example using numbers from their (i) e.g. reduction is

%6410028

1028

(iii) 25 [%] B1

[1]

(iv) 125 000

25% of 125 000

31 250 cao

B1

M1

A1

[3]

FT their iii (eg may be 95%) and their 125 000 (or 250 000) for M1

May be rounded

37500 from 6250+31250 scores B1M1A0


9


(v) Negative correlation

B1

[1]

Or description of this eg ‘higher internet use [corresponds] lower smoking’

Not a wrong description eg ‘households with more internet smoke less’

Ignore weak/strong

(vi) One relevant and correct comment B1

[1]

Eg ‘Correlation does not imply causation’

Eg ‘No obvious reason why her statement should be true’

Allow reasonable arguments such as ‘one addiction replaced by another’ or

‘improved (internet) access to information / health risks of smoking’

Accept ‘may be other factors [causing reduction]’

4 (i) (A) Total UK population is 65 million B1 Answer in range 60 to 70 million

(B) Ages 0 to 80

Age 16+ is about 80% of the total

52 million

0.31 × their 52 million

≈ 16 million

M1

M1

A1

M1

A1 FT

[6]

Soi consideration of overall age with adults 50%

Proportion of adults in range 70% to 85%

If elderly specifically excluded allow proportion to be as low as 50%

Adults in range 42 to 60 million (elderly excluded: allow 30 million)

Calculation must be 31% or 30% or ⅓ of either adults or whole population. M0

any other % adults.

Expect final answer in range 12 million to 20 million if working correct

Elderly excluded: accept 9 million

M0M0A0M1A1FT is possible.

(ii) 1.2 billion ÷ their (i)[=75]

Comparison with time

About once a week OR a bit more than once a week

oe.

M1

M1

A1 FT

[3]

Must be their final answer [for adults taking selfies] from i

To give better than per year: eg 12 months, 52 weeks or 365 days. Can be

awarded independently of first M1

FT from their i is dependent on M1M1. Do not accept “once a week” with no

working

(iii) Zero oe B1

[1]


10


5 (i) Length for one medal in range 50 cm to 120cm oe

Multiply by 30

Ans 30 times value for 1 medal

M1

M1

A1 FT

[3]

20-48 inches

Dep on 2nd

M mark only (M0M1A1 possible); need units for A1

SC2 for answer in range 15m to 36m with no working.

5 (ii)

12 – 7 and 6 – 1

Same number of turns [each roll] but each turn has

longer length on 12 cm roll oe

Roll A has more ribbon from correct working

M1

M1

A1

[3]

Widths method:

Soi ‘same width’ of ribbon. M1M0A1 possible if not fully explained.

Alternative method: cross-sectional areas

M1 2 2(6 3.5 )[ 23.75 ] [=74.6] oe

M1 2 2(3 0.5 )[ 8.75 ] [27.5] oe condone -12

A1 Roll A has more from correct working

6 (i) 1002 or 100 47

53% oe

[to] 98% oe

M1

A1

A1

[3]

For subtracting either from 100

Allow 0.53

Allow 0.98

(ii)

Working can be shown in a two way table instead of a tree diagram

+ve -ve total

Liar 85 15 100

Not liar 10 190 200

total 95 205 300

M1

Tree diagram with appropriate labelling or table with labels or equivalent

numbers clearly shown

A1 Correct partial frequencies or probabilities on ONE set of branches


11


A1 Correct partial frequencies or probabilities on ALL branches

Probabilities can be fractions/decimals/percentages

Calculations with no tree or table: Allow M1A1 once two outcomes have been

identified in calculation. Final A1 requires all 4 outcomes.

(A)

Adding the 2 relevant outcomes

95

10 oe 10.5% cao no FT

M1

A1

Frequencies:

their 85 + their 10 [=95]

Probabilities:

[=

(B)

Adding the 2 relevant outcomes

M1

Frequencies:

their 15 + their 190 [=205]

Probabilities:

[=

205

15 oe 7.3% cao no FT A1

[7]

(iii) (A) 200 B1

[1]

(B)

2

400

10

M1

A1

[2]

(C) their 200 at axis of symmetry

Scale going up in their sd soi

B1 FT

B1 FT

[2]

FT their (A)

FT their (B)


12


(iv) Polygraph seems to be doing better than 50%

detection oe

Over 2 or over 3 sd from mean oe

B1

B1

[2]

Correct conclusion oe: eg the [50%] assumption was not right / he was incorrect

Reason oe: eg 5.7 sd above mean or ‘very unlikely result’ FT reasoning here

from their sd in B

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Oxford Cambridge and RSA Examinations

Level 3 Certificate

Quantitative Reasoning (MEI) OCR Level 3 Certificate Quantitative Reasoning (MEI) H866

OCR Report to Centres June 2018

OCR Report to Centres – June 2018

9

H866/02 Critical Maths

1. General Comments: The vast majority of candidates were well prepared for the examination, not just in terms of their knowledge and understanding but also in terms of their willingness to tackle problems in a variety of contexts. Most candidates showed a good understanding of mathematical and statistical concepts, skills and techniques in their responses. Many were able to confidently apply mathematical and statistical thinking and reasoning to evaluate quantitative information and explain a wide range of real-life problems. Most candidates set out their working in a way, which made it, clear what they had done. This is especially important in problem solving or in questions where candidates are asked to show that a given answer is true. However, there was still scope for improvement in this area for some candidates. Accuracy was also an issue for some candidates; as there were questions that clearly indicated the use of rounded or approximate values and others where exact values were required throughout calculations. 2. Comments on Individual Questions: 1(i) Most candidates were successful in reading from the diagram to find reasonable estimates of the percentages. Using the fact that the two answers had to add to 100% allowed others to gain a follow through mark. 1(ii) Most candidates were familiar with the concept of a percentage point change and completed this successfully. A common wrong answer was to find the percentage change. 1(iii) Most candidates were able to use the figures in the question to calculate and compare the relevant numbers of households and to show that there were more households in “ownership” in 2011. A few candidates lost marks because they thought that 21.7 million was something other than 21 700 000. A minority attempted a qualitative argument but rarely said anything relevant to the situation – those who did were able to score one of the four marks available by providing an argument that there might have been more households in “ownership” in 2011. 1(iv) Many candidates successfully explained that each bar represented 100% and so would not change in length. Some candidates said that the bars represented percentages but did not complete the explanation of why the total bar length remained the same over the years. 2(i) Nearly all candidates were able to successfully apply the given rules to find the scores.


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2(ii) Many candidates were successful in finding a possible set of scores where the “eliminate extremes” rule gave a higher score than the “average all” rule. A small number of candidates did not use whole number scores in the range 0 to 10, as specified in the question, and so lost marks. Other candidates lost marks by either not writing down their set of scores clearly or making errors in finding either of the two combined scores. 2(iii) The two most commonly used strategies for finding a set of scores where the “eliminate extremes” rule gave the same score as the “average all” rule were either to use a set of identical scores or to use a set of scores which formed an arithmetic sequence but there were other ways of finding a suitable set of scores. Some candidates rounded scores and claimed they had equal scores using both rules when this was only true to one decimal place. 2(iv) Many candidates identified the advantages correctly and succinctly. 3(i) Many candidates found it difficult to work with a percentage over 100. In addition to this, some attempted to increase the wrong value by 180%. 3(ii) There were some good clear explanations which generally either explained that it was not possible to decrease a value by over 100% or that a percentage increase and the reverse percentage decrease have different percentage values because they refer to different numerical starting values. Some candidates claimed that percentages over 100 are not possible showing a lack of familiarity with large percentages. Others claimed that probability depends on the individual – this suggestion is not consistent with the context of the question, which is about an average cohort of people. 3(iii) Most candidates got this correct. The most common wrong answer was 95% which suggests that they were focussed on mutually exclusive outcomes rather than reading the question carefully to understand the question. 3(iv) Many candidates did this correctly showing clear working. Some added on extra 5%, showing that they had not fully understood the context but they got some of the marks for the question. A few candidates showed long, detailed non-calculator methods for calculating 25% of 125 000. Those who worked more efficiently either used a calculator to find 25% or divided by 4. 3(v) Most candidates described the negative correlation, for example, “as the percentage of households with internet access goes up the percentage of adult smokers goes down”; this was a correct answer so got the mark. A minority said there was no relationship and so did not score. 3(vi) Many candidates scored here by saying that there could have been other factors involved or by saying that the internet could have helped reduce smoking due to wider availability of information about the risks of smoking. Fewer candidates scored by simply saying that correlation does not imply causation, which was sufficient. Others referred to percentage decrease in smoking, claiming that this was evidence for causation but this is not the case. 4(i) Most candidates knew the approximate size of the UK population but a sizeable minority had no idea – suggested wrong answers ranged from hundreds of thousands to billions. Nearly all candidates wrote down a figure for the total UK population and so were able to gain follow through marks for method. Some candidates either assumed that the whole population was adults or did not notice the distinction between the total population and the adult population – this reduced the number of marks they had access to. It was pleasing to see some candidates making good estimates of the proportion of adults in the population and showing their reasoning


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but a few had over half the population being children – this is not realistic for the UK. Virtually all candidates gained some marks for this part question, only stronger candidates scored full marks. 4(ii) Candidates who had an answer for part (i) were able to gain full marks for follow through in this part and some did so. Marks were lost by candidates who were content to find the average number of selfies per person who admitted to taking them but did not address the “how often” which the question asked. 4(iii) Only the strongest candidates got this mark. Most candidates who attempted this part divided one number by another number, which is not a suitable method for finding the median. It was pleasing to see that a very small number of candidates not only got this right but were able to explain their reasoning clearly. 5(i) Most candidates took the approach of estimating the length of ribbon needed for one medal then multiplying by 30 with most of them getting full marks. The main errors were either not having a ribbon long enough to go over the head or finding the length for one ribbon and stopping there. A few candidates did not know how many cm were in a metre – leaving the final answer in cm was accepted. 5(ii) There were two methods of tackling this question – both were seen and most candidates gained at least one mark with many gaining two or three. The main sources of error were using an incorrect formula for area of a circle, incorrect rounding or assuming that an equal number of turns of ribbon implied equal total length. 6(i) Candidates who had a good understanding of the pre-release article and were able to use it to answer the given question usually gained full marks – with some slips in calculation leading to 2 out of 3. Other candidates did not seem to be able to relate the information in the passage to the question asked and gained no marks. 6(ii) A small number of candidates gained full marks; they had usually produced a fully labelled tree diagram using representative frequencies with labels on appropriate branches including “pass” or “fail” – this enabled them to use the information in the tree diagram correctly to produce the answers to parts (A) and (B). Other candidates gained the marks for producing a correct tree diagram but were not able to find the probabilities asked for. There were some

rounding errors caused by assuming that 13

is 33% - candidates who worked with the given

values of 13

and 23

found that this gave easier numbers to work with. It was rare for candidates

who did not use either a labelled tree diagram or a labelled table to score more than 2 marks. 6(iii) Nearly all candidates were able to find the correct mean but fewer were successful with the standard deviation. Some candidates were able to relate their mean and standard deviation to the Normal curve in part (C) and so score full marks for this part, but others numbered the axis so it ran from 0 to 400. 6(iv) Only the strongest candidates scored full marks here but some picked up one of the two available marks. Some candidates did not realise the relevance of the mean and standard deviation in this part of the question and so said that John was fairly close when, according to their calculated values of mean and standard deviation, the value was well over 10 standard deviations from the mean.

Published: 15 August 2018 Version 1.0 1

Level 3 Certificate, Level 3 Extended Project and FSMQ raw mark grade boundaries June 2018 series

Level 3 Certificate Mathematics - Quantitative Methods (MEI)Max Mark a b c d e u

G244 A 01 Introduction to Quantitative Methods with Coursework (WrittenPaper) Raw 72 58 50 43 36 28 0

G244 A 02 Introduction to Quantitative Methods with Coursework(Coursework) Raw 18 14 12 10 8 7 0

UMS 100 80 70 60 50 40 0Overall 90 72 62 53 44 35 0

Level 3 Certificate Mathematics - Quantitative Reasoning (MEI)Max Mark a b c d e u

H866 01 Introduction to quantitative reasoning Raw 72 56 49 42 35 28 0H866 02 Critical maths Raw 60 44 39 34 29 24 0

*To create the overall boundaries, component 02 is weighted to give marks out of 72 Overall 144 109 96 83 70 57 0

Level 3 Certificate Mathematics - Quantitative Problem Solving (MEI)Max Mark a b c d e u

H867 01 Introduction to quantitative reasoning Raw 72 56 49 42 35 28 0H867 02 Statistical problem solving Raw 60 40 36 32 28 24 0

*To create the overall boundaries, component 02 is weighted to give marks out of 72 Overall 144 104 92 80 69 57 0

Advanced Free Standing Mathematics Qualification (FSMQ)Max Mark a b c d e u

6993 01 Additional Mathematics Raw 100 56 50 44 38 33 0

Intermediate Free Standing Mathematics Qualification (FSMQ)Max Mark a b c d e u

6989 01 Foundations of Advanced Mathematics (MEI) Raw 40 35 30 25 20 16 0

For more information about results and grade calculations, see https://www.ocr.org.uk/students/getting-your-results/

Documents

Oxford Cambridge and RSA Level 3 Certificate Quantitative ...€¦ · Average all rule • Find the mean of all five scores. Eliminate extremes rule • Cross out the highest and