General Certificate of Education Advanced Subsidiary Examination January 2012

Mathematics

MS/SS1B

Unit Statistics 1B

Statistics Unit Statistics 1B Tuesday 17 January 2012

d

9.00 am to 10.30 am

For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.

e

s

n

Time allowed * 1 hour 30 minutes

e

Instructions * Use black ink or black ball-point pen. Pencil should only be used for drawing. * Fill in the boxes at the top of this page. * Answer all questions. * Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. * You must answer the questions in the spaces provided. Do not write outside the box around each page. * Show all necessary working; otherwise marks for method may be lost. * Do all rough work in this book. Cross through any work that you do not want to be marked. * The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures.

d

n

o

C

Information * The marks for questions are shown in brackets. * The maximum mark for this paper is 75. * Unit Statistics 1B has a written paper only. Advice * Unless stated otherwise, you may quote formulae, without proof, from the booklet. * You do not necessarily need to use all the space provided.

P47436/Jan12/MS/SS1B 6/6/6/

MS/SS1B

2

Giles, a keen gardener, rents a council allotment.

1

During early April 2011, he planted 27 seed potatoes. When he harvested his potato crop during the following August, he counted the number of new potatoes that he obtained from each seed potato. He recorded his results as follows. Number of new potatoes Frequency

46

7

8

9

10

11

5 12

2

2

1

4

8

6

4

(a)

Calculate values for the median and the interquartile range of these data.

(3 marks)

(b)

Advise Giles on how to record his corresponding data for 2012 so that it would then be possible to calculate the mean number of new potatoes per seed potato. (1 mark)

Dr Hanna has a special clinic for her older patients. She asked a medical student, Lenny, to select a random sample of 25 of her male patients, aged between 55 and 65 years, and, from their clinical records, to list their heights, weights and waist measurements.

2

Lenny was then asked to calculate three values of the product moment correlation coefficient based upon his collected data. His results were: (a)

0.365 between height and waist measurement;

(b)

1.16 between height and weight;

(c)

0.583 between weight and waist measurement. For each of Lenny’s three calculated values, state whether the value is definitely correct, probably correct, probably incorrect or definitely incorrect. (3 marks)

(02)

P47436/Jan12/MS/SS1B

3

During June 2011, the volume, X litres, of unleaded petrol purchased per visit at a supermarket’s filling station by private-car customers could be modelled by a normal distribution with a mean of 32 and a standard deviation of 10 .

3

Determine:

(a) (i)

PðX < 40Þ ;

(ii) PðX > 25Þ ; (iii) Pð25 < X < 40Þ .

(7 marks)

(b)

Given that during June 2011 unleaded petrol cost £1.34 per litre, calculate the probability that the unleaded petrol bill for a visit during June 2011 by a private-car customer exceeded £65 . (3 marks)

(c)

Give two reasons, in context, why the model Nð32, 102 Þ is unlikely to be valid for a visit by any customer purchasing fuel at this filling station during June 2011. (2 marks)

The records at a passport office show that, on average, 15 per cent of photographs that accompany applications for passport renewals are unusable.

4

Assume that exactly one photograph accompanies each application. Determine the probability that in a random sample of 40 applications:

(a) (i)

exactly 6 photographs are unusable;

(ii) at most 5 photographs are unusable; (iii) more than 5 but fewer than 10 photographs are unusable.

(7 marks)

(b)

Calculate the mean and the standard deviation for the number of photographs that are unusable in a random sample of 32 applications. (3 marks)

(c)

Mr Stickler processes 32 applications each day. His records for the previous 10 days show that the numbers of photographs that he deemed unusable were 8

6

10

7

9

7

8

9

6

7

By calculating the mean and the standard deviation of these values, comment, with reasons, on the suitability of the Bð32, 0.15Þ model for the number of photographs deemed unusable each day by Mr Stickler. (4 marks)

s

(03)

Turn over

P47436/Jan12/MS/SS1B

4

An experiment was undertaken to collect information on the burning of a specific type of wood as a source of energy. At given fixed levels of the wood’s moisture content, x per cent, its corresponding calorific value, y MWh=tonne, on burning was determined. The results are shown in the table.

5

x

5

10

15

20

25

30

35

40

45

50

55

60

65

y

5.2

4.7

4.3

4.0

3.2

2.8

2.5

2.2

1.8

1.5

1.3

1.0

0.6

(a)

Explain why calorific value is the response variable.

(1 mark)

(b)

Calculate the equation of the least squares regression line of y on x, giving your answer in the form y ¼ a þ bx . (5 marks)

(c)

Interpret, in context, your values for a and b.

(d)

Use your equation to estimate the wood’s calorific value when it has a moisture content of 27 per cent. (2 marks)

(e)

Calculate the value of the residual for the point ð35, 2.5Þ.

(f)

Given that the values of the 13 residuals lie between 0.28 and þ0.23 , comment on (1 mark) the likely accuracy of your estimate in part (d).

(g) (i)

Give a general reason why your equation should not be used to estimate the wood’s calorific value when it has a moisture content of 80 per cent. (1 mark)

(3 marks)

(2 marks)

(ii) Give a specific reason, based on the context of this question and with numerical

support, why your equation cannot be used to estimate the wood’s calorific value when it has a moisture content of 80 per cent. (2 marks)

Twins Alec and Eric are members of the same local cricket club and play for the club’s under 18 team.

6

The probability that Alec is selected to play in any particular game is 0.85 . The probability that Eric is selected to play in any particular game is 0.60 . The probability that both Alec and Eric are selected to play in any particular game is 0.55 . By using a table, or otherwise:

(a) (i)

show that the probability that neither twin is selected for a particular game is 0.10 ;

(ii) find the probability that at least one of the twins is selected for a particular game; (iii) find the probability that exactly one of the twins is selected for a particular game.

(5 marks)

(04)

P47436/Jan12/MS/SS1B

5

The probability that the twins’ younger brother, Cedric, is selected for a particular game is:

(b)

0.30 given that both of the twins have been selected; 0.75 given that exactly one of the twins has been selected; 0.40 given that neither of the twins has been selected. Calculate the probability that, for a particular game: (i)

all three brothers are selected;

(ii) at least two of the three brothers are selected.

(6 marks)

A random sample of 50 full-time university employees was selected as part of a higher education salary survey.

7

The annual salary in thousands of pounds, x, of each employee was recorded, with the following summarised results. X X x ¼ 2290:0 and ðx  xÞ2 ¼ 28 225:50 Also recorded was the fact that 6 of the 50 salaries exceeded £60 000 . (a) (i)

Calculate values for x and s, where s 2 denotes the unbiased estimate of s 2 . (3 marks)

(ii) Hence show why the annual salary, X , of a full-time university employee is unlikely

to be normally distributed. Give numerical support for your answer. (b) (i)

(2 marks)

Indicate why the mean annual salary, X , of a random sample of 50 full-time university employees may be assumed to be normally distributed. (2 marks)

(ii) Hence construct a 99% confidence interval for the mean annual salary of full-time

university employees. (c)

(4 marks)

It is claimed that the annual salaries of full-time university employees have an average which exceeds £55 000 and that more than 25% of such salaries exceed £60 000 . Comment on each of these two claims.

(3 marks)

Copyright Ó 2012 AQA and its licensors. All rights reserved.

(05)

P47436/Jan12/MS/SS1B

Version 1.0

General Certificate of Education (A-level) January 2012

Mathematics

MS/SS1B

(Specification 6360) Statistics 1B

Final

Mark Scheme

Mark schemes are prepared by the Principal Examiner and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all examiners participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every examiner understands and applies it in the same correct way. As preparation for standardisation each examiner analyses a number of students’ scripts: alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, examiners encounter unusual answers which have not been raised they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.

Further copies of this Mark Scheme are available from: aqa.org.uk Copyright © 2012 AQA and its licensors. All rights reserved. Copyright AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 3644723) and a registered charity (registered charity number 1073334). Registered address: AQA, Devas Street, Manchester M15 6EX.

Key to mark scheme abbreviations M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA c sf dp

mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent 2 or 1 (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach candidate significant figure(s) decimal place(s)

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.

MS/SS1B - AQA GCE Mark Scheme (VP) 2012 January Series

MS/SS1B Q 1 Median = 10 (a)

Solution

Marks

Total

Comments

B1

CAO

B1

CAO; either May be implied by IQR = 2

Upper quartile = 11 Lower quartile = 9 Interquartile range = 2

B1

3

Do not group results

B1

1

(b)

Illustrations for B1: Use all values Replace ≤6 by or use (0), 1, ..., 6 Replace ≥12 by or use 12, 13, ... Record exact values/frequencies

Q 2 (a)

(b)

(c)

Solution

CAO; do not award if seen to be not based on 11 and 9

OE statement that implies non grouping or recording of all separate observed values Illustrations for B0: Record max and/or min values Construct frequency table Use 1, 2 or 12, 13

Total

4

Marks

Total

Comments

Probably correct

B1

CAO; accept minimum of PC or Pc or pC or pc

Definitely incorrect

B1

CAO; accept minimum of DI or Di or dI or di

Probably incorrect

B1

3

Notes: Ignore reasoning in all parts, unless it includes 2 of the 4 statements in which case  B0 If answers not labelled, then assume above order

CAO; accept minimum of PI or Pi or pI or pi Definitely wrong, etc  B0 Likely correct, etc  B0

Total

3

MS/SS1B - AQA GCE Mark Scheme (VP) 2012 January Series

MS/SS1B (cont) Q 3(a) (i)

Solution Volume, X ~ N(32, 102)

40  32   P(X < 40) = P  Z   10  

(ii)

Marks

Total

M1

Standardising 40 with 32 and 10; allow (32 – 40)

= P(Z < 0.8)

A1

CAO; ignore inequality and sign May be implied by a correct answer

= 0.788

A1

3

= P(Z < +0.7)

M1

= 0.758

A1

P(25 < X < 40) =

(0.78814)

Area change May be implied by a correct answer or an answer > 0.5 2

(i) – (1 – (ii)

M1

= 0.78814 – (1 – 0.75804) = 0.546 Note: If (ii) is 0.242, then (0.788 – 0 242) = 0.546  M0 A0

A1

AWRT

(0.75804)

OE; allow new start ignoring (i) & (ii) Allow even if incorrect standardising providing 0 < answer < 1 May be implied by a correct answer 2

AWRT

(0.54618)

P(B > £65) = 48.5  32   P Z   10  

M1

Attempt to change from B to X using (48 to 49), 32 and 10 or Attempt to work with distribution of B using 65, (42.8 to 42.9) and 13.4

= P(Z > 1.65) = 1 – P(Z < 1.65)

m1

Area change May be implied by a correct answer or an answer < 0.5

= 1 – 0.95053 = 0.049 to 0.05(0)

A1

or 65  42.88   P Z   13.4  

(c)

AWRT

P(X > 25) = P(Z > –0.7)

(iii)

(b)

Comments

Other fuels Other vehicles with an example (not other cars) Other types of customer Minimum purchase (policy) Purchases in integer/fixed £s Customers filling fuel cans

3

AWFW

(0.04947)

B2,1

2

Size of car/engine/fuel tank  B0 Price of fuel  B0 Customer paying capacity  B0 Must be two clearly different valid reasons for award of B2 Drivers and vehicles related  B1 eg lorry drivers & lorries

Total

12

MS/SS1B - AQA GCE Mark Scheme (VP) 2012 January Series

MS/SS1B (cont) Q 4(a) (i)

(ii) (iii)

Solution U ~ B(40, 0.15)

Marks M1

P(U = 6) = 0.6067 – 0.4325 or  40  6 34 =    0.15   0.85  6   = 0.174 P(U ≤ 5)

M1 Can be implied by a correct answer

= 0.432 to 0.433

A1

3

AWRT

(0.1742)

B1

1

AWFW

(0.4325)

See supplementary sheet for individual probabilities

MINUS 0.4325 or 0.2633

(p1)

(p2)

Mean or  = 32 × 0.15

Accept 3 dp rounding or truncation but allow 0.97 p2 – p1  M0 M0 A0 (1 – p2) – p1  M0 M0 A0 p1 – (1 – p2)  M1 M0 A0 (1 – p2) – (1 – p1)  M1 M1 (A1) only providing result > 0 Accept 3 dp rounding or truncation

M1

M1

= 0.5(00) to 0.501

(c)

Comments Used somewhere in (a) Accept 3 dp rounding or truncation

P(5 < U < 10) = 0.9328 or 0.9701

(b)

Total

A1

3

AWFW

(0.5003)

B1

CAO

(V or 2 =) 32  0.15  0.85 or (SD or  =) 32  0.15  0.85

M1

Either numerical expression; ignore terminology May be implied by 4.08 CAO seen or 2.02 AWRT seen

(SD or ) = 2.02

A1

Mean = 7.7

B1

CAO

SD = 1.26 to 1.34

B1

AWFW

= 4.8

3

(Sample) mean is bigger / greater / different or 7.7/32 = 0.24 > 0.15 and (Sample) SD is smaller / less / different

Bdep1

So model appears unsuitable

Bdep1

4

Total

14

AWRT (2.0199) Do not award if labelled V or 2

  x  77 

 x

2

 609 

Both; dependent on all previous 5 marks of B1 M1 A1 B1 B1 Can be scored for incorrect (b) re-done correctly in (c) Means & SDs different  Bdep0 OE; dependent on Bdep1

MS/SS1B - AQA GCE Mark Scheme (VP) 2012 January Series

MS/SS1B (cont) Q Solution See supplementary sheet for alternative solutions 5

Marks

Total

B1

1

Comments

and additional guidelines to parts (b), (d) and (e)

(a)

Calorific value depends upon moisture content Moisture (content) is set/are fixed values

(b)

b (gradient) = –0.076 b (gradient) = –0.07 to –0.08

B2 (B1)

a (intercept) = a (intercept) =

B2 (B1)

Must be in context; not “it”, etc Use of x and y  B0

AWRT; including –ve sign AWFW; including –ve sign

(–0.07582)

Treat rounding of correct answers as ISW

Thus

(c)

5.35 to 5.36 5.1 to 5.6

y = (5.35 to 5.36) –0.076x

a: calorific value of wood with zero/no moisture or dry maximum calorific value

B1

b: each 1(%) rise in moisture content reduces calorific value by 0.076 MWh/tonne

B2

As x increases y decreases

(d)

BF1

y27 = 3.28 to 3.32

AWFW AWFW 5

3

2

(B1)

(e)

r(35, 2.5) = –0.21 to –0.19 = 0.1 to 0.3

B2 (B1)

2

(f)

Good/reasonable/accurate/correct/etc B1

1

AWFW; including –ve sign (–0.20000) AWFW; ignore sign

OE; ignore reasoning Very good (B1)

Extrapolation/outside (observed) range (of x)

B1

y80 = –0.5 to –1

B1

Negative value for calorific value is impossible or More energy needed than is generated

AWFW (3.30659) AWFW; even if by interpolation from original data giving likely values of 3 or 3.04

Accept more positive qualifying adjectives

(ii)

In context and with values; F on b b ≥ 0  B0 Negative relationship/correlation

= 2.5 to 3.5

(g)(i)

F on a and b even if rounded

OE; a ≤ 0  B0

(B1)

B2

(5.35385)

1

OE

AWFW

Bdep1

2

Total

17

Not good (B0)

(–0.71209)

OE; dependent on B1 Must be in context; negative value impossible  Bdep0

MS/SS1B - AQA GCE Mark Scheme (VP) 2012 January Series

MS/SS1B (cont) Q Solution See supplementary sheet for alternative solutions 6

Marks

Total

Comments

to parts (a)(i) and (b)(ii)

(a)(i)

Table Method (2- way with either R or C totals)

E E Total

A 0.55 0.30 0.85

A 0.05 0.10 0.15

Total 0.60 0.40 1.00

B1

0.15 or 0.4; CAO; allow fractions

B1

0.05 and 0.3; CAO; allow fractions

Bdep1

3

0.1; AG so dependent on B1 B1

(ii)

P(≥1) = 0.9 or 9/10

B1

1

CAO

(iii)

P(1) = 0.3 + 0.05 = 1 – (0.55 + 0.10) = 0.35 or 35/100 or 7/20

B1

1

CAO

P(3) = 0.55 × 0.30

B1

= 0.165 or 165/1000 or 33/200

B1

0.55 × (1 – 0.3) or 0.385

M1

(b)(i)

(ii)

or or

OE; implied by correct answer 2

(0.3 × 0.75) or 0.225 (0.05 × 0.75) or 0.0375 (0.35 × 0.75) or 0.2625

M1

At least one of these expressions or values

(0.385 + 0.2625) + 0.165

B1

OE; implied by correct answer

= 0.812 to 0.813 or

CAO

8125 1625 325 65 13 or or or or 10000 2000 400 80 16

AWFW A1

4 CAO

Total

11

(0.8125)

MS/SS1B - AQA GCE Mark Scheme (VP) 2012 January Series

MS/SS1B (cont) Q Solution 7(a) (i) 2290 x = 45.8 or 45800 50

28225.5  s   49 or 50 2

Marks

 s 

or

28225.5 49 or 50

s = 24(.0) or 24000 to 24001

Total

Comments

B1

CAO

M1

Ignore notation

A1

3

AWRT/AWFW

( = 23.75942)

SCs: (for no seen working) M1 A1 for 24.0 or 24000 to 24001 M1 A0 for 24 or 23700 to 23800 (ii)

(b)(i)

See supplementary sheet for alternative solutions

x  ns =  45.8  n  24.0  < 0 SC: Accept quoted values of (–4 to –1) (n = 2) or (–28.5 to –23.5) (n = 3) (both AWFW) and

M1

negative salaries are impossible

A1

Large sample or n > 25 or 30 or n = 50 so

B1

CLT applies

Bdep1 99% (0.99)  z = 2.57 to 2.58

(ii)

s n

CI for  is

x  z 

Thus

45.8  2.5758 

24.0 50

Hence 45.8  (8.7 to 8.8) or 45800  (8700 to 8800) OR (37.(0) to 37.1, 54.5 to 54.6) or (37000 to 37100, 54500 to 54600) (c)

(24.00064)

Allow (45 to 47) and any multiple of (23.5 to 24.5) which gives value < 0 Must clearly state the value of a numerical expression 2

OE; must be in context Negative values impossible  A0 OE

2

Must indicate CLT; dependent on B1 Indication that other than sample mean is normally distributed  Bdep0

B1

AWFW

M1

Used with ( x & s) from (a)(i) and z(1.64 to 2.58) &  n with n > 1

AF1

F on ( x & s) with  50 or 49 & z(1.64 to 1.65 or 2.32 to 2.33 or 2.57 to 2.58)

A1

4

(2.5758)

CAO/AWFW (8.74) Ignore (absence of) quoted units AWFW

See supplementary sheet for additional illustrations

Clear correct comparison of 55 or 55000 with c’s UCL or CI

B1

Accept 55000 compared with c’s 54.5 to 54.6 (ie different units)

(6/50 or 0.12 or 12%) 60 | N(45.8, 24.02)) = P(Z > 0.59)

M1

= 0.27 to 0.28

A1

2

Standardising 60 using 45.8 & 24.0

2

In addition to probability within range, must compare calculated value to 6/50 = 0.12 OE

Additional comment illustrations

It/(claimed) mean/(claimed) value > UCL/CI

B0

Must indicate 55 or 55000

99% have (mean) weights between CLs so ...

B0

Any comparison of 60 (£60 000) with UCL/CI

B0

Value of 60 does not refer to mean

P(X > 60 | N(45.8, 24.02)) = P(Z > 0.59) = (0.27 to 0.28) > 6/50 = 0.12

B0

Assumes salaries ~ N; cf (a)(ii)



Scaled mark unit grade boundaries - January 2012 exams A-level Max. Scaled Mark

Scaled Mark Grade Boundaries and A* Conversion Points A* A B C D E

Code

Title

LAW02 LAW03

LAW UNIT 2 LAW UNIT 3

94 80

69

MD01 MFP1 MM1A MM1B MPC1

MATHEMATICS UNIT MD01 MATHEMATICS UNIT MFP1 MATHEMATICS UNIT MM1A MATHEMATICS UNIT MM1B MATHEMATICS UNIT MPC1

75 75 100 75 75

-

MS1A MATHEMATICS UNIT MS1A MS/SS1A/W MATHEMATICS UNIT S1A - WRITTEN MS/SS1A/C MATHEMATICS UNIT S1A - COURSEWORK

-

73 63

66 57

59 51

52 45

46 40

62 56 50 44 67 60 53 46 no candidates were entered for this unit 59 52 46 40 61 55 49 43

39 39 34 37

100 75 25

-

74 54 20

65

56

47

38 28 10

MS1B MD02 MFP2 MM2B MPC2 MS2B MFP3 MPC3 MFP4 MPC4

MATHEMATICS UNIT MS1B MATHEMATICS UNIT MD02 MATHEMATICS UNIT MFP2 MATHEMATICS UNIT MM2B MATHEMATICS UNIT MPC2 MATHEMATICS UNIT MS2B MATHEMATICS UNIT MFP3 MATHEMATICS UNIT MPC3 MATHEMATICS UNIT MFP4 MATHEMATICS UNIT MPC4

75 75 75 75 75 75 75 75 75 75

69 59 69 69 67 64 60 63

56 64 52 63 66 63 60 57 54 57

49 57 45 55 59 55 52 50 48 51

42 50 38 47 52 47 44 43 42 45

36 44 31 39 46 40 37 37 37 39

30 38 25 32 40 33 30 31 32 33

MEST1 MEST2 MEST3 MEST4

MEDIA STUDIES UNIT 1 MEDIA STUDIES UNIT 2 MEDIA STUDIES UNIT 3 MEDIA STUDIES UNIT 4

80 80 80 80

67 74

55 63 57 68

47 54 47 56

40 45 37 45

33 36 27 34

26 28 18 23

PHIL1

PHILOSOPHY UNIT 1

90

-

55

49

43

37

32