Oxford Cambridge and RSA Level 3 Certificate Quantitative ...mei.org.uk/files/papers/H867_02_SPS_2017_June.pdf · The study covers three families of birds: owls, hawks and crows

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INFORMATION• The total mark for this paper is 60.• The marks for each question are shown in brackets [ ].• This document consists of 20 pages.• Final answers should be given to a degree of accuracy appropriate to the context.

© OCR 2017 [601/4782/9]DC (LK/SW) 145720/2

Last name

First name

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Oxford Cambridge and RSA

Level 3 CertificateQuantitative Problem Solving (MEI)H867/02 Statistical Problem Solving

Wednesday 24 May 2017 – MorningTime allowed: 2 hours

You must have:• the Insert (inserted)• the Statistical Tables (ST1) (inserted)

You may use:• a scientific or graphical calculator

OCR is an exempt CharityTurn over

* H 8 6 7 0 2 *

2

© OCR 2017

Answer all the questions.

Section A (30 marks)

1 A town council is trying to save money by closing its tourist information centre. It costs £45 000 a year to run it.

David is trying to estimate the cost to businesses in the town of not having it.

He identifies the following businesses that might suffer as a result.

• 35 small shops• 10 large shops• 16 guest houses• 6 hotels• 33 restaurants.

He selects a sample of 20 businesses and asks them the value of the trade, to the nearest £1000, that they think they would lose in a year if the tourist information centre was closed.

(i) (A) How many of each type of business should he select for his sample to be representative? [3]

(B) State which of the following terms best describes your sampling method.

Opportunity, simple random, stratified, quota, cluster, self-selected [1]

(ii) All of the businesses in the sample reply to David’s question. This table gives a summary of their replies.

Amount (£1000) Frequency Amount (£1000) Frequency

0 3 6 1

1 2 7 0

2 5 8 2

3 1 9 0

4 0 10 2

5 4 210 0

Use these data to estimate the total cost to businesses of closing the tourist information centre. [3]

(iii) It is suggested that the answer to part (ii) may be an overestimate. Give one reason why this might be the case. [1]

3

Turn over© OCR 2017

1 (i)(A)

1 (i)(B)

1 (ii)

1 (iii)

4

© OCR 2017

2 Simon is worried about his blood pressure. He buys an instrument to measure it; it gives two figures, the systolic and diastolic pressures. These are referred to as S and D in this question.

The table below gives Simon’s figures for ten days chosen at random. There are also columns for working out Spearman’s rank correlation coefficient.

(i) Complete the table and calculate Spearman’s rank correlation coefficient. [4]

(ii) Carry out a hypothesis test at the 5% significance level of whether there is positive association between systolic and diastolic pressures. State your null and alternative hypotheses, and your conclusion. [4]

2 (i)

Systolic Diastolic S rank D rank d d2

158 95 6 2 4 16

178 88 2 4½ –2.5 6.25

174 97 3 1 2 4

162 85 5 7 –2 4

170 88 4 4½ –0.5 0.25

156 84 8

128 87 10 6 4 16

179 93 1 3 –2 4

157 83 7

154 79 9

Σ

5


2 (ii)

6

© OCR 2017

Simon goes to see his doctor who prescribes some medicine. Simon then takes his blood pressure every day. He records his systolic readings for 100 days and uses a spreadsheet to draw this graph.

200190180170160150140

Syst

olic

pre

ssur

e

130120110100

1 100Day

(iii) Write down two things that this graph tells Simon about his blood pressure. [2]

(iv) Simon shows his doctor the graph. State, with brief reasons, what the doctor can conclude from the graph about Simon’s diastolic pressure

(A) on any particular day,

(B) over the whole period? [2]

2 (iii)

7


2 (iv)(A)

2 (iv)(B)

8

© OCR 2017

3 There is concern about the use of a certain type of rat poison. Some of the dead rats are eaten by birds which are then also poisoned.

A team of scientists are investigating whether this is happening in a particular area and whether it is affecting all types of birds to the same extent. The study covers three families of birds: owls, hawks and crows.

The scientists collect dead birds and analyse their livers. They classify the level of rat poison in them.

• Low (or none). The rat poison was unlikely to be the cause of death.• Medium. The rat poison was a possible cause of death.• High. The rat poison was almost certainly the cause of death.

At the end of one year their findings are summarised in Table 3.1.

Observed frequency, fo Low Medium High Total

Owls 6 16 26 48

Hawks 10 26 12 48

Crows 36 12 16 64

Total 52 54 54 160

Table 3.1

The scientists carry out a χ2 test using these data.

(i) Complete Table 3.2 for the expected frequencies, fe. [2]

(ii) Table 3.3 shows the contributions of the individual cells to the test statistic X2.

Complete the empty cells and work out the value of X2. [3]

(iii) State the result of the χ2 test at the 1% significance level. [3]

3 (i)

Expected frequency, fe Low Medium High Total

Owls 15.6 16.2 16.2 48

Hawks

Crows 21.6

Total 52 54 54

Table 3.2

9


3 (ii)

Low Medium High

Owls.

( . ).15 6

6 15 65 9077

2-= 0.0025 5.9284

Hawks

Crows

Table 3.3

3 (iii)

(iv) Answer these two questions in terms of the situation being investigated.

(A) What should the scientists conclude from the hypothesis test?

(B) What else might the data suggest to them? [2]

3 (iv)(A)

3 (iv)(B)

10

© OCR 2017

BLANK PAGE

PLEASE DO NOT WRITE ON THIS PAGE

11


Section B (30 marks)

The questions in this section are based on the pre-release data set. A hard copy of this is provided with this examination paper.

4 (a) Calculate the total GDP for the Cayman Islands in millions of US$.

Give your answer to the nearest million US$. [3]

(b) During the year in which the data were collected the death rate in Iceland was 6.20 per 1000.

Estimate the change in Iceland’s population that year, giving your answer to a reasonable level of accuracy.

State one assumption that you have made. [4]

4 (a)

4 (b)

12

© OCR 2017

5 The graph in Fig. 5.1 is a Normal probability plot for the life expectancy in the 43 mainland countries in Sub-Saharan Africa for which data is available.

(i) Draw rings round the points on the graph corresponding to Chad and South Africa. [1]

5 (i)

2.5

2

–2

1.5

–1.5

1

–1

0.5

–0.5

046 48 50 52 54 56 58 60 62 64

Life expectancy

z

66

Fig. 5.1

(ii) State which features of the graph indicate the following points about the distribution.

(A) The Normal distribution is a possible model.

(B) The Normal distribution is not a perfect model.

(C) The mean of the life expectancies in these countries is about 57.3 years. [3]

(iii) The standard deviation of these data is 4.8 years. Show that none of the values of life expectancy is more than 2 standard deviations from the mean.

State, with justification, how many values you would expect to be more than 2 standard deviations from the mean if the data followed a perfect Normal distribution. [3]

13


5 (ii)(A)

5 (ii)(B)

5 (ii)(C)

5 (iii)

14

© OCR 2017

The frequency chart shows the life expectancy for the 38 mainland European countries.

6

4

2

065 70 75 80

Age

Frequency

85 90

(iv) (A) State which one of the following terms best describes this distribution.

Positively skewed Bimodal Normal

(B) Which country in Europe has the highest life expectancy? [2]

(v) Write two statements comparing this distribution with that for the mainland Sub-Saharan African countries. [2]

(vi) In the pre-release data set, European countries are classified as “Europe (Western)” or “Europe (Eastern)”.

Look through the figures for these two groups.

Without doing any calculations, comment on how they influence the graph above. [1]

5 (iv)(A)

5 (iv)(B)

15


5 (v)

5 (vi)

16

© OCR 2017

6 Aisha is investigating access to health care for ordinary people in countries of different wealth. To model the

situation she devises a variable which she calls M; it is given by the formula M xy

= where

• x is the GDP per capita, measured in 10 000 US$,• y is the number of medical doctors per 1000 population.

Aisha starts with a pilot study based on the 13 members of OPEC, the Organisation of Petroleum Exporting Countries. They are given in the spreadsheet below.

(i) (A) Fill in the missing entries in column D.

(B) What formula could Aisha have used in cell D7? [2]

6 (i)(A)

A B C D

1 Country x, GDP per capita(in 10 000 US$)

y, Medical doctors(per 1000 people)

M xy

=

2 Algeria 0.75 1.21 1.61

3 Angola 0.63 0.17

4 Ecuador 1.06 1.69

5 Indonesia 0.52 0.20

6 Iran 1.28 0.89 0.70

7 Iraq 0.71 0.61 0.86

8 Kuwait 4.21 1.79 0.43

9 Libya 1.13 1.90 1.68

10 Nigeria 0.28 0.40 1.43

11 Qatar 10.21 2.76 0.27

12 Saudi Arabia 3.13 0.94 0.30

13 UAR 2.99 1.93 0.65

14 Venezuela 1.36 1.94 1.43

6 (i)(B)

17


(ii) Give a brief statement for each of the three figures for Nigeria in the spreadsheet interpreting it in comparison with those for the other OPEC countries. [3]

(iii) Aisha thinks that the quantity M will be a useful measure for her study.

Give one reason why this might be the case. [1]

6 (ii) Column B

Column C

Column D

6 (iii)

18

© OCR 2017

Aisha decides to move from the pilot study to one covering all the countries of the world. She obtains data for the number of doctors per 1000 people for a total of 194 countries. She then uses a spreadsheet to draw the scatter diagram below.

(iv) On the scatter diagram draw, by eye, a line of best fit passing through the origin. Write the corresponding value of M beside it.

Circle the point on the scatter diagram corresponding to Saudi Arabia. [3]

6 (iv)

0.000

1

2

3

4

5

6

7

8

2.00 4.00 6.00 8.00 10.00 12.00GDP per capita (×104 US$)

Doctors /1000 people

19

© OCR 2017

(v) Aisha uses the spreadsheet to calculate the product moment correlation coefficient between x and y for all 194 countries. Its value is 0.58.

Make two comments indicating how this value, together with the scatter diagram, is relevant to Aisha’s investigation. [2]

6 (v)

END OF QUESTION PAPER

20

© OCR 2017

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Level 3 CertificateQuantitative Problem Solving (MEI)H867/02 Statistical Problem Solving

Insert

Wednesday 24 May 2017 – MorningTime allowed: 2 hours

*6835425514*

OCR is an exempt CharityTurn over

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MEI EXAMINATION FORMULAE ANDSTATISTICAL TABLES (ST1)

For use with:

H867/02 STATISTICAL PROBLEM SOLVING

ST1CST262

July 2014

STATISTICS: HYPOTHESIS TESTS

Description Test statistic

Spearman rank correlation test rs = 1 −6Ðd2

i

n�n2 − 1�

#2 test X2 =Ð�fo − fe�

2

fe

2

CRITICAL VALUES FOR CORRELATION COEFFICIENTS

Critical values for the product moment correlation coefficient, r

5% 212% 1% 1

2% 1-Tail Test 5% 212% 1% 1

2%

10% 5% 2% 1% 2-Tail Test 10% 5% 2% 1%

n n

1 – – – – 31 0.3009 0.3550 0.4158 0.4556

2 – – – – 32 0.2960 0.3494 0.4093 0.4487

3 0.9877 0.9969 0.9995 0.9999 33 0.2913 0.3440 0.4032 0.4421

4 0.9000 0.9500 0.9800 0.9900 34 0.2869 0.3388 0.3972 0.4357

5 0.8054 0.8783 0.9343 0.9587 35 0.2826 0.3338 0.3916 0.4926

6 0.7293 0.8114 0.8822 0.9172 36 0.2785 0.3291 0.3862 0.4238

7 0.6694 0.7545 0.8329 0.8745 37 0.2746 0.3246 0.3810 0.4182

8 0.6215 0.7067 0.7887 0.8343 38 0.2709 0.3202 0.3760 0.4128

9 0.5822 0.6664 0.7498 0.7977 39 0.2673 0.3160 0.3712 0.4076

10 0.5494 0.6319 0.7155 0.7646 40 0.2638 0.3120 0.3665 0.4026

11 0.5214 0.6021 0.6851 0.7348 41 0.2605 0.3081 0.3621 0.3978

12 0.4973 0.5760 0.6581 0.7079 42 0.2573 0.3044 0.3578 0.3932

13 0.4762 0.5529 0.6339 0.6835 43 0.2542 0.3008 0.3536 0.3887

14 0.4575 0.5324 0.6120 0.6614 44 0.2512 0.2973 0.3496 0.3843

15 0.4409 0.5140 0.5923 0.6411 45 0.2483 0.2940 0.3457 0.3801

16 0.4259 0.4973 0.5742 0.6226 46 0.2455 0.2907 0.3420 0.3761

17 0.4124 0.4821 0.5577 0.6055 47 0.2429 0.2876 0.3384 0.3721

18 0.4000 0.4683 0.5425 0.5897 48 0.2403 0.2845 0.3348 0.3683

19 0.3887 0.4555 0.5285 0.5751 49 0.2377 0.2816 0.3314 0.3646

20 0.3783 0.4438 0.5155 0.5614 50 0.2353 0.2787 0.3281 0.3610

21 0.3687 0.4329 0.5034 0.5487 51 0.2329 0.2759 0.3249 0.3575

22 0.3598 0.4227 0.4921 0.5368 52 0.2306 0.2732 0.3218 0.3542

23 0.3515 0.4132 0.4815 0.5256 53 0.2284 0.2706 0.3188 0.3509

24 0.3438 0.4044 0.4716 0.5151 54 0.2262 0.2681 0.3158 0.3477

25 0.3365 0.3961 0.4622 0.5052 55 0.2241 0.2656 0.3129 0.3445

26 0.3297 0.3882 0.4534 0.4958 56 0.2221 0.2632 0.3102 0.3415

27 0.3233 0.3809 0.4451 0.4869 57 0.2201 0.2609 0.3074 0.3385

28 0.3172 0.3739 0.4372 0.4785 58 0.2181 0.2586 0.3048 0.3357

29 0.3115 0.3673 0.4297 0.4705 59 0.2162 0.2564 0.3022 0.3328

30 0.3061 0.3610 0.4226 0.4629 60 0.2144 0.2542 0.2997 0.3301

3

Critical values for Spearman’s rank correlation coefficient, rs

5% 212% 1% 1

2% 1-Tail Test 5% 212% 1% 1

2%

10% 5% 2% 1% 2-Tail Test 10% 5% 2% 1%

n n

1 – – – – 31 0.3012 0.3560 0.4185 0.4593

2 – – – – 32 0.2962 0.3504 0.4117 0.4523

3 – – – – 33 0.2914 0.3449 0.4054 0.4455

4 1.0000 – – – 34 0.2871 0.3396 0.3995 0.4390

5 0.9000 1.0000 1.0000 – 35 0.2829 0.3347 0.3936 0.4328

6 0.8286 0.8857 0.9429 1.0000 36 0.2788 0.3300 0.3882 0.4268

7 0.7143 0.7857 0.8929 0.9286 37 0.2748 0.3253 0.3829 0.4211

8 0.6429 0.7381 0.8333 0.8810 38 0.2710 0.3209 0.3778 0.4155

9 0.6000 0.7000 0.7833 0.8333 39 0.2674 0.3168 0.3729 0.4103

10 0.5636 0.6485 0.7455 0.7939 40 0.2640 0.3128 0.3681 0.4051

11 0.5364 0.6182 0.7091 0.7545 41 0.2606 0.3087 0.3636 0.4002

12 0.5035 0.5874 0.6783 0.7273 42 0.2574 0.3051 0.3594 0.3955

13 0.4835 0.5604 0.6484 0.7033 43 0.2543 0.3014 0.3550 0.3908

14 0.4637 0.5385 0.6264 0.6791 44 0.2513 0.2978 0.3511 0.3865

15 0.4464 0.5214 0.6036 0.6536 45 0.2484 0.2945 0.3470 0.3822

16 0.4294 0.5029 0.5824 0.6353 46 0.2456 0.2913 0.3433 0.3781

17 0.4142 0.4877 0.5662 0.6176 47 0.2429 0.2880 0.3396 0.3741

18 0.4014 0.4716 0.5501 0.5996 48 0.2403 0.2850 0.3361 0.3702

19 0.3912 0.4596 0.5351 0.5842 49 0.2378 0.2820 0.3326 0.3664

20 0.3805 0.4466 0.5218 0.5699 50 0.2353 0.2791 0.3293 0.3628

21 0.3701 0.4364 0.5091 0.5558 51 0.2329 0.2764 0.3260 0.3592

22 0.3608 0.4252 0.4975 0.5438 52 0.2307 0.2736 0.3228 0.3558

23 0.3528 0.4160 0.4862 0.5316 53 0.2284 0.2710 0.3198 0.3524

24 0.3443 0.4070 0.4757 0.5209 54 0.2262 0.2685 0.3168 0.3492

25 0.3369 0.3977 0.4662 0.5108 55 0.2242 0.2659 0.3139 0.3460

26 0.3306 0.3901 0.4571 0.5009 56 0.2221 0.2636 0.3111 0.3429

27 0.3242 0.3828 0.4487 0.4915 57 0.2201 0.2612 0.3083 0.3400

28 0.3180 0.3755 0.4401 0.4828 58 0.2181 0.2589 0.3057 0.3370

29 0.3118 0.3685 0.4325 0.4749 59 0.2162 0.2567 0.3030 0.3342

30 0.3063 0.3624 0.4251 0.4670 60 0.2144 0.2545 0.3005 0.3314

4

THE NORMAL DISTRIBUTION AND ITS INVERSE

The Normal distribution: values of Φ(Ï) = p

The table gives the probability, p, of a

random variable distributed as N(0, 1)

being less than Ï.N(0, 1)

0 Ï

p

(add)

Ï .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 1 2 3 4 5 6 7 8 9

0.0 .5000 5040 5080 5120 5160 5199 5239 5279 5319 5359 4 8 12 16 20 24 28 32 36

0.1 .5398 5438 5478 5517 5557 5596 5636 5675 5714 5753 4 8 12 16 20 24 28 32 35

0.2 .5793 5832 5871 5910 5948 5987 6026 6064 6103 6141 4 8 12 15 19 23 27 31 35

0.3 .6179 6217 6255 6293 6331 6368 6406 6443 6480 6517 4 8 11 15 19 23 26 30 34

0.4 .6554 6591 6628 6664 6700 6736 6772 6808 6844 6879 4 7 11 14 18 22 25 29 32

0.5 .6915 6950 6985 7019 7054 7088 7123 7157 7190 7224 3 7 10 14 17 21 24 27 31

0.6 .7257 7291 7324 7357 7389 7422 7454 7486 7517 7549 3 6 10 13 16 19 23 26 29

0.7 .7580 7611 7642 7673 7704 7734 7764 7794 7823 7852 3 6 9 12 15 18 21 24 27

0.8 .7881 7910 7939 7967 7995 8023 8051 8078 8106 8133 3 6 8 11 14 17 19 22 25

0.9 .8159 8186 8212 8238 8264 8289 8315 8340 8365 8389 3 5 8 10 13 15 18 20 23

1.0 .8413 8438 8461 8485 8508 8531 8554 8577 8599 8621 2 5 7 9 12 14 16 18 21

1.1 .8643 8665 8686 8708 8729 8749 8770 8790 8810 8830 2 4 6 8 10 12 14 16 19

1.2 .8849 8869 8888 8907 8925 8944 8962 8980 8997 9015 2 4 6 7 9 11 13 15 16

1.3 .9032 9049 9066 9082 9099 9115 9131 9147 9162 9177 2 3 5 6 8 10 11 13 14

1.4 .9192 9207 9222 9236 9251 9265 9279 9292 9306 9319 1 3 4 6 7 8 10 11 13

1.5 .9332 9345 9357 9370 9382 9394 9406 9418 9429 9441 1 2 4 5 6 7 8 10 11

1.6 .9452 9463 9474 9484 9495 9505 9515 9525 9535 9545 1 2 3 4 5 6 7 8 9

1.7 .9554 9564 9573 9582 9591 9599 9608 9616 9625 9633 1 2 3 3 4 5 6 7 8

1.8 .9641 9649 9656 9664 9671 9678 9686 9693 9699 9706 1 1 2 3 4 4 5 6 6

1.9 .9713 9719 9726 9732 9738 9744 9750 9756 9761 9767 1 1 2 2 3 4 4 5 5

2.0 .9772 9778 9783 9788 9793 9798 9803 9808 9812 9817 0 1 1 2 2 3 3 4 4

2.1 .9821 9826 9830 9834 9838 9842 9846 9850 9854 9857 0 1 1 2 2 2 3 3 4

2.2 .9861 9864 9868 9871 9875 9878 9881 9884 9887 9890 0 1 1 1 2 2 2 3 3

2.3 .9893 9896 9898 9901 9904 9906 9909 9911 9913 9916 0 1 1 1 1 2 2 2 2

2.4 .9918 9920 9922 9925 9927 9929 9931 9932 9934 9936 0 0 1 1 1 1 1 2 2

2.5 .9938 9940 9941 9943 9945 9946 9948 9949 9951 9952

2.6 .9953 9955 9956 9957 9959 9960 9961 9962 9963 9964

2.7 .9965 9966 9967 9968 9969 9970 9971 9972 9973 9974

2.8 .9974 9975 9976 9977 9977 9978 9979 9979 9980 9981

2.9 .9981 9982 9982 9983 9984 9984 9985 9985 9986 9986 differences

3.0 .9987 9987 9987 9988 9988 9989 9989 9989 9990 9990 untrustworthy

3.1 .9990 9991 9991 9991 9992 9992 9992 9992 9993 9993

3.2 .9993 9993 9994 9994 9994 9994 9994 9995 9995 9995

3.3 .9995 9995 9996 9996 9996 9996 9996 9996 9996 9997

3.4 .9997 9997 9997 9997 9997 9997 9997 9997 9997 9998

5

The Inverse Normal function: values of Φ–1

(p) = Ï

p .000 .001 .002 .003 .004 .005 .006 .007 .008 .009

.50 .0000 .0025 .0050 .0075 .0100 .0125 .0150 .0175 .0201 .0226

.51 .0251 .0276 .0301 .0326 .0351 .0376 .0401 .0426 .0451 .0476

.52 .0502 .0527 .0552 .0577 .0602 .0627 .0652 .0677 .0702 .0728

.53 .0753 .0778 .0803 .0828 .0853 .0878 .0904 .0929 .0954 .0979

.54 .1004 .1030 .1055 .1080 .1105 .1130 .1156 .1181 .1206 .1231

.55 .1257 .1282 .1307 .1332 .1358 .1383 .1408 .1434 .1459 .1484

.56 .1510 .1535 .1560 .1586 .1611 .1637 .1662 .1687 .1713 .1738

.57 .1764 .1789 .1815 .1840 .1866 .1891 .1917 .1942 .1968 .1993

.58 .2019 .2045 .2070 .2096 .2121 .2147 .2173 .2198 .2224 .2250

.59 .2275 .2301 .2327 .2353 .2378 .2404 .2430 .2456 .2482 .2508

.60 .2533 .2559 .2585 .2611 .2637 .2663 .2689 .2715 .2741 .2767

.61 .2793 .2819 .2845 .2871 .2898 .2924 .2950 .2976 .3002 .3029

.62 .3055 .3081 .3107 .3134 .3160 .3186 .3213 .3239 .3266 .3292

.63 .3319 .3345 .3372 .3398 .3425 .3451 .3478 .3505 .3531 .3558

.64 .3585 .3611 .3638 .3665 .3692 .3719 .3745 .3772 .3799 .3826

.65 .3853 .3880 .3907 .3934 .3961 .3989 .4016 .4043 .4070 .4097

.66 .4125 .4152 .4179 .4207 .4234 .4261 .4289 .4316 .4344 .4372

.67 .4399 .4427 .4454 .4482 .4510 .4538 .4565 .4593 .4621 .4649

.68 .4677 .4705 .4733 .4761 .4789 .4817 .4845 .4874 .4902 .4930

.69 .4959 .4987 .5015 .5044 .5072 .5101 .5129 .5158 .5187 .5215

.70 .5244 .5273 .5302 .5330 .5359 .5388 .5417 .5446 .5476 .5505

.71 .5534 .5563 .5592 .5622 .5651 .5681 .5710 .5740 .5769 .5799

.72 .5828 .5858 .5888 .5918 .5948 .5978 .6008 .6038 .6068 .6098

.73 .6128 .6158 .6189 .6219 .6250 .6280 .6311 .6341 .6372 .6403

.74 .6433 .6464 .6495 .6526 .6557 .6588 .6620 .6651 .6682 .6713

.75 .6745 .6776 .6808 .6840 .6871 .6903 .6935 .6967 .6999 .7031

.76 .7063 .7095 .7128 .7160 .7192 .7225 .7257 .7290 .7323 .7356

.77 .7388 .7421 .7454 .7488 .7521 .7554 .7588 .7621 .7655 .7688

.78 .7722 .7756 .7790 .7824 .7858 .7892 .7926 .7961 .7995 .8030

.79 .8064 .8099 .8134 .8169 .8204 .8239 .8274 .8310 .8345 .838l

.80 .8416 .8452 .8488 .8524 .8560 .8596 .8633 .8669 .8705 .8742

.81 .8779 .8816 .8853 .8890 .8927 .8965 .9002 .9040 .9078 .9116

.82 .9154 .9192 .9230 .9269 .9307 .9346 .9385 .9424 .9463 .9502

.83 .9542 .9581 .9621 .9661 .9701 .9741 .9782 .9822 .9863 .9904

.84 .9945 .9986 1.003 1.007 1.011 1.015 1.019 1.024 1.028 1.032

.85 1.036 1.041 1.045 1.049 1.054 1.058 1.063 1.067 1.071 1.076

.86 1.080 1.085 1.089 1.094 1.099 1.103 1.108 1.112 1.117 1.122

.87 1.126 1.131 1.136 1.141 1.146 1.150 1.155 1.160 1.165 1.170

.88 1.175 1.180 1.185 1.190 1.195 1.200 1.206 1.211 1.216 1.221

.89 1.227 1.232 1.237 1.243 1.248 l.254 1.259 1.265 1.270 1.276

.90 1.282 1.287 1.293 1.299 1.305 1.311 1.317 1.323 1.329 1.335

.91 1.341 1.347 1.353 1.360 1.366 1.372 1.379 1.385 1.392 1.398

.92 1.405 1.412 1.419 1.426 1.433 1.440 1.447 1.454 1.461 1.468

.93 1.476 1.483 1.491 1.499 1.506 1.514 1.522 1.530 1.538 1.546

.94 1.555 1.563 1.572 1.581 1.589 1.598 1.607 1.616 1.626 1.635

.95 1.645 1.655 1.665 1.675 1.685 1.695 1.706 1.717 1.728 1.739

.96 1.751 1.762 1.774 1.787 1.799 1.812 1.825 1.838 1.852 1.866

.97 1.881 1.896 1.911 1.927 1.943 1.960 1.977 1.995 2.014 2.034

.98 2.054 2.075 2.097 2.120 2.144 2.170 2.197 2.226 2.257 2.290

.99 2.326 2.366 2.409 2.457 2.512 2.576 2.652 2.748 2.878 3.090

6

PERCENTAGE POINTS OF THE #2 DISTRIBUTION

p%

#2

p%

#2

p% 99 97.5 95 90 10 5 2.5 1 0.5

. = 1 .0001 .0010 .0039 .0158 2.706 3.841 5.024 6.635 7.879

2 .0201 .0506 0.103 0.211 4.605 5.991 7.378 9.210 10.60

3 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.34 12.84

4 0.297 0.484 0.711 1.064 7.779 9.488 11.14 13.28 14.86

5 0.554 0.831 1.145 1.610 9.236 11.07 12.83 15.09 16.75

6 0.872 1.237 1.635 2.204 10.64 12.59 14.45 16.81 18.55

7 1.239 1.690 2.167 2.833 12.02 14.07 16.01 18.48 20.28

8 1.646 2.180 2.733 3.490 13.36 15.51 17.53 20.09 21.95

9 2.088 2.700 3.325 4.168 14.68 16.92 19.02 21.67 23.59

10 2.558 3.247 3.940 4.865 15.99 18.31 20.48 23.21 25.19

11 3.053 3.816 4.575 5.578 17.28 19.68 21.92 24.72 26.76

12 3.571 4.404 5.226 6.304 18.55 21.03 23.34 26.22 28.30

13 4.107 5.009 5.892 7.042 19.81 22.36 24.74 27.69 29.82

14 4.660 5.629 6.571 7.790 21.06 23.68 26.12 29.14 31.32

15 5.229 6.262 7.261 8.547 22.31 25.00 27.49 30.58 32.80

16 5.812 6.908 7.962 9.312 23.54 26.30 28.85 32.00 34.27

17 6.408 7.564 8.672 10.09 24.77 27.59 30.19 33.41 35.72

18 7.015 8.231 9.390 10.86 25.99 28.87 31.53 34.81 37.16

19 7.633 8.907 10.12 11.65 27.20 30.14 32.85 36.19 38.58

20 8.260 9.591 10.85 12.44 28.41 31.41 34.17 37.57 40.00

21 8.897 10.28 11.59 13.24 29.62 32.67 35.48 38.93 41.40

22 9.542 10.98 12.34 14.04 30.81 33.92 36.78 40.29 42.80

23 10.20 11.69 13.09 14.85 32.01 35.17 38.08 41.64 44.18

24 10.86 12.40 13.85 15.66 33.20 36.42 39.36 42.98 45.56

25 11.52 13.12 14.61 16.47 34.38 37.65 40.65 44.31 46.93

26 12.20 13.84 15.38 17.29 35.56 38.89 41.92 45.64 48.29

27 12.88 14.57 16.15 18.11 36.74 40.11 43.19 46.96 49.64

28 13.56 15.31 16.93 18.94 37.92 41.34 44.46 48.28 50.99

29 14.26 16.05 17.71 19.77 39.09 42.56 45.72 49.59 52.34

30 14.95 16.79 18.49 20.60 40.26 43.77 46.98 50.89 53.67

35 18.51 20.57 22.47 24.80 46.06 49.80 53.20 57.34 60.27

40 22.16 24.43 26.51 29.05 51.81 55.76 59.34 63.69 66.77

50 29.71 32.36 34.76 37.69 63.17 67.50 71.42 76.15 79.49

100 70.06 74.22 77.93 82.36 118.5 124.3 129.6 135.8 140.2

7

8

Oxford Cambridge and RSA Examinations

Level 3 Certificate

Quantitative reasoning (MEI)

Unit H867/02 Statistical Problem Solving

OCR Level 3 Certificate Quantitative Reasoning

Mark Schemes for June 2017

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 will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2017

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

H867/02 Mark Scheme June 2017

4

Subject-specific Marking Instructions

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.

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.

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.

gh 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 units. 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 AO Level

1 (i) (A) There are 35 + 10 + 16 + 6 + 33 = 100 businesses B1 100 seen or implied 2 E

For example

135 7

5 so he selects 7small shops

B1 For two correct answers (need not

be rounded) 1 E

He selects 7 small shops, 2 large shops, 3 guest houses, 1 hotel and 7

restaurants B1 All correct including sensible

rounding 2 E

(i) (B) The sample is stratified. B1 1 E

[4]

(ii) 0 3 + 1 2 + .,,, + 10 2 = 77 B1 77 or 77 000 seen 2 E

77 5 = 385 M1 1 E

385 1000 = 385 000 so £385 000 A1 ft their ‘77’ 1 E

[3]

(iii) Some of the responses may be overstated. The businesses will benefit from the

tourist information centre staying open.

They rounded up to the nearest 1000.

B1 Any reasonable comment that

refers to overestimate rather than

general uncertainty.

3 A

[1]


8

2 (i)

Systolic Diastolic S rank D rank d d2

158 95 6 2 4 16

178 88 2 4½ -2.5 6.25

174 97 3 1 2 4

162 85 5 7 -2 4

170 88 4 4½ -0.5 0.25

156 84 8 8 0 0

128 87 10 6 4 16

179 93 1 3 -2 4

157 83 7 9 -2 4

154 79 9 10 -1 1

Σ 55.5

B1

B1

Correct ranking

All correct

1

1

E

E

2

s 2

6 6 55.51 1

10 100 11

dr

n n

M1 (Allow calculation based on

use of pmcc on ranks)

Has to use n = 10.

2 E

s 0.6636...r A1 FT from their value of

2d

The rS calculation can be seen

in part ii.

2 C

[4]


9

2 (ii) H0: There is no association between S and D ‘association’ or ‘correlation’

H1: There is positive association between S and D B1 Both H0 and H1 correct. (Must be

correctly labelled or in the correct

order.)

2 C

(1-tail test) 2 C

The critical value (for n = 10 at the 5% level) is 0.5636 B1 0.5636 seen

0.6636 > 0.5636 (so the test is significant) M1 Comparison seen or implied 2 C

The evidence supports the alternative hypothesis that there is positive association

(between systolic and diastolic blood pressure). A1 Correct conclusion, with some

interpretation (e.g. “there is

positive association”, need not

mention blood pressure), from

correct CV, but ft their rS from i.

3 C

[4]

2 (iii) The general trend is down B1 Accept "It tells Simon about

trends in his blood pressure" oe

Accept “Medicines are working”

Accept numerical values from

graph.

3 E

There is considerable fluctuation B1 3 C

[2]


10

2 (iv) (A) No accurate information on any particular day but, because S and D are

correlated, we have some approximate information. B1 In either (A) or (B) there must be a

statement that the diastolic

pressure is correlated with the

systolic from the graph. Award 1

mark for two correct statements

without this reason.

The graph provides and upper

bound for D.

Can’t refer to the values from the

table.

3 A

(B) As S and D are correlated, the doctor can conclude that D decreases / there

is considerable fluctuation in D.

B1 Either statement

Can’t refer to the values from the

table.

3 A

[2]


11

3 (i)

Expected frequency, fe Low Medium High Total

Owls 15.6 16.2 16.2 48

Hawks 15.6 16.2 16.2 48

Crows 20.8 21.6 21.6 64

Total 52 54 54 160

B1

B1

If the numbers are rounded to

the nearest integer, award max

1 mark.

Hawks row

Crows row

2

1

E

E

[2]

3 (ii)

X2 Low Medium High

Owls 5.9077 0.0025 5.9284

Hawks 2.0103 5.9284 1.0889

Crows 11.1077 4.2667 1.4519

B1

B1

Accept answers given to 1dp

(rounded or truncated)

2 correct values

All correct

2

1

C

A

X2 = 37.69(25) B1 Total (ft their table) 1 A

[3]


12

3 (iii) 3 1 3 1 4 B1 Can be implied by 13.28 2 A

Critical value at the 1% significance level is 13.28 M1 2 A

Since 37.69 > 13.28, the test is significant. A1 Comparison can be in words. 1 A

[3]

3 (iv) (A) There is (strong) evidence that some types of bird are more at risk from the rat

poison than others. B1 Answers for (A) and (B) can

be interchanged.

Must refer to different types of

birds.

3 E

(B) The rat poison is indeed killing birds. B1 Accept other sensible answers

e.g. Crows are least affected;

More crows died

3 C

[2]


13

4 (a) Cayman Islands: Population 54 914, GDP per capita 43 800 US$ B1 2 E

Total GDP = 54 914 43 800 (= 2 405 233 200 US$)

M1 For multiplication (must be

popxGDP) 2 E

= 2405 million US$ A1 Must be rounded correctly 1 C

[3]

4 (b) (Iceland: Population 317 351, Birth rate 13.09) 2 E

Deaths: 317351 6.20 1000 1968 (to the nearest whole number)

M1 Either births or deaths;

condone use of incorrect

country.

3 C

Births: 317351 13.09 1000 4154 (to the nearest whole number) A1 Cao both births and deaths

Increase (= 4154 – 1968) = 2186 (or 2187)

To the nearest 100 the increase is 2200 A1 Don’t allow decimal answers.

Accept answers from early

rounding (art 2100).

1 C

It is assumed that immigration and emigration can be ignored.

B1 Allow any other sensible

modelling assumption 3 A

[4]


14

ALTERNATIVE

(Iceland: Population 317 351, Birth rate 13.09)

Difference between birth and death rates: 13.09 – 6.20 (= 6.89) M1

Change = 371 351 × 6.89 ÷ 1000 (= 2186) A1

To the nearest 100 the increase is 2200 A1 Don’t allow decimal answers.

Accept answers from early

rounding (art 2100).

It is assumed that immigration and emigration can be ignored.

B1 Allow any other sensible

modelling assumption


15

5 (i)

B1

Both points required but need

not be labelled (ignore

incorrect labelling)

2

C

[1]

5 (ii) (A) Most points lie on or near the straight line

B1 Condone reference to “line of

best fit” 2 E

(B) The points are not exactly on the line B1 2 E

(C) The line crosses the horizontal axis at 57.3

B1 Accept "The median is about

57.3" oe 1 E

[3]


16

5 (iii) 57.3 - 2 × 4.8 = 47.7 z-values for lowest and highest

values: -1.6375 (Chad) and

+1.7604 (Ghana)

57.3 + 2 × 4.8 = 66.9

None of the points on the graph have life expectance < 47.7 or > 66.9

B1 Some comparison needed (z-

scores are sufficient) 2 E

In the Normal distribution 5% of the points are more than 2 standard deviations from

the mean M1 5% or 95% seen 3 C

5% of 43 = 2.15

So about 2 values would be expected to be more than 2 standard deviations from the

mean. A1 2 seen 3 C

[3]

5 (iv) (A) Bimodal B1 2 E

(B) Monaco B1 1 E

[2]

5 (v) The life expectancy in Europe is much higher than in mainland Sub-Saharan Africa B1 Any two sensible, different

comments (e.g. comparison of

ranges)

3 E

The distribution is nowhere near Normal for Europe but quite close for Africa B1 3 E

[2]

5 (vi) Most East European countries have life expectancy in the 70s whereas for most West

European countries it is in the 80s. This explains the bimodal distribution. B1 3 A

[1]


17

Question Answer Marks Guidance AO Level

6 (i) (A)

A B C D

1 Country x, GDP per capita

(in 10 000 US$)

y, Medical doctors

(per 1000 people)

yM

x

2 Algeria 0.75 1.21 1.61

3 Angola 0.63 0.17 0.27

4 Ecuador 1.06 1.69 1.59

5 Indonesia 0.52 0.20 0.38

6 Iran 1.28 0.89 0.70

7 Iraq 0.71 0.61 0.86

8 Kuwait 4.21 1.79 0.43

9 Libya 1.13 1.90 1.68

10 Nigeria 0.28 0.40 1.43

11 Qatar 10.21 2.76 0.27

12 Saudi Arabia 3.13 0.94 0.30

13 UAR 2.99 1.93 0.65

14 Venezuela 1.36 1.94 1.43

B1

Cao (to 2dp)

1

E

(B) The formula in cell D7 is = C7/B7

B1 Allow “=C2/B2 copied

down” 1 E

[2]


18

6 (ii) Column B: Nigeria is the poorest of the OPEC countries as measured by per capita

GDP B1 oe 2 E

Column C: Nigeria has one of the lowest (third lowest) numbers of doctors per 1000

people B1 oe 2 E

Column D: Nigeria is spending more of its money on doctors than most other OPEC

countries

B1 This mark requires an

element of interpretation. 3 A

[3]

(iii) M gives an indication of how much of a country's wealth is being spent on health for

ordinary people. B1 Any reasonable answer,

eg "M shows up

differences between

countries."

3 A

[1]


19

6 (iv)

M1

A1

B1

Reasonable

line

M must

follow

through as

gradient of

the line

drawn

2

3

1

E

A

C

[3]

M = 1

Saudi Arabia GDP per capita (× 10

4 US$)

Doctors / 1000 people


20

(v) There is a definite correlation between x and y. / It shows whether there is a correlation.

It is sensible to look at the two variables together.

B1 Any sensible comments 3 A

The value also shows that the correlation is not perfect.

This is shown in the scatter diagram.

Many points are far from the line of best fit and so could be interesting for

the investigation.

B1

3 A

[2]

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Published: 16 August 2017 Version 1.0 1

Unit level raw mark and UMS grade boundaries June 2017 series

For more information about results and grade calculations, see www.ocr.org.uk/ocr-for/learners-and-parents/getting-your-results

AS GCE / Advanced GCE / AS GCE Double Award / Advanced GCE Double Award

GCE Mathematics (MEI) Max Mark a b c d e u

4751 01 C1 – MEI Introduction to advanced mathematics (AS) Raw 72 63 58 53 49 45 0 UMS 100 80 70 60 50 40 0

4752 01 C2 – MEI Concepts for advanced mathematics (AS) Raw 72 55 49 44 39 34 0 UMS 100 80 70 60 50 40 0

4753 01 (C3) MEI Methods for Advanced Mathematics withCoursework: Written Paper Raw 72 54 49 45 41 36 0

4753 02 (C3) MEI Methods for Advanced Mathematics withCoursework: Coursework Raw 18 15 13 11 9 8 0

4753 82 (C3) MEI Methods for Advanced Mathematics withCoursework: Carried Forward Coursework Mark Raw 18 15 13 11 9 8 0

UMS 100 80 70 60 50 40 0 4754 01 C4 – MEI Applications of advanced mathematics (A2) Raw 90 67 61 55 49 43 0

UMS 100 80 70 60 50 40 0

4755 01 FP1 – MEI Further concepts for advanced mathematics(AS) Raw 72 57 52 47 42 38 0

UMS 100 80 70 60 50 40 0

4756 01 FP2 – MEI Further methods for advanced mathematics(A2) Raw 72 65 58 52 46 40 0

UMS 100 80 70 60 50 40 0

4757 01 FP3 – MEI Further applications of advanced mathematics(A2) Raw 72 64 56 48 41 34 0

UMS 100 80 70 60 50 40 0

4758 01 (DE) MEI Differential Equations with Coursework: WrittenPaper Raw 72 63 56 50 44 37 0

4758 02 (DE) MEI Differential Equations with Coursework:Coursework Raw 18 15 13 11 9 8 0

4758 82 (DE) MEI Differential Equations with Coursework: CarriedForward Coursework Mark Raw 18 15 13 11 9 8 0

UMS 100 80 70 60 50 40 0 4761 01 M1 – MEI Mechanics 1 (AS) Raw 72 57 49 41 34 27 0

UMS 100 80 70 60 50 40 0 4762 01 M2 – MEI Mechanics 2 (A2) Raw 72 56 48 41 34 27 0



UMS 100 80 70 60 50 40 0 4766 01 S1 – MEI Statistics 1 (AS) Raw 72 61 55 49 43 37 0

UMS 100 80 70 60 50 40 0 4767 01 S2 – MEI Statistics 2 (A2) Raw 72 56 50 45 40 35 0



UMS 100 80 70 60 50 40 0 4771 01 D1 – MEI Decision mathematics 1 (AS) Raw 72 52 46 41 36 31 0

UMS 100 80 70 60 50 40 0 4772 01 D2 – MEI Decision mathematics 2 (A2) Raw 72 53 48 43 39 35 0

UMS 100 80 70 60 50 40 0 4773 01 DC – MEI Decision mathematics computation (A2) Raw 72 46 40 34 29 24 0

UMS 100 80 70 60 50 40 0

4776 01 (NM) MEI Numerical Methods with Coursework: WrittenPaper Raw 72 58 53 48 43 37 0

4776 02 (NM) MEI Numerical Methods with Coursework:Coursework Raw 18 14 12 10 8 7 0

4776 82 (NM) MEI Numerical Methods with Coursework: CarriedForward Coursework Mark Raw 18 14 12 10 8 7 0

UMS 100 80 70 60 50 40 0 4777 01 NC – MEI Numerical computation (A2) Raw 72 55 48 41 34 27 0

http://www.ocr.org.uk/ocr-for/learners-and-parents/getting-your-results


UMS 100 80 70 60 50 40 0 4798 01 FPT - Further pure mathematics with technology (A2) Raw 72 57 49 41 33 26 0

UMS 100 80 70 60 50 40 0

GCE Statistics (MEI) Max Mark a b c d e u G241 01 Statistics 1 MEI (Z1) Raw 72 61 55 49 43 37 0

UMS 100 80 70 60 50 40 0 G242 01 Statistics 2 MEI (Z2) Raw 72 55 48 41 34 27 0

UMS 100 80 70 60 50 40 0 G243 01 Statistics 3 MEI (Z3) Raw 72 56 48 41 34 27 0

UMS 100 80 70 60 50 40 0

GCE Quantitative Methods (MEI) Max Mark a b c d e u G244 01 Introduction to Quantitative Methods MEI Raw 72 58 50 43 36 28 0 G244 02 Introduction to Quantitative Methods MEI Raw 18 14 12 10 8 7 0

UMS 100 80 70 60 50 40 0 G245 01 Statistics 1 MEI Raw 72 61 55 49 43 37 0

UMS 100 80 70 60 50 40 0 G246 01 Decision 1 MEI Raw 72 52 46 41 36 31 0

UMS 100 80 70 60 50 40 0


Level 3 Certificate and FSMQ raw mark grade boundaries June 2017 series

Level 3 Certificate Mathematics for EngineeringMax Mark a* a b c d e u

H860 01 Mathematics for EngineeringH860 02 Mathematics for Engineering

Level 3 Certificate Mathematical Techniques and Applications for EngineersMax Mark a* a b c d e u

H865 01 Component 1 Raw 60 48 42 36 30 24 18 0

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

H866 01 Introduction to quantitative reasoning Raw 72 54 47 40 34 28 0H866 02 Critical maths Raw 60* 48 42 36 30 24 0

*Component 02 is weighted to give marks out of 72 Overall 144 112 97 83 70 57 0

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

H867 01 Introduction to quantitative reasoning Raw 72 54 47 40 34 28 0H867 02 Statistical problem solving Raw 60* 41 36 31 27 23 0

*Component 02 is weighted to give marks out of 72 Overall 144 103 90 77 66 56 0

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

6993 01 Additional Mathematics Raw 100 72 63 55 47 39 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

This unit has no entries in June 2017

For more information about results and grade calculations, see www.ocr.org.uk/ocr-for/learners-and-parents/getting-your-results

Documents

Oxford Cambridge and RSA Level 3 Certificate Quantitative ...mei.org.uk/files/papers/H867_02_SPS_2017_June.pdf · The study covers three families of birds: owls, hawks and crows