General Certificate of Education June 2006 Advanced Subsidiary Examination

MATHEMATICS Unit Statistics 1B

MS/SS1B

STATISTICS Unit Statistics 1B Wednesday 24 May 2006 1.30 pm to 3.00 pm For this paper you must have: * an 8-page answer book * the blue AQA booklet of formulae and statistical tables You may use a graphics calculator.

Time allowed: 1 hour 30 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MS/SS1B. Answer all questions. Show all necessary working; otherwise marks for method may be lost. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. * *

* * *

Information The maximum mark for this paper is 75. The marks for questions are shown in brackets. Unit Statistics 1B has a written paper only.

* * *

Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet.

*

P85535/Jun06/MS/SS1B 6/6/6/

MS/SS1B

2

Answer all questions.

1 The table shows, for each of a random sample of 8 paperback fiction books, the number of pages, x, and the recommended retail price, £y, to the nearest 10p.

(a)

x

223

276

374

433

564

612

704

766

y

6.50

4.00

5.50

8.00

4.50

5.00

8.00

5.50

(i) Calculate the value of the product moment correlation coefficient between x and y. (3 marks) (ii) Interpret your value in the context of this question.

(2 marks)

(iii) Suggest one other variable, in addition to the number of pages, which may affect the recommended retail price of a paperback fiction book. (1 mark) (b) The same 8 books were later included in a book sale. The value of the product moment correlation coefficient between the number of pages and the sale price was 0.959, correct to three decimal places. What can be concluded from this value?

(2 marks)

2 The heights of sunflowers may be assumed to be normally distributed with a mean of 185 cm and a standard deviation of 10 cm. (a) Determine the probability that the height of a randomly selected sunflower: (i) is less than 200 cm;

(3 marks)

(ii) is more than 175 cm;

(3 marks)

(iii) is between 175 cm and 200 cm.

(2 marks)

(b) Determine the probability that the mean height of a random sample of 4 sunflowers is more than 190 cm. (4 marks)

P85535/Jun06/MS/SS1B

3

3 A new car tyre is fitted to a wheel. The tyre is inflated to its recommended pressure of 265 kPa and the wheel left unused. At 3-month intervals thereafter, the tyre pressure is measured with the following results: Time after fitting (x months) Tyre pressure ( y kPa) (a)

0

3

6

9

12

15

18

21

24

265

250

240

235

225

215

210

195

180

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

(4 marks)

(ii) Interpret in context the value for the gradient of your line.

(2 marks)

(iii) Comment on the value for the intercept with the y-axis of your line.

(2 marks)

(b) The tyre manufacturer states that, when one of these new tyres is fitted to the wheel of a car and then inflated to 265 kPa, a suitable regression equation is of the form y ˆ 265 ‡ bx The manufacturer also states that, as the car is used, the tyre pressure will decrease at twice the rate of that found in part (a). (i) Suggest a suitable value for b.

(2 marks)

(ii) One of these new tyres is fitted to the wheel of a car and inflated to 265 kPa. The car is then used for 8 months, after which the tyre pressure is checked for the first time. Show that, accepting the manufacturer's statements, the tyre pressure can be expected to have fallen below its minimum safety value of 220 kPa. (2 marks)

Turn over for the next question

P85535/Jun06/MS/SS1B

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4

4 The weights of packets of sultanas may be assumed to be normally distributed with a standard deviation of 6 grams. The weights of a random sample of 10 packets were as follows: 498 (a)

496

499

511

503

505

510

509

513

508

(i) Construct a 99% confidence interval for the mean weight of packets of sultanas, giving the limits to one decimal place. (5 marks) (ii) State why, in calculating your confidence interval, use of the Central Limit Theorem was not necessary. (1 mark) (iii) On each packet it states `Contents 500 grams'. Comment on this statement using both the given sample and your confidence interval. (3 marks)

(b) Given that the mean weight of all packets of sultanas is 500 grams, state the probability that a 99% confidence interval for the mean, calculated from a random sample of packets, will not contain 500 grams. (1 mark)

P85535/Jun06/MS/SS1B

5

5 Kirk and Les regularly play each other at darts. (a) The probability that Kirk wins any game is 0.3, and the outcome of each game is independent of the outcome of every other game. Find the probability that, in a match of 15 games, Kirk wins: (i) exactly 5 games;

(3 marks)

(ii) fewer than half of the games;

(3 marks)

(iii) more than 2 but fewer than 7 games.

(3 marks)

(b) Kirk attends darts coaching sessions for three months. He then claims that he has a probability of 0.4 of winning any game, and that the outcome of each game is independent of the outcome of every other game. (i) Assuming this claim to be true, calculate the mean and standard deviation for the number of games won by Kirk in a match of 15 games. (3 marks) (ii) To assess Kirk's claim, Les keeps a record of the number of games won by Kirk in a series of 10 matches, each of 15 games, with the following results: 8

5

6

3

9

12

4

2

6

Calculate the mean and standard deviation of these values. (iii) Hence comment on the validity of Kirk's claim.

5 (2 marks) (3 marks)

Turn over for the next question

P85535/Jun06/MS/SS1B

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6

6 A housing estate consists of 320 houses: 120 detached and 200 semi-detached. The numbers of children living in these houses are shown in the table. Number of children None

One

Two

At least three

Total

Detached house

24

32

41

23

120

Semi-detached house

40

37

88

35

200

Total

64

69

129

58

320

A house on the estate is selected at random. D denotes the event `the house is detached'. R denotes the event `no children live in the house'. S denotes the event `one child lives in the house'. T denotes the event `two children live in the house'. (D 0 denotes the event `not D'.) (a) Find:

(b)

(i) P…D†;

(1 mark)

(ii) P…D Ç R†;

(1 mark)

(iii) P…D È T†;

(2 marks)

(iv) P…D j R†;

(2 marks)

(v) P…R j D 0 †.

(3 marks)

(i) Name two of the events D, R, S and T that are mutually exclusive.

(1 mark)

(ii) Determine whether the events D and R are independent. Justify your answer. (2 marks) (c) Define, in the context of this question, the event: (i) D 0 È T;

(2 marks)

(ii) D Ç …R È S†:

(2 marks)

END OF QUESTIONS

P85535/Jun06/MS/SS1B

MS1B – AQA GCE Mark Scheme, 2006 June series

Key To Mark Scheme And Abbreviations Used In Marking M

mark is for method

m or dM A B E

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

or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA

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

MC MR RA FW ISW FIW BOD WR FB NOS G c sf dp

mis-copy mis-read required accuracy further work ignore subsequent work from incorrect work given benefit of doubt work replaced by candidate formulae book not on scheme graph 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. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be provided on the mark scheme. 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.

2

AQA GCE Mark Scheme, 2006 June series – MS1B

MS1B Q Solution 1(a)(i) r = 0.143 to 0.1432 or r = 0.142 to 0.144 or r = 0.1 to 0.2 Attempt at or Attempt at

Marks B3

Total AWFW

B2

AWFW

B1

AWRT

Σx Σx2 Σy Σy2 Σxy Sxx Syy Sxy

3952, 2228282 47.00, 292.0000 23517.50 M1

Attempt at a correct formula for r

m1

r = 0.143 to 0.1432

A1

(ii) Little/weak/no correlation/relationship/association between number of pages and (retail) price (iii) Size (page, thickness), author, ranking, publicity/marketing, cover design, recommendations on back, publisher, font, popularity, quality, print-run, etc (b) (Very) strong/almost exact positive/perfect correlation/relationship/ association between number of pages and sale/new price Sale price appears to be determined by number of pages Total

Comments

275994, 15.875, 299.5

3

AWFW or equivalent; but not poor

B1 B1

2

context

B1

1

or any sensible variable but not pictures, coloured pictures, age, words, weight, mass

or equivalent

B1 B1

2

B2

or equivalent 8

3

context

MS1B – AQA GCE Mark Scheme, 2006 June series

MS1B (cont) Q Solution 2(a) Height, X ~ N(185, 102) (i) P(X < 200) = P ⎛ Z < 200 − 185 ⎞ ⎜ ⎟ 10 ⎝ ⎠

Marks

Total

standardising (199.5, 200 or 200.5) with 185 and ( 10 , 10 or 102) and/or (185 − x)

M1

= P(Z < 1.5) = Φ(1.5) = 0.933

A1 A1

(ii) P(X > 175) = P ⎛ Z > 175 − 185 ⎞ ⎜ ⎟ 10 ⎝ ⎠

3

M1

= P(Z > –1) = P(Z < 1) = 0.841

m1 A1

(iii) P(175 < X < 200) = (i) – [1 – (ii)] = 0.93319 – [1 – 0.84134] = 0.774 to 0.775 (b) Mean of X = 185

102 = 25 4 190 − 185 ⎞ ⎛ P ( X > 190 ) = P ⎜ Z > ⎟ 5 ⎝ ⎠ = P(Z > 1) = 1 – Φ(1) = 0.159

Variance of X =

4

CAO; ignore sign AWRT (0.93319) standardising (174.5, 175 or 175.5) with 185 and ( 10 , 10 or 102) and/or (185 − x)

3

M1 A1

Comments

area change AWRT

(0.84134)

or equivalent 2

AWFW (0.77453) on (i) and (ii) providing > 0

B1

CAO; may be implied by use in standardising

B1

CAO; or equivalent

M1

standardising 190 with 185 and 5 and/or (185 − 190)

A1

4

Total

12

AWRT (0.15866) on (a)(ii) if used

AQA GCE Mark Scheme, 2006 June series – MS1B

MS1B (cont) Q Solution 3(a)(i) Gradient, b = –3.24 to –3.26 b = –3.2 to –3.3

Marks B2 B1

Total

AWFW AWFW

Intercept, a = 262 to 264

B2

AWFW

a = 260 to 270

B1

AWFW

Attempt at Σx Σx2 Σy Σxy or Attempt at Sxx Sxy Attempt at a correct formula for b b = –3.24 to –3.26 a = 262 to 264

Comments (–3.25)

( 262.88! )

108, 1836, 2015, 22425 M1 m1 A1 A1

540, –1755 4

AWRT AWFW

2

or equivalent or better

Accept a & b interchanged only if identified correctly in (b) and (c) (ii) Gradient, b: Decrease in pressure per month Change in pressure

B2 B1

(iii) Intercept, a: Initial pressure or pressure at x = 0 Reference to 265, actual or expected value

B1 B1

(b)(i) Value for b = 2 × [gradient or b from (a)(i)] = –6.4 to –6.6

(ii) P8 = 265 – 6.5 × 8

or equivalent; not y-intercept 2

M1 A1

accept 2b; ignore sign 2

M1

= 212 to 214 (< 220)

A1 Total

2 12

5

AWFW (–6.5) from (a)(i) but must be < 0 must use 265 and x = 8 and 2 × [b (< 0) from (a)(i)] or [from (b)(i) (< 0)] AWFW AG

MS1B – AQA GCE Mark Scheme, 2006 June series

MS1B (cont) Q Solution 4(a)(i) Mean, x = 505.2

Marks B1

99% ⇒ z = 2.57 to 2.58 or 99% ⇒ t = 3.25

Total

Comments CAO; stated or implied

B1

AWFW

(2.5758)

B1

AWRT

(3.250)

M1

use of; must have ÷ n with n > 1

(Knowledge of the t-distribution is not required in this unit)

CI for µ is

x ± ( z or t ) ×

(σ

or s )

(

n

6

Thus

505.2 ± 2.5758 ×

or

5.65 ⎞ ⎛ 5.96 505.2 ± 3.25 × ⎜ or ⎟ 9 ⎠ ⎝ 10

10

Hence 505.2 ± 4.9 or (500.3, 510.1)

A1

on x and z only

A1

on x only

)

use of t ⇒ 505.2 ± 6.1

(ii) Weights of packets can be assumed to be normally distributed (iii) Given sample: 3 in 10/ some of packets have weights below 500 grams

A1

5

AWRT

B1

1

accept ‘population of weights’; not ‘sample of weights’ or ‘it’

B1

Confidence interval: CI > 500 Conclusion: Statement does not appear justified

or equivalent

B1

(b) 0.01 or 1% Total

6

on CI in (a)(i)

B1 dep

3

or equivalent dependent on both B1 and B1

B1

1 10

CAO; or equivalent

AQA GCE Mark Scheme, 2006 June series – MS1B

MS1B (cont) Q

Solution

Marks M1

5(a) B(15, 0.3) (i) P(K = 5) = P(K ≤ 5) – P(K ≤ 4) ⎛15 ⎞ 5 10 P(K = 5) = ⎜ ⎟ ( 0.3) ( 0.7 ) ⎝5⎠ = 0.7216 – 0.5155 = 0.2055 to 0.2065

Total

use of in (a)

M1 A1

Comments

may be implied 3

AWFW

(0.2061)

(ii) (Fewer than) half ⇒ 7 or 7½ or 8 Thus require P(K ≤ 7 or < 8) = 0.9495 to 0.9505

B1 M1 A1

3

stated or implied used or implied by correct answer AWFW (0.9500)

(iii) P(2 < K < 7) = 0.8689 or 0.9500 minus 0.1268 or 0.2969 = 0.7415 to 0.7425 or B(15, 0.3) expressions stated for at least 3 terms within 2 ≤ K ≤ 7 Answer

M1 M1 A1

3

AWFW

(b)(i) Mean, µ = np = 15 × 0.4 = 6

M1

Variance, σ = np (1 − p ) = 6 × 0.6 = 3.6 Standard deviation =

A1

3.6 = 1.89 to 1.9

or implied by a correct answer

A2 B1 M1

2

CAO use of σ 2 = np (1 − p ) 3

AWFW; or equivalent

2

CAO (Σx = 60) CSO if evidence of np(1–p) or 1.9 AWFW; or equivalent. (Σx2 = 440)

3

on 2 means; accept 156 = 0.4 if not contradicted by x in (ii) dependent on 2 correct SDs dependent on 2 correct SDs

B1

(ii) Mean, x = 6

Standard deviation, s or σ = 2.82 to 2.99 (iii) Means are same/equal

B1 B1

Standard deviations are different Reason to doubt validity of Kirk’s claim

(0.7421)

B1 dep B1 dep

Total

17

7

MS1B – AQA GCE Mark Scheme, 2006 June series

MS1B (cont) Q 6

Solution D(D)

|

0(R)

1(S)

2(T)

≥3

|

|

24

32

41

SD(D′) |

40

37

88

35 | 200

T

64

69

129

58 | 320

|

(a)(i) P(D) =

120 3 or or 0.375 320 8

120 + 88 129 + 24 + 32 + 23 = 320 320 208 13 or or 0.65 = 320 20

P(D ∪ T) =

24 3 or or 0.375 64 8

or

R and T

1

CAO; or equivalent

B1

1

CSO; or equivalent

2

CAO; or equivalent

M1

A1

or

M0 if independence assumed

2

M1

S and T

(ii) P(D) = 0.375 = P(D | R) or (i) = (iv)

A1

3

CAO; or equivalent

B1

1

not D and D′ P(D) × P(R) = 0.375 × 0.2 = 0.075 = P(D ∩ R) or (ii) or P(R | D) = P(R) = 0.2, etc

M1

so YES (c)(i) A semi-detached house or two children (or both) (ii) A detached house and/with less than two children

A1

2

B1 B1

2

B1 B1 Total TOTAL

8

CAO; or equivalent

numerator allow independence assumed denominator

M1

40 1 = or or 0.2 200 5

R and S

B1

A1

40 ′ P R ∩ D ( ) ( 320 ) (v) P(R | D′) = = 200 P ( D′ ) ( 320 )

(b)(i)

Comments

M1

24 ( 320 ) (iv) P(D | R) = P ( D ∩ R ) = ( ii ) = 64 P(R) P(R) ( 320 ) =

Total

23 | 120

(ii) P(D ∩ R) = 24 or 3 or 0.075 320 40 (iii)

Marks T_

2 16 75

CAO or equivalent CAO (0 or 1 must not include ‘both’)

AQA June Examinations 2006 Scaled Mark Unit Grade Boundaries (GCE Specifications) Unit Code

Unit Title

MD01 MD02 MFP1 MFP2 MFP3 MFP4 MM03 MM04 MM05 MM1A MM1B MM2A MM2B MPC1 MPC2 MPC3 MPC4 MS03 MS04 MS1A MS1B MS2A MS2B

Scaled Mark Grade Boundaries B C D

Maximum Scaled Mark

A

GCE MATHEMATICS UNIT D01

75

61

54

47

40

33

GCE MATHEMATICS UNIT D02 GCE MATHEMATICS UNIT FP1 GCE MATHEMATICS UNIT FP2 GCE MATHEMATICS UNIT FP3 GCE MATHEMATICS UNIT FP4 GCE MATHEMATICS UNIT M03 GCE MATHEMATICS UNIT M04 GCE MATHEMATICS UNIT M05 GCE MATHEMATICS UNIT M1A GCE MATHEMATICS UNIT M1B GCE MATHEMATICS UNIT M2A GCE MATHEMATICS UNIT M2B GCE MATHEMATICS UNIT PC1 GCE MATHEMATICS UNIT PC2 GCE MATHEMATICS UNIT PC3 GCE MATHEMATICS UNIT PC4 GCE MATHEMATICS UNIT S03 GCE MATHEMATICS UNIT S04 GCE MATHEMATICS UNIT S1A GCE MATHEMATICS UNIT S1B GCE MATHEMATICS UNIT S2A GCE MATHEMATICS UNIT S2B

75 75 75 75 75 75 75 75 100 75 100 75 75 75 75 75 75 75 100 75 100 75

64 60 58 62 59 59 59 57 79 61 81 62 61 60 60 61 61 61 80 60 78 60

56 52 51 54 51 51 51 49 69 53 71 54 53 53 53 54 53 53 70 52 68 52

48 44 44 46 44 44 43 41 59 45 61 46 45 46 46 47 45 45 60 44 58 44

41 36 37 39 37 37 36 33 49 37 51 38 38 39 39 40 38 38 50 37 49 37

34 29 30 32 30 30 29 26 40 30 41 31 31 33 32 33 31 31 41 30 40 30

E