General Certificate of Education Advanced Subsidiary Examination June 2011

Mathematics

MS/SS1B

Unit Statistics 1B

Statistics Unit Statistics 1B Friday 20 May 2011

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1.30 pm to 3.00 pm

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

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Time allowed * 1 hour 30 minutes

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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.

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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.

P38946/Jun11/MS/SS1B 6/6/6/

MS/SS1B

2

The number of matches in each of a sample of 85 boxes is summarised in the table.

1

Number of matches

Number of boxes

Less than 239

1

239–243

1

244–246

2

247

3

248

4

249

6

250

10

251

13

252

16

253

20

254

5

255–259

3

More than 259

1

Total

85

For these data:

(a) (i)

state the modal value;

(ii) determine values for the median and the interquartile range.

(1 mark) (3 marks)

Given that, on investigation, the 2 extreme values in the above table are 227 and 271 :

(b)

(i)

calculate the range;

(ii) calculate estimates of the mean and the standard deviation. (c)

(02)

(1 mark) (4 marks)

For the numbers of matches in the 85 boxes, suggest, with a reason, the most appropriate measure of spread. (2 marks)

P38946/Jun11/MS/SS1B

3

The diameter, D millimetres, of an American pool ball may be modelled by a normal random variable with mean 57.15 and standard deviation 0.04 .

2

Determine:

(a) (i)

PðD < 57:2Þ ;

(3 marks)

(ii) Pð57:1 < D < 57:2Þ .

(2 marks)

A box contains 16 of these pool balls. Given that the balls may be regarded as a random sample, determine the probability that:

(b)

(i)

all 16 balls have diameters less than 57.2 mm;

(2 marks)

(ii) the mean diameter of the 16 balls is greater than 57.16 mm.

(4 marks)

During a particular summer holiday, Rick worked in a fish and chip shop at a seaside resort.

3 (a)

He suspected that the shop’s takings, £y, on a weekday were dependent upon the forecast of that day’s maximum temperature, x C, in the resort, made at 6.00 pm on the previous day. To investigate this suspicion, he recorded values of x and y for a random sample of 7 weekdays during July.

(i)

x

23

18

27

19

25

20

22

y

4290

3188

5106

3829

5057

4264

4485

Calculate the equation of the least squares regression line of y on x.

(4 marks)

(ii) Estimate the shop’s takings on a weekday during July when the maximum

temperature was forecast to be 24 C.

(2 marks)

(iii) Explain why your equation may not be suitable for estimating the shop’s takings on a

weekday during February.

(1 mark)

(iv) Describe, in the context of this question, a variable other than the maximum

temperature, x, that may affect y.

(1 mark)

Seren, who also worked in the fish and chip shop, investigated the possible linear relationship between the shop’s takings, £z, recorded in £000s, and each of two other explanatory variables, v and w.

(b)

(i)

She calculated correctly that the regression line of z on v had a z-intercept of 1 and a gradient of 0.15 . Draw this line, for values of v from 0 to 40, on Figure 1 on page 4. s

(03)

Turn over

P38946/Jun11/MS/SS1B

4 (ii) She also calculated correctly that the regression line of z on w had a z-intercept of 5

and a gradient of 0.40 . Draw this line, for values of w from 0 to 10, on Figure 2 below.

(3 marks)

Figure 1 z~ 6– 5– 4– 3– 2– 1– 10

20

30

40

~

0– 0

v

1 –

Figure 2 z~ 6– 5– 4– 3– 2– 1– 2

4

6

8

10

~

0– 0

w

1 –

(04)

P38946/Jun11/MS/SS1B

5

Rice that can be cooked in microwave ovens is sold in packets which the manufacturer claims contain a mean weight of more than 250 grams of rice.

4

The weight of rice in a packet may be modelled by a normal distribution. A consumer organisation’s researcher weighed the contents, x grams, of each of a random sample of 50 packets. Her summarised results are: X x ¼ 251:1 and ðx  xÞ2 ¼ 184:5 (a)

Show that, correct to two decimal places, s ¼ 1:94 , where s 2 denotes the unbiased estimate of the population variance. (1 mark)

(b) (i)

Construct a 96% confidence interval for the mean weight of rice in a packet, giving the limits to one decimal place. (4 marks)

(ii) Hence comment on the manufacturer’s claim.

(2 marks)

The statement ‘250 grams’ is printed on each packet.

(c)

Explain, with reference to the values of x and s, why the consumer organisation may consider this statement to be dubious. (2 marks)

Emma visits her local supermarket every Thursday to do her weekly shopping.

5 (a)

The event that she buys orange juice is denoted by J , and the event that she buys bottled water is denoted by W . At each visit, Emma may buy neither, or one, or both of these items. (i)

Complete the table of probabilities, printed below, for these events, where J ’ and W ’ denote the events ‘not J ’ and ‘not W ’ respectively. (3 marks)

(ii) Hence, or otherwise, find the probability that, on any given Thursday, Emma buys

either orange juice or bottled water but not both.

(2 marks)

(iii) Show that: (A) the events J and W are not mutually exclusive; (B) the events J and W are not independent.

Turn over

s

(05)

(3 marks)

P38946/Jun11/MS/SS1B

6

Rhys visits the supermarket every Saturday to do his weekly shopping. Items that he may buy are milk, cheese and yogurt.

(b)

The The The The

probability, PðMÞ , that he buys milk on any given Saturday is 0.85 . probability, PðCÞ , that he buys cheese on any given Saturday is 0.60 . probability, PðY Þ , that he buys yogurt on any given Saturday is 0.55 . events M, C and Y may be assumed to be independent.

Calculate the probability that, on any given Saturday, Rhys buys: none of the 3 items;

(2 marks)

(ii) exactly 2 of the 3 items.

(3 marks)

(i)

J

J’

0.65

W W’

Total

0.15

Total

0.30

1.00

An amateur tennis club purchases tennis balls that have been used previously in professional tournaments.

6

The probability that each such ball fails a standard bounce test is 0.15 . The club purchases boxes each containing 10 of these tennis balls. Assume that the 10 balls in any box represent a random sample. Determine the probability that the number of balls in a box which fail the bounce test is:

(a)

at most 2 ;

(1 mark)

(ii) at least 2 ;

(2 marks)

(iii) more than 1 but fewer than 5 .

(3 marks)

(i)

Determine the probability that, in 5 boxes, the total number of balls which fail the bounce test is:

(b)

(i)

more than 5 ;

(ii) at least 5 but at most 10 .

(06)

(2 marks) (3 marks)

P38946/Jun11/MS/SS1B

7

Three airport management trainees, Ryan, Sunil and Tim, were each instructed to select a random sample of 12 suitcases from those waiting to be loaded onto aircraft.

7 (a)

Each trainee also had to measure the volume, x, and the weight, y, of each of the 12 suitcases in his sample, and then calculate the value of the product moment correlation coefficient, r, between x and y. * *

Ryan obtained a value of 0.843 . Sunil obtained a value of þ0.007 .

Explain why neither of these two values is likely to be correct.

(2 marks)

Peggy, a supervisor with many years’ experience, measured the volume, x cubic feet, and the weight, y pounds, of each suitcase in a random sample of 6 suitcases, and then obtained a value of 0.612 for r.

(b)

*

*

Ryan and Sunil each claimed that Peggy’s value was different from their values because she had measured the volumes in cubic feet and the weights in pounds, whereas they had measured the volumes in cubic metres and the weights in kilograms. Tim claimed that Peggy’s value was almost exactly half his calculated value because she had used a sample of size 6 whereas he had used one of size 12 .

Explain why neither of these two claims is valid.

(2 marks)

Quentin, a manager, recorded the volumes, v, and the weights, w, of a random sample of 8 suitcases as follows.

(c)

(i)

v

28.1

19.7

46.4

23.6

31.1

17.5

35.8

13.8

w

14.9

12.1

21.1

18.0

19.8

19.2

16.2

14.7

Calculate the value of r between v and w.

(ii) Interpret your value in the context of this question.

(3 marks) (2 marks)

END OF QUESTIONS

Copyright ª 2011 AQA and its licensors. All rights reserved.

(07)

P38946/Jun11/MS/SS1B

MS/SS1B Q 1 (a)(i)

Solution

(ii)

Marks

Total

Comments

Mode = 253

B1

1

Median = 252

B1

CAO

B1

CAO; either May be implied by IQR = 3

CAO

Upper quartile = 253 Lower quartile = 250 Interquartile range = 3

B1

3

CAO; do not award if seen to be not based on 253 and 250

Range = 271 – 227 = 44

B1

1

CAO; do not award if seen to be not based on 271 and 227

(ii) Mean, x = 251 to 251.4 Award B1 if divisor seen not to be 85 but answer in range

B2

(b)(i)

AWFW

Note: If B0 then can award M1 for attempt at  fx  85 seen Standard deviation,

 fx

= 21352

x = 251.2

Ignore notation and condone incorrect midpoints (eg upper or lower limits used)

s or  = 4.21 to 4.28

B2

4

AWFW  = 4.217

 fx 2 = 5365134 s = 4.242

Award B1 if divisor seen not to be 84 or 85 but answer in range (c) Interquartile range (IQR)

B1

Not affected by unknown/large/small/extreme/ outlying/227 & 271 values

Bdep1

Named 2

Or equivalent Dependent on previous B1

Only negative comments on other measures  Bdep0

More than one named  B0 Bdep0 Range  B0 Bdep0

OR Standard deviation (s or  )

Named

(B1)

Uses all data values

Or equivalent Dependent on previous (B1)

(Bdep1)

Only negative comments on other measures  Bdep0

Total

11

MS/SS1B (cont) Q 2

Solution

Marks

Diameter, D ~ N(57.15, 0.042) 57.2  57.15   P(D < 57.2) = P  Z   0.04  

(a)(i)

Total

Comments

M1

Standardising 57.2 with 57.15 and 0.04; allow (57.15 – 57.2)

= P(Z < 1.25)

A1

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

= 0.894 to 0.895

A1

= p – (1 – p)

M1

= 2 × 0.89435 – 1 = 0.788 to 0.79(0)

A1

3

AWFW

(0.89435)

(ii) P(57.1 < D < 57.2) Allow even if incorrect standardising providing p – (1 – p) seen May be implied by a correct answer 2

(b)(i) P(16 balls < 57.2) = p16

0 < p < 1

M1

= [(a)(i)]16 = (0.89435)16 = 0.166 to 0.17(0)

A1

(ii)

with

2

= P(Z > 1) = 1 – P(Z < 1)

= 1 – 0.84134 = = 0.158 to 0.159

AWFW

B1

M1

Standardising 57.16 with 57.15 and 0.01 or equivalent; allow (57.15 – 57.16)

m1

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

A1

4

Notes: Ignore partial/incomplete attempts at (ii) in (i) if followed by correct method

AWFW (0.15866) (1 – answer)  B1 M1 max Mark two complete answers in (i) as two attempts so (0 + 2)/2  1max

Answer to (i) or (ii) repeated

Mark as per scheme; thus (2 max, 0) or (0, 4 max) Total

(0.16754)

CAO Stated or used (see Notes below) CAO If only seen in (b)(i), allow just B1

Sd of D16 = 0.04/16 = 0.01 57.16  57.15   P  D16  57.16  = P  Z   0.01  

(0.78870)

Any probability to power 16 or 1 – p16; do not allow multiplying factors If only seen in (b)(ii), allow just M1

Variance of D16 = 0.042/16 = 0.0001 or

AWFW

11

MS/SS1B (cont) Q 3

Solution

Marks

b (gradient) = 191 b (gradient) = 190 to 192

(a)

Total

B2 (B1)

Comments CAO AWFW Treat rounding of correct answers as ISW

a (intercept) = 115 a (intercept) = 93 to 137

B2 (B1)

4

CAO AWFW

OR Attempt at

 x  x 2  y &  xy

or Attempt at S xx & S xy

 y 

154 3452 30219 & 677042 (133170091) (all 4 attempted)

2

(M1)

S 

12224 & 64 (2714668) (both attempted)

yy

Attempt at correct formula for b (gradient) b (gradient) = 191 a (intercept) = 115

(m1) (A1) (A1)

CAO CAO If a and b are not identified anywhere in question, then: 190 to 192  B1 93 to 137  B1

Accept a & b interchanged only if identified and used correctly in (ii)

(ii) y24 = 115 + 191 × 24

= £4699 or £4700 = £4650 to £4750 SC: (4290 + 5057)/2 = 4673 to 4674  B1 If B0 but clear evidence of correct use of c’s equation with x = 24

(iii)

(Maximum) temperature (in February) is likely to be/will be lower/different

B2 (B1)

2

B1

Rainfall amount/wind strength/sunshine hours/ daylight hours/opening times/day of week/ visitor numbers/public holidays/school holidays/ local attractions/etc

1

B1

1

Allow if at least 1 variable correctly identified

Total

(£4699)

(M1)

Must imply a temperature comparison with July

(iv)

Either; ignore units AWFW

8

Or equivalent; must be clear indication that (max) temperature is less than/different Extrapolation/not July/not summer/winter/etc  B0 Or equivalent Accept any sensible reason; do not penalise for dubious ‘variable name’ so, for example, accept ‘rainfall’ Minimum/average temp/etc  B0 Quality or price of food/ staff/etc  B0

MS/SS1B (cont) Q 3

Solution

Marks

Total

Comments

(b) Any line (straight, freehand, curve) from (0, –1) on Figure 1 or from (0, 5) on Figure 2

B1

Accept clear marking of (0, –1) or (0, 5) with no line

(i) Straight, not freehand, line from (0, –1) to (40, 5) on F1 only; allow line extensions and only very minor inaccuracies in points plotted

B1

(10, 0.5)

(ii) Straight , not freehand, line from (0, 5) to (10, 1) on F2 only; allow line extensions and only very minor inaccuracies in points plotted

B1

3

Notes: Both lines on F1  B1 B1 B0 max Both lines on F2  B1 B0 B1 max >1 undeleted line on either F1 or F2  2 max Total

3

(2, 4.2)

(20, 2)

(4, 3.4)

(30, 3.5)

(6, 2.6)

(8, 1.8)

MS/SS1B (cont) Q 4 (a)

Solution 184.5 49

or

Marks

50 1.92  49

B1

Total

1

= 1.94 (b) (i)

96% (0.96)  z = 2.05 to 2.06

s n

CI for  is

x  z 

Thus

251.1  2.0537 

Hence or

1.94 50 or 49

251.1  0.6

Comments Fully correct expression or equivalent must be seen Note: s  184.5 50  1.939  B0 AG

B1

AWFW

M1

Used with 251.1 and 1.94 correctly Must have n with n > 1

AF1

F on z only

Adep1

4

(250.5, 251.7)

(2.0537)

CAO/AWRT Dependent on AF1 but not on z so can be gained using an incorrect z AWRT

(ii) Claim is  > 250 Clear correct comparison of 250 with LCL or CI so Claim is supported/reasonable/correct/true/etc Must be consistent with c’s comparison (c)

x  ns = 251.1  n  1.94 < 250 SC: Quoted values of 249.2, 247.2 or 245.3 (AWRT)  M1

BF1 Bdep1

F on CI 2

Dependent on BF1 Allow any multiple of 1.94 Must clearly indicate the value of a numerical expression giving a result less than 250

M1

so Some individual packets are likely to/will contain less than 250 grams

A1 Total

(250 < LCL or CI)

2 9

Or equivalent

MS/SS1B (cont) Q 5 (a)(i)

Solution

W W Total

J 0.55 0.15 0.70

J 0.10 0.20 0.30

Marks Total 0.65 0.35 1.00

Total

B1

0.35 and 0.7; CAO

B1

0.55; CAO

B1

3

Notes: Use of Venn or tree diagrams without table completion  B0 B0 B0 Printed table not completed but constructed and completed on Page 12/13  B1 B1 B1 max (ii) P(purchases exactly one) = P W  J    0.15

(iii) (A)

Only c’s equivalent to 0.10 shown and added to 0.15 Can be implied by correct answer

M1

= 0.10 + 0.15 = 0.25 or 25/100 or 5/20 or 1/4

A1

P W  J  = 0.8 &/ P W   P  J  = 1.35

B1

2

&/ P W  = 0.65 or

Bdep1

P  J | W   0.55 0.65 = 0.85

Do not accept use of W  and/or J  AWRT 3

&/ P  J  = 0.70 or

CAO

Any one of these three seen Ignore contradictions, explanations & justifications

B1

P W | J   0.55 0.70 = 0.79

0.1 and 0.2; CAO Accept fractional answers Do not accept percentages

or P W  J  = 0.55 (>0); accept if indicated in a Venn diagram or P W   P  J  = 1.35 >0 or impossible (B)

Comments

P W   P  J  = 0.45 to 0.46

Any one of these three seen Ignore contradictions, explanations & justifications AWFW

&/ P W  J  = 0.55 (b) (i)

Do not allow multiplying factors in (b)

P(0) = 0.15 × 0.40 × 0.45

B1

= 0.027 or 27/1000 (ii) P(2) = 0.85 × 0.60 × 0.45 = 0.2295 + 0.85 × 0.40 × 0.55 = 0.1870 + 0.15 × 0.60 × 0.55 = 0.0495 or

B1

2

For either method: At least two bold expressions correct Only one bold expression correct Can be implied by correct answer For second method: Must have ‘1 –’ for any marks

M2 (M1)

= 1 – (0.027 + 0.2265 + 0.2805) = 0.466 or 466/1000 or 233/500 Total

A1

Can be implied by correct answer or 1 – (0.2265 + 0.466 + 0.2805) CAO

3 13

CAO; do not imply this from (i)

MS/SS1B (cont) Q 6 (a) (i)

Solution

X ~ B(10, 0.15) P(X  2) = 0.82(0)

Marks

Total

B1

1

Comments

AWRT

(0.8202)

(ii) P(X ≥ 2) = 1 – P(X  1) = 1 – (0.5443 or 0.8202)

M1

= 0.455 to 0.456

A1

(iii) P(1 < X < 5) = 0.9901 or 0.9986

(p1)

M1

minus 0.5443 or 0.1969

(p2)

M1

= 0.445 to 0.446

A1

OR B(10, 0.15) expressions stated for at least 3 terms within 1  X  5 gives probability = 0.445 to 0.446 (b)

Requires ‘1 –’ Accept 3/2 dp rounding or truncation Can be implied by 0.455 to 0.456 but not by 0.179 to 0.18(0) 2

AWFW

(0.4557)

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

AWFW

(0.4458)

(M1)

Can be implied by a correct answer

(A2)

AWFW

(0.4458)

Normal approximation  0 marks

Y ~ B(50, 0.15)

(i) P(Y > 5) = 1 – P(Y  5) = 1 – (0.2194 or 0.1121)

M1

= 0.78(0) to 0.781

A1

(ii) P(5 ≤ Y ≤ 10) = 0.8801 or 0.7911

(p1)

M1

minus 0.1121 or 0.2194

(p2)

M1

= 0.768

A1

OR B(50, 0.15) expressions stated for at least 3 terms within 4  Y  10 gives probability = 0.768 Total

2

Requires ‘1 –’ Accept 3 dp rounding or truncation Can be implied by 0.78(0) to 0.781 but not by 0.888 to 0.89 AWFW (0.7806) Accept 2/3 dp rounding or truncation p2 – p1  M0 M0 A0 (1 – p2) – p1  M0 M0 A0 p1 – (1 – p2)  M1 M0 A0 only providing result > 0 Accept 3 dp rounding or truncation

3

AWRT

(0.7680)

(M1)

Can be implied by a correct answer

(A2)

AWRT 11

(0.7680)

MS/SS1B (cont) Q 7

Solution

Marks

(a) Ryan: Value indicates that as volume increases then weight decreases

B1

Sunil: Value indicates no correlation/relationship/ association/link between volume and weight

B1

SC: If B0 B0: Would expect weight to increase with volume or Would expect strong(er) positive correlation between weight and volume (b) Ryan & Sunil: r is not affected by units/(linear) scaling Tim: r is not affected by sample size or 2 × 0.612 > 1  impossibility (c) (i)

r = 0.541 to 0.543 r = 0.54 to 0.55 r = 0.5 to 0.6

Total

Comments

Or equivalent in context

2

(B1)

Or equivalent in context

Or equivalent in context

B1

Or equivalent

B1

2

Either; or equivalent

B3 (B2) (B1)

3

AWFW AWFW AWFW

(0.54186)

OR Attempt at

216 6633.16 136 2376.84 3795.5 (all 5 attempted) Accept notation of x and y 801.16 64.84 & 123.5 (all 3 attempted)

 v  v2  w  w2 &  vw (M1)

or Attempt at Svv Sww & Svw Attempt at substitution into correct corresponding formula for r r = 0.541 to 0.543

(m1) (A1)

AWFW Dependent on 0.5 ≤ r ≤ 0.6 Or equivalent; must qualify strength and state positive Bdep0 for very strong/strong/high/ good/average/medium/reasonable/ poor/very weak/little/etc

(ii) (Quite or fairly) weak/some/moderate positive (linear) correlation/relationship/ association/link (but not ‘ trend’) between volumes and weights of suitcases

Bdep1

B1 Total TOTAL

&

2 9 75

Context; providing 0 < r < 1



Scaled mark unit grade boundaries - June 2011 exams A-level

Code

Title

Max. Scaled Mark

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

MS1B MD02 MFP2 MM2B MPC2 MS2B XMCA2 MFP3 MM03 MPC3 MS03 MFP4 MM04 MPC4 MS04 MM05

GCE MATHEMATICS UNIT S1B GCE MATHEMATICS UNIT D02 GCE MATHEMATICS UNIT FP2 GCE MATHEMATICS UNIT M2B GCE MATHEMATICS UNIT PC2 GCE MATHEMATICS UNIT S2B GCE MATHEMATICS UNIT XMCA2 GCE MATHEMATICS UNIT FP3 GCE MATHEMATICS UNIT M03 GCE MATHEMATICS UNIT PC3 GCE MATHEMATICS UNIT S03 GCE MATHEMATICS UNIT FP4 GCE MATHEMATICS UNIT M04 GCE MATHEMATICS UNIT PC4 GCE MATHEMATICS UNIT S04 GCE MATHEMATICS UNIT M05

75 75 75 75 75 75 125 75 75 75 75 75 75 75 75 75

69 62 68 62 88 69 67 68 68 68 63 58 67 62

59 64 55 62 54 54 76 64 59 59 62 61 57 51 60 55

52 56 48 55 47 46 66 55 51 52 54 53 51 46 52 48

46 49 41 48 41 38 57 46 43 46 46 46 45 41 44 41

40 42 34 41 35 30 48 38 36 40 39 39 39 36 37 34

34 35 28 34 29 23 39 30 29 34 32 32 33 31 30 28

MEST1 MEST2 MEST3 MEST4

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

80 80 80 80

70 74

55 63 61 68

47 54 50 56

40 45 39 45

33 36 28 34

26 28 18 23

MHEB1 MHEB2

GCE MODERN HEBREW UNIT 1 GCE MODERN HEBREW UNIT 2

100 100

80

61 71

54 62

47 54

40 46

34 38

MUSC1 MUS2A MUS2B MUS2C

GCE MUSIC UNIT 1 GCE MUSIC UNIT 2A GCE MUSIC UNIT 2B GCE MUSIC UNIT 2C

80 60 60 60

-

57 45 49 49

51 40 43 44

45 35 37 39

39 30 32 34

34 26 27 29