Mark Scheme 4766 June 2005 - Mathematics in …mei.org.uk/files/papers/All packs/MS_MEI_S1.pdfMark Scheme 4766 January 2006. Q 1 (i) The range = 55 – 15 = 40 The interquartile range

Mark Scheme 4766June 2005

Statistics 1 (4766) Qn Answer Mk Comment 1 (i) (ii)

Mean = 657/20 = 32.85

Variance = 191

(22839 - 20

657 2

) = 66.13

Standard deviation = 8.13 32.85 + 2(8.13) = 49.11 none of the 3 values exceed this so no outliers

B1 cao M1 A1 cao M1 ft A1 ft

Calculation of 49.11

2 (i) (ii) (iii)

Length of journey

020406080

100120

0 2 4 6 8 10

Leng t h o f jo urney

Median = 1.7 miles Lower quartile = 0.8 miles Upper quartile = 3 miles Interquartile range = 2.2 miles The graph exhibits positive skewness

G1 G1 G1 B1 M1 M1 A1 ft E1

For calculating 38,68,89,103,112,120 Plotting end points Heights inc (0,0)

Statistics 1 (4766) June 2005 Final Mark Scheme 3 (i) (ii) (iii)

P(X = 4) = 401

(4)(5) = 21

(Answer given)

E = (2+12+36+80)( )X 401

So E = 3.25 ( )X

Var = (2+24+108+320)( )X 401

- 3.25²

= 11.35 – 10.5625 = 0.7875

Expected number of weeks = 406

x45

= 6.75 weeks

B1 M1 A1 cao M1 M1 dep A1 cao M1 A1

Calculation must be seen Sum of rp Sum of r²p -3.25² Use of np

4 (i) (ii) (iii)

Number of choices = = 20 ⎟⎟⎠

⎞⎜⎜⎝

⎛36

Number of ways = ⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛58

47

36

= 563520 ×× = 39200 Number of ways of choosing 12 questions

= = 293930 ⎟⎟⎠

⎞⎜⎜⎝

⎛1221

Probability of choosing correct number from each section = 39200/293930 = 0.133

M1 A1 M1 M1 A1 cao M1 M1 ft A1 cao

For ⎟⎠

⎞⎜⎝

⎛36

Correct 3 terms Multiplied

For ⎟⎟⎠

⎞⎜⎜⎝

⎛1221

5 (i) (ii) (iii)

1 2 3 4 5 6 1 1 2 3 4 5 6 2 2 2 6 4 10 6 3 3 6 3 12 15 6 4 4 4 12 4 20 12 5 5 10 15 20 5 30 6 6 6 6 12 30 6

(A) P(LCM > 6) = 1/3 (B) P(LCM = 5n) = 11/36 (C) P(LCM > 6 ∩ LCM = 5n) = 2/9

31 x

3611 ≠

92

Hence events are not independent

B1 B1 B1 M1 A1 cao M1 E1

All correct Use of diagram Use of definition

6 (i) (ii)

P(First) 0.9 P 0.9 0.1 F(2nd) P 0.9 0.1 0.5 P(2nd) F 0.5 F(Leave) 0.1 0.5 P 0.9 F P(2nd) 0.5 0.1 F (Leave) F(Leave)

G1 G1 M1

Probabilities Outcomes

(A) (B) (iii) (iv) (v) (vi)

P(First team) = = 0.729 39.0 P(Second team) =

5.09.01.05.01.09.01.09.09.0 ××+××+×× = 0.081 + 0.045 + 0.045 = 0.171 P(asked to leave) = 1 -0.729 – 0.171 = 0.1 P(Leave after two games given leaves)

= 1.0

5.01.0 × = 21

P(at least one is asked to leave)

39.01−= = 0.271 P(Pass a total of 7 games) =P(First, Second, Second) + P(First, First, Leave after three games) = + 2171.0729.03 ×× 05.0729.03 2 ×× = 0.064 + 0.080 = 0.144

A1 M1 M1 A1 B1 M1 ft A1 cao M1 ft M1 A1 cao M1 M1 ft M1 ft M1 A1 cao

1 correct triple 3 correct triples added Denominator Calc’n of 0.9 1 – ( )³ Attempts both 0.729(0.171)² 0.05(0.729)² multiply by 3

7 (i) (ii)

⎟⎠⎞

⎜⎝⎛

61,15~ BX

( )0=XP = 15

65⎟⎠⎞

⎜⎝⎛ = 0.065

( )114

65

61

415

4 ⎟⎠⎞

⎜⎝⎛×⎟

⎠⎞

⎜⎝⎛×⎟⎟

⎠

⎞⎜⎜⎝

⎛==XP

= 0.142 (or 0.9102-0.7685)

M1 A1 cao M1 M1 A1 cao

15

65⎟⎠⎞

⎜⎝⎛

114

65

61

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

multiply by ⎟⎠

⎞⎜⎝

⎛4

15

(iii) (iv) (A) (B) (v)

)3( >XP = )3(1 ≤− XP = 1 – 0.7685 = 0.232 Let p = probability of a six on any throw

61:0 =pH

61:1 <pH

⎟⎠⎞

⎜⎝⎛

61,15~ BX

( 0=XP ) = 0.065 0.065 < 0.1 and so reject 0HConclude that there is sufficient evidence at the 10% level that the dice are biased against sixes. Let p = probability of a six on any throw

61:0 =pH

61:1 >pH

⎟⎠⎞

⎜⎝⎛

61,15~ BX

( )5≥XP = = 1 – 0.910 = 0.09 ( 41 ≤− XP )0.09 < 0.1 and so reject 0HConclude that there is sufficient evidence at the 10% level that the dice are biased in favour of sixes. Conclusions contradictory. Even if null hypothesis is true, it will be rejected 10% of the time purely by chance. Or other sensible comments.

M1 A1 B1 B1 M1 M1 dep E1 dep B1 M1 M1 dep E1 dep E1 E1

Definition of p Both hypotheses 0.065 Comparison Both hypotheses 0.09 Comparison Contradictory By chance

Mark Scheme 4766January 2006

Q 1 (i)

The range = 55 – 15 = 40 The interquartile range = 35 – 26 = 9

B1 CAO B1 CAO 2

(ii)

35 + 1.5 x 9 = 48.5 26 – 1.5 x 9 = 12.5 Any value > 48.5 is an outlier (so 55 will be an outlier),

M1 for 48.5 oe M1 for 12.5 oe A1 (FT their IQR in (i)) 3

(iii) One valid comment such as eg: Positively skewed Middle 50% of data is closely bunched

E1

1 TOTAL 6 2 (i)

Impossible because if 3 letters are correct, the fourth must be also.

E1 1

(ii)

There is only one way to place letters correctly. There are 4! = 24 ways to arrange 4 letters. OR:

41 x

31 x 1

2 NOTE: ANSWER GIVEN

E1 E1

B1 for 41 x

31 B1 for x 1

2

2 (iii)

E( X ) = 1 x 31 + 2 x

41 + 4 x

241

= 1

E( X² ) = 1 x 31 + 4 x

41 + 16 x

241

= 2

So Var( X ) = 2 – 12 = 1

M1 For ∑xp (at least 2 non-zero terms correct) A1 CAO M1 for 2x p∑ (at least 2 non-zero terms correct) M1dep for – their E( X )² A1 FT their E(X) provided Var( X ) > 0 5

TOTAL 8

3 (i)

X ~ B(10,0.2) P(X < 4) = P(X ≤ 3) = 0.8791 OR attempt to sum P(X = 0,1,2,3) using X ~ B(10,0.2) can score M1, A1

M1 for X ≤ 3 A1 2

(ii) Let p = the probability that a bowl is imperfect

2.0:0 =pH 2.0:1 <pH X ~ B(20,0.2) P(X ≤ 3) = 0.2061 0.2061 > 5% Cannot reject 0H and so insufficient evidence to claim a reduction. OR using critical region method: CR is {0} B1, 2 not in CR M1, A1 as above

B1 Definition of p B1, B1 B1 for 0.2061 seen M1 for this comparison A1 dep for comment in context

3 3

TOTAL 8 4 (i)

The company could increase the mean weight. The company could decrease the standard deviation.

B1 CAO B1

2 (ii)

Sample mean = 11409/25 = 456.36

2114095206937 325.7625xxS = − =

Sample s.d = 325.7624

=3.68

B1 M1 for Sxx A1

3

TOTAL 5 5 (i)

P( A ∩ B ) = 0.4

B1 CAO 1

(ii)

P( C U D ) = 0.6 B1 CAO 1

(iii)

Events B and C are mutually exclusive.

B1 CAO 1

(iv) P( B ) = 0.6, P( D ) = 0.4 and P(B ∩ D) = 0.2 0.6 x 0.4 ≠ 0.2 (so B and D not independent)

B1 for P(B ∩ D) = 0.2 soi E1 2

TOTAL 5 6 (i) Number of selections = ⎟

⎠

⎞⎜⎝

⎛7

12 = 792 M1 for ⎟

⎠

⎞⎜⎝

⎛7

12 A1 CAO

2 (ii) Number of arrangements = 7! = 5040 M1 for 7!, A1 CAO 2 TOTAL 4

7 (i)

Mean score = (2x8 + 3x7 + 4x6 + 5 + 4)/11 = 6.36

M1 for /11fx∑ A1 CAO 2

(ii)

Mean GCSE Score

G1 Linear sensible scales G1 fds of 8, 28, 38, 26, 6 or 4k, 14k, 19k, 13k, 3k for sensible values of k either on script or on graph. G1 (dep on reasonable attempt at fd) Appropriate label for vertical scale eg ‘Frequency density’, ‘frequency per ½ unit’, ‘students per mean GCSE score’. (allow Key) 3

(iii)

Mid point, x

f fx fx²

5 8 40 200 5.75 14 80.5 462.875 6.25 19 118.75 742.18756.75 13 87.75 592.31257.5 6 45 337.5

60 372 2334.875 Sample mean = 372/60 = 6.2

23722334.875 28.47560xxS = − =

Sample s.d = 28.47559

= 0.695

B1 mid points B1FT fx∑ and 2fx∑ B1 CAO M1 for their Sxx A1 CAO

5 (iv)

Prediction of score = 13 x 7.4 – 46 = 50.2 So predicted AS grade would be B

M1 For 13 x 7.4 – 46 A1 dep on 50.2 (or 50) seen 2

(v)

Prediction of score = 13 x 5.5 – 46 = 25.5 So predicted grade would be D/E (allow D or E) Because score roughly halfway from 20 to 30, OR (for D) closer to D than E OR (for E) past E but not up to D boundary

M1 For 13 x 5.5 – 46 A1 dep on 25.5 (or 26 or 25) seen E1 For explanation of conversion – logical statement/argument that supports their choice. 3

(vi) Mean = 13 x 6.2 – 46 = 34.6 Standard deviation = 13 x 0.695 = 9.035

B1 FT their 6.2 M1 for 13 x their 0.695 A1 FT 3

TOTAL 18

8 (i)

P( all jam )

= 103

114

125

××

= 221

= 0.04545

M1 5 x 4 x 3 or 53⎛ ⎞⎜ ⎟⎝ ⎠

in

numerator

M1 12 x 11 x 10 or 123

⎛ ⎞⎜ ⎟⎝ ⎠

in

denominator A1 CAO

3

(ii) P( all same )

=103

114

125

×× + 4 3 212 11 10

× × + 3 2 112 11 10

× ×

= 1 1 122 55 220

+ + = 443

= 0.06818

M1 Sum of 3 reasonable triples or combinations M1 Triples or combinations correct A1 CAO 3

(iii)

P(all different)

= 103

114

125

6 ×××

= 113

=0.2727

M1 5,4,3 M1 ×6 three fractions or 12

3⎛ ⎞⎜ ⎟⎝ ⎠

denom. A1 CAO

3 (iv) P(all jam given all same) =

122

344

= 32

M1 Their (i) in numerator M1 Their (ii) in denominator A1 CAO 3

(v)

P(all jam exactly twice)

= 2 35 1 21

2 22 22⎛ ⎞ ⎛ ⎞ ⎛ ⎞× ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠= 0.01797

M1 for 52⎛ ⎞⎜ ⎟⎝ ⎠

x …

M1 for their p2 q3 A1 CAO

3 (vi) P(all jam at least once)

5211 0.207522

⎛ ⎞= − =⎜ ⎟⎝ ⎠

M1 for their q5 M1 indep for 1 – 5th power A1 CAO 3

TOTAL 18


4766 Mark Scheme June 2006

( ) 3.0P =∩→ BA

Q1 (i)

G1 Labelled linear scales G1 Height of lines

2 (ii) Negative (skewness) B1 1 (iii) Σfx = 123 so mean = 123/25 = 4.92 o.e.

2123681 75.8425xxS = − =

M.s.d = 75.84 3.034

25=

B1 M1 for Sxx attempted

A1 FT their 4.92 3

(iv)

Total for 25 days is 123 and totals for 31 days is 155.

Hence total for next 6 days is 32 and so mean = 5.33

M1 31 x 5 – 25xtheir 4.92

A1 FT their 123 2

TOTAL 8 Q2(i)

o.e.

M1 Product of these fractions

A1 2 (ii)

B1FT either 0.4 or 0.2 in correct place

B1FT all correct and labelled

2 (iii) P(B|A) ≠ P(B), 3/7 ≠ 0.5

Unequal so not independent

E1 Correct comparison

E1dep for ‘not independent’

2 (iv) 3/7 < 0.5

so Isobel is less likely to score when her parents attend

E1 for comparison

E1dep

2 TOTAL 8

.1 .2 .3 .4

A B

0

1

2

3

4

5

6

7

8

1 2 3 4 5 6 7

Number Correct

Freq

uenc

y

( ) ( ) ( ) 7 3P P P |10 7

A B A B A∩ = = ×


Q3(i)

P(X = 1) = 7k, P(X = 2) = 12k, P(X = 3) = 15k, P(X = 4) = 16k

50k = 1 so k = 1/50

M1 for addition of four multiples of k

A1 ANSWER GIVEN

2

(ii) E(X) = 1 x 7k + 2 x 12k + 3 x 15k + 4 x 16k = 140k = 2.8

OR E(X) = 1 x 7/50 + 2 x 12/50 + 3 x 15/50 + 4 x 16/50 = 140 /50 = 2.8 oe

Var(X) = 1 x 7k + 4 x 12k + 9 x 15k + 16 x 16k -7.84 = 1.08

OR Var(X) = 1 x 7/50 + 4 x 12/50 + 9 x 15/50 + 16 x 16/50 -7.84

= 8.92 – 7.84 = 1.08

M1 for Σxp (at least 3 terms correct) A1 CAO M1 Σx2p (at least 3 terms correct) M1dep for – their E( X )² NB provided Var( X ) > 0 A1 FT their E(X)

5

TOTAL 7 Q4(i)

4 x 5 x 3 = 60 M1 for 4 x 5 x 3 A1 CAO

2

(ii) (A) ( )4

2 = 6

(B) ( )( )( )32

52

42 = 180

B1 ANSWER GIVEN B1 CAO

2 (iii) (A) 1/5

(B) 52

32

54

43

=××

B1 CAO

M1 for 3 4 24 5 3

× ×

A1

3

TOTAL 7 Q5 (i)

( ) ( ) 2952.013.087.02P 232 =××==X

M1 0.872 x 0.13

M1 ( )32 x p2q with p+q=1

A1 CAO

3 (ii) In 50 throws expect 50 (0.2952) = 14.76 times B1 FT 1 (iii) P (two 20’s twice) = ( )4

2 2597.07048.02952.0 22 =××

M1 22 7048.02952.0 × A1 FT their 0.2952

2 TOTAL 6


0.9

0.1

Genuine

Fake

0.95

0.05

0. 2

0. 8

Positive

Negative

Positive

Negative

Q6 (i)

G1 for left hand set of branches fully correct including labels and probabilities G1 for right hand set of branches fully correct

2

(ii) P (test is positive) = (0.9)(0.95) + (0.1)(0.2) = 0.875 M1 Two correct pairs added A1 CAO

2

(iii) P (test is correct) = (0.9)(0.95) + (0.1)(0.8) = 0.935 M1 Two correct pairs added A1 CAO

2

(iv) P (Genuine|Positive)

= 0.855/0.875

= 0.977

M1 Numerator M1 Denominator A1 CAO

3 (v) P (Fake|Negative) = 0.08/0.125 = 0.64

M1 Numerator M1 Denominator A1 CAO

3 (vi) EITHER: A positive test means that the painting is almost

certain to be genuine so no need for a further test.

However, more than a third of those paintings with a negative result are genuine so a further test is needed.

NOTE: Allow sensible alternative answers

E1FT E1FT

2

(vii) P (all 3 genuine) = (0.9 x 0.05 x 0.96)3

= (0.045 x 0.96)3

= (0.0432)3

= 0.0000806

M1 for 0.9 x 0.05 (=0.045) M1 for complete correct triple product M1indep for cubing A1 CAO

4

TOTAL 18


Q7 (i)

X ~ B(20, 0.1)

1920( ) P( = 1) 0.1 0.9 0.2702

1

A X ⎛ ⎞= × × =⎜ ⎟

⎝ ⎠

OR from tables 0.3917 0.1216 0.2701− = ( ) P( 1) 1 0.1216 0.8784B X ≥ = − =

M1 0.1 x 0.919

M1 ( )201 x pq19

A1 CAO OR: M2 for 0.3917 – 0.1216 A1 CAO M1 P(X=0) provided that P(X≥1)=1–P(X≤1) not seen M1 1-P(X=0) A1 CAO

3

3

(ii) EITHER: 1 – 0.9n ≥ 0.8 0.9n ≤ 0.2 Minimum n = 16 OR (using trial and improvement): Trial with 0.915 or 0.916 or 0.917 1 – 0.915 = 0.7941 < 0.8 and 1 – 0.916 = 0.8147 > 0.8 Minimum n = 16 NOTE: n = 16 unsupported scores SC1 only

M1 for 0.9n

M1 for inequality A1 CAO M1 M1 A1 CAO

3 (iii) ( ) A Let p = probability of a randomly selected rock

containing a fossil (for population) H0: p = 0.1 H1: p < 0.1 ( ) B Let X ~ B(30, 0.1) P(X ≤ 0) = 0.0424 < 5% P(X ≤ 1) = 0.0424 + 0.1413 = 0.1837 > 5% So critical region consists only of 0.

( ) C 2 does not lie in the critical region. So there is insufficient evidence to reject the null hypothesis and we conclude that it seems that 10% of rocks in this area contain fossils.

B1 for definition of p B1 for H0 B1 for H1 M1 for attempt to find P(X ≤ 0) or P(X ≤ 1) using binomial M1 for both attempted M1 for comparison of either of the above with 5% A1 for critical region dep on both comparisons (NB Answer given) M1 for comparison A1 for conclusion in context

3

4

2 TOTAL 18

63

Mark Scheme 4766January 2007

4766 Mark Scheme January 2007

64

GENERAL INSTRUCTIONS Marks in the mark scheme are explicitly designated as M, A, B, E or G. M marks ("method") are for an attempt to use a correct method (not merely for stating the method). A marks ("accuracy") are for accurate answers and can only be earned if corresponding M mark(s) have been earned. Candidates are expected to give answers to a sensible level of accuracy in the context of the problem in hand. The level of accuracy quoted in the mark scheme will sometimes deliberately be greater than is required, when this facilitates marking. B marks are independent of all others. They are usually awarded for a single correct answer. E marks ("explanation") are for explanation and/or interpretation. These will frequently be sub divisible depending on the thoroughness of the candidate's answer. G marks ("graph") are for completing a graph or diagram correctly.

• Insert part marks in right-hand margin in line with the mark scheme. For fully correct parts tick the answer. For partially complete parts indicate clearly in the body of the script where the marks have been gained or lost, in line with the mark scheme.

• Please indicate incorrect working by ringing or underlining as appropriate.

• Insert total in right-hand margin, ringed, at end of question, in line with the mark scheme.

• Numerical answers which are not exact should be given to at least the accuracy shown.

Approximate answers to a greater accuracy may be condoned.

• Probabilities should be given as fractions, decimals or percentages.

• FOLLOW-THROUGH MARKING SHOULD NORMALLY BE USED WHEREVER POSSIBLE. There will, however, be an occasional designation of 'c.a.o.' for "correct answer only".

• Full credit MUST be given when correct alternative methods of solution are used. If errors

occur in such methods, the marks awarded should correspond as nearly as possible to equivalent work using the method in the mark scheme.

• The following notation should be used where applicable:

FT Follow-through marking

BOD Benefit of doubt

ISW Ignore subsequent working


65

Q1 (i)

Mean = 127.6/13 = 9.8 Median = 8.6 Midrange = 14.5

M1 for 127.6/13 soi A1 CAO B1 CAO B1 CAO

4 (ii)

Mean slightly inflated due to the outlier Median good since it is not affected by the outlier Midrange poor as it is highly inflated due to the outlier

B1 B1 B1

3 TOTAL 7 Q2 (i)

0

24

6

8

10

12

14

16

0 1 2 3 4 5 6

Number of absentees

Freq

uenc

y

G1 labelled linear scales on both axes G1 heights

2

(ii) Mean =

9950

= 1.98

Sxx = 299315

50− (= 118.98)

rmsd = 118.98

50= 1.54

NB full marks for correct results from recommended method which is use of calculator functions

B1 for mean M1 for attempt at Sxx

A1 CAO

3

(iii) New mean = 30 – 1.98 = 28.02

New rmsd = 1.54 (unchanged)

B1 FT their mean B1 FT their rmsd

2

TOTAL 7

Q3 (i)

time freq width f dens 0- 34 5 6.8 5- 153 5 30.6 10- 188 10 18.8 20- 73 10 7.3 30- 27 10 2.7 40- 5 20 0.25

M1 for fds A1 CAO Accept any suitable unit for fd such as eg freq per 5 mins. G1 linear scales on both axes and label G1 width of bars G1 height of bars

5

(ii) Positive skewness B1 CAO (indep) 1 TOTAL 6


66

Q4(i)

r 1 2 3 4 5 6

P(X = r) k 3k 5k 7k 9k 11k

36k = 1 , so k = 136

B1 for 3k, 5k, 7k, 9k M1 for sum of six multiples of k = 1 A1 CAO MUST BE FRACTION IN SIMPLEST FORM

3

(ii) E(X) = 1 3 5 7 9 11 1611 2 3 4 5 6 4.4736 36 36 36 36 36 36

× + × + × + × + × + × = =

M1 for Σ rp

A1 CAO

2

(iii) P(X=16) =

3166

⎛ ⎞× ⎜ ⎟⎝ ⎠

6 1

216 36= =

M1 for 6 ×

M1 indep for 31

6⎛ ⎞⎜ ⎟⎝ ⎠

A1 CAO

3

TOTAL 8

Q5(i)

P(jacket and tie) = 0.4 × 0.3 = 0.12

M1 for multiplying A1 CAO

2

(ii)

G1 for two intersecting circles labelled G1 for 0.12 and either 0.28 or 0.08 G1 for remaining probabilities Note FT their 0.12 provided < 0.2

3

(iii) (A) P(jacket or tie) =P(J) + P(T) – P(J∩T)

= 0.4 + 0.2 – 0.12 = 0.48

OR = 0.28 + 0.12 + 0.08 = 0.48

(B) P(no jacket or no tie) = 0.52 + 0.28 + 0.08 = 0.88

OR 0.6 + 0.8 – 0.52 = 0.88

OR 1 – 0.12 = 0.88

B1 FT B2 FT Note FT their 0.12 provided < 0.2

3

TOTAL 8

.28 0.12

0.52

Jacket Tie

0.08


67

Q6 (i)

Median = 3370

Q1 = 3050 Q3 = 3700

Inter-quartile range = 3700 – 3050 = 650

B1 B1 for Q3 or Q1 B1 for IQR

3

(ii) Lower limit 3050 – 1.5 × 650 = 2075

Upper limit 3700 + 1.5 × 650 = 4675

Approx 40 babies below 2075 and 5 above 4675

so total 45

B1 B1 M1 (for either) A1

4 (iii) Decision based on convincing argument:

eg ‘no, because there is nothing to suggest that they are not genuine data items and these data may influence health care provision’

E2 for convincing argument

2

(iv) All babies below 2600 grams in weight B2 CAO 2

(v) (A) X ~ B(17, 0.12)

P(X = 2) = 2 15170.12 0.88

2⎛ ⎞

× ×⎜ ⎟⎝ ⎠

= 0.2878

(B) P(X > 2)

= 1 – (0.2878 + 16170.12 0.88

1⎛ ⎞

× ×⎜ ⎟⎝ ⎠

+ 0.8817)

= 1 – (0.2878 + 0.2638 + 0.1138) = 0.335

M1 172

⎛ ⎞×⎜ ⎟

⎝ ⎠p2 × q15

M1 indep 2 150.12 0.88× A1 CAO M1 for P(X=1)+ P(X=0) M1 for 1 – P(X ≤ 2) A1 CAO

3

3

(vi) Expected number of occasions is 33.5

B1 FT 1 TOTAL 18


68

Q7 (i)

(A) P(both) =22 4

3 9⎛ ⎞ =⎜ ⎟⎝ ⎠

(B) P(one) =2 1 423 3 9

× × =

(C) P(neither) =21 1

3 9⎛ ⎞ =⎜ ⎟⎝ ⎠

B1 CAO B1 CAO B1 CAO

3 (ii) Independence necessary because otherwise, the probability

of one seed germinating would change according to whether or not the other one germinates. May not be valid as the two seeds would have similar growing conditions eg temperature, moisture, etc. NB Allow valid alternatives

E1 E1

2

(iii) Expected number = 2 ×

23

=43

(= 1.33)

E(X2) = 1 4 4 200 1 49 9 9 9

× + × + × =

Var(X) = 220 4 4

9 3 9⎛ ⎞− =⎜ ⎟⎝ ⎠

= 0.444

NB use of npq scores M1 for product, A1CAO

B1 FT M1 for E(X2) A1 CAO

3

(iv) Expect

8200 177.8 plants9

× =

So expect 0.85 × 177.8 = 151 onions

M1 for 200 × 89

M1 dep for × 0.85 A1 CAO

3

(v) Let X ~ B(18, p) Let p = probability of germination (for population) H0: p = 0.90 H1: p < 0.90

P(X ≤ 14) = 0.0982 > 5% So not enough evidence to reject H0 Conclude that there is not enough evidence to indicate that the germination rate is below 90%.

Note: use of critical region method scores M1 for region {0,1,2,…, 13} M1 for 14 does not lie in critical region then A1 E1 as per scheme

B1 for definition of p B1 for H0 B1 for H1 M1 for probability M1 dep for comparison A1 E1 for conclusion in context

7

TOTAL 18


Q1 (i) 84⎛ ⎞⎜ ⎟⎝ ⎠

ways to select = 70

M1 for 84⎛ ⎞⎜ ⎟⎝ ⎠

A1 CAO

2

(ii) 4! = 24

B1 CAO

1

TOTAL 3

Q2 (i)

Amount 0- <20 20- <50 50- <100 100- <200

Frequency 800 480 400 200

B1 for amounts B1 for frequencies

2

(ii) Total ≈ 10 800 35 480 75 400 150 200 £84800× + × + × + × =

M1 for their midpoints × their frequencies A1 CAO

2

TOTAL 4

Q3 (i) Mean =

302656

= 54.0

Sxx = 23026178890

56− = 15378

s = 15378

55= 16.7

B1 for mean

M1 for attempt at Sxx

A1 CAO

3

(ii) x + 2s = 54.0 + 2×16.7 = 87.4 So 93 is an outlier

M1 for their x +2×their sA1 FT for 87.4 and comment

2

(iii) New mean = 1.2 ×54.0 – 10 = 54.8 New s = 1.2 ×16.7 = 20.1

B1 FT M1A1 FT

3

TOTAL 8 Q4 (i) (A) P(at least one) =

36 1850 25

= = 0.72

(B) P(exactly one) =9 6 5 20 2

50 50 5+ +

= = = 0.4

B1 aef M1 for (9+6+5)/50 A1 aef

3

(ii) P(not paper | aluminium) =

1324

M1 for denominator 24 or 24/50 or 0.48 A1 CAO

2

(iii) P(one kitchen waste) =

18 32 576250 49 1225

× × = = 0.470

M1 for both fractions M1 for 2 × product of both, or sum of 2 pairs A1

3

TOTAL 8

Q5 (i) 11th value is 4,12th value is 4 so median is 4

Interquartile range = 5 – 2 = 3

B1 M1 for either quartile A1 CAO

3 (ii) No, not valid

any two valid reasons such as : • the sample is only for two years, which may not be

representative • the data only refer to the local area, not the whole of

Britain • even if decreasing it may have nothing to do with global

warming • more days with rain does not imply more total rainfall • a five year timescale may not be enough to show a long

term trend

B1 E1 E1

3

TOTAL 6

Q6 (i)

74

4 3 2 1 1Either P(all 4 correct)7 6 5 4 35

1 1or P(all 4 correct)35C

= × × × =

= =

M1 for fractions, or 7C4 seen A1 NB answer given

2

(ii) E(X) =

4 18 12 1 80 21 2 3 4 2 2.35 35 35 35 35 7

× + × + × + × = = = 29

E(X2) = 4 18 12 1 2001 4 9 16 5.71435 35 35 35 35

× + × + × + × = =

Var(X) = 2200 80 24

35 35 49⎛ ⎞− =⎜ ⎟⎝ ⎠

= 0.490 (to 3 s.f.)

M1 for Σrp (at least 3 terms correct)

A1 CAO

M1 for Σ x2p (at least 3 terms correct) M1dep for – their E( X )² A1 FT their E(X) provided Var( X ) > 0

5

TOTAL 7

Section B Q7 (i)

Positive result

Doubtful result

Negative result

0.03

Has the disease

Clear

0.95

Has the disease

Clear

Has the disease

Clear

0.06

0.91

0.05

0.10

0.90

0.01

0.99

G1 probabilities of result G1 probabilities of disease G1 probabilities of clear G1 labels

4

(ii) P(negative and clear) = 0.91 × 0.99

= 0.9009

M1 for their 0.91 × 0.99

A1 CAO

2

(iii) P(has disease) = 0.03 × 0.95 + 0.06 × 0.10 + 0.91 × 0.01

= 0.0285 + 0.006 + 0.0091

= 0.0436

M1 three products M1dep sum of three products A1 FT their tree

3

(iv) P(negative | has disease)

= P(negative has disease) 0.0091 0.2087

P(has disease) 0.0436= =

and

M1 for their 0.01 × 0.91 or 0.0091 on its own or as numerator M1 indep for their 0.0436 as denominator A1 FT their tree

3

(v) Thus the test result is not very reliable. A relatively large proportion of people who have the disease will test negative.

E1 FT for idea of ‘not reliable’ or ‘could be improved’, etc E1 FT

2

(vi)

P(negative or doubtful and declared clear) =0.91 0.06 0.10 0.02 0.06 0.90 1 =0.91 0.00012 0.054 0.96412

+ × × + × ×+ + =

M1 for their 0.91 + M1 for either triplet M1 for second triplet A1 CAO

4 TOTAL 18

Q8 (i)

X ~ B(17, 0.2) P(X ≥ 4) = 1 – P(X ≤ 3)

= 1 – 0.5489 = 0.4511

B1 for 0.5489 M1 for 1 – their 0.5489 A1 CAO

3

(ii) E(X) = np = 17 × 0.2 = 3.4 M1 for product A1 CAO

2

(iii) P(X = 2) = 0.3096 – 0.1182 = 0.1914 P(X = 3) = 0.5489 – 0.3096 = 0.2393 P(X = 4) = 0.7582 – 0. 5489 = 0.2093 So 3 applicants is most likely

B1 for 0.2393 B1 for 0.2093 A1 CAO dep on both B1s

3

(iv) (A) Let p = probability of a randomly selected maths graduate applicant being successful (for population) H0: p = 0.2 H1: p > 0.2 (B) H1 has this form as the suggestion is that mathematics graduates are more likely to be successful.

B1 for definition of p in context B1 for H0B1 for H1E1

4

(v) Let X ~ B(17, 0.2) P(X ≥ 6) = 1 – P(X ≤ 5) = 1 – 0.8943 = 0.1057 > 5% P(X ≥ 7) = 1 – P(X ≤ 6) = 1 – 0.9623 = 0.0377 < 5% So critical region is {7,8,9,10,11,12,13,14,15,16,17}

B1 for 0.1057 B1 for 0.0377 M1 for at least one comparison with 5% A1 CAO for critical region dep on M1 and at least one B1

4

(vi) Because P(X ≥ 6) = 0.1057 > 10% Either: comment that 6 is still outside the critical region Or comparison P(X ≥ 7) = 0.0377 < 10%

E1 E1

2

TOTAL 18


46

4766 Statistics 1

Q1 (i)

Mode = 7 Median = 12.5

B1 cao B1 cao

2

(ii) Positive or positively skewed E1 1 (iii)

(A) Median

(B) There is a large outlier or possible outlier of 58 / figure of 58.

Just ‘outlier’ on its own without reference to either 58 or large scores E0

Accept the large outlier affects the mean (more) E1

E1 cao E1indep

2

(iv)

There are 14.75 × 28 = 413 messages So total cost = 413 × 10 pence = £41.30

M1 for 14.75 × 28 but 413 can also imply the mark A1cao

2

TOTAL 7 Q2 (i)

43! 4 6 24

3⎛ ⎞

× = × =⎜ ⎟⎝ ⎠

codes or 4P3 = 24 (M2 for 4P3 )

Or 4 × 3 × 2 = 24

M1 for 4 M1 for ×6 A1

3

(ii) 43 = 64 codes

M1 for 43

A1 cao

2

TOTAL 5

Q3 (i)

Probability = 0.3 × 0.8 = 0.24

M1 for 0.8 from (1-0.2) A1

2

(ii)

Either: P(AUB) = P(A) + P(B) – P(A∩B)

= 0.3 + 0.2 – 0.3 × 0.2

= 0.5 – 0.06 = 0.44

Or: P(AUB) = 0.7 × 0.2 + 0.3 × 0.8 + 0.3 × 0.2

= 0.14 + 0.24 + 0.06 = 0.44

Or: P(AUB) = 1 – P(A′ ∩B′ )

= 1 – 0.7 × 0.8 = 1 – 0.56 = 0.44

M1 for adding 0.3 and 0.2 M1 for subtraction of ( 0.3 × 0.2) A1 cao M1 either of first terms M1 for last term A1 M1 for 0.7 × 0.8 or 0.56 M1 for complete method as seen A1

3

(iii) P(A|B) =

P( ) 0.06 6 0.136P( ) 0.44 44

∩= = =

A BB

M1 for numerator of their 0.06 only M1 for ‘their 0.44’ in denominator A1 FT (must be valid p)

3

TOTAL 8


47

Q4 (i)

E(X) = 1 0.2 2 0.16 3 0.128 4 0.512 2.952× + × + × + × = Division by 4 or other spurious value at end loses A mark E(X2) = 1 0.2 4 0.16 9 0.128 16 0.512 10.184× + × + × + × = Var(X) = 10.184 – 2.9522 = 1.47 (to 3 s.f.)

M1 for Σ rp (at least 3 terms correct) A1 cao M1 for Σ x2p at least 3 terms correct M1 for E(X2) – E(X) 2 Provided ans > 0 A1 FT their E(X) but not a wrong E(X2)

5

(ii) Expected cost = 2.952 × £45000 = £133000 (3sf)

B1 FT ( no extra multiples / divisors introduced at this stage)

1

(iii)

0

0.2

0.4

0.6

1 2 3 4

r

P(X

= r)

G1 labelled linear scales G1 height of lines

2

TOTAL 8

Q5(i)

Impossible because the competition would have finished as soon as Sophie had won the first 2 matches

E1

1

(ii) SS, JSS, JSJSS B1, B1, B1 (-1 each error or omission)

3

(iii) 0.72 + 0.3 × 0.72 + 0.7 × 0.3 × 0.72 = 0.7399 or 0.74(0) { 0.49 + 0.147 + 0.1029 = 0.7399}

M1 for any correct termM1 for any other correct term M1 for sum of all three correct terms A1 cao

4

TOTAL 8


48

Section B Q6 (i) Mean =

180.612

= 15.05 or 15.1

Sxx = 2180.63107.56

12− or 3107.56 – 12(their 15.05)2 =

(389.53)

s = 389.53

11= 5.95 or better

NB Accept answers seen without working (from calculator)

B1 for mean M1 for attempt at Sxx

A1 cao

3

(ii) x + 2s = 15.05 + 2 × 5.95 = 26.95 x – 2s = 15.05 – 2 × 5.95 = 3.15 So no outliers

M1 for attempt at either M1 for both A1 for limits and conclusion FT their mean and sd

3

(iii)

New mean = 1.8 × 15.05 + 32 = 59.1 New s = 1.8 × 5.95 = 10.7

B1FT M1 A1FT

3 (iv) New York has a higher mean or ‘ is on average’ higher (oe)

New York has greater spread /range /variation or SD (oe)

E1FT using 0F ( x dep) E1FT using 0F (σ dep)

2 (v) (vi)

NB all G marks dep on attempt at cumulative frequencies. NB All G marks dep on attempt at cumulative frequencies Line on graph at cf = 43.2(soi) or used 90th percentile = 166

B1 for all correct cumulative frequencies (may be implied from graph). Ignore cf of 0 at this stage G1 for linear scales (linear from 70 to 190) ignore x < 70 vertical: 0 to 50 but not beyond 100 (no inequality scales) G1 for labels G1 for points plotted as (UCB, their cf). Ignore (70,0) at this stage. No mid –point or LCB plots. G1 for joining all of ‘their points’(line or smooth curve) AND now including (70,0) M1 for use of 43.2 A1FT but dep on 3rd G mark earned

5

2

TOTAL 18

Upper bound (70) 100 110 120 150 170 190Cumulative frequency (0) 6 14 24 35 45 48

0

10

20

30

40

50

0 50 100 150 200

Hours

Cum

ulat

ive

freq

uenc

y


49

Q7 (i)

X ~ B(12, 0.05)

1112( ) P( = 1) 0.05 0.95 0.3413

1

A X ⎛ ⎞= × × =⎜ ⎟

⎝ ⎠

OR from tables 0.8816 0.5404 0.3412− = ( ) P( 2) 1 0.8816 0.1184B X ≥ = − = ( ) Expected number E( ) 12 0.05 0.6= = × =C X np

M1 0.05 × 0.9511

M1 ( )121 × pq11 (p+q) =

1 A1 cao OR: M1 for 0.8816 seen and M1 for subtraction of 0.5404 A1 cao M1 for 1 – P(X ≤ 1) A1 cao M1 for 12 0.05× A1 cao (= 0.6 seen)

3

2

2

(ii) (iii)

Either: 1 – 0.95n ≤ ⅓ 0.95n ≥ ⅔ n ≤ log ⅔ /log0.95, so n ≤ 7.90 Maximum n = 7 Or: (using tables with p = 0.05): n = 7 leads to P(X ≥ 1) = 1 - P(X = 0) = 1 – 0.6983 = 0.3017 ( < ⅓ ) or 0.6983 ( > 2/3) n = 8 leads to P(X ≥ 1) = 1 - P(X = 0) = 1 – 0.6634 = 0.3366 ( > ⅓ ) or 0.6634 ( < 2/3 ) Maximum n = 7 (total accuracy needed for tables) Or: (using trial and improvement): 1 – 0.957 = 0.3017 ( < ⅓) or 0.957 = 0.6983 ( > 2/3) 1 – 0.958 = 0.3366 ( > ⅓) or 0.968 = 0.6634 ( < 2/3)

Maximum n = 7 (3 sf accuracy for calculations) NOTE: n = 7 unsupported scores SC1 only

Let X ~ B(60, p) Let p = probability of a bag being faulty H0: p = 0.05 H1: p < 0.05

P(X ≤ 1) = 0.9560 + 60 × 0.05 × 0.9559 = 0.1916 > 10% So not enough evidence to reject H0 Conclude that there is not enough evidence to indicate that the new process reduces the failure rate or scientist incorrect/ wrong.

M1 for equation in n M1 for use of logs A1 cao M1indep M1indep A1 cao dep on both M’s M1indep (as above) M1indep (as above) A1 cao dep on both M’s B1 for definition of p B1 for H0 B1 for H1 M1 A1 for probability M1 for comparison A1 E1

3

8

TOTAL 18


60

4766 Statistics 1 Q1 (i)

Mean = 7.35 (or better) Standard deviation: 3.69 – 3.70 (awfw) Allow s2 = 13.62 to 13.68 Allow rmsd = 3.64 – 3.66 (awfw) After B0, B0 scored then if at least 4 correct mid-points seen or used.{1.5, 4, 6, 8.5, 15}

Attempt of their mean =44

fx∑, with 301 ≤ fx ≤ 346 and fx

strictly from mid-points not class widths or top/lower boundaries.

B2cao fx∑ = 323.5 B2cao fx∑ 2 = 2964.25 (B1) for variance s.o.i.o (B1) for rmsd (B1) mid-points (B1) 6.84 ≤ mean ≤ 7.86

4

(ii) Upper limit = 7.35 + 2 × 3.69 = 14.73 or ‘their sensible mean’ + 2 × ‘their sensible s.d.’ So there could be one or more outliers

M1 ( with s.d. < mean) E1dep on B2, B2 earned and comment

2

TOTAL 6 Q2 (i)

P(W) × P(C) = 0.20 × 0.17 = 0.034

P(W∩C) = 0.06 (given in the question)

Not equal so not independent (Allow 0.20 × 0.17 ≠ 0.06 or

≠ p (W ∩ C) so not independent).

M1 for multiplying or 0.034 seen A1 (numerical justification needed)

2

(ii)

The last two G marks are independent of the labels

G1 for two overlapping circles labelled G1 for 0.06 and either 0.14 or 0.11 in the correct places G1 for all 4 correct probs in the correct places (including the 0.69) NB No credit for Karnaugh maps here

3

(iii)

P(W|C) = P( ) 0.06 6 0.353

P(C) 0.17 17∩ = = =W C

(awrt 0.35)

M1 for 0.06 / 0.17 A1 cao

2

0.14

0.06

0.69

W C

0.11


61

(iv) Children are more likely than adults to be able to speak

Welsh or ‘proportionally more children speak Welsh than

adults’

Do not accept: ‘more Welsh children speak Welsh than

adults’

E1FT Once the correct idea is seen, apply ISW

1

TOTAL 8

Q3 (i)

( ) 0.5 + 0.35 + 1 so 0.15( ) 0 0.5 1 0.35 2 3 0.67 so 2 3 0.32( ) from above 2 2 0.30 so 0.02, 0.13

A p qp q

B p qp q

C p qq p

+ =+ =

× + × + + =+ =

+ == =

B1 p + q in a correct equation before they reach p + q =0.15 B1 2p + 3q in a correct equation before they reach 2p + 3q = 0.32 (B1) for any 1 correct answer B2 for both correct answers

1 1 2

(ii) E(X2) = 0 0.5 1 0.35 4 0.13 9 0.02 1.05× + × + × + × = Var(X) = ‘their 1.05’ – 0.672 = 0.6011 (awrt 0.6) (M1, M1 can be earned with their p+

and q + but not A mark)

M1 Σx2p (at least 2 non zero terms correct) M1dep for (– 0.67²), provided Var( X ) > 0 A1 cao (No n or n-1 divisors)

3

TOTAL 7 Q4 (i)

X ~ B(8, 0.05) 8( ) P( = 0) 0.95 0.6634 = =A X 0.663 or better

Or using tables P( = 0) = 0.6634 X

78( ) P( = 1) 0.05 0.95 0.2793

1P( 1) 1 (0.6634 0.2793) 0.0573

⎛ ⎞= × × =⎜ ⎟⎝ ⎠

> = − + =

B X

X

Or using tables P(X > 1 ) = 1 – 0.9428 = 0.0572

M1 0.958 A1 CAO Or B2 (tables) M1 for P(X = 1)(allow 0.28 or better) M1 for 1 – P(X ≤ 1) must have both probabilities A1cao (0.0572 – 0.0573) M1 for P(X ≤ 1) 0.9428 M1 for 1 – P(X ≤ 1) A1 cao (must end in…2)

2

3

(ii) Expected number of days = 250 × 0.0572 = 14.3 awrt

M1 for 250 x prob(B) A1 FT but no rounding at end

2

TOTAL 7


62

Q5 (i)

Let p = probability of remembering or naming all items (for population) (whilst listening to music.) H0: p = 0.35 H1: p > 0.35

H1 has this form since the student believes that the probability will be increased/ improved/ got better /gone up.

B1 for definition of p B1 for H0 B1 for H1 E1dep on p>0.35 in H0 In words not just because p > 0.35

4

(ii) Let X ~ B(15, 0.35)

Either: P(X ≥ 8) = 1 - 0.8868 = 0.1132 > 5% Or 0.8868 < 95% So not enough evidence to reject H0 (Accept Ho ) Conclude that there is not enough evidence to indicate that the probability of remembering all of the items is improved / improved/ got better /gone up. (when listening to music.) -------------------------------------------------------------------------- Or: Critical region for the test is {9,10,11,12,13,14,15} 8 does not lie in the critical region. So not enough evidence to reject H0

Conclude that there is not enough evidence to indicate that the probability of remembering all of the items is improved / improved/ got better /gone up. (when listening to music.) ------------------------------------------------------------------------- Or: The smallest critical region that 8 could fall into is {8, 9, 10, 11, 12, 13, 14, and 15}. The size of this region is 0.1132 0.1132 > 5% So not enough evidence to reject H0

Conclude that there is not enough evidence to indicate that the probability of remembering all of the items is improved (when listening to music)

Either: M1 for probability (0.1132) M1dep for comparison A1dep E1dep on all previous marks for conclusion in context ----------------------------- Or: M1 for correct CR(no omissions or additions) M1dep for 8 does not lie in CR A1dep E1dep on all previous marks for conclusion in context ----------------------------- Or: M1 for CR{8,9,…15}and size = 0.1132 M1 dep for comparison A1dep E1dep on all previous marks for conclusion in context

4

TOTAL 8


63

Section B

Q6 (i)

(A) P(both rest of UK) = 0.20 × 0.20

= 0.04 (B) Either: All 5 case P(at least one England) = (0.79 x 0.20) + (0.79 x 0.01) + (0.20 x 0.79) + (0.01 x 0.79) + (0.79 x 0.79) = 0.158 + 0.0079 + 0.158 + 0.0079 + 0.6241 = 0.9559 Or P(at least one England) = 1 – P(neither England) = 1 – (0.21 × 0.21) = 1 – 0.0441 = 0.9559 or listing all = 1 –{ ( 0.2 x 0.2) + (0.2 x 0.01) + (0.01 x 0.20) + (0.01x 0.01)} = 1 – (**) = 1 – { 0.04 + 0.002 + 0.002 + 0.0001) = 1 – 0.0441 = 0.9559 Or: All 3 case P(at least one England) = = 0.79 × 0.21 + 0.21 × 0.79 + 0.792 = 0.1659 + 0.1659 + 0.6241 = 0.9559 ________________________________________________ (C)Either 0.79 x 0.79 + 0.79 x 0.2 + 0.2 x 0.79 + 0.2 x 0.2 = 0.9801 Or 0.99 × 0.99 = 0.9801 Or 1 – { 0.79 x 0.01 + 0.2 x 0.01 + 0.01 x 0.79 + 0.01 x 0.02 + 0.012 } = 1 – 0.0199 = 0.9801

M1 for multiplying

A1cao M1 for any correct term (3case or 5case) M1 for correct sum of all 3 (or of all 5) with no extras A1cao (condone 0.96 www) Or M1 for 0.21 × 0.21 or for (**) fully enumerated or 0.0441 seen M1dep for 1 – (1st part) A1cao See above for 3 case ___________________ M1 for sight of all 4 correct terms summed A1 cao (condone 0.98 www) or M1 for 0.99 x 0.99 A1cao Or M1 for everything 1 – {…..} A1cao

2

3

2

(ii) P(both the rest of the UK | neither overseas)

P(the rest of the UK neither overseas)P(neither overseas)

0.04 0.04080.9801

and=

= =

{Watch for:)()(

CanswerAanswer

as evidence of method (p <1)}

M1 for numerator of 0.04 or ‘their answer to (i)(A)’ M1 for denominator of 0.9801 or ‘their answer to (i) (C)’ A1 FT (0 < p < 1) 0.041 at least

3


64

(iii) (A) Probability = 1 – 0.795 = 1 – 0.3077 = 0.6923 (accept awrt 0.69) see additional notes for alternative solution (B) 1 – 0.79n > 0.9 EITHER: 1 – 0.79n > 0.9 or 0.79n < 0.1 (condone = and ≥ throughout) but not reverse inequality

n >79010

.log.log

, so n > 9.768…

Minimum n = 10 Accept n ≥ 10 -------------------------------------------------------------------------- OR (using trial and improvement): Trial with 0.799 or 0.7910 1 – 0.799 = 0.8801 (< 0.9) or 0.799 = 0.1198 (> 0.1) 1 – 0.7910 = 0.9053 ( > 0.9 ) or 0.7910 = 0.09468 (< 0.1) Minimum n = 10 Accept n ≥ 10 -------------------------------------------------------------------------- NOTE: n = 10 unsupported scores SC1 only

M1 for 0.795 or 0.3077... M1 for 1 – 0.795 dep A1 CAO M1 for equation/inequality in n (accept either statement opposite) M1(indep) for process

of using logs i.e. ba

loglog

A1 CAO ----------------------------- M1(indep) for sight of 0.8801 or 0.1198 M1( indep) for sight of 0.9053 or 0.09468 A1 dep on both M’s cao -----------------------------

3

3

TOTAL 16


65

Q7 (i)

Positive

B1

1

(ii)

Number of people = 20 × 33 ( 000) + 5 × 58 (000 )

= 660 ( 000 ) + 290 (000) = 950 000

M1 first term M1(indep) second term A1 cao NB answer of 950 scores M2A0

3

(iii)

(A) a = 1810 + 340 = 2150

(B) Median = age of 1 385 (000th ) person or 1385.5 (000)

Age 30, cf = 1 240 (000); age 40, cf = 1 810 (000)

Estimate median = (30) + 570145

× 10

Median = 32.5 years (32.54...) If no working shown then 32.54 or

better is needed to gain the M1A1. If 32.5 seen with no previous working allow SC1

M1 for sum A1 cao 2150 or 2150 thousand but not 215000 B1 for 1 385 (000) or 1385.5 M1 for attempt to

interpolate kk

570145 × 10

(2.54 or better suggests this) A1 cao min 1dp

2 3

(iv)

Frequency densities: 56, 65, 77, 59, 45, 17

(accept 45.33 and 17.43 for 45 and 17)

B1 for any one correct B1 for all correct (soi by listing or from histogram) Note: all G marks below dep on attempt at frequency density, NOT frequency G1 Linear scales on both axes (no inequalities) G1 Heights FT their listed fds or all must be correct. Also widths. All blocks joined G1 Appropriate label for vertical scale eg ‘Frequency density (thousands)’, ‘frequency (thousands) per 10 years’, ‘thousands of people per 10 years’. (allow key). OR f.d.

5


66

(v)

Any two suitable comments such as:

Outer London has a greater proportion (or %) of people

under 20 (or almost equal proportion)

The modal group in Inner London is 20-30 but in Outer London it is 30-40 Outer London has a greater proportion (14%) of aged 65+ All populations in each age group are higher in Outer London Outer London has a more evenly spread distribution or balanced distribution (ages) o.e.

E1 E1

2 (vi) Mean increase ↑

median unchanged (-) midrange increase ↑ standard deviation increase ↑ interquartile range unchanged. ( - )

Any one correct B1 Any two correct B2 Any three correct B3 All five correct B4

4

TOTAL 20


47

4766 Statistics 1

Section A Q1 (i)

(With ∑ fx = 7500 and ∑ f = 10000 then arriving at the

mean) (i) £0.75 scores (B1, B1) (ii) 75p scores (B1, B1) (iii) 0.75p scores (B1, B0) (incorrect units) (iv) £75 scores (B1, B0) (incorrect units)

After B0, B0 then sight of 100007500

scores SC1. SC1or an answer

in the range £0.74 - £0.76 or 74p – 76p (both inclusive) scores SC1 (units essential to gain this mark) Standard Deviation: (CARE NEEDED here with close proximity of answers)

• 50.2(0) using divisor 9999 scores B2 (50.20148921) • 50.198 (= 50.2) using divisor 10000 scores B1(rmsd) • If divisor is not shown (or calc used) and only an answer

of 50.2 (i.e. not coming from 50.198) is seen then award B2 on b.o.d. (default)

After B0 scored then an attempt at Sxx as evident by either

Sxx = 27500(5000 200000 25000000)

10000+ + − (= 25199375)

or Sxx = (5000 + 200000 + 25000000) – 10000(0.75)2 scores (M1) or M1ft ‘their 75002’ or ‘their 0.752’ NB The structure must be correct in both above cases with a max of 1 slip only after applying the f.t.

B1 for numerical mean (0.75 or 75 seen) B1dep for correct units attached

B2 correct s.d.

(B1) correct rmsd

(B2) default

∑ fx 2 = 25,205,000

Beware ∑ x 2 =25, 010, 100

After B0 scored then

(M1) or M1f.t. for

attempt at Sxx NB full marks for correct results from recommended method which is use of calculator functions

4


48

(ii) P(Two £10 or two £100)

= 50 49 20 19

10000 9999 10000 9999× + ×

= 0.0000245 + 0.0000038 = (0.00002450245 + 0.00000380038) = 0.000028(3) o.e. = (0.00002830283)

After M0, M0 then ×10000

50 10000

50 + ×

1000020

10000

20 o.e.

Scores SC1 (ignore final answer but SC1 may be implied by sight of 2.9 × 10 – 5

o.e.)

Similarly, ×10000

50 10000

49 + ×

1000020

10000

19 scores SC1

M1 for either correct product seen (ignore any multipliers) M1 sum of both correct (ignore any multipliers) A1 CAO (as opposite with no rounding) (SC1 case #1) (SC1 case #2) CARE answer is also 2.83 × 10 – 5

3

TOTAL 7 Q2 (i)

1 1 1 1 1 1 1Either P(all correct)6 5 4 3 2 1 720

1 1or P(all correct) 0.001396! 720

= × × × × × =

= = =

M1 for 6! Or 720 (sioc) or product of fractions A1 CAO (accept 0.0014)

2

(ii) 3 2 1 1Either P(picks T, O, M)6 5 4 20

1 1 1 1or P(picks T, O, M) 3!6 5 4 201 1or P(picks T, O, M)6 203

= × × =

= × × × =

= =⎛ ⎞⎜ ⎟⎝ ⎠

M1 for denominators M1 for numerators or 3! A1 CAO Or M1 for 6

3⎛ ⎞⎜ ⎟⎝ ⎠

or 20 sioc

M1 for 1/ 63

⎛ ⎞⎜ ⎟⎝ ⎠

A1 CAO

3

TOTAL 5 Q3 (i)

p = 0.55

B1 cao

1

(ii) E(X) = 0 0.55 1 0.1 2 0.05 3 0.05 4 0.25 1.35× + × + × + × + × = E(X2) = 0 0.55 1 0.1 4 0.05 9 0.05 16 0.25× + × + × + × + × = 0 + 0.1 + 0.2 + 0.45 + 4 = (4.75) Var(X) = ‘their’ 4.75 – 1.352 = 2.9275 awfw (2.9275 – 2.93)

M1 for Σrp (at least 3 non zero terms correct) A1 CAO(no ‘n’ or ‘n-1’ divisors) M1 for Σr2p (at least 3 non zero terms correct) M1dep for – their E( X )² provided Var( X ) > 0 A1 cao (no ‘n’ or ‘n-1’ divisors)

5

(iii) P(At least 2 both times) = (0.05+0.05+0.25)2 = 0.1225 o.e. M1 for (0.05+0.05+0.25)2 or 0.352 seen A1cao: awfw (0.1225 -0.123) or 49/400

2


49

TOTAL 8


50

Q4 (i)

X ~ B(50, 0.03)

4950( ) P( = 1) 0.03 0.97 0.3372

1

⎛ ⎞= × × =⎜ ⎟

⎝ ⎠A X

50( ) P( = 0) = 0.97 0.2181( 1) 1 0.2181 0.3372 0.4447

=> = − − =

B XP X

M1 0.03 × 0.9749 or 0.0067(4)….

M1 501

⎛ ⎞⎜ ⎟⎝ ⎠

× pq49 (p+q

=1) A1 CAO (awfw 0. 337 to 0. 3372) or 0.34(2s.f.) or 0.34(2d.p.) but not just 0.34 B1 for 0.9750 or 0.2181 (awfw 0.218 to 0.2181) M1 for 1 – ( ‘their’ p (X = 0) + ‘their’ p(X = 1)) must have both probabilities A1 CAO (awfw 0.4447 to 0.445)

3

3

(ii) Expected number = np = 240 × 0.3372 = 80.88 – 80.93 = (81) Condone 240 × 0.34 = 81.6 = (82) but for M1 A1f.t.

M1 for 240× prob (A) A1FT

2

TOTAL 8 Q5 (i)

P(R) × P(L) = 0.36 × 0.25 = 0.09 ≠ P(R∩L) Not equal so not independent. (Allow 0.36 × 0.25 ≠ 0.2or 0.09 ≠ 0.2 or ≠ p(R ∩ L) so not independent)

M1 for 0.36 × 0.25 or 0.09 seen A1 (numerical justification needed)

2

(ii) G1 for two overlapping circles labelled G1 for 0.2 and either 0.16 or 0.05 in the correct places G1 for all 4 correct probs in the correct places (including the 0.59) The last two G marks are independent of the labels

3

(iii) ( ) 0.2 5( | ) 0.556

( ) 0.36 9P L RP L R

P R∩

= = = = (awrt 0.56)

This is the probability that Anna is late given that it is raining. (must be in context) Condone ‘if ’ or ‘when’ or ‘on a rainy day’ for ‘given that’ but not the words ‘and’ or ‘because’ or ‘due to’

M1 for 0.2/0.36 o.e. A1 cao E1 (indep of M1A1) Order/structure must be correct i.e. no reverse statement

3

TOTAL 8

.166

0.2

0.59

R L

0.05


51

Section B Q6 (i)

Median = 4.06 – 4.075 (inclusive) Q1 = 3.8 Q3 = 4.3 Inter-quartile range = 4.3 – 3.8 = 0.5

B1cao B1 for Q1 (cao) B1 for Q3 (cao) B1 ft for IQR must be using t-values not locations to earn this mark

4

(ii) Lower limit ‘ their 3.8’ – 1.5 × ‘their 0.5’ = (3.05) Upper limit ‘ their 4.3’ + 1.5 × ‘their 0.5’ = (5.05) Very few if any temperatures below 3.05 (but not zero) None above 5.05 ‘So few, if any outliers’ scores SC1

B1ft: must have -1.5 B1ft: must have +1.5 E1ft dep on -1.5 and Q1 E1ft dep on+1.5 and Q3 Again, must be using t-values NOT locations to earn these 4 marks

4

(iii) Valid argument such as ‘Probably not, because there is nothing to suggest that they are not genuine data items; (they do not appear to form a separate pool of data.’) Accept: exclude outlier – ‘measuring equipment was wrong’ or ‘there was a power cut’ or ref to hot / cold day [Allow suitable valid alternative arguments]

E1

1

(iv) Missing frequencies 25, 125, 50 B1, B1, B1 (all cao)

3

(v) Mean = (3.2×25 + 3.6×125 + 4.0×243 + 4.4×157 + 4.8×50)/600 = 2432.8/600 = 4.05(47)

M1 for at least 4 midpoints correct and being used in attempt to find ∑ ft A1cao: awfw (4.05 – 4.055) ISW or rounding

2

(vi) New mean = 1.8 × ‘their 4.05(47)’ + 32 = 39.29(84) to 39.3 New s = 1.8 × 0.379 = 0.682

B1 FT M1 for 1.8 × 0.379 A1 CAO awfw (0.68 – 0.6822)

3

TOTAL 17


52

Q7 (i)

X ~ B(10, 0.8)

(A) Either 8 210P( = 8) 0.8 0.2 0.3020

8⎛ ⎞

= × × =⎜ ⎟⎝ ⎠

X (awrt)

or P(X = 8) = P(X ≤ 8) – P(X ≤ 7) = 0.6242 – 0.3222 = 0.3020 (B) Either P(X ≥ 8) = 1 – P(X ≤ 7) = 1 – 0.3222 = 0.6778

or P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.3020 + 0.2684 + 0.1074 = 0.6778

M1 0.88 × 0.22 or 0.00671…

M1 ( )108 × p8 q2; (p +q

=1) Or 45 × p8 q2; (p +q =1) A1 CAO (0.302) not 0.3 OR: M2 for 0.6242 – 0.3222 A1 CAO M1 for 1 – 0.3222 (s.o.i.) A1 CAO awfw 0.677 – 0.678 or M1 for sum of ‘their’ p(X=8) plus correct expressions for p(x=9) and p(X=10) A1 CAO awfw 0.677 – 0.678

3

2

(ii) Let X ~ B(18, p) Let p = probability of delivery (within 24 hours) (for population) H0: p = 0.8 H1: p < 0.8

P(X ≤ 12) = 0.1329 > 5% ref: [pp =0.0816] So not enough evidence to reject H0 Conclude that there is not enough evidence to indicate that less than 80% of orders will be delivered within 24 hours

Note: use of critical region method scores M1 for region {0,1,2,…,9, 10} M1dep for 12 does not lie in critical region then A1dep E1dep as per scheme

B1 for definition of p B1 for H0 B1 for H1 M1 for probability 0.1329 M1dep strictly for comparison of 0.1329 with 5% (seen or clearly implied) A1dep on both M’s E1dep on M1,M1,A1 for conclusion in context

7


53

(iii) Let X ~ B(18, 0.8) H1: p ≠ 0.8 LOWER TAIL P(X ≤ 10) = 0.0163 < 2.5% P(X ≤ 11) = 0.0513 > 2.5% UPPER TAIL P(X ≥ 17) = 1 – P(X ≤ 16) = 1 – 0.9009 = 0.0991 > 2.5% P(X ≥ 18) = 1 – P(X ≤ 17) = 1 – 0.9820 = 0.0180 < 2.5% So critical region is {0,1,2,3,4,5,6,7,8,9,10,18} o.e.

Condone X ≤ 10 and X ≥ 18 or X = 18 but not p(X ≤ 10) and

p(X ≥ 18)

Correct CR without supportive working scores SC2 max after

the 1st B1 (SC1 for each fully correct tail of CR)

B1 for H1 B1 for 0.0163 or 0.0513 seen M1dep for either correct comparison with 2.5% (not 5%) (seen or clearly implied) A1dep for correct lower tail CR (must have zero) B1 for 0.0991 or 0.0180 seen M1dep for either correct comparison with 2.5% (not 5%) (seen or clearly implied) A1dep for correct upper tail CR

7

TOTAL 19


67

4766 Statistics 1

Q1 (i) Median = 2

Mode = 1

B1 CAO B1 CAO

2

(ii)

0

10

20

30

40

50

60

1 2 3 4

Number o f P eople

S1 labelled linear scales on both axes H1 heights

2

(iii) Positive B1 1 TOTAL 5 Q2 (i)

255

⎛ ⎞⎜ ⎟⎝ ⎠

different teams = 53130

M1 for 255

⎛ ⎞⎜ ⎟⎝ ⎠

A1 CAO

2

(ii) 14 11364 55 20020

3 2⎛ ⎞ ⎛ ⎞

× = × =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

M1 for either combination M1 for product of both A1 CAO

3

TOTAL 5

Q3 (i) Mean =

12612

= 10.5

Sxx = 21261582

12− = 259

s = 25911

= 4.85

B1 for mean M1 for attempt at Sxx A1 CAO

3

(ii) New mean = 500 + 100 ×10.5 = 1550

New s = 100 ×4.85 = 485

B1 ANSWER GIVEN M1A1FT

3

(iii) On average Marlene sells more cars than Dwayne.

Marlene has less variation in monthly sales than Dwayne.

E1 E1FT

2

TOTAL 8


68

Q4 (i)

E(X) = 25 because the distribution is symmetrical. Allow correct calculation of Σrp

E1 ANSWER GIVEN

1

(ii) E(X2) = 2 2 2 210 0.2 20 0.3 30 0.3 40 0.2× + × + × + × = 730 Var(X) = 730 – 252 = 105

M1 for Σr2p (at least 3 terms correct) M1dep for – 25² A1 CAO

3

TOTAL 4 Q5 (i)

Distance freq width f dens

0- 360 50 7.200 50- 400 50 8.000 100- 307 100 3.070

200-400 133 200 0.665

M1 for fds A1 CAO Accept any suitable unit for fd such as eg freq per 50 miles. L1 linear scales on both axes and label W1 width of bars H1 height of bars

5

(ii) Median = 600th distance Estimate = 50 + 240/400 × 50 = 50 + 30 = 80

B1 for 600th M1 for attempt to interpolate A1 CAO

3

TOTAL 8

Q6 (i) (A) P(at most one) =

83 0.83100

=

(B) P(exactly two) =10 2 1 13 0.13

100 100+ + = =

B1 aef M1 for (10+2+1)/100 A1 aef

1

2

(ii) P(all at least one) =

53 52 51 140556 0.145100 99 98 970200

× × = =

M1 for 53100

×

M1dep for product of next 2 correct fractions A1 CAO

3

TOTAL 6


69

Q7 (i)

a = 0.8, b = 0.85, c = 0.9.

B1 for any one B1 for the other two

2

(ii) P(Not delayed) = 0.8 × 0.85 × 0.9 = 0.612

P(Delayed) = 1 – 0.8 × 0.85 × 0.9 = 1 – 0.612 = 0.388

M1 for product A1 CAO M1 for 1 – P(delayed) A1FT

4

(iii) P(just one problem)

= 0.2×0.85×0.9 + 0.8×0.15×0.9 + 0.8×0.85×0.1

= 0.153 + 0.108 + 0.068 = 0.329

B1 one product correct M1 three products M1 sum of 3 products A1 CAO

4

(iv) P(Just one problem | delay)

= P(Just one problem and delay) 0.329

P(Delay) 0.388= = 0.848

M1 for numerator M1 for denominator A1FT

3

(v) P(Delayed | No technical problems)

Either = 0.15 + 0.85 × 0.1 = 0.235

Or = 1 – 0.9 × 0.85 = 1 – 0.765 = 0.235

Or = 0.15 × 0.1 + 0.15 × 0.9 + 0.85 × 0.1 = 0.235

Or (using conditional probability formula)

P(Delayed and no technical problems)P(No technical problems)

0.8 0.15 0.1 0.8 0.15 0.9 0.8 0.85 0.10.8

0.188 0.2350.8

× × + × × + × ×=

= =

M1 for 0.15 + M1 for second term A1CAO M1 for product M1 for 1 – product A1CAO M1 for all 3 products M1 for sum of all 3 products A1CAO M1 for numerator M1 for denominator A1CAO

3

(vi) Expected number = 110 × 0.388 = 42.7 M1 for product A1FT

2

TOTAL 18


70

Q8 (i)

X ~ B(15, 0.2)

3 1215( ) P( = 3) 0.2 0.8 0.2501

3

⎛ ⎞= × × =⎜ ⎟⎝ ⎠

A X

OR from tables 0.6482 0.3980 0.2502− =

( ) P( 3) 1 0.3980 0.6020≥ = − =B X

(C) E(X) = np = 15 × 0.2 = 3.0

M1 0.23 × 0.812

M1 ( )153 × p3 q12

A1 CAO OR: M2 for 0.6482 – 0.3980 A1 CAO M1 P(X≤2) M1 1-P(X≤2) A1 CAO M1 for product A1 CAO

3 3 2

(ii) (A) Let p = probability of a randomly selected child eating at least 5 a day H0: p = 0.2 H1: p > 0.2 (B) H1 has this form as the proportion who eat at least 5 a day is expected to increase.

B1 for definition of p in context B1 for H0 B1 for H1 E1

4

(iii) Let X ~ B(15, 0.2) P(X ≥ 5) = 1 – P(X ≤ 4) = 1 – 0.8358 = 0.1642 > 10% P(X ≥ 6) = 1 – P(X ≤ 5) = 1 – 0.9389 = 0.0611 < 10% So critical region is {6,7,8,9,10,11,12,13,14,15}

7 lies in the critical region, so we reject null hypothesis and we conclude that there is evidence to suggest that the proportion who eat at least five a day has increased.

B1 for 0.1642 B1 for 0.0611 M1 for at least one comparison with 10% A1 CAO for critical region dep on M1 and at least one B1 M1 dep for comparison A1 dep for decision and conclusion in context

6

TOTAL 18


46

4766 Statistics 1

1 (i) 5 2 6 3 4 7 8 7 1 2 2 3 4 5 5 7 9 8 1 Key 6 3 represents 63 mph

G1 stem G1 leaves CAO G1 sorted G1 key [4]

(ii) Median = 72 Midrange = 66.5

B1 FT B1 CAO [2]

(iii)

EITHER: Median since midrange is affected by outlier (52) OR: Median since the lack of symmetry renders the midrange less representative

E1 for median E1 for explanation [2]

TOTAL [8]

2 (i) (A) P(X = 10) = P(5 then 5) = 0.4×0.25 = 0.1

(B) P(X = 30) = P(10 and 20) = 0.4×0.25 + 0.2×0.5 = 0.2

B1 ANSWER GIVEN M1 for full

calculation A1 ANSWER GIVEN

[1]

[2] (ii) E(X) = 10 0.1 15 0.4 20 0.1 25 0.2 30 0.2 20× + × + × + × + × =

E(X2) = 100 0.1 225 0.4 400 0.1 625 0.2 900 0.2 445× + × + × + × + × = Var(X) = 445 – 202 = 45

M1 for Σrp (at least 3 terms correct)

A1 CAO M1 for Σr2p (at least 3

terms correct) M1 dep for – their

E (X )² A1 FT their E(X)

provided Var( X ) > 0

[5]

TOTAL [8]

3 (i) G1 for two labelled

intersecting circles G1 for at least 2

correct probabilities

G1 for remaining

probabilities [3]

(ii) P(G) × P(R) = 0.24 × 0.13 = 0.0312 ≠ P(G∩R) or ≠ 0.06 So not independent.

M1 for 0.24 × 0.13 A1 [2]

0.18 0.06

0.69

G R

0.07


47

(iii)

( ) 0.06 1( | ) 0.25( ) 0.24 4

P R GP R GP G

∩= = = =

M1 for numerator M1 for denominator A1 CAO

[3]

TOTAL [8]

4 (i)

P(20 correct) = 20 10300.6 0.4 0.1152

20

× × =

M1 0.620 × 0.410

M1 ( )3020 × p20 q10

A1 CAO

[3]

(ii) Expected number = 100 × 0.1152 = 11.52

M1 A1 FT (Must not

round to whole number)

[2]

TOTAL [5]

5 (i) P(Guess correctly) = 0.14 = 0.0001 B1 CAO [1]

(ii)

P(Guess correctly) = 1 14! 24

= M1 A1 CAO

[2]

TOTAL [3]

6 (i) 20 × 19 × 18 = 6840 M1

A1 [2]

(ii) 203 – 20 = 7980 M1 for figures – 20

A1 [2]

TOTAL [4]


48

7 (i) 10 × 2 = 20. M1 for 10 × 2

A1 CAO [2]

(ii)

Mean = 10 65 35 75 55 85 20 95 9850 82.08120 120

× + × + × + × = =

It is an estimate because the data are grouped.

M1 for midpoints M1 for double pairs A1 CAO E1 indep

[4]

(iii)

10 × 652 + 35 × 752 + 55 × 852 + 20 × 952 (= 817000)

M1 for Σfx2

Sxx =

29850817000120

− (= 8479.17)

s = 8479.17119

= 8.44

M1 for valid attempt at Sxx A1 CAO [3]

(iv) x – 2s = 82.08 – 2×8.44= 65.2

x + 2s = 82.08 + 2×8.44= 98.96 So there are probably some outliers.

M1 FT for x – 2s M1 FT for x + 2s A1 for both E1 dep on A1

[4]

(v) Negative. E1

[1]

(vi) Upper bound 60 70 80 90 100

Cumulative frequency 0 10 45 100 120

C1 for cumulative frequencies

S1 for scales L1 for labels ‘Length and CF’ P1 for points J1 for joining points

dep on P1 All dep on attempt at cumulative frequency.

[5]

TOTAL [19]


49

8 (i) (A) P(Low on all 3 days) = 0.53 = 0.125 or 1/8

(B) P(Low on at least 1 day) = 1 – 0.53 = 1 – 0.125 = 0.875 (C) P(One low, one medium, one high) = 6 × 0.5 × 0.35 × 0.15 = 0.1575

M1 for 0.53

A1 CAO M1 for 1 – 0.53 A1 CAO M1 for product of probabilities 0.5 × 0.35 × 0.15 or 21/800 M1 × 6 or × 3! or 3P3 A1 CAO

[2]

[2]

[3]

(ii)

X ~ B(10, 0.15) (A) P(No days) = 0.8510 = 0.1969 Or from tables P(No days) = 0.1969

(B) Either 1 910P(1 day) 0.15 0.85 0.3474

1

= × × =

or from tables P(1 day) = P(X ≤ 1) – P(X ≤ 0) = 0.5443 – 0.1969 = 0.3474

M1 A1 M1 0.151 × 0.859 M1 ( )10

1 × p1 q9

A1 CAO OR: M2 for 0.5443 –

0.1969 A1 CAO

[2]

[3]

(iii) Let X ~ B(20, 0.5)

Either: P(X ≥ 15) = 1 – 0.9793 = 0.0207 < 5% Or: Critical region is {15,16,17,18,19,20} 15 lies in the critical region. So there is sufficient evidence to reject H0 Conclude that there is enough evidence to indicate that the probability of low pollution levels is higher on the new street. H1 has this form as she believes that the probability of a low pollution level is greater in this street.

Either: B1 for correct probability of 0.0207 M1 for comparison Or: B1 for CR, M1 for comparison A1 CAO dep on B1M1 E1 for conclusion in

context E1 indep

[5]

TOTAL [17]

Oxford Cambridge and RSA Examinations

GCE

Mathematics (MEI) Advanced Subsidiary GCE 4766

Statistics 1

Mark Scheme for June 2010


Q1 (i)

Positive skewness

B1

1

(ii) Inter-quartile range = 10.3 – 8.0 = 2.3 Lower limit 8.0 – 1.5 2.3 = 4.55 Upper limit 10.3 + 1.5 2.3 = 13.75 Lowest value is 7 so no outliers at lower end Highest value is 17.6 so at least one outlier at upper end.

B1 M1 for 8.0 – 1.5 2.3 M1 for 10.3 + 1.5 2.3 A1 A1

5

(iií) Any suitable answers Eg minimum wage means no very low values Highest wage earner may be a supervisor or manager or specialist worker or more highly trained worker

E1 one comment relating to low earners E1 one comment relating to high earners

2 TOTAL 8 Q2 (i)

4k + 6k + 6k + 4k = 1 20k = 1 k = 0.05

M1 A1 NB Answer given

2 (ii) E(X) = 1 0.2 2 0.3 3 0.3 4 0.2 2.5

(or by inspection) E(X2) = 1 0.2 4 0.3 9 0.3 16 0.2 7.3 Var(X) = 7.3 – 2.52 = 1.05

M1 for Σrp (at least 3 terms correct) A1 CAO M1 for Σr2p (at least 3 terms correct) M1dep for – their E( X )² A1 FT their E(X) provided Var( X ) > 0

5

TOTAL 7 Q3 (i)

Lifetime (x hours) Frequency Width FD

0 < x ≤ 20 24 20 1.2 20 < x ≤ 30 13 10 1.3 30 < x ≤ 50 14 20 0.7 50 < x ≤ 65 21 15 1.4

65 < x ≤ 100 18 35 0.51

M1 for fds A1 CAO Accept any suitable unit for fd such as eg freq per 10 hours. L1 linear scales on both axes and label on vert axis W1 width of bars H1 height of bars

5

3


(ii) Median lies in third class interval (30 < x ≤ 50) Median = 45.5th lifetime (which lies beyond 37 but not as far as 51)

B1 CAO E1 dep on B1

2

TOTAL 7 Q4 (i)

1 × 15

= 15

M1 A1

2

(ii) 4 3 2 1 241 0.03845 5 5 5 625

M1 For

54

541 justor

M1 dep for fully correct product A1

3

(iii) 1 – 0.0384 = 0.9616 or 601/625 B1 1 TOTAL 6 Q5 (i)

Mean =

0 37 1 23 2 11 3 3 4 0 5 1 59 0.78775 75

Sxx = 2

2 2 2 2 2 2 590 37 1 23 2 11 3 3 4 0 5 1 72.5975

s = 72.5974

= 0.99

M1 A1 M1 for Σfx2 s.o.i. M1 dep for good attempt at Sxx BUT NOTE M1M0 if their Sxx < 0 A1 CAO

5

(ii) New mean = 0.787 × £1.04 = £0.818 or 81.8 pence

New s = 0.99 × £1.04 = £1.03 or 103 pence

B1 ft their mean B1 ft their s B1 for correct units dep on at least 1 correct (ft)

3

TOTAL 8 Section B

Q6 (i)

X ~ B(18, 0.1) 2 1618

( ) P(2 faulty tiles) 0.1 0.9 0.28352

A

OR from tables 0.7338 0.4503 0.2835 ( ) P(More than 2 faulty tiles) 1 0.7338 0.2662 B

M1 0.12 0.916 M1 p18

22 q16

A1 CAO OR: M2 for 0.7338 – 0.4503 A1 CAO M1 P(X≤2) M1 dep for 1-P(X≤2) A1 CAO

3

3 4


5

0.95

0.05 Late

0.6

0.4

On time

On time

On time 0.95

0.05 Late

Late

1000

1100

On time 0.95

Late

On time

0.05

0.95

0.05 Late

1200

On time

On time

Late

Late

0.6

0.4

0.6

0.4

(C) E(X) = np = 18 0.1 = 1.8 M1 for product 18 0.1 A1 CAO

2

(ii) (A) Let p = probability that a randomly selected tile is faulty H0: p = 0.1 H1: p > 0.1

B1 for definition of p in context B1 for H0 B1 for H1

3

(B) H1 has this form as the manufacturer believes that the number of faulty tiles may increase.

E1

1

(iii) Let X ~ B(18, 0.1) P(X ≥ 4) = 1 – P(X ≤ 3) = 1 – 0.9018 = 0.0982 > 5% P(X ≥ 5) = 1 – P(X ≤ 4) = 1 – 0.9718 = 0.0282 < 5% So critical region is {5,6,7,8,9,10,11,12,13,14,15,16,17,18}

B1 for 0.0982 B1 for 0.0282 M1 for at least one comparison with 5% A1 CAO for critical region dep on M1 and at least one B1

4

(iv) 4 does not lie in the critical region, (so there is insufficient evidence to reject the null hypothesis and we conclude that there is not enough evidence to suggest that the number of faulty tiles has increased.

M1 for comparison A1 for conclusion in context

2 TOTAL 18Q7 (i)

G1 first set of branches G1 indep second set of branches G1 indep third set of branches G1 labels

4


6

(ii)

(A) P(all on time) = 0.953 = 0.8574 (B) P(just one on time) = 0.95×0.05×0.4 + 0.05×0.6×0.05 + 0.05×0.4×0. 6 = 0.019 + 0.0015 + 0.012 = 0.0325 (C) P(1200 is on time) = 0.95×0.95×0.95 +0.95×0.05×0.6 + 0.05×0.6×0.95 + 0.05×0.4×0. 6 = 0.857375+0.0285+0.0285+0.012= 0.926375

M1 for 0.953 A1 CAO M1 first term M1 second term M1 third term A1 CAO M1 any two terms M1 third term M1 fourth term A1 CAO

2

4

4

(iii) P(1000 on time given 1200 on time) = P(1000 on time and 1200 on time) / P(1200 on time) = 0.95 0.95 0.95 0.95 0.05 0.6 0.885875 0.9563

0.926375 0.926375

M1 either term of numerator M1 full numerator M1 denominator A1 CAO

4

Total 18

GCE


Advanced Subsidiary GCE

Unit 4766: Statistics 1

Mathematics (MEI)

Mark Scheme for January 2011


SECTION A Q1 (i)

Mode = 960 (grams) Median = 1020 (grams) N.B. 96 and 102 gets SC1

B1 CAO B1 CAO

2

Ignore units and working

(ii) Positive E1 1 Not right skewed Not positive correlation

TOTAL 3 Q2 (i) P(product of two scores < 10) = 13

16 = 0.8125

B1 1

Allow 0.813 or 0.812

(ii)

P(even) P(< 10) = 0.5 1316

= 1332

= 0.40625

P(even ∩ < 10) = 616

= 0.375

So not independent.

M1 for 0.5 1316

or 1332

FT their answer to (i)

M1 for 616

A1

3

Do not allow these embedded in probability formulae Also allow P(even|<10) = 6/13≠ P(even) = 1/2 Or P(<10|even) = 6/8≠ P(<10) = 13/16 Or P(even|<10) = 6/13≠ P(even|<10’) = 2/3 Or P(<10|even) = 6/8≠ P(<10|even’) = 7/8 For all of these alternatives allow M2 for both probabilities. (M1 not available except if they correctly state both probabilities EG P(even|<10) and P(even) and get one correct) If they do not state what probabilities they are finding, give M2 for one of the above pairs of probabilities with ≠ symbol

TOTAL 4

1


Q3 (i)

133

ways of choosing the men = 286

M1 for seen 133

A1

2

Accept 13C3 or 13!/(3!10!) or equivalent for M1 No marks for permutations

(ii) 133

× = 286 × 120 = 34320 103

M1 for product A1 FT their 286

2

For permutations 1716 × 720 = 1235520 allow SC1 406 (from 286 + 120) scores SC1 (without further working)

(iii) 236

=100947

34320/100947 = 1040/3059 = 0.340 (allow 0.34)

M1 for denominator of 23

6

A1 FT

2

FT their 34320 Or 6C3×13/23×12/22×11/21×10/20×9/19×8/18 = 0.340 scores M1 for product of fractions and A1 for 6C3× and correct evaluation For permutations 1235520/72681840=0.017 scores SC1 Allow full marks for fractional answers, even if unsimplified 406/100947 = 0.00402 gets M1A1 with or without working

TOTAL 6

2


Q4 (i)

2k + 6k + 12k + 20k + 30k = 1, 70k = 1

k = 170

M1 A1 NB ANSWER GIVEN

2

For five multiples of k (at least four correct multiples) Do not need to sum or =1 for M1 Condone omission of either 70k = 1 or k = 1/70 but not both Condone omission of k: 2+6+12+20+30=70 Allow substitution of k = 1/70 into formula and getting at least four of 2/70, 6/70, 12/70, 20/70, 30/70 for M1 and 2/70+6/70+12/70+20/70+30/70 = 1 for A1

(ii) E(X) = 2 6 12 20 301 2 3 4 5 470 70 70 70 70

E(X2) =

2 6 12 20 30 12041 4 9 16 25 17.270 70 70 70 70 70

Var(X) = 17.2 – 42 = 1.2

M1 for Σrp (at least 3 terms correct) A1 CAO M1 for Σr2p (at least 3 terms correct) M1dep for - their E(X )² A1 FT their E(X) but not an error in E(X2) provided Var( X ) > 0

5

280/70 scores M1A0 USE of E(X-µ) gets M1 for attempt at (x-µ) should see (-3) , (-2) , (-1) , 0 , 1 (if E(X) correct but FT their E(X)) (all 5 correct for M1), then M1 for Σp(x-µ)

2 2

2 2 2 2 2

2 (at least 3 terms correct with their probabilities) Allow all M marks with their probabilities, (unless not between 0 and 1, see below for all probs 1/70). Division by 5 or other spurious value at end gives max M1A1M1M1A0, or M1A0M1M1A0 if E(X) also divided by 5. Unsupported correct answers get 5 marks. SC2 for use of 1/70 for all probabilities leading to E(X) = 3/14 and Var(X) = 145/196 = 0.74

TOTAL 7

3


Q5 (i)

P(Wet and bus) = 0.4 0.7

= 0.28

M1 for multiplying probabilities

A1 CAO

2

Fractional answer = 7/25 (Allow 28/100)

(ii) P(Walk or bike) = 0.6 0.5 + 0.6 0.4 + 0.4 0.2 + 0.4 0.1 or 0.3+0.24+0.08+0.04 = 0.66

M1 for any two correct pairs M1 for sum of all four correct terms With no extra terms for second M1 A1 CAO

3

Or = 0.6 0.9 + 0.4 0.3 gets M1 for either term = 0.54 + 0.12 gets M1 for sum of both A1 CAO Or = 1 – 0.6×0.1 – 0.4×0.7 = 0.66. M1 for 1 – one correct term, M1 for complete correct expression and A1 for correct evaluation.

(iii) P(Dry given walk or bike) P(Dry and walk or bike)

P(Walk or bike)

= 0.6 0.90.66 = 0.54

0.66=

119 = 0.818

M1 for numerator leading to 0.54 M1 for denominator Ft their P(Walk or bike) from (ii) provided between 0 and 1 A1 FT

3

Allow 0.82, not 0.819 More accurate answer =0.81818 Fractional answer = 54/66 = 27/33 = 9/11 Condone answer of 0.8181 Do not give final A1 if ans ≥ 1

TOTAL 8

4


Q6 (i) (A) P(Avoided air travel) = 7

100= 0.07

(B) P(At least two) =11 2 1 4 18 9100 100 50

= 0.18

B1 aef isw M1 for (11+2+1+4)/100 A1 aef isw

1

2

For M1 terms must be added must be as above or better with no extra terms (added or subtracted) for M1 Must simplify to 18/100 or 9/50 or 0.18 for A1 SC1 for 18/58 Or 1 – (14+26+0+42)/100 = 0.18 gets M1A1

(ii) P(Reduced car use | Avoided air travel) = 67

= 0.857

M1 for denominator 7 or 7/100 or 0.07 FT their (i)A A1 CAO

2

Allow 0.86

(iii) P(None have avoided air travel) = 93 92 91 0.8025100 99 98

M1 for 93/100× (triple product) M1 for product of remaining fractions A1

3

Fuller answer 0.802511, so allow 0.803 without working, but 0.80 or 0.8 only with working . (93/100)3 scores M1M0A0 which gives answer 0.804357 so watch for this. M0M0A0 for binomial probability including 0.93100 but 3C0×0.070×0.933 still scores M1 (k/100)3 for values of k other than 93 scores M0M0A0

( 1) ( 2)100 99 98

k k k for values of k other than 93 scores

M1M0A0 Correct working but then multiplied or divided by some factor scores M1M0A0 93P3 / 100P3 = 0.803 93P3 seen M1 divided by 100P3 M1 0.803 A1 93C / 100

3 C3 = 0.803 Allow unsimplified fractional answer 778596/970200 =9269/11550

TOTAL 8

5


SECTION B Q7 (i)

Income Frequency Width FD

0 ≤ x ≤ 20 238 20 11.9 20 < x ≤ 40 365 20 18.25 40 < x ≤ 60 142 20 7.1 60 < x≤ 100 128 40 3.2

100 < x ≤ 200 45 100 0.�5

INCORRECT DIAGRAMS: Frequency diagrams can get M0, A0, L0, W1, H0 MAXIMUM Thus frequency density = frequency × width, frequency/midpoint etc gets MAX M0A0L0W1H0

M1 for fds A1 CAO Accept any suitable unit for fd such as eg freq per £1000. L1 linear scale and label on vertical axis W1 linear scale on horizontal axis and correct width of bars H1 height of bars

5

At least 4 fds correct for M1 M1 can be also be gained from freq per 10K - 119, 182.5, 71, 32, 4.5 (at least 4 correct) and A1 for all correct Accept any suitable unit for fd, eg freq per £10K, BUT NOT FD per £1000 Allow fds correct to at least one dp If fd not explicitly given, M1 A1 can be gained from all heights correct (within one square) on histogram (and M1A0 if at least 4 correct) Allow restart although given fd wrong For L1, label required on vert axis in relation to first M1 mark ie fd or frequency density or if relevant freq/£10K, freq/£k etc (NOT fd/£10K) Accept f/w or f/cw (freq/width or freq/class width) Ignore horizontal label L1 can also be gained from an accurate key – may see 1 square = 36.5 or 23.8 or 14.2 For W1, must be drawn at 0, 20, 40 etc NOT 19.5 or 20.5 etc NO GAPS ALLOWED Must have linear scale. No inequality labels on their own such as 0≤I<20, 20≤I<40 etc but allow if a clear horizontal linear scale is also given. FT of heights dep on M1 all must agree with their fds If fds not given and one height is wrong then max M1A0L1W1H0 – visual check only (within one square) –no need to measure precisely

6


(ii)

Mean = 918

4515012880142503653023810

= 918

37420 = 40.8

M1 for midpoints M1 for midpoints ×frequencies with divisor 918 A1 CAO

3

At least three midpoints correct for M1 (seen in (ii) or in table in (i)) No marks if not using midpoints Second M1 for sight of at least 3 double pairs seen out of 10 × 238 + 30 × 365 + 50 × 142 + 80 × 128 + 150 × 45 with divisor 918 Numerator = 2380+10950+7100+10240+6750 Use of LCB or UCB for midpoints here scores 0 For answer 40.76 or 40.8 or 41 mark as B3 37420/918 o.e. scores M1M1A0 NB Accept answers seen without working in part (ii) or (iii) (from calculator) Use of ‘not quite right’ midpoints such as 10.5, 30.5, etc can get M0M1A0 here and SC3 in (iii) Watch for incorrect method 238/10+365/30+142/50+128/80+45/150=40.71 Allow max 4 sf in final answer Also accept £40760, £40800 etc

(iii) ∑fx = 238×10 + 365×30 + 142×50 + 128×80 + 45×1502 2 2 2 2 2 = 2539000 Or 238×100 + 365×900 + 142×2500 + 128×6400 + 45×22500 = 2539000 Or 2380×10 + 10950×300 + 7100×50 + 10240×80 + 13500×150 =2539000

Sxx = 2374202539000

918 = 1013666

s = 1013666917

= 33.2

M1 for at least 3 multiples fx2 A1 for Σfx2

M1 for attempt at Sxx Dep on first M1 BUT NOTE M1M0 if their Sxx < 0 A1 CAO If using LCB or UCB

4

For A1, all midpoints and frequencies correct Or Sxx = 2539000 – 918 × 40.76 = 1013855, s=33.25. Using mean 40.8 leads to 1010861 , s=

2

33.20, Using mean = 41 leads to Sxx = 995844 and s = 32.95 M1M1 for ∑f(x-xbar) M1 for first three terms, M1 for all 5 terms

2

238 × (10-40.76) + 365 × (30-40.76) + 142 × (50-40.76) + 128 × (80-40.76) + 45 × (150-40.76) (= 1013666) A1 for

2 2

2 2 2

Sxx = 1013666 A1 for final answer

7


consistently then allow SC2 if working is fully correct but SC0 otherwise but no marks in part (ii)

For answer 33.25 or 33.3 or 33.2 (www) can just mark as B4 - these may be from calculator without working Allow 33 with correct working rmsd = √(1013666/918) (=33.23) gets M1A1M1A0 (if seen) WATCH FOR DIVISOR OF 918 Allow max 4 sf in final answer Allow £33200 etc

(iv) ( x – 2s = 40.76 – 2 33.25 = –25.74 ) x + 2s = 40.76 + 2 33.25 = 107.26 Comment that there are almost certainly some outliers. Appropriate comment such as

‘No, since there is nothing to indicate that these high earners represent a separate population.’

M1 for x + 2s or x – 2s

A1 for 107.26 (FT) E1 E1 Dep on upper limit in range 106 - 108

4

FT any positive mean and positive sd for M1 Only follow through numerical values, not variables such as s, so if a candidate does not find s but then writes here ‘limit is 40.76+ 2 standard deviation’, do NOT award M1 (This rule of not following through variables applies in all situations) Award E0E0 if their upper limit > 200 Allow ‘Must be some outliers’ Allow any comments that implies that there are outliers No marks in (iv) unless using x + 2s or x – 2s

(v) New mean = 1.15 × 40.76 = 46.87 New variance = 1.152 × 33.252 = 1462

For misread 1.5 in place of 1.15 For 1.5 × 40.76 = 61.1 and 1.52 × 33.252 = 2490 allow SC2 if all present but SC0 otherwise

B1 FT M1A1 FT

3

FT their mean (if given to ≥ 2 s.f.) FT their s (if given to ≥ 2 s.f.) provided their s>0 If RMSD found in part (i) rather than s, then FT their RMSD For new SD = 38.24 found instead of variance give M1A0 even if called variance (and FT their s) M0A0 for 1.15 x 33.252= 1271 Allow max 4 sf in final answers Min 2 sf If candidate ‘starts again’ only award marks for CAO

TOTAL 19

8


Q8 (i)

E(X) = np = 12 0.2 = 2.4 Do not allow subsequent rounding.

M1 for product A1 CAO

2

If wrong n used consistently throughout, allow M marks only. NB If they round to 2, even if they have obtained 2.4 first they get M1A0. For answer of ‘2.4 or 2 if rounded up’ allow M1A0 Answer of 2 without working gets M0A0. If they attempt E(X) by summing products xp give no marks unless answer is fully correct.

(ii)

X ~ B(12, 0.2)

2 1012( ) P(Wins exactly 2) 0.2 0.8 0.2835

2

A

OR from tables 0.5583 0.2749 0.2834

M1 0.22 0.810

M1 122 p2 q10

A1 CAO OR: M2 for 0.5583 – 0.2749 A1 CAO

3

With p + q = 1 Also for 66 × 0.004295 Allow answers within the range 0.283 to 0.284 with or without working or 0.28 to 0.283 if working shown See tables at the website http://www.mei.org.uk/files/pdf/formula_book_mf2.pdf

( ) P(Wins at least 2) 1 0.2749 0.7251 B

M1 P(X≤1) M1 1-P(X≤1) A1 CAO

3

M1 0.2749 seen M1 1 – 0.2749 seen Allow 0.725 to 0.73 but not 0.72. Point probability method: P(1) = 12×0.2×0.811 = 0.2062 , P(0) = 0.812 = 0.0687 So P(X≤1) = 0.2749 gets M1 then mark as per scheme SC1 for 1 – P(X≤2) = 1 – 0.5583 = 0.4417 For misread of tables value of 0.2749 , allow 0 in (A) but MAX M1M1 in (B) For P(X>1) = P(X=2) + P(X=3) + P(X=4) +.... allow M1 for 0.2835+0.2362+0.1329+0.0532+0.0155 and second M1 for 0.0033+0.0005+0.0001 and A1 for 0.725 or better M0M0A0 for 1 - P(X=1) = 1 – 0.2062 = 0.7938

9


(iii) Let p = probability that Ali wins a game H0: p = 0.2 H1: p > 0.2 H1 has this form as Ali claims that he is better at winning games than Mark is. EITHER Probability method: P(X ≥ 7) = 1 – P(X ≤ 6) = 1 – 0.9133 = 0.0867 > 5% So not significant, so there is not enough evidence to reject the null hypothesis and we conclude that there is not enough evidence to suggest that Ali is better at winning games than Mark. Must include ‘not enough evidence’ or something similar for E1. ‘Not enough evidence’ can be seen in the either for the A mark or the E mark. Do not allow final conclusions for E1 such as : 'there is evidence to suggest that Ali is no better at winning games than Mark' or 'Mark and Ali have equal probabilities of winning games'

B1 for definition of p in context B1 for H0 B1 for H1 E1 B1 for P(X ≥ 7) B1 for 0.0867 Or 1 – 0.9133 seen M1 for comparison with 5% dep on B1 for 0.0867 A1 for not significant or ‘accept H0’ or ‘cannot reject H0’ or ‘reject H1’ E1 dep on M1A1 Do not award first B1 for poor symbolic notation such as P(X = 7) = 0.0867 This comment applies to all methods

4

5

Minimum needed for B1 is p = probability that Ali wins. Allow p = P(Ali wins) for B1 Definition of p must include word probability (or chance or proportion or percentage or likelihood but NOT possibility). Preferably as a separate comment. However can be at end of H as long as it is a clear definition ‘ p = the probability that Ali wins a game, NOT just a sentence ‘probability is 0.2’

0

H : p(Ali wins) = 0.2, H : p(Ali wins) > 0.2 gets B0B1B1Allow p=20%, allow θ or π and ρ but not x. However allow any single symbol

0 1

if defined / Allow H = p=0.2, Allow H : p=0 0

210

Do not allow H : P(X=x) = 0.2, H : P(0 1 X=x) > 0.2 Do not allow H : =0.2, =20%, P(0.2), p(0.2), p(x)=0.2, x=0.2 (unless x correctly defined as a probability)

0

Do not allow H :p≥0.2, 1Do not allow H and H reversed for B marks but can still get E1

0 1

Allow NH and AH in place of H and H 0 1For hypotheses given in words allow Maximum B0B1B1E1 Hypotheses in words must include probability (or chance or proportion or percentage) and the figure 0.2 oe. Zero for use of point prob - P(X = 7) = 0.0546

10


OR Critical region method: Let X ~ B(20, 0.2) P(X ≥ 7) = 1 – P(X ≤ 6) = 1 – 0.9133 = 0.0867 > 5% P(X ≥ 8) = 1 – P(X ≤ 7) = 1 – 0.9679 = 0.0321 < 5% So critical region is {8,9,10,11,12,13,14,15,16,17,18,19,20}

7 does not lie in the critical region, so not significant, So there is not enough evidence to reject the null hypothesis and we conclude that there is not enough evidence to suggest that Ali is better at winning games than Mark.

B1 for 0.0867 B1 for 0.0321 M1 for at least one comparison with 5% A1 CAO for critical region and not significant or ‘accept H0’ or ‘cannot reject H0’ or ‘reject H1’ dep on M1 and at least one B1 E1 dep on M1A1

Allow any form of statement of CR eg X ≥8, 8 to 20, 8 or above, X > 8, {8, …}, annotated number line, etc but not P(X ≥8) {8,9,10,11,12} gets max B2M1A0 – tables stop at 8. NB USE OF POINT PROBABILITIES gets B0B0M0A0 Use of complementary probabilities Providing there is sight of 95%, allow B1 for 0.9133, B1 for 0.9679, M1 for comparison with 95% A1CAO for correct CR See additional notes below the scheme for other possibilities PLEASE CHECK THAT THERE IS NO EXTRA WORKING ON THE SECOND PAGE IN THE ANSWER BOOKLET

TOTAL 17 NOTE RE OVER-SPECIFICATION OF ANSWERS If answers are grossly over-specified (see instruction 8), deduct the final answer mark in every case, except where there are more than two over-specified answers in a single question (only likely in question 7) in which case deduct a mark in only the first two cases of over-specification in that question. Probabilities should also be rounded to a sensible degree of accuracy. ADDITIONAL NOTES RE Q8 PART iii Use of n = 12 P(X ≥ 7) = 1 – P(X ≤ 6) = 1 – 0.9961 = 0.0039 < 5% So significant or reject H0 etc, so there evidence to suggest that Ali is better at winning games than Mark. Gets B1 for P(X ≥ 7) B1 for 0.0039 M1 for comparison with 5% dep on B1 for 0.0039 A1 for significant E1 for evidence to suggest that Ali is better at winning games than Mark. Then award MR -1 so maximum of 4 possible Comparison with 95% method B1 for P(X ≤ 6) B1 for 0.9133

11


M1 for comparison with 95% dep on B1 A1 for not significant or ‘accept H0’ or ‘cannot reject H0’ E1 Smallest critical region method: Either: Smallest critical region that 7 could fall into gets B1 and has size 0.0867 gets B1, This is > 5% gets M1, A1, E1 as per scheme NB These marks only awarded if 7 used, not other values. Use of k method with no probabilities quoted: P(X ≥ 7) = 1 – P(X ≤ 6) > 5% P(X ≥ 8) = 1 – P(X ≤ 7) < 5% These may be seen in terms of k or n. Either k = 8 or k – 1 = 7 so k = 8 gets SC1 so CR is {8,9,10,11,12,13,14,15, 16, 17, 18, 19, 20} gets another SC1and conclusion gets another SC1 Use of k method with one probability quoted: 1 - 0.9679 < 5% or 0.0321 < 5% gets B0B1M1 P(X ≤ k - 1) = P(X ≤ 7) so k – 1 = 7 so k = 8 (or just k = 8 ) so CR is {8,9,10,11,12,13,14,15, 16, 17, 18, 19, 20} and conclusion gets A1E1 Two tailed test with H1: p≠0.2 Hyp gets max B1B1B0E0 P(X ≥ 7) = 0.0867 gets B1B1comparison with 2.5% gets M1 (must be 2.5%) Final marks A0E0 Two tailed test done but with correct H1: p>0.2 Hyp gets max B1B1B1E1

12


13

if compare with 5% ignore work on lower tail and mark upper tail as per scheme so can score full marks if compare with 2.5% no marks B0B0M0A0E0 One tailed test with H1: p<0.2 Hyp gets max B1B1B0E0 no further marks B0B0M0A0E0 Lower tailed test with H1: p>0.2 Hyp gets max B1B1B0E0 no further marks B0B0M0A0E0 Line diagram method B1 for squiggly line between 7 and 8 or on 8 exclusively (ie just one line), B1dep for arrow pointing to right, M1 0.0321 seen on diagram from squiggly line or from 8, A1E1 for correct conclusion Bar chart method B1 for line clearly on boundary between 7 and 8 or within 8 block exclusively (ie just one line),, B1dep for arrow pointing to right, M1 0.0321 seen on diagram from boundary line or from 8, A1E1 for correct conclusion Using P(Not faulty) method H0: p=0.8, H1: p<0.8, where p represents the prob that Ali loses a game Ali claims that the proportion of games that he loses is less than 80% gets B1B1B1E1 P(X≤13)=0.0867>5% So not significant, so there is not enough evidence to reject the null hypothesis and we conclude that there is not enough evidence to suggest that Ali is better at winning games than Mark. Gets B1B1M1A1E1


GCE

Mathematics (MEI) Advanced Subsidiary GCE

Unit 4766: Statistics 1



1. 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.

2. For answers scoring no marks, you must either award NR (no response) or 0, as follows:

Award NR (no response) if: Nothing is written at all in the answer space There is a comment which does not in any way relate to the question being asked (“can’t do”, “don’t know”, etc.) There is any sort of mark that is not an attempt at the question (a dash, a question mark, etc.)

The hash key [#] on your keyboard will enter NR.

Award 0 if: There is an attempt that earns no credit. This could, for example, include the candidate copying all or some of the question, or any working

that does not earn any marks, whether crossed out or not.

3. The following abbreviations may be used in this mark scheme. M1 method mark (M2, etc, is also used) A1 accuracy mark B1 independent mark 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 ft follow through isw ignore subsequent working oe or equivalent rot rounded or truncated sc special case soi seen or implied www without wrong working

1


4. Annotating scripts. The following annotations are available:

and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working (after correct answer obtained) 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 is also available to highlight any particular points on a script.

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6. Write a brief report on the performance of the candidates. Your Team Leader will tell you when this is required. The Assistant Examiner’s Report Form (AERF) can be found on the Cambridge Assessment Support Portal. This should contain notes on particular strengths displayed, as well as common errors or weaknesses. Constructive criticisms of the question paper/mark scheme are also appreciated.

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8. The schedule of dates for the marking of this paper is displayed under ‘OCR Subject Specific Details’ on the Cambridge Assessment Support Portal. It is vitally important that you meet these requirements. If you experience problems that mean you may not be able to meet the deadline then you must contact your Team Leader without delay.

2


SECTION A Q1 (i)

1000 × 0.013 = 13 Or 0.2 × 65 = 13 Or 0.2 × 5 × 13= 13

M1 A1 M1 for 0.2 × 65

2

Allow with or without working For MR 1000 × 0. 13 = 130 Allow M1A0 Allow M1A0 if extra terms added eg 1000 × 0.004 SC1 for 1000 × 0.014 = 14 For whole calculation

(ii) Positive B1 1 Allow +ve but NOT skewed to the right Do not allow ‘positive correlation’

(iii) Minimum value = 1500 Maximum value = 2500

B1 Without wrong working B1 Without wrong working

2 Exact answers only unless good explanation such as eg no road has length zero so min is eg 1501 SC1 for lower answer between 1499 and 1501 and upper between 2499 and 2501 Allow answer given as inequality

TOTAL 5 Q2 (i)

1 1 1 1 1 1Either P(alphabetic order)5 4 3 2 1 120

1 1or P(alphabetic order) 0.008335! 120

M1 for 5! or 120 or 5P5 seen or product of correct fractions A1 CAO

2

Allow 0.0083 or 1/120 but not 0.008

(ii) 2 1 1Either P(picks Austen and Bronte)5 4 10

1 1 1or P(picks Austen and Bronte) 25 4 10

1 1or P(picks Austen and Bronte)5 102

M1 for denominators M1 for 2× dep on correct denominators A1 CAO Or M1 for

or 10 5

2 5M1 for 1/ 2

A1 CAO

3

1/5P2 scores M1 also 1/20 oe scores M1 even if followed by further incorrect working

52

seen as part of a binomial expression gets

M0M0A0

TOTAL 5

3


Q3 (i)

P(X = 0) = 0.756 = 0.178 M1 for 0.756 A1 CAO

2

Or from tables 0.1780 Or 729/4096 Allow 0.18 with working

(ii) E(X) = np = 50 0.178 = 8.9

M1 for product A1 FT

2

FT their answer to (i) providing it’s a probability NB A0 if subsequently rounded

TOTAL 4 Q4 (i)

G1 labelled linear scales on both axes G1 heights

2

Accept r or x for horizontal label and p or better for vertical including probability distribution Visual check only Allow G1G0 for points rather than lines Bars must not be wider than gaps for second G1 Condone vertical scale 1, 2, 3, 4, 5 and Probability (×) 1/18 as label BOD for height of r = 0 on vertical axis

(ii) (A) If X = 1, possible scores are (1,2), (2,3), (3,4), (4,5), (5,6) and (2,1), (3,2), (4,3), (5,4), (6,5)

(All are equally likely) so probability = 1036

= 518

(B) If X = 0, possible scores are (1,1), (2,2), (3,3), (4,4), (5,5),

(6,6) so probability = 636

= 16

M1 A1 B1

2

1

Also M1 for a clear correct sample space seen with the ten 1’s identified by means of circles or ticks oe soi. Must be convincing. No additional values such as 0,1 and 1,0 Do not allow ‘ just 10 ways you can have a difference of 1 so 10/36’ or equivalent SC1 for possible scores are (1,2), (2,3), (3,4), (4,5), (5,6) so probability = 2 × 5 × 1/36 with no explanation for 2 × Also B1 for a clear correct sample space seen with the six 0’s identified by means of circles or ticks oe soi. Must be convincing. No additional values. Allow both dice must be the same so probability = 6/36 = 1/6. Allow 1 × 1/6= 1/6 BOD

(iii) Mean value of X = 1 5 2 1 1 1 170 1 2 3 4 5 1 1.9 46 18 9 6 9 18 18

M1 for Σrp (at least 3 terms correct) A1 CAO

2

Or 35/18 Division by 6 or other spurious factor gets MAX M1A0

TOTAL 7

4


Q5 (i)

G1 for two labelled intersecting circles G1 for at least 2 correct probabilities. G1 for remaining correct probabilities

3

Allow labels such as P(W) and P(F) Allow other sensible shapes in place of circles

(ii) P(W) P(F) = 0.14 0.41 = 0.0574 P(W∩F) = 0.11 So not independent.

M1 for 0.41 0.14 A1 Condone dependent Must have full method www Must have either P(W∩F) or 0.11

2

Answer of 0.574 gets Max M1A0 Omission of 0.0574 gets M1A0 Max Or: P(W|F) = 0.11/0.41 = 0.268 ≠ P(W) (= 0.14) M1 for full working P(F|W) = 0.11/0.14 = 0.786 ≠ P(F) (= 0.41) M1 for full working No marks without correct working

(iii) ( ) 0.11 11( | ) 0.268( ) 0.41 41

P W FP W FP F

This is the probability that a randomly selected respondent works (part time), given that the respondent is female.

M1 for correct fraction A1 E1 For E1 must be in context – not just talking about events F and W

3

Allow 0.27 with working Allow 11/41 as final answer Condone ‘if’ or ‘when’ for ‘given that’ but not the words ‘and’ or ‘because’ or ‘due to’ for E1. E1 (independent of M1): the order/structure must be correct i.e. no reverse statement Allow ‘The probability that a randomly selected female respondent works part time’ oe

TOTAL 8

.03 0.11

0.56

W F

0.30

5


Q6 (i) Mean = 1 10 2 40 3 15 4 5 155 2.214

70 70

Sxx = 2

2 2 2 2 1551 10 2 40 3 15 4 5 385 343.21 41.7970

s = 41.7969

= 0.778

M1 A1 CAO M1 for Σfx2 s.o.i. M1 for attempt at Sxx Dep on first M1 A1 CAO If 0.778 or better seen ignore previous incorrect working (calculator answer) Allow final answer to 2 sig fig (www)

5

For M1 allow sight of at least 3 double pairs seen from 1 10 2 40 3 15 4 5 with divisor 70. Allow answer of 155/70 or 2.2 or 2.21 or 31/14 oe For 155/70 = eg 2.3 , allow A1 isw M1 for 2 52 2 21 10 2 40 3 15 4 with at least three correct terms Using exact mean leads to Sxx = 41.79 , s=0.778, Using mean 2.214 leads to Sxx = 41.87 , s=0.779, Using mean 2.21 leads to Sxx = 43.11 and s = 0.790 Using mean 2.2 leads to Sxx = 46.2 and s = 0.818 Using mean 2 leads to Sxx = 105 and s = 1.233 All the above get M1M1A1 except the last one which gets M1M1A0 RMSD(divisor n rather than n – 1) =√(41.79/70) = 0.772 gets M1M1A0 Alternative method, award M1for at least 3 terms of and second M1 for all 4 terms of

2 2 2

21 2.214 10 2 2.214 40 3 2.214 15

4 2.214 5( 41.79)

NB Allow full credit for correct answers without working (calculator used)

(ii) Mean would decrease

Standard deviation would increase

B1

B1

2

Do not accept increase/decrease seen on their own – must be linked to mean and SD. Allow eg ‘It would skew the mean towards zero’ And eg ‘ It would stretch the SD’ SC1 for justified argument that standard deviation might either increase or decrease according to number with no eggs (n≤496 increase, n≥497 decrease)

TOTAL 7

6


SECTION B Q7 (i)

X ~ B(20, 0.15)

(A) Either 1 1920P( = 1) 0.15 0.85 0.1368

1

X

or P(X = 1) = P(X ≤ 1) – P(X ≤ 0) = 0.1756 – 0.0388 = 0.1368 (B) P(X ≥ 2) = 1 – P(X ≤ 1) = 1 – 0.1756 = 0.8244

M1 0.151 0.8519

M1 201 p1 q19

A1 CAO OR: M2 for 0.1756 – 0.0388 A1 CAO M1 for 1 – their 0.1756 A1 CAO

3

2

With p + q = 1 Allow answer 0.137 with or without working or 0.14 if correct working shown See tables at the website http://www.mei.org.uk/files/pdf/formula_book_mf2.pdf For misread of tables 0.3917 – 0.1216 = 0.2701 allow M1M1A0 also for 0.1304 – 0.0261 = 0.1043

Provided 0.1756 comes from P(X=0) +P(X=1) Allow answer 0.824 with or without working or 0.82 if correct working shown Point probability method: P(1) = 0.1368 , P(0) = 0.0388 So 1 - P(X≤1) = 1 - 0.1756 gets M1 then mark as per scheme M0A0 for 1 – P(X ≤ 1) = 1 – 0.4049 = 0.5951 For misread of tables 1 – 0.3917 = 0.6083 allow M1A1 also for 1 – 0. 1304 = 0.8696 provided consistent with part (A) OR M1A0 if formula used in part (A)

7

http://www.mei.org.uk/files/pdf/formula_book_mf2.pdf


(ii) Let X ~ B(n, p) Let p = probability of a ‘no-show’ (for population) H0: p = 0.15 H1: p < 0.15

H1 has this form because the hospital management hopes to reduce the proportion of no-shows.

B1 for definition of p B1 for H0 B1 for H1 E1 Allow correct answer even if H1 wrong

4

Allow p = P(no-show) for B1 Definition of p must include word probability (or chance or proportion or percentage or likelihood but NOT possibility). Preferably as a separate comment. However can be at end of H0

as long as it is a clear definition ‘ p = the probability of no-show, NOT just a sentence ‘probability is 0.15’ H0 : p(no-show) = 0.15, H1 : p(no-show) < 0.15 gets B0B1B1 Allow p=15%, allow θ or π and ρ but not x. However allow any single symbol if defined Allow H0 = p=0.15, Do not allow H0 : P(X=x) = 0.15, H1 : P(X=x) < 0.15 Do not allow H0: =0.15, =15%, P(0.15), p(0.15), p(x)=0.15, x=0.15 (unless x correctly defined as a probability) Do not allow H1:p≤0.15, Do not allow H0 and H1 reversed for B marks but can still get E1 Allow NH and AH in place of H0 and H1 For hypotheses given in words allow Maximum B0B1B1E1 Hypotheses in words must include probability (or chance or proportion or percentage) and the figure 0.15 oe.

(iii) P(X 1) = 0.1756 > 5% So not enough evidence to reject H0. Not significant. Conclude that there is not enough evidence to indicate that the proportion of no-shows has decreased.

M1 for probability seen, but not in calculation for point probability M1 dep for comparison A1

4

Zero for use of point prob - P(X = 1) = 0.1368 Do NOT FT wrong H1 Allow accept H0, or reject H1 Full marks only available if ‘not enough evidence to…’ oe mentioned somewhere Do not allow ‘enough evidence to reject H1’ for final mark but can still get 3/4 Upper end comparison: 1 – 0.1756 = 0.8244 < 95% gets

8


Note: use of critical region method scores M1 for region {0} M1 for 1 does not lie in critical region, then A1 E1 as per scheme

E1 dep for conclusion in context.

M2 then A1E1 as per scheme Line diagram method M1 for squiggly line between 0 and 1 with arrow pointing to left, M1 0.0388 seen on diagram from squiggly line or from 0, A1E1 for correct conclusion Bar chart method M1 for line clearly on boundary between 0 and 1 and arrow pointing to left, M1 0.0388 seen on diagram from boundary line or from 0, A1E1 for correct conclusion

(iv) 6 < 8 So there is sufficient evidence to reject H0 Conclude that there is enough evidence to indicate that the proportion of no-shows appears to have decreased.

M1 for comparison seen A1 E1 for conclusion in context

3

Allow ‘6 lies in the CR’ Do NOT insist on ‘not enough evidence’ here Do not FT wrong H1:p>0.15 but may get M1 In part (iv) ignore any interchanged H0 and H1 seen in part (ii)

(v) For n ≤ 18, P(X 0) > 0.05 so the critical region is empty.

E1 for P(X 0) > 0.05 E1 indep for critical region is empty

2

E1 also for sight of 0.0536 Condone P(X = 0) > 0.05 or all probabilities or values, (but not outcomes) in table (for n ≤ 18) > 0.05 Or ‘There is no critical region’ For second E1 accept ‘H0 would always be accepted’ Do NOT FT wrong H1 Use professional judgement - allow other convincing answers

TOTAL 18

9


Q8 (i)

Upper Bound 9.1 9.3 9.5 9.7 9.9 10.1Cumulative frequency 0 5 12 27 43 50

B1 for cumulative frequencies G1 for scales G1 for labels G1 for points (Provided plotted at correct UCB positions) G1 for joining points All G’s dep on attempt at cumulative frequency but not cumulative fx’s or other spurious values.

5

May be implied from graph. Condone omission of 0 at this stage. Linear horizontal scale. Linear vertical scale: 0 to 50 (no inequality scales - Not even <9.1, <9.3, <9.5 …) Heating quality or x and Cumulative frequency or just CF or similar but not just frequency or fd nor cumulative fd Plotted as (UCB, their cf). Ignore (9.1,0) at this stage. No midpoint or LCB plots. Plotted within ½ small square For joining all of ‘their points’ (line or smooth curve) AND now including (9.1,0) dep on previous G1 Mid point or LCB plots may score first three marks Can get up to 3/5 for cum freq bars Allow full credit if axes reversed correctly Lines of best fit could attract max 4 out of 5.

(ii) Median = 9.67

B1 FT Allow answers between 9.66 and 9.68 without checking curve. Otherwise check curve.

3

Based on 25th to 26th value on a cumulative frequency graph ft their mid-point plot (not LCB’s) approx 9.57 for m.p. plot Allow 9.56 to 9.58 without checking B0 for interpolation

CF

x

10


Q1 = 9.51 Q3 = 9.83 Inter-quartile range = 9.83 – 9.51 = 0.32

B1 FT for Q3 or Q1 B1 FT for IQR providing both Q1 and Q3 are correct Allow answers between 9.50 and 9.52 and between 9.82 and 9.84 without checking curve. Otherwise check curve.

Based on 12th to 13th and 37th to 38th values on a cumulative frequency graph ft their mid -point plot (not LCB’s) approx Q1 = 9.42; Q3 = 9.73 Allow 9.41 to 9.43 and 9.72 to 9.74 without checking B0 for interpolation Allow correct IQR from graph if quartiles not stated Lines of best fit: B0 B0 B0 here.

(iii) Lower limit 9.51 – 1.5 × 0.32 = 9.03 Upper limit 9.83 + 1.5 × 0.32 = 10.31 Thus there are no outliers in the sample.

B1 FT their Q1, IQR B1 FT their Q3, IQR E1 NB E mark dep on both B marks

3

Any use of median 1.5 IQR scores B0 B0 E0 If FT leads to limits above 9.1 or below 10.1 then E0 No marks for ± 2 or 3 IQR In this part FT their values from (ii) if sensibly obtained (eg from LCB plot) or lines of best fit, but not from location ie 12.5, 37.5 or cumulative fx’s or similar. For use of mean ± 2s, Mean = 9.652, s = 0.235, Limits 9.182, 10.122 gets M1 for correct lower limit, M1 for correct upper limit, zero otherwise, but E0 since there could be outliers using this definition

(iv) (A) P(All 3 more than 9.5) = 38 37 36 0.430450 49 48

(=50616/117600 = 2109/4900)

M1 for 38/50 × (triple product) M1 for product of remaining fractions A1 CAO

3

(38/50)3 which gives answer 0.4389 scores M1M0A0 so watch for this. M0M0A0 for binomial probability including 0.76100 but 3C0×0.240×0.763 still scores M1 (k/50)3 for values of k other than 38 scores M0M0A0

( 1) ( 2)50 49 48k k k

for values of k other than 38 scores M1M0A0 Correct working but then multiplied or divided by some factor scores M1M0A0

11


12

(B) P(At least 2 more than 9.5) 38 37 123 0.430450 49 48

= 3 × 0.1435 + 0.4304 = 0.4304 + 0.4304 = 0.861 (=101232/117600 = 4218/4900 = 2109/2450) OR P(At least 2 more than 9.5) = 1 – (P(0) + P(1))

= 12 11 10 12 11 381 350 49 48 50 49 48

= 1 – [0.01122 + 0.12796] = 1 – 0.13918 = 0.861

M1 for product of 3 correct fractions seen M1 for 3 × a sensible triple or sum of 3 sensible triples M1 indep for + 0.4304 FT (providing it is a probability) A1 CAO M1 for 12/50×11/49×38/48 M1 for 3 × a sensible triple or sum of 3 sensible triples M1 dep on both previous M1’s for 1 –[0.01122 + 0.12796] A1 CAO

4

Accept 0.43 with working and 0.430 without working

17600

r gets first two M1M1’s

SC1 for

Or /3 3

= 2109/4900 = 0.4304 38 50

Allow unsimplified fraction as final answer 50616/1 O

38 12 50/

2 1 3

=0.4304

38 38 12350 50 50

or other sensible triple and SC2

if this + their 0.4304 (= 0.8549) 8/4900, but only M3A0

se of 1 – method ‘with replacement’

Allow 0.86 or 2109/2450 or 421for other unsimplified fractions U

SC1 for 12 12 383 50 50 50

SC2 for whole of 1 - 12 12 38 12 12 12350 50 50 50 50 50

(= 1 – (0.1313 + 0.0138) = 1 – 0.1451 = 0.8549)

TOTAL 18 NOTE RE OVER-SPECIFICATION OF ANSWERS If answers are grossly over-specified (see instruction 8), deduct the final answer mark in every case. Probabilities should also be rounded to a sensible degree of accuracy. In general final non probability answers should not be given to more than 4 significant figures. Allow probabilities given to 5 sig fig. In general accept answers which are correct to 3 significant figures when given to 4 or 5 significant figures.


Question Answer Marks Guidance 1 (i)

P(All blue) =4828

4929

5030

= 0.2071

M1 For

5030

(as part of a

triple product)

(30/50)3 = 0.216 scores M1M0A0

( 1) ( 2)50 49 48k k k

for values of k other than 30 scores M1M0A0 Zero for binomial unless simplifies to (3/5)3

M1 For product of other two fractions

Correct working but then multiplied or divided by some factor scores M1M0A0

OR 30

3 = 4060/19600 = 29/140 = 0.2071 50

/3

M2 for the complete method

A1 CAO SC2 for P(All red) = 0.0582

Accept 0.21 with working and 0.207 without working Allow unsimplified fraction as final answer 24360/117600 oe

[3] 1 (ii)

P(All red) = 4818

4919

5020

= 0.0582 or 3 = 0.0582 20 50

/3

M1 For P(All red) SC2 for 1 – (30/50)3 – (20/50)3 = 1 – 0.216 – 0.064 = 0.72, providing consistent with (i) . If not consistent with (i) M0M0A0

P(At least one of each colour) = 1 – (0.2071 + 0.0582) = 0.7347

or 29 57 260 13 361 1 1

140 980 980 49 49

M1 For 1 – (0.2071 + 0.0582)

A1 CAO Allow 0.73 with working OR [3] Allow unsimplified fraction as

final answer 86400/117600 oe P(2b,1r)+P(1b,2r) (M1)

For either 4820

4929

5030

or 4830

4919

5020

Allow M1 for 3×(30/50)2×(20/50) or 3×(30/50) ×(20/50) 2 and second M1 for sum of both if = 0.72 If not consistent with (i) M0M0A0

5


Question Answer Marks Guidance

= 4820

4929

50303 +

4830

4919

50203

(M1) For sum of both or for 3× either

NB M2 also for 30 20 4850 49 48

even if not multiplied by 3 = 3 × 0.1480 + 3 × 0.0969 = 0.7347 (A1) CAO Allow 0.73 or better with

working OR Either

or 30 20 50/

2 1 3

30 20 50/

1 2 3

(M1)

(M1) For sum of both (A1) CAO 2 (i) 9C3 × 5C3 = 84 × 10 = 840 M1 For either 9C3 or 5C3 Zero for permutations M1 For product of both

correct combinations

A1 CAO [3] 2 (ii) Total number of ways of answering 6 from 14 is 14C6 = 3003 M1 For 14C6 seen in part (ii)

Probability =14340

3003840

= 0.27972 = 0.280 M1 For their 840/ 3003 or

their 840/14C6

A1 FT their 840 Allow full marks for unsimplified fractional answers

[3] OR 6C3 × 5/14 × 4/13 × 3/12 × 9/11 × 8/10 × 7/9 = 0.280 (M1) For product of fractions SC1 for 6C3 × (5/14)3 × (9/14)3 =

0.2420 (M1) For 6C3× correct product (A1)

6


Question Answer Marks Guidance 3 (i) X ~ B(30, 0.85) M1 For 0.8529 × 0.151 =

0.0013466

P(X = 29) = × 0.8529 × 0.151 = 30 × 0.0013466 = 0.0404

2930 M1

For × p29 × q1

2930 With p + q = 1

A1 CAO Allow 0.04 www [3] If further working (EG P(X=29)

–P(X=28)) give M2A0 3 (ii) P(X = 30) = 0.8530 = 0.0076 M1 For 0.8530 P(X ≥ 29) = 0.0404 + 0.0076 = 0.0480 M1 For P(X = 29) + P(X =

30) (not necessarily correct, but both attempts at binomial, including coefficient in (i))

Allow eg 0.04+0.0076=0.0476 Allow 0.05 with working

A1 CAO [3] 3 (iii) Expected number = 10 × 0.0480 = 0.480 M1 For 10 × their (ii) provided (ii) between 0 and 1 A1 FT their (ii) but if answer

to (ii) leads to a whole number for (iii) give M1A0

Do not allow answer rounded to 0 or 1.

[2]

7


Question Answer Marks Guidance 4 (i) (A) P(third selected) = 0.922 × 0.08 = 0.0677 M1 For 0.922 Or = 1058/15625 M1 For p2 × q With p + q = 1

With no extra terms A1 CAO Allow 0.068 but not 0.067 nor

0.07 [3] SC1 for ‘without replacement’ method 92/100×91/99×8/98

=0.0690 4 (i) (B) P (second) + P(third) M1 For 0.92 × 0.08 With no extra terms = (0.92 × 0.08) + (0.922 × 0.08) A1 FT their 0.0677 Allow 0.141 to 0.142 and = 0.0736 + 0.0677 = 0.1413 allow 0.14 with working = 2208/15625 [2] SC1 for answer of 0.143 from ‘without replacement’ method 4 (ii) P(At least one of first 20) = 1 - P(None of first 20) M1 0.9220 Accept answer of 0.81 or better

from P(1) + P(2) + ... , or SC2 if all correct working shown but wrong answer No marks for ‘without replacement’ method’

= 1 – 0.9220 = 1 – 0.1887 = 0.8113 M1 1 – 0.9220 A1 CAO Allow 0.81 with working but

not 0.812 [3]

8


Question Answer Marks Guidance 5 Let p = probability that a randomly selected frame is faulty B1 For definition of p in context

Minimum needed for B1 is p = probability that frame/bike is faulty. Do not allow is p = probability that it is faulty Allow p = P(frame faulty) Definition of p must include word probability (or chance or proportion or percentage or likelihood but NOT possibility). Preferably as a separate comment. However can be at end of H0

as long as it is a clear definition ‘ p = the probability that frame is faulty, NOT just a sentence ‘probability is 0.05’ Do NOT allow ‘p = the probability that faulty frames have increased’

H0: p = 0.05 B1 H0 : p(frame faulty) = 0.05, H1 : p(frame faulty) > 0.05 gets B0B1B1 Allow p=5%, allow θ or π and ρ but not x. However allow any single symbol if defined Allow H0 = p=0.05, Allow H0 : p=1/20 Do not allow H0 : P(X=x) = 0.05, H1 : P(X=x) > 0.05 Do not allow H0: =0.05, =5%, P(0.05), p(0052), p(x)=0.05, x=0.05 (unless x correctly defined as a probability) Do not allow H1:p≥0.05, Do not allow H0 and H1 reversed Allow NH and AH in place of H0 and H1 For hypotheses given in words allow Maximum B0B1B1 Hypotheses in words must include probability (or chance or proportion or percentage) and the figure 0.05 oe.

H1: p > 0.05 B1 P(X ≥ 4) B1 For notation P(X ≥ 4) or

1- P(X ≤ 3) This mark may be implied by 0.0109 as long as no incorrect notation.

No further marks if point probs used - P(X = 4) = 0.0094 DO NOT FT wrong H1 But if H1 is p ≥ 0.05 allow the rest of the marks if earned so max 7/8

= 1- P(X ≤ 3) = 1 – 0.9891 = 0.0109 B1* For 0.0109, indep of previous mark

Or for 1 – 0.9891

9


Question Answer Marks Guidance 0.0109 < 0.05 M1*

dep For comparison with 5%

So reject H0 A1* or significant or ‘accept H1’

There is evidence to suggest that the proportion of faulty frames has increased.

E1* Dep on

A1

Must include ‘sufficient evidence’ or something similar such as ‘to suggest that’ ie an element of doubt for E1. ‘Sufficient evidence’ or similar can be seen in the either the A mark or the E mark.

[8] OR Critical region method: No marks if CR not justified Let X ~ B(18, 0.05)

P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – 0.9419 = 0.0581 > 5%

(B1)

For 0.0581

Do not insist on correct notation as candidates have to work out two probabilities for full marks

P(X ≥ 4) = 1 – P(X ≤ 3) = 1 – 0.9891 = 0.0109 < 5%

(B1) For 0.0109

(M1)

For at least one correct comparison with 5%

So critical region is {4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}

4 lies in the critical region, so significant,

(A1) CAO for critical region and significant oe

Condone {4,5 ..}, X ≥ 4, oe but not P(X ≥ 4)

There is evidence to suggest that the proportion of faulty frames has increased.

(E1)

10


Question Answer Marks Guidance 6 (i) M1 At least 4 fds correct for M1

M1 can be also be gained from freq per 1000 – 14, 44, 52, 18, 3.5 (at least 4 correct) and A1 for all correct or freq per 500 - 7, 22, 26, 9, 1.75 Accept any suitable unit for fd, eg freq per 1000, BUT NOT FD per 1000 Allow fds correct to at least three dp If fd not explicitly given, M1 A1 can be gained from all heights correct (within one square) on histogram (and M1A0 if at least 4 correct) Allow restart with correct heights if given fd wrong

A1 For fd’s all correct

Engine size Frequency Group width Frequency density 500 ≤ x ≤ 1000 7 500 0.014

1000 < x ≤ 1500 22 500 0.044 1500 < x ≤ 2000 26 500 0.052 2000 < x ≤ 3000 18 1000 0.018 3000 < x ≤ 5000 7 2000 0.0035

G1(L1) linear scales on both axes and label on vertical axis Label required on vert axis IN RELATION to first M1 mark ie fd or frequency density or if relevant freq/1000, etc (NOT fd/1000, but allow fd×1000, etc) Accept f/w or f/cw (freq/width or freq/class width) Ignore horizontal label and allow horizontal scale to start at 500 Can also be gained from an accurate key

INCORRECT DIAGRAMS: Frequency diagrams can get M0, A0, G0, G1, G0 MAXIMUM Thus frequency density = frequency × width, frequency/midpoint etc gets MAX M0A0G0G1G0 Frequency polygons MAX M1A1G0G0G0

G1(W1) Width of bars Must be drawn at 500, 1000etc NOT 499.5 or 500.5 etc NO GAPS ALLOWED Must have linear scale. No inequality labels on their own such as 500≤S<1000, etc but allow if a clear horizontal linear scale is also given.

11


Question Answer Marks Guidance G1(H1)

Height of bars FT of heights dep on at least 3 heights correct and all must agree with their fds If fds not given and one height is wrong then max M1A0G1G1G0 – visual check only (within one square) –no need to measure precisely

[5] 6 (ii) Do not know exact highest and lowest values so cannot tell what the

midrange is. OR No and a counterexample to show it may not be 2750 OR (500 + 5000) / 2 = 2750. But very unlikely to be absolutely correct but probably close to the true value. Some element of doubt needed. Allow ‘Likely to be correct’

E1

[1]

Allow comment such as 'Highest value could be 5000 and lowest could be 500 therefore midrange could be 2750' NO mark if incorrect calculation Sight of 1750 AND 3000 (min and max of midrange) scores E1

6 (iii) Mean =

80)74000()182500()261750()221250()7750(

M1 For midpoints (at least 3 correct)

No marks for mean or sd unless using midpoints Answer must NOT be left as improper fraction

1891

80151250

A1 CAO

Accept correct answers for mean (1890 or 1891) and sd (850 or 846 or 845.5) from calculator even if eg wrong Sxx given

2 2 2 2 2 2(750 7) (1250 22) (1750 26) (2500 18) (4000 7)x f = 3937500 + 34375000 + 79625000 + 112500000 + 112000000

M1 For sum of at least 3 correct multiples fx2 Allow M1 for anything which rounds to 342400000

=342437500

5648046980

151250-342437500S2

xx

846714943

7956480469

s A1

Only penalise once in part (iii) for over specification, even if mean and standard deviation both over specified. Allow SC1 for RMSD 840.2 or 840 from calculator

Only an estimate since the data are grouped. E1 indep

Or for any mention of midpoints or ‘don’t have actual data’ or ‘data are not exact’ oe

[5]

12


Question Answer Marks Guidance 6 (iv) x – 2s = 1891 – (2 846) = 199

Allow 200 M1 For either.

FT any positive mean and their positive sd/rmsd for M1 Only follow through numerical values, not variables such as s, so if a candidate does not find s but then writes here ‘limit is 40.76+ 2 standard deviation’, do NOT award M1 No marks in (iv) unless using x + 2s or x – 2s

x + 2s = 1891 + (2 846) = 3583 Allow 3580 or 3600

A1 For both (FT) Do NOT penalise over specification here as it is not the final answer

So there are probably some outliers E1 Must include an element of doubt Dep on upper limit in range 3000 – 5000 Allow comments such as ‘any value over 3583 is an outlier’ Ignore comments about possible outliers at lower end.

[3] 6 (v) Number of cars over 2000 cm3 = 25/80 × 2.5 million = 781250 M1 For 25/80× 2.5 million or (18+7) /80× 2.5 million So duty raised = 781250 × £1000 = £781 million M1

indep For something × £1000 even if this is the first step

A1 CAO NB £781250000 is over specified so only 2/3 [3] 6 (vi) Because the numbers of cars sold with engine size greater than 2000 E1 Allow any other reasonable suggestion cm3 might be reduced due to the additional duty. Condone ‘sample may not be representative’ [1]

Allow ‘sample is not of NEW cars’

13


Question Answer Marks Guidance 7 (i) P(X = 0) = 0.4 × 0.54 = 0.025 NB ANSWER GIVEN M1 For 0.54 A1 [2] 7 (ii) P(X = 1) = (0.6 × 0.54) + (4 × 0.4 × 0.5× 0.53) M1* For 0.6 × 0.54 seen as a single term (not multiplied or divided

by anything) = 0.0375 + 0.1 = 0.1375 NB ANSWER GIVEN M1* For 4 × 0.4 × 0.54 Allow 4 × 0.025

Watch out for incorrect methods such as (0.4/4) 0.1 MUST be justified

M1* dep

For sum of both , dep on both M1’s

A1 [4] 7 (iii) G1 For labelled linear scales on both axes Dep on attempt at vertical line chart. Accept P on vertical

axis G1 For heights – visual check only but last bar taller than first

and fifth taller than second and fourth taller than third. Lines must be thin (gap width > line width). All correct. Zero if vertical scale not linear Everything correct but joined up tops G0G1 MAX Everything correct but f poly G0G1 MAX Everything correct but bar chart G0G1 MAX Curve only (no vertical lines) gets G0G0 Best fit line G0G0 Allow transposed diagram

[2]

14


Question Answer Marks Guidance 7 (iv) ‘Negative’ or ‘very slight negative’ E1 E0 for symmetrical [1] but E1 for (very slight) negative skewness even if also

mention symmetrical Ignore any reference to unimodal

7 (v) E(X) = (0×0.025) + (1×0.1375) + (2×0.3) + (3×0.325) + (4×0.175) M1 For Σrp (at least 3 terms correct) + (5×0.0375) A1 CAO = 2.6 E(X2) = (0×0.025) + (1×0.1375) + (4×0.3) + (9×0.325) + 16×0.175) M1* For Σr2p (at least 3 terms correct) + (25×0.0375) = 0 + 0.1375 + 1.2 + 2.925 + 2.8 + 0.9375 = 8 Var (X) = 8 – 2.62 M1*

dep for – their E( X )²

= 1.24 A1 FT their E(X) provided Var( X ) > 0 [5] USE of E(X-µ)2 gets M1 for attempt at (x-µ)2 should see (-

2.6)2, (-1.6)2, (-0.6)2, 0.42, 1.42, 2.42 (if E(X) correct but FT their E(X)) (all 5 correct for M1), then M1 for Σp(x-µ)2 (at least 3 terms correct) Division by 5 or other spurious value at end gives max M1A1M1M1A0, or M1A0M1M1A0 if E(X) also divided by 5. Unsupported correct answers get 5 marks.

7 (vi) P(Total of 3) = (3×0.325×0.0252) + (6×0.3×0.1375×0.025) + M1 For decimal part of first term 0.325×0.0252 0.13753 = 3 × 0.000203 + 6 × 0.001031 + 0.002600=

0.000609 + 0.006188 + 0.002600 = 0.00940 M1 For decimal part of second term 0.3×0.1375×0.025

(= 3×13/64000 + 6×33/32000 + 1331/512000) M1 For third term – ignore extra coefficient All M marks above depend on triple probability products

A1 CAO: AWRT 0.0094. Allow 0.009 with working. [4] NOTE RE OVER-SPECIFICATION OF ANSWERS If answers are grossly over-specified, deduct the final answer mark in every case. Probabilities should also be rounded to a sensible degree of accuracy. In general final non probability answers should not be given to more than 4 significant figures. Allow probabilities given to 5 sig fig. In general accept answers which are correct to 3 significant figures when given to 4 or 5 significant figures. If answer given as a fraction and as an over-specified decimal – ignore decimal and mark fraction.

15


ADDITIONAL NOTES RE Q5 Comparison with 95% method If 95% seen anywhere then B1 for P(X ≤ 3) B1 for 0.9891 M1* for comparison with 95% dep on B1 A1* for significant oe E1* Smallest critical region method: Either: Smallest critical region that 4 could fall into is {4,5,6,7,8,9,10,11,12,13,14,15, 16, 17, 18} gets B1 and has size 0.0109 gets B1, This is < 5% gets M1*, A1*, E1* as per scheme NB These marks only awarded if 4 used, not other values. Use of k method with no probabilities quoted: P(X ≥ 3) = 1 – P(X ≤ 2) > 5% P(X ≥ 4) = 1 – P(X ≤ 3) < 5% These may be seen in terms of k or n. Either k = 4 or k – 1 = 3 so k = 4 gets SC1 so CR is {4,5,6,7,8,9,10,11,12,13,14,15, 16, 17, 18} gets another SC1and conclusion gets another SC1 Use of k method with one probability quoted: 1 - 0.9891 < 5% or 0.0109 < 5% gets B0B1M1 P(X ≤ k - 1) = P(X ≤ 3) so k – 1 = 3 so k = 4 (or just k = 8 ) so CR is {4,5,6,7,8,9,10,11,12,13,14,15, 16, 17, 18} and conclusion gets A1E1

16


17

Two tailed test done but with correct H1: p>0.05 Hyp gets max B1B1B1 if compare with 5% ignore work on lower tail and mark upper tail as per scheme but withhold A1E1 if compare with 2.5% no marks B0B0M0A0E0 Line diagram method B1 for squiggly line between 3 and 4 or on 4 exclusively (ie just one line), B1dep for arrow pointing to right, M1 0.0109 seen on diagram from squiggly line or from 4, A1E1 for correct conclusion Bar chart method B1 for line clearly on boundary between 3 and 4 or within 4 block exclusively (ie just one line), B1dep for arrow pointing to right, M1 0.0109 seen on diagram from boundary line or from 8, A1E1 for correct conclusion. Using P(Not faulty) method H0: p=0.95, H1: p<0.95 where p represents the prob that a frame is faulty gets B1B1B1. P(X≤14)=0.0109 < 5% So significant, etc gets B1B1M1A1E1 NB If H0: p=0.5, H1: p>0.5, etc seen, but then revert to 0.05 in working allow marks for correct subsequent working. However if 0.5 used consistently throughout, then max B1 for definition of p and possibly B1 for notation P(X ≥ 4).


6

Question Answer Marks Guidance 1 (i) Positive B1 CAO [1] 1 (ii) Mean = 5.064 allow 5.1 with working 126.6/25 or 5.06 without B1 SD = 1.324 allow 1.3 with working or 1.32 without B2 Allow B1 for RMSD =

1.297 or var =1.753 or MSD = 1.683

Also allow B1 for Sxx = 42.08 or for Σx2 = 683 SC1 for both mean = 50.64 and SD = 13.24 (even if over-specified)

[3] 1 (iii) x – 2s = 5.064 – 21.324 = 2.416 B1FT FT their mean and sd For use of quartiles and IQR

Q1 = 3.95; Q3 = 6.0; IQR = 2.05 3.95 – 1.5(2.05) gets M1 Allow other sensible definitions of quartiles

x + 2s = 5.064 + 21.324 = 7.712 M1

for x + 2s but withhold final E mark if their limits mean that there are no outliers.

6.0 + 1.5(2.05) gets M1

A1FT For upper limit Limits 0.875 and 9.075 So there is an outlier. E1 Incorrect statement such as

7.6 and 8.1 are outliers gets E0 Do not award E1 if calculation error in upper limit

So there are no outliers NB do not penalise over-specification here as not the final answer but just used for comparison. FT from SC1

[4] 2 (i)

r 2 3 4 5 P(X = r) 3k 8k 15k 24k

B1

For correct table (ito k or correct probabilities 0.06, 0.16, 0.30, 0.48)

3k + 8k + 15k + 24k = 1 M1 For their four multiples of k added and =1.

k = 0.02 A1 or k = 1/50 (with or without working)

Allow M1A1 even if done in part (ii) – link part (ii) to part (i)

[3]


7

Question Answer Marks Guidance 2 (ii) E(X) = (2 0.06) (3 0.16) (4 0.30) (5 0.48) 4.2

M1

For Σrp (at least 3 terms correct Provided 4 reasonable probabilities seen.

If probs wrong but sum = 1 allow full marks here. If sum ≠ 1 allow max M1A0M1 M0A0 (provided all probabilities between 0 and 1)

or 21/5

A1 cao Or ito k NB E(X) = 210k , E(X2) = 924k gets M1A0M1M0A0. E(X) = 210k , Var (X) = 924k – (210k)2 gets M1A0M1M1A0.

E(X2) = (4 0.06) (9 0.16) (16 0.30) (25 0.48) 18.48 M1 For Σr2p (at least 3 terms correct)

Var(X) = 18.48 – 4.22 M1 dep for – their E(X)² = 0.84 = 21/25 A1 FT their E(X) provided

Var(X) > 0 (and of course E(X2) is correct)

Use of E(X – µ)2 gets M1 for attempt at (x – µ)2 should see (–2.2)2, (–1.2)2, (–0.2)2, 0.82, (if E(X) wrong FT their E(X)) (all 4 correct for M1), then M1 for Σp(x – µ)2 (at least 3 terms correct with their probabilities) Division by 4 or other spurious value at end gives max M1A1M1M1A0, or M1A0M1M1A0 if E(X) also divided by 4. Unsupported correct answers get 5 marks

[5] 3 (i)

( ) ( | ) ( ) 0.4 0.07 0.028P L W P L W P W

M1

For ( | ) ( )P L W P W

A1 cao [2]


8

Question Answer Marks Guidance 3 (ii)

B1 For two labelled intersecting circles

B1 For at least 2 correct probabilities.

FT their 0.028 provided < 0.038

B1 For remaining probabilities

[3] 3 (iii)

( ) 0.028, ( ) ( ) 0.038 0.07 0.00266P L W P L P W

M1

For correct use of

( ) ( )P L P W If P(L) wrong, max M1A0E0. No marks if P(W) wrong

Or EG P( | ) 0.4,P( ) 0.038L W L Not equal so not independent M1 is for comparing with some attempt at numbers P( | ) with P( )L W L , A1 for 0.038 If P(L) wrong, max M1A0E0

A1 For 0.00266 Not equal so not independent E1*

dep on M1

Allow ‘they are dependent’ Do not award E1 if P( )wrongL W

[3] 4 (i)

113

M1

Seen

= 165 A1 Cao [2]

0.01 0.028

0.92

L W

0.042


9

Question Answer Marks Guidance 4 (ii) 5 6 5 6

2 1 3 0 60 10 70 1411 11 165 165 165 333 3

= 0.424

M1

For intention to add correct two fractional terms

Or For attempt at correct two terms

M1 For numerator of first term For prod of 3 correct fractions =4/33 Alternative

1 – P(1 or 0) = 1 – 5 6 5 6 5 4311 10 9 11 10 9

= 5 4111 33

= 1433

M1 for 1 – P(1 or 0) , M1 for first product, M1 for ×3, M1 for

second product, A1

M1 For numerator of sec term Do not penalise omission

of 60

For whole expression ie 5 4 6 43

11 10 9 11

(= 3 × 0.1212…)

M1 For correct denominator For attempt at 5 4 3 2

11 10 9 33

A1 cao cao [5] Use of binomial can get max first M1 5 (i) 25 1 25

6 6 216

(= 0.116)

M1 For 5/6 (or 1 – 1/6) seen

If extra term or whole number factor present give M1M0A0

M1 For whole product A1 cao Allow 0.12 with working [3] 5 (ii) 1051 1 0.1615 0.8385

6

M1

For (5/6)10 (without extra terms)

Allow 0.838 or 0.839 without working and 0.84 with working. For addition P(X = 1) +…+ P(X = 10) give M1A1 for 0.84 or better, otherwise M0A0

A1 cao [2]


10

Question Answer Marks Guidance 6 (i) 4 + ½ of 18 = 4 + 9 =13 M1 For ½ of 18 A1 cao 13/100 gets M1A0 [2] 6 (ii) (Median) = 50.5th value M1

For 50.5 seen SC2 for use of 50th value leading to

Est = 140 + (25 /29 × 5) = 144.3 (SC1 if over-specified)

Est = 140 + 25.5 5

29

or = 140 + 50.5 25 554 25

M1 For attempt to find this value or Est = 145 – 3.5 5

29

= 144.4

= 144.4 A1 NB no marks for mean = 144.35 [3] NB Watch for over-specification


11

Question Answer Marks Guidance 6 (iii)

Height FrequencyGroup width

Frequency density

125 ≤ x ≤ 140 25 15 1.67 140 < x ≤ 145 29 5 5.80 145 < x ≤ 150 24 5 4.80 150 < x ≤ 160 18 10 1.80 160 < x ≤ 170 4 10 0.40

M1

For fd’s - at least 3 correct Accept any suitable unit for fd such as eg freq per cm.

M1 can be also be gained from freq per 10 – 16.7, 58, 48, 18, 4 (at least 3 correct) or freq per 5 – 8.35, 29, 24, 9, 2

A1 correct to at least one dp allow 1.66 but not 1.6 for first fd

for all correct. If fd not explicitly given, M1 A1 can be gained from all heights correct (within one square) on histogram (and M1A0 if at least 3 correct)

G1

linear scales on both axes and label on vertical axis

Linear scale and label on vertical axis IN RELATION to first M1 mark ie fd or frequency density or if relevant freq/10, etc (NOT eg fd/10). However allow scale given as fd×10, or similar Accept f/w or f/cw (freq/width or freq/class width) Can also be gained from an accurate key G0 if correct label but not fd’s.

W1

width of bars Must be drawn at 125, 140 etc NOT 124.5 or 125.5 etc NO GAPS ALLOWED Must have linear scale. No inequality labels on their own such as 125≤S<140, etc but allow if a clear horizontal linear scale is also given. Ignore horizontal label.

H1 height of bars Height of bars – must be linear vertical scale. FT of heights dep on at least 3 heights correct and all must agree with their


12

Question Answer Marks Guidance fds If fds not given and at least 3 heights correct then max M1A0G1W1H0 Allow restart with correct heights if given fd wrong (for last three marks only)

[5] 6 (iv) 4 boys 0.6 × 15 M1 For 0.6 × 15 Or 45 × 0.2 = 9 (number of squares

and 0.2 per square) = 9 girls A1 For 9 girls So 5 more girls A1 cao [3] 6 (v)

Frequencies and midpoints for girls are Height 132.5 142.5 147.5 155 167.5 Frequency 18 23 31 19 9

B1

For at least three frequencies correct

B1 At least three midpoints correct

No further marks if not using midpoints

So mean = (132.5 18) (142.5 23) (147.5 31) (155 19) (167.5 9)

100

= (2385) (3277.5) (4572.5) (2945) (1507.5)

100

M1 For attempt at ∑xf For sight of at least 3 xf pairs M1*

Dep on M1

For division by 100

= 146.9 (Exact answer 146.875)

A1 Cao NB Watch for over-specification

Allow answer 146.9 or 147 but not 150 NB Accept answers seen without working (from calculator) Use of ‘not quite right’ midpoints such as 132.49 or 132.51 etc can get B1B0M1M1A0

[5]


13

Question Answer Marks Guidance 7 (i) (A) X ~ B(10, 0.35)

P(5 accessing internet) = 5 5100.35 0.65

5

M1 or 5 50.35 0.65

M1 For5 510

5p q

With p + q = 1 Also for 252 × 0.0006094

= 0.1536 A1 cao Allow 0.15 or better NB 0.153 gets A0 See tables at the website http://www.mei.org.uk/files/pdf/formula_book_mf2.pdf

OR OR from tables = 0.9051 – 0.7515 = 0.1536 M2 For 0.9051 – 0.7515 A1 cao [3] 7 (i) (B) P(X ≥ 5) = 1 – P(X ≤ 4) =1 – 0.7515 M1 For 0.7515 = 0.2485 A1 cao Accept 0.25 or better – allow 0.248 or

0.249 Calculation of individual probabilities gets B2 if fully correct 0.25 or better, otherwise B0.

[2] 7 (i) (C) E(X) = np = 10 × 0.35 M1 For 10 × 0.35 If any indication of rounding to 3 or 4

allow M1A0 = 3.5 A1 cao [2]


14

Question Answer Marks Guidance 7 (ii) Let X ~ B(20, 0.35) Let p = probability of a customer using the internet (for

population) B1 For definition of p in

context Minimum needed for B1 is p = probability of using internet. Allow p = P(using internet) Definition of p must include word probability (or chance or proportion or percentage or likelihood but NOT possibility). Preferably as a separate comment. However can be at end of H0 as long as it is a clear definition ‘p = the probability of using internet’, Do NOT allow ‘p = the probability of using internet is different’

H0: p = 0.35 B1 For H0 Allow p=35%, allow only p or θ or π or ρ. However allow any single symbol if defined (including x) Allow H0 = p=0.35, Allow H0 : p=7/20or p=35/100 Allow NH and AH in place of H0 and H1 Do not allow H0 : P(X=x) = 0.35 Do not allow H0: =0.35, =35%, P(0.35), p(x)=0.35, x=0.35 (unless x correctly defined as a probability) Do not allow H0 and H1 reversed For hypotheses given in words allow Maximum B0B1B1 Hypotheses in words must include probability (or chance or proportion or percentage) and the figure 0.35 oe Thus eg H0 : p(using internet) = 0.35, H1 : p(using internet) ≠ 0.35 gets B0B1B1


15

Question Answer Marks Guidance H1: p 0.35 B1 For H1 Allow ‘p < 0.35 or p > 0.35’in place of

p 0.35 H1 has this form because the test is to investigate whether the

proportion is different, (rather than lower or higher). E1 Do not allow if H1 wrong.

P(X ≥ 10) B1 For notation P(X ≥ 10) or P(X > 9) or 1 – P(X ≤ 9) (as long as no incorrect notation)

This mark may be implied by 0.1218 as long as no incorrect notation. No further marks if point probs used - P(X = 10) = 0.0686 (do not even give the notation mark for correct notation) DO NOT FT wrong H1, but see extra notes

= 1 – 0.8782 = 0.1218 B1* For 0.1218 Allow 0.12 Or for 1 – 0.8782 Indep of previous mark

> 2.5% M1* dep

For comparison with 2.5%

So not significant. A1* Allow ‘accept H0’ or ‘reject H1’ Conclude that there is not enough evidence to indicate that the

probability is different. (Must state ‘probability’, not just ‘p’) E1*

dep on A1

Must include ‘sufficient evidence’ or something similar such as ‘to suggest that’ ie an element of doubt either in the A or E mark.

ALTERNATIVE METHOD FOR FINAL 5 MARKS

Critical region method LOWER TAIL P(X ≤ 2) = 0.0121 < 2.5% P(X ≤ 3) = 0.0444 > 2.5%

B1

For either probability

Do not insist on correct notation as candidates have to work out two probabilities for full marks. If only upper tail of CR given (or only upper tail justified), allow max 4/5 for final 5 marks.

UPPER TAIL P(X ≥ 11) = 1 – P(X ≤ 10) = 1 – 0.9468 = 0.0532 > 2.5% P(X ≥ 12) = 1 – P(X ≤ 11) = 1 – 0.9804 = 0.0196 < 2.5%

B1

For either probability


16

Question Answer Marks Guidance So critical region is {0,1,2,12,13,14,15,16,17,18,19,20} M1*

dep cao dep on at least one correct comparison with 2.5%

No marks if CR not justified Condone {0,1,2, 12, … 20}, X ≤ 2, X ≥ 12, oe but not P(X ≤ 2) etc

So not significant A1* Conclude that there is not enough evidence to indicate that the

probability is different. E1*

dep on A1

NB If CR found correctly then P(X = 10) subsequently found but cand says ’10 not in CR’ then allow up to all last five marks. If do not say ’10 not in CR’ allow none of last five marks

[9] 7 (iii) 0.0022 < 2.5% So reject Ho,

Significant. B1 For either reject Ho or

significant, dep on correct comparison

Conclude that there is enough evidence to indicate that the probability is different.

E1* dep

Dep on good attempt at correct hypotheses in part (ii)

If they have H1: p>0.35, allow SC1 if all correct including comparison with 5%.

[2]


17

APPENDIX NOTE RE OVER-SPECIFICATION OF ANSWERS If answers are grossly over-specified, deduct the final answer mark in every case. Probabilities should also be rounded to a sensible degree of accuracy. In general final non-probability answers should not be given to more than 4 significant figures. Allow probabilities given to 5 sig fig. Additional notes re Q7 part ii Comparison with 97.5% method If 97.5% seen anywhere then B1 for P(X ≤ 9) B1 for 0.8782 M1* for comparison with 97.5% dep on second B1 A1* for not significant oe E1* Smallest critical region method: Smallest critical region that 10 could fall into is {10,11,12,13,14,15, 16, 17, 18,19,20} gets B1 and has size 0.1218 gets B1, This is > 2.5% gets M1*, A1*, E1* as per scheme NB These marks only awarded if 10 used, not other values. Use of k method with no probabilities quoted: This gets zero marks. Use of k method with one probability quoted: Mark as per scheme Line diagram method and Bar chart method No marks unless correct probabilities shown on diagram, then mark as per scheme. Upper tailed test done with H1: p>0.35 Hyp gets max B1B1B0E0 If compare with 5% give SC2 for P(X ≥ 10) = 1 – 0.8782 = 0.1218 > 5% and SC1 for final conclusion (must be ‘larger than’ not ‘different from’) If compare with 2.5% no further marks B0B0M0A0E0 Lower tailed test done with H1: p<0.35 No marks out of last 5.


6


Mean = 24940100

= 249.4g or 249g B1

Ignore units CAO

NB 249.40 gets B0 for over-specification

Sxx =

2249406240780100

= 20744 M1

For Sxx

M1 for 6240780 - 100 their mean2 BUT NOTE M0 if their Sxx < 0

s = 20744

99 = 209.53 = 14.4751 =14.5g

A1

CAO ignore units For s2 of 210 (or better) allow M1A0 with or without working For RMSD of 14.4 (or better) allow M1A0 provided working seen For RMSD2 of 207 (or better) allow M1A0 provided working seen Allow 14.48 but NOT 14.47

[3]

1 (ii) New mean = (0.9 × 249.4) – 15 = 209.5g B1 FT their mean provided answer is positive

If candidate ‘starts again’ only award marks for CAO Allow 209

New sd = 0.9 × 14.48 = 13.03g M1 A1

FT their sd FT Allow 13.0 to 13.1

Or for 0.92×14.52 Deduct at most 1 mark overall in whole question for over-specification of Mean and 1 mark overall for SD

[3]


7

Question Answer Marks Guidance 2 (i) 5 4 5 300 53

10 9 8 720 12 = (0.4167)

M1 For 5/10 × 4/9 Correct working but then multiplied or divided by some factor scores M1M1M0A0

M1 For × 5/8 Zero for binomial Allow M2 for equivalent triple such as 5 5 4

10 9 8

M1 For 3 × triple product Or 3 separate equal triplets added A1 CAO (Fully simplified) Answer must be a fraction

[4] Or

5 52 1 10 5 5

10 120 123

M1* For

5 52 1

Seen

M1* For

103

Seen

M1* dep

For whole fraction Correct working but then multiplied or divided by some factor scores M1M1M0A0

A1

CAO (Fully simplified)

2 (ii) 3 47 5 5412 12 12

M1FT For first probability Allow 4C3

M1FT For (5/12)4 M1FT For sum of both correct

probabilities Provided sum <1

0.169 0.030 0.199

Or 875 625 13755184 20736 6912

A1 CAO Do not allow 0.2, unless fuller answer seen first

Alternative for 1- (P(0)+P(1)+P(2)) allow M1FT for two ‘correct’ probs, M1 for sum of three ‘correct’, M1 for 1 – answer, A1 CAO

[4]


8

Question Answer Marks Guidance 3 (i) X ~ B(50, 0.1) M1 For 5 450.1 0.9

P(5 underweight) = 5 4550

0.1 0.95

= 0.1849

M1 For 5 4550

5p q

With p + q = 1 Also for 2118760 × 8.73 × 10-8

A1 CAO Allow 0.185 or better NB 0.18 gets A0

[3]

3 (ii) X ~ B(20, 0.1) P(X ≥ 1) = 1 – P(X = 0) M1 For 0.1216 Allow M1 for 0.920 = 1 – 0.1216 = 0.8784 A1 CAO Allow 0.878 or better [2] See tables at the website

http://www.mei.org.uk/files/pdf/formula_book_mf2.pdf

3 (iii) E(X) = 48 × 0.8784 = 42.16 (= 42.2) M1 A1

FT their probability from part (ii)

If any indication of rounding to 42 or 43 or to another integer on FT allow M1A0 SC1 for 48 ´ their p giving an integer answer. NB 0.6083 in (ii) leads to 29.20

[2]


9


P(X = 15) 3 2 16 5 4

M1 For product of three correct

fractions

= 6 1 0.05

120 20

Or 6 3

1 1 0.05C 20

Or 3! 3! 1 0.056! 20

A1

NB ANSWER GIVEN NB 1 – (0.45 + 0.45 + 0.05) = 0.05 scores M0A0

Full marks for 1 1 1 63! 0.056 5 4 120

Allow 3 ´ 2 in place of 3! SC1 for

1 1 1 66 0.056 5 4 120

[2] 4 (ii) E(X) = (15 × 0.05) + (1010 × 0.45) + (2005 × 0.45) + (3000 ×

0.05) M1

For Σrp (at least 3 terms correct)

= 1507.5 so 1508 (4sf) A1 CAO Allow 1507, 1510, 1507.5, 1507.50 or 3015/2

E(X2) = (152 × 0.05) + (10102 × 0.45) + (20052 × 0.45) + (30002 × 0.05) = 2718067.5

M1 For Σr2p (at least 3 terms correct)

Use of E(X-µ)2 gets M1 for attempt at (x-µ)2 should see (-1492.5)2, (-497.5)2, 497.52, 1492.52, (if E(X) wrong FT their E(X)) (all 4 correct for M1), then M1 for Σp(x-µ)2 (at least 3 terms correct with their probabilities) Division by 4 or other spurious value at end gives max M1A1M1M1A0, or M1A0M1M1A0 if E(X) also divided by 4. Unsupported correct answers get 5 marks

Var (X) = 2718067.5 – 21507.5 M1 dep for – their E(X)²

= 445511.25 so 445500 (4sf) A1

FT their E(X) provided Var(X) > 0 (and of course E(X2) is correct)

Allow 446000

[5]


10

Question Answer Marks Guidance 5 (i) Because if people cannot make a correct identification, then the

probability that they guess correctly will be 0.5 For ‘equally likely to guess right or wrong’ or ‘two outcomes with equal probability’ or ‘50:50 chance of success’ or ‘right one in two occasions on average’ or ‘two (equally likely) outcomes’ etc

E1

E1

For idea of a guess or ‘chosen at random’ For idea of two outcomes

NB The question includes the sentence ‘She suspects that people do no better than they would by guessing.’, so this on its own does not get the mark for the idea of a guess

[2] 5 (ii) ‘Because people may do better than they would by guessing’

or similar B1 For idea of selecting

correctly /identifying /knowing

No marks if answer implies that it is because there are over half in the sample who make a correct identification

[1] 5 (iii) P(X ≥ 13) = 1– P(X ≤ 12) = 1 – 0.8684 = 0.1316

NB PLEASE ANNOTATE THE TOP AND BOTTOM OF THE EXTRA PAGE IF NOT USED

M1 For notation P(X ≥ 13) or P(X > 12) or 1– P(X ≤ 12)

Notation P(X = 13) scores M0. If they have the correct P(X ≥ 13) then give M1 and ignore any further incorrect notation.

B1* For 0.1316 Or for 1 – 0.8684 indep of previous mark

0.1316 > 0.05 M1* dep

For comparison with 5%

So not significant A1*

Allow ‘accept H0’ or ‘reject H1’

There is insufficient evidence to suggest that people can make a correct identification.

E1* dep

NB Point probabilities score zero.

Must include ‘insufficient evidence’ or something similar such as ‘to suggest that’ ie an element of doubt either in the A or E mark. Must be in context to gain E1 mark. Do not allow ‘sufficient evidence to suggest proportion making correct identification is 0.5’ or similar


11

Question Answer Marks Guidance ALTERNATIVE METHOD – follow method above unless

some mention of CR seen

Must see some reference to CR to gain any marks

Critical region method UPPER TAIL P(X ≥ 14) = 1 – P(X ≤ 13) = 1 – 0.9423 = 0.0577 > 5% P(X ≥ 15) = 1 – P(X ≤ 14) = 1 – 0.9793 = 0.0207 < 5%

B1

M1*

For either probability For a correct comparison with 5%

Do not insist on correct notation as candidates have to work out two probabilities for full marks. Allow comparison in form of statement ‘critical region at 5% level is …’

So critical region is {15,16,17,18,19,20} M1* dep

cao dep on the two correct probabilities

No marks if CR not justified Condone {15, … 20}, X ≥ 15, oe but not P(X ≥ 15,) etc

13 not in CR so not significant A1* Must include ‘13 not in CR’


There is insufficient evidence to indicate that people can make a correct identification.

E1* dep on A1

Ignore any work on lower critical region

NB If CR found correctly, then P(X=13) subsequently found, but cand says ‘13 not in CR’ then allow up to all five marks. If do not say ’13 not in CR’ allow no marks

[5]


12

Question Answer Marks Guidance 6 (i) Median = 3.32 kg B1

For Q1 allow 2.82 to 2.84 For Q3 allow 3.70 to 3.72 If no quartiles given allow B0B1 for

Q1 (= 6.5th value) = 2.83 Q3 (= 19.5th value) = 3.71 B1 For Q1 or Q3 Inter-quartile range = 3.71 – 2.83 = 0.88 B1 For IQR dep on both

quartiles correct [3] IQR in range 0.86 to 0.90 (ii)

G1

G1

G1

For reasonably linear scale shown. For boxes in approximately correct positions, with median just to right of centre For whiskers in approximately correct positions in proportion to the box FT their median and quartiles if sensible – guidance above is only for correct values

Dep on attempt at box and whisker plot with at least a box and one whisker. Condone lack of label. Do not award unless RH whisker significantly shorter than LH whisker Allow LH whisker going to 2.5 and outlier marked at 1.39

[3] 6 (iii) Lower limit 2.83 – (1.5 × 0.88) = 1.51 B1 For 1.51 FT Any use of median 1.5 × IQR scores

B0 B0 E0 No marks for ± 2 or 3 × IQR In this part FT their values from (i)or (ii) if sensibly obtained but not from location ie 6.5, 19.5

Upper limit 3.71 + (1.5 × 0.88) = 5.03 B1 For 5.03 FT Do not penalise over-specification as not the final answer

Exactly one baby weighs less than 1.51 kg and none weigh over 5.03 kg so there is exactly one outlier.

E1* Dep on their 1.51 and 5.03

Do not allow unless FT leads to upper limit above 4.34 and lower limit between 1.39 and 2.50


13

Question Answer Marks Guidance ‘Nothing to suggest that this baby is not a genuine data value

so she should not be excluded’ or ‘This baby is premature and therefore should be excluded’.

E1* Dep

Any sensible comment in context

For use of mean ± 2sd allow B1 For 3.27 + 2 × 0.62= 4.51 B1 For 3.27 - 2 × 0.62= 2.03 Then E1E1 as per scheme

[4] 6 (iv) Median = 3.5 kg B1 Q1 = 50th value = 3.12 Q3 = 150th value = 3.84 B1 For Q1 or Q3 For Q1 allow 3.11 to 3.13

For Q3 allow 3.83 to 3.85 Inter-quartile range = 3.84 – 3.12 = 0.72 B1 For IQR FT their quartiles Dep on both quartiles correct [3] If no quartiles given allow B0B1 for

IQR in range 0.70 to 0.74 6 (v) Female babies have lower weight than male babies on the

whole E1 FT

Allow ‘on average’ or similar in place of ‘on the whole’

Do not allow lower median

Female babies have higher weight variation than male babies E1 FT

Allow ‘more spread’ or similar but not ‘higher range’ Condone less consistent

Do not allow higher IQR, but SC1 for both lower median and higher IQR, making clear which is which

[2] 6 (vi) Male babies must weigh more than 4.34 kg Approx 10 male babies weigh more than this. M1* For 10 or 9 or 8 Or 200 – 190, 200 –191 or 200 –192

Probability = 10 9 90 9 0.00226200 199 39800 3980

or 9 8 72 0.00181200 199 39800

or 8 7 56 7 0.00141200 199 39800 4975

M1* dep

A1

For first fraction multiplied by any other different fraction (Not a binomial probability) CAO Allow their answer to min of 2 sf

Allow any of these answers For spurious factors, eg 2 × correct answer allow M1M1A0 SC1 for n/200 × (n-1)/199

[3]


14

0.1

0.9 Miss

0.05

0.95

Hit

Hit

Hit 0.2

0.8 Miss

Miss

First

Second Hit 0.2

0.8 Miss

Hit 0.2

0.8 Miss

Third

Hit

Hit

Miss

Miss

0.05

0.95

0.05

0.95


G1

G1

G1

G1

For first set of branches For second set of branches (indep) For third set of branches (indep) For labels

All probabilities correct All probabilities correct All probabilities correct All correct labels for ‘Hit’ and ‘Miss’, ‘H’ and ‘M’ etc. Condone omission of First, Second, Third. Do not allow misreads here as all FT

[4] 7 (ii) A P(Hits with at least one) = 1 – P(misses with all) M1* For 0.9 × 0.95 × 0.95 FT their tree for both M marks, = 1 – (0.9 × 0.95 × 0.95) = 1 – 0.81225 = 0.18775 M1*

dep For 1 – ans provided three terms

A1 CAO 0.188 or better. Condone 0.1877 Allow 751/4000

ALTERNATIVE METHOD only if there is an attempt to add 7 probabilities

At least three correct triple products M1 Attempt to add 7 triple products M1 (not necessarily correct triple products) A1 CAO FURTHER ALTERNATIVE METHOD 0.1 + 0.9 × 0.05 M1 Above probability + 0.9 × 0.95 × 0.05 M1 A1 CAO [3]


15

Question Answer Marks Guidance 7 (ii) B P(Hits with exactly one) M1 For two correct products FT their tree for all three M marks, = (0.1 × 0.8 × 0.95) + (0.9 × 0.05 × 0.8) + (0.9 × 0.95 × 0.05) M1 For all three correct

products provided three terms

= 0.076 + 0.036 + 0.04275 19 9 171

250 250 4000

M1 For sum of all three correct products

6194000

= 0.15475 A1 CAO Allow 0.155 or better

[4] 7 (iii) P(Hits with exactly one given hits with at least one)

=

P Hits with exactly one and hits with at least oneP Hits with at least one

If answer to (B) > than answer to (A) then max M1M0A0

= 0.15475

0.18775

M1 M1

For numerator FT For denominator FT

Both must be part of a fraction

= 0.8242 A1 CAO Allow 0.824 or better or 619/751 [3] 7 (iv) P(Hits three times overall) =

(0.1 × 0.2 × 0.2) + (0.9 × 0.95 × 0.95 × 0.05 × 0.2 × 0.2) M1 For 0.1 × 0.2 × 0.2 or 0.004

or 1/250 FT their tree for all three M marks

M1 For 0.9 × 0.95 × 0.95 × 0.05 × 0.2 × 0.2

provided three terms in first product and six in second product. Last three probs must be 0.05 × 0.2 × 0.2 unless they extend their tree

= 0.004 + 0.0016245 M1* Dep on both prev M1’s

For sum of both With no extras

= 0.0056245 A1 CAO Allow 0.00562 or 0.00563 or 0.0056 [4]


16

NOTE RE OVER-SPECIFICATION OF ANSWERS If answers are grossly over-specified, deduct the final answer mark in every case. Probabilities should also be rounded to a sensible degree of accuracy. In general final non probability answers should not be given to more than 4 significant figures. Allow probabilities given to 5 sig fig. PLEASE HIGHLIGHT ANY OVER-SPECIFICATION Please note that there are no G or E marks in scoris, so use B instead

NB PLEASE ANNOTATE EVERY ADDITIONAL ANSWER SHEET EVEN IF FULL MARKS AWARDED OR THE PAGE IS BLANK


17

Additional notes re Q5 part iii Comparison with 95% method If 95% seen anywhere then M1 for P(X ≤ 12) B1 for 0.8684 M1* for comparison with 95% dep on second B1 A1* for not significant oe E1* Comparison with 95% CR method If 95% seen anywhere then B1 for 0.9423 or 0.9793 M1 for correct comparison with 95% M1dep for correct CR provided both probs correct then follow mark scheme for CR method Smallest critical region method: Smallest critical region that 13 could fall into is {13, 14, 15, 16, 17, 18, 19, 20} gets B1 and has size 0.1316 gets B1, This is > 5% gets M1*, A1*, E1* as per scheme NB These marks only awarded if 13 used, not other values. Use of k method with no probabilities quoted: This gets zero marks. Use of k method with one probability quoted: Mark as per scheme Line diagram method and Bar chart method No marks unless correct probabilities shown on diagram, then mark as per scheme..


1. Annotations and abbreviations

Annotation in scoris Meaning

Blank Page – this annotation must be used on all blank pages within an answer booklet (structured or unstructured) and on each page of an additional object where there is no candidate response.

and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in mark scheme

Meaning

E1 Mark for explaining U1 Mark for correct units G1 Mark for a correct feature on a graph M1 dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working


4

2. Subject-specific Marking Instructions for GCE Mathematics (MEI) Statistics strand a Annotations should be used whenever appropriate during your marking.

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

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

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


5

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

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

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

f Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy. 3 significant figures may often be the norm for this, but this always needs to be considered in the context of the problem in hand. For example, in quoting probabilities from Normal tables, we generally expect some evidence of interpolation and so quotation to 4 decimal places will often be appropriate. But even this does not always apply – quotations of the standard critical points for significance tests such as 1.96, 1.645, 2.576 (maybe even 2.58 – but not 2.57) will commonly suffice, especially if the calculated value of a test statistic is nowhere near any of these values. Sensible discretion must be exercised in such cases. Discretion must also be exercised in the case of small variations in the degree of accuracy to which an answer is given. For example, if 3 significant figures are expected (either because of an explicit instruction or because the general context of a problem demands it) but only 2 are given, loss of an accuracy ("A") mark is likely to be appropriate; but if 4 significant figures are given, this should not normally be penalised. Likewise, answers which are slightly deviant from what is expected in a very minor manner (for example a Normal probability


6

given, after an attempt at interpolation, as 0.6418 whereas 0.6417 was expected) should not be penalised. However, answers which are grossly over- or under-specified should normally result in the loss of a mark. This includes cases such as, for example, insistence that the value of a test statistic is (say) 2.128888446667 merely because that is the value that happened to come off the candidate's calculator. Note that this applies to answers that are given as final stages of calculations; intermediate working should usually be carried out, and quoted, to a greater degree of accuracy to avoid the danger of premature approximation. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader.

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

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 Genuine misreading (of numbers or symbols, occasionally even of text) occurs. If this results in the object and/or difficulty of the question being considerably changed, it is likely that all the marks for that question, or section of the question, will be lost. However, misreads are often such that the object and/or difficulty remain substantially unaltered; these cases are considered below. The simple rule is that all method ("M") marks [and of course all independent ("B") marks] remain accessible but at least some accuracy ("A") marks do not. It is difficult to legislate in an overall sense beyond this global statement because misreads, even when the object and/or difficulty remains unchanged, can vary greatly in their effects. For example, a misread of 1.02 as 10.2 (perhaps as a quoted value of a sample mean) may well be catastrophic; whereas a misread of 1.6748 as 1.6746 may have so slight an effect as to be almost unnoticeable in the candidate's work. A misread should normally attract some penalty, though this would often be only 1 mark and should rarely if ever be more than 2. Commonly in sections of questions where there is a numerical answer either at the end of the section or to be obtained and commented on (eg the value of a test statistic), this answer will have an "A" mark that may actually be designated as "cao" [correct answer only]. This should be interpreted strictly – if the misread has led to failure to obtain this value, then this "A" mark must be withheld even if all method marks have been earned. It will also often be the case that such a mark is implicitly "cao" even if not explicitly designated as such.


7

On the other hand, we commonly allow "fresh starts" within a question or part of question. For example, a follow-through of the candidate's value of a test statistic is generally allowed (and often explicitly stated as such within the marking scheme), so that the candidate may exhibit knowledge of how to compare it with a critical value and draw conclusions. Such "fresh starts" are not affected by any earlier misreads. A misread may be of a symbol rather than a number – for example, an algebraic symbol in a mathematical expression. Such misreads are more likely to bring about a considerable change in the object and/or difficulty of the question; but, if they do not, they should be treated as far as possible in the same way as numerical misreads, mutatis mutandis. This also applied to misreads of text, which are fairly rare but can cause major problems in fair marking. The situation regarding any particular cases that arise while you are marking for which you feel you need detailed guidance should be discussed with your Team Leader. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.


1


1 (i) Upper Bound 20 30 40 50 60 70 80 90

Cumulative Freq 0 10 40 82 105 114 119 120

B1 Cumulative frequencies

All correct

G1

For plotted points

(Provided plotted at correct

UCB positions)

Plotted within ½ small square

If cf not given then allow G1 for

good attempt at cf. e.g. if they

have 0,10,40,72,95,104,109,110

G1

For joining points

(within ½ a square)

G1 For scales

G1 For labels

All marks dep on good

attempt at cumulative

frequency, but not

cumulative fx’s or other

spurious values.

[5]


2

0.2

0.3

Accept

Retest

0.5

0. 3

0. 5 Reject

Accept

Reject

0.2

Reject

Accept

Retest

0.4

0.6


1 (ii)

Median = 45 B1

Allow answers between 44

and 46 without checking

curve. Otherwise check

curve.

No marks if not using

diagram.

Based on 60th value

ft their curve (not LCB’s) Allow

40 for m.p. plot without

checking graph

B0 for interpolation

If max value wrong (eg 110) FT

their max value for all 3 marks

Q1 = 37 Q3 = 53 B1

For Q3 or Q1

Allow Q1 between 37 and

38 without checking

Allow Q3 between 52 and

54 without checking

Based on 30th and 90

th values

ft their curve (not LCB’s) Allow

Q1 = 32; Q3 = 48 without

checking graph

Inter-quartile range = 53 – 37 = 16 B1

For IQR providing both Q1

and Q3 are correct

B0 for interpolation

B2 for correct IQR from graph if

quartiles not stated but indicated

on graph

Allow from mid-point plot

Must be good attempt at

cumulative frequency in part (i)

to score any marks here

Lines of best fit: B0 B0 B0 here.

Also cumulative frequency bars:

B0 B0 B0 here

[3]

2 (i)

Alternative version of tree

diagram for Q2(i)


3

0.2

0.5

Accept

Reject

0.3 Retest

0. 5

0. 3 Retest

Accept

0.2

Reject

Accept Reject

0.4

0.6


2 (i) Do a vertical scan and give:

Allow labels such as A, R,

F(Fail) etc

G1 First column All probabilities correct

G1 Second column All probabilities correct

G1

Final column

Do not award if first two

branches missing

Branches two and three

should come out of ‘retest’

All probabilities correct

If any labels missing or incorrect

allow max 2/3

Do not allow misreads here as all

FT (eg 0.3 and 0.5 reversed)

[3]

2 (ii)

P(Accepted) = 0.2 + (0.3 × 0.2) + (0.3 × 0.3 × 0.4) M1 For second or third product

FT their tree provided correct

numbers of terms and correct

structure of 3, 3, 2 branches.

= 0.2 + 0.06 + 0.036 = 0.296 A1 CAO Allow 37/125 oe

[2]

2 (iii) P(At least one retest given accepted)

=P(At least one retest and accepted)

P(Accepted) M1

For numerator

FT their tree provided correct

numbers of terms and correct

structure of 3, 3, 2 branches. for

both M1’s

= 0.3 0.2 0.3 0.3 0.4

0.296=

0.096

0.296

M1 For denominator Both must be part of a fraction

Allow 12/125 oe

= 0.324 A1

FT their 0.296 and 0.096

Allow 0.32 with working Allow 12/37 oe

[3]


4

0.286

0.099

0.729

L

R

0.121 0.051


3 (i)

Because P(L | R) P(L) E1

If two or more methods

given and only one correct,

do not award the mark

Allow 0.45 ≠ 0.15

Either P(L ∩ R) (= 0.099) ≠ P(L)

× P(R), provided 0.099 in (ii)

or 0.099 ≠ 0.15 × 0.22 (= 0.033)

Look out for complement

methods, etc

[1]

3 (ii) P(L ∩ R) = P(L | R) × P(R) = 0.45 × 0.22 M1 For product Allow if done correctly in part(i)

= 0.099 A1 CAO Allow 99/1000

[2]

3 (iii)

G1

For two labelled

intersecting circles,

provided no incorrect

labelling.

Condone labels such as P(L) etc

Allow other shapes in place of

circles

No need for ‘box’

G1

For at least 2 correct

probabilities.

FT their P(L ∩ R) from part

(ii) provided ≤ 0.15

FT from 0.033 in (ii) gives

0.117, 0.033, 0.187, 0.663

In general 0.15 – x, x, 0.22 – x,

0.63 + x

May also see 0.0825, 0.0675,

0.1525, 0.6975

G1

For remaining probabilities.

FT their P(L ∩ R) providing

probabilities between 0 and

1.

[3]


5


4 (i)

P(All four are girls) = 16 15 14 13

30 29 28 27

M1 For

16

30

OR 16 30 1820 52

/ 0.06644 4 27405 783

M1 for either term in correct

position in a fraction

M1 for correct fraction

A1 CAO

Allow full marks for

unsimplified fractional answers

SC2 for

14 13 12 110.0365

30 29 28 27

M1

For product of other three

correct fractions

Without extra terms

= 0.0664 A1

CAO

Allow 0.066 with working

but not 0.07

[3] SC2 for 14 30 143

/ 0.03654 4 3915


6


4 (ii)

P(All four are boys) = 14 13 12 11

0.036530 29 28 27

M1 For P(All four are boys)

Without extra terms

OR 14 30 143/ 0.0365

4 4 3915

M1 for this then as per scheme.

P(At least one girl and at least one boy)

= 1 – (0.0664 + 0.0365) M1

For 1 – (0.0664 + 0.0365)

FT their ‘sensible'

probabilities

= 0.897 A1

CAO

Allow answer 0.8975 from

use of 0.066

NB Watch for (1 – 0.0365)×(1 –

0.0664) = 0.9635×0.9336 =

0.8995 Gets just M1 for 0.0365

[3]

Accept 0.90 work working, but

not 0.9

OR P(3b,1g) + P(2b,2g) +P(1b,3g)

= 14 13 12 16

430 29 28 27

+14 13 16 15

630 29 28 27

+

14 16 15 14 208 52 564 4 6 4

30 29 28 27 3915 783 783

4 0.0531 6 0.0664 4 0.0715

M1 For any one product, even if

coefficient missing

Or 14

C3×16

C1/30

C4 + 14

C2×16

C2/30

C4 +14

C1×16

C3/30

C4

= 0.2125 + 0.3985 + 0.2861

= 0.897

M1 for any one term

= 0.2125 + 0.3985 + 0.2861 M1

For sum of all three (all

correct)

M1 for sum of all three (all

correct) = 0.897 3512

3915 A1 CAO A1 CAO

[3]


7


5 (i) k + 0.01 + k + 0.04 + k + 0.09 + k + 0.16 + k + 0.25 = 1

5k + 0.55 = 1

M1 For equation in k

k = 0.09 A1 NB Answer Given

Allow substitution of k = 0.09 to

show probabilities add to 1 with

convincing working

r 1 2 3 4 5

P(X= r) 0.1 0.13 0.18 0.25 0.34

B1 Complete correct table Must tabulate probabilities,

though may be seen in part(ii)

[3]

5 (ii)

E(X) = (1 0.1) (2 0.13) (3 0.18) (4 0.25) (5 0.34) M1

For Σrp (at least 3 terms

correct Provided 5

reasonable probabilities

seen.

If probs wrong but sum = 1

allow max M1A0M1M1A1. If

sum ≠ 1 allow max

M1A0M1M0A0 (provided all

probabilities ≥0 and <1)

No marks if all probs =0.2

.

Use of E(X-µ)2 gets M1 for

attempt at (x-µ)2 should see (-

2.6)2, (-1.6)

2, (-0.6)

2, 0.4

2, 1.4

2,

(if E(X) wrong FT their E(X))

(all 5 correct for M1), then M1

for Σp(x-µ)2 (at least 3 terms

correct with their probabilities)

Division by 5 or other spurious

value at end and/or rooting final

answer gives max

M1A1M1M1A0, or

M1A0M1M1A0 if E(X) also

divided by 5.

Unsupported correct answers get

5 marks (Probably from

calculator)

= 3.6 A1 CAO

E(X2) =

(1 0.1) (4 0.13) (9 0.18) (16 0.25) (25 0.34) 14.74 M1*

For Σr2p (at least 3 terms

correct)

Var(X) = 14.74 – 3.62

M1*

dep

for – their (E[X])²

FT their E(X) provided Var(

X ) > 0

= 1.78 A1 CAO

[5]


8


6 (i)

Weight Frequency Group Width Frequency density

30 ≤ w < 50 11 20 0.55

50 ≤ w < 60 10 10 1

60 ≤ w < 70 18 10 1.8

70 ≤ w < 80 14 10 1.4

80 ≤ w < 90 7 10 0.7

M1

For fd’s - at least 3 correct

Accept any suitable unit for

fd such as eg freq per 10g.

M1 can be also be gained from

freq per 10 – 5.5, 10, 18, 14, 7

(at least 3 correct) or similar.

If fd not explicitly given, M1

A1 can be gained from all

heights correct (within half a

square) on histogram (and M1A0

if at least 3 correct)

A1

G1

linear scales on both axes

and labels

Vertical scale starting from

zero (not broken - but can

get final mark for heights if

broken)

Linear scale and label on vertical

axis IN RELATION to first M1

mark ie fd or frequency density

or if relevant freq/10, etc (NOT

eg fd/10).

However allow scale given as

fd×10, or similar.

Accept f/w or f/cw (freq/width or

freq/class width)

Ignore horizontal label

Can also be gained from an

accurate key

G0 if correct label but not fd’s.

G1 width of bars

Must be drawn at 30, 50 etc

NOT 29.5 or 30.5 etc NO GAPS

ALLOWED

Must have linear scale.

No inequality labels on their own

such as 30 ≤W<50, 50 ≤W<60

etc but allow if 30, 50, 60 etc

occur at the correct boundary

position. See additional notes.

Allow this mark even if not

using fd’s


9


G1 height of bars Height of bars – must be linear

vertical scale.

FT of heights dep on at least 3

heights correct and all must

agree with their fds

If fds not given and at least 3

heights correct then max

M1A0G1G1G0

Allow restart with correct

heights if given fd wrong (for

last three marks only)

[5]


10


6 (ii) Mean = (40 11) (55 10) (65 18) (75 14) (85 7)

60

3805

60

M1

For midpoints

Products are 440, 550,

1170, 1050, 595

For midpoints (at least 3 correct)

No marks for mean or sd unless

using midpoints

= 63.4 (or 63.42) A1 CAO

(exact answer 63.41666…)

Answer must NOT be left as

improper fraction as this is an

estimate

Accept correct answers for mean

and sd from calculator even if eg

wrong Sxx given

x

2f =

2 2 2 2 2(40 11) (55 10) (65 18) (75 14) (85 7)

253225

23805

253225- 11924.660

xxS M1

For attempt at xxS

Should include sum of at

least 3 correct multiples fx2

– Ʃx2/n

Allow M1 for anything which

rounds to 11900

11924.6202.11 14.2

59s

A1

At least 1dp required

Use of mean 63.4 leading to

answer of 14.29199.. with

Sxx = 12051.4 gets full

credit.

63.42 leads to 14.2014…

Do not FT their incorrect

mean

(exact answer14.2166…)

Allow SC1 for RMSD 14.1

(14.0976…) from calculator.

Only penalise once in part (ii) for

over specification, even if mean

and standard deviation both over

specified.

If using (x – x )2 method, B2 if

14.2 or better (14.3 if use of

63.4), otherwise B0

[4]


11


6 (iii)

x – 2s = 63.4 – (2 14.2) = 35 M1

For either

No marks in (iii) unless

using x + 2s or x – 2s

Only follow through numerical

values, not variables such as s, so

if a candidate does not find s but

then writes here ‘limit is 63.4 +

2 standard deviation’, do NOT

award M1

x + 2s = 63.4 + (2 14.2) = 91.8 A1 For both (FT)

Do not penalise for over-

specification

So there are probably some outliers at the lower end, but none at the

upper end E1

Must include an element of

doubt and must mention

both ends

Must have correct limits to get

this mark

[3]

6 (iv) Mean =

3624.5

50 = 72.5g (or exact answer 72.49g) B1 CAO Ignore units

xxS =

23624.5265416

50= 2676 M1 For xxS

M1 for 265416 - 50 their

mean2

BUT NOTE M0 if their Sxx < 0

s = 2676

49 = 54.61 = 7.39g A1

CAO ignore units

Allow 7.4 but NOT 7.3

(unless RMSD with

working)

For s2 of 54.6 (or better) allow

M1A0 with or without working.

For RMSD of 7.3 (or better)

allow M1A0 provided working

seen

For RMSD2 of 53.5 (or better)

allow M1A0 provided working

seen

[3]


12


6 (v)

Variety A have lower average than Variety B oe E1

FT their means

Do not condone lower

central tendency or lower

mean

Allow ‘on the whole’ or similar

in place of ‘average’.

Variety A have higher variation than Variety B oe E1 FT their sd

Allow ‘more spread’ or similar

but not ‘higher range’ or ‘higher

variance’

Condone less consistent.

[2]

7 (i) (A) X ~ B(15, 0.85)

P(exactly 12 germinate) = 12 3

150.85 0.15

12

M1 For 12 30.85 0.15

M1 For

12 315

12p q

= 0.2184 A1 CAO

OR OR

from tables: 0.3958 – 0.1773 M2 For 0.3958 – 0.1773

= 0.2185 A1 CAO

[3]

7 (i) (B) P(X <12) = P(X ≤ 11) = 0.1773 M1

For P(X ≤ 11) or P(≤11)

(With no extras)

A1

CAO (as final answer)

May see alternative method:

0.3958 – 0.2185 = 0.1773

0.3958 - their wrong answer

to part (i) scores M1A0

[2]


13


7 (ii) Let p = probability of a seed germinating (for the population) B1 For definition of p See below for additional notes

H0: p = 0.85 B1 For H0

H1: p < 0.85 B1 For H1

H1 has this form because the test is to investigate whether the

proportion of seeds which germinate is lower. E1

Dep on < 0.85 used in H1

Do not allow just

‘Germination rate will be

lower’ or similar.

For use of 0.15 as P(not

germinating), contact team

leader

E0 for simply stating H1 in

words

[4]

7 (iii) Let X ~ B(20, 0.85)

P(X 13) = 0.0219

M1*

For probability (provided

not as part of finding

P(X = 13)) Ignore notation

No further marks if point probs

used - P(X = 13) = 0.0160

DO NOT FT wrong H1, but see

extra notes


Must include ‘sufficient

evidence’ or something similar

such as ‘to suggest that’ ie an

element of doubt either in the A

or E mark.

0.0219 > 1%

M1*

dep For comparison

So not enough evidence to reject H0.

Not significant. A1

* For not significant oe

Conclude that there is not enough evidence to indicate that the

proportion of seeds which have germinated has decreased.

E1*

dep

For conclusion in context

Must mention decrease, not

just change

ALTERNATIVE METHOD – follow method above unless some

mention of CR seen

Critical region method

No marks if CR not justified

Condone {0,1,2,…, 12}, X ≤ 12,

oe but not P(X ≤ 12) etc

LOWER TAIL

P(X ≤ 13) = 0.0219 > 1%

P(X ≤ 12) = 0.0059 < 1%

M1 For either probability Could get M1A0A1E1 if poor

notation for CR

Do not allow just ’13 not in CR’

- Must say ‘not significant’ or

accept H0 or similar

So critical region is {0,1,2,3,4,5,6,7,8,9,10,11,12} A1

cao dep on at least one

correct comparison with 1%

13 not in CR so not significant A1*

There is insufficient evidence to indicate that the proportion of seeds

which have germinated has decreased.

E1*

dep

[4]


14


7 (iv) 33 < 35 M1 For comparison Allow ‘33 lies in the CR’

So there is sufficient evidence to reject H0 A1*

Must include ‘sufficient

evidence’ or something similar

such as ‘to suggest that’ ie an

element of doubt either in the A

or E mark.

Do not FT wrong H1: In part (iv)

ignore any interchanged H0 and

H1 seen in part (ii)

Conclude that there is enough evidence to indicate that the proportion

of seeds which have germinated has decreased.

E1*

dep

For conclusion in context

Must mention decrease, not

just change

If use a calculator to find

P(X ≤33) = 0.000661 and

compare with 1% then B2 for

P(X ≤ 33) = 0.000661 < 0.01 so

reject H0 then final E1 as per

scheme.

[3]

7 (v) For n = 3 , P(X 0) = 0.0034 < 0.01 M1 For P(X 0) = 0.0034 Allow 0.003

For n = 2 , P(X 0) = 0.0225 > 0.01 M1 For P(X 0) = 0.0225

So the least value of n for which the critical region is not empty and

thus H0 could be rejected is 3. A1 CAO

Condone just ‘n = 3’ for final A

mark dep on both M marks

If wrong H1 allow max M2A0 if

correct probabilities seen.

ALTERNATIVE METHOD using logs

0.15n < 0.01

n > log 0.01 / log 0.15

n > 2.427

Least n = 3

M1

M1

A1

[3]

GCE

Mathematics (MEI) Unit 4766: Statistics 1 Advanced Subsidiary GCE



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 2015

4766 Mark Scheme June 2015 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


3

4766 Mark Scheme June 2015 Subject-specific Marking Instructions for GCE Mathematics (MEI) Statistics strand

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


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

4


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.



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. 3 significant figures may often be the norm for this, but this always needs to be considered in the context of the problem in hand. For example, in quoting probabilities from Normal tables, we generally expect some evidence of interpolation and so quotation to 4 decimal places will often be appropriate. But even this does not always apply – quotations of the standard critical points for significance tests such as 1.96, 1.645, 2.576 (maybe even 2.58 – but not 2.57) will commonly suffice, especially if the calculated value of a test statistic is nowhere near any of these values. Sensible discretion must be exercised in such cases. Discretion must also be exercised in the case of small variations in the degree of accuracy to which an answer is given. For example, if 3 significant figures are expected (either because of an explicit instruction or because the general context of a problem demands it) but only 2 are given, loss of an accuracy ("A") mark is likely to be appropriate; but if 4 significant figures are given, this should not normally be penalised. Likewise, answers which are slightly deviant from what is expected in a very

5


minor manner (for example a Normal probability given, after an attempt at interpolation, as 0.6418 whereas 0.6417 was expected) should not be penalised. However, answers which are grossly over- or under-specified should normally result in the loss of a mark. This includes cases such as, for example, insistence that the value of a test statistic is (say) 2.128888446667 merely because that is the value that happened to come off the candidate's calculator. Note that this applies to answers that are given as final stages of calculations; intermediate working should usually be carried out, and quoted, to a greater degree of accuracy to avoid the danger of premature approximation. 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.




6



7



Mean = 59972

40 = 1499

Condone full answer of 1499.3 (despite over-specification rule) B1 CAO Ignore units

NB Allow 1500 NB Answer must be decimal

Sxx =

2599729676702840

− = 6851008 M1

For Sxx

M1 for 96767028 - 40 × their mean2 BUT NOTE M0 if their Sxx < 0

s =

685100839

= 175667 = 419

NB Full answer is 419. 1263 (but only allow to 4sf due to over-specification rule)

A1

CAO ignore units

For s2 of 176000 (or better) allow M1A0 with or without working For RMSD of 414 (or better) allow M1A0 provided working seen For RMSD2 of 171000 (or better) allow M1A0 provided working seen For use of 1499: Sxx = 6886988, s2 = 176589, s = 420.225, RMSD = 414.9 For use of 1500: Sxx = 6767028, s2 = 173513.5, s = 416.549, RMSD = 411.3 Give same credit to answers as for correct answers

[3]

1 (ii) New mean = (0.163 × 1499) + 14.5 = £258.84 (No penalty for giving to 5sf as this is an exact sum of money)

B1 FT their mean provided answer is positive

If candidate ‘starts again’ only award marks for CAO Allow £259 or £259.00 from 1500 or £258.89 from 1499.3 Condone 258.8 and 258.9 Accept answers rounded to 3 sf or more eg £258.80, £258.90

New sd = 0.163 × 419 M1 FT their sd for M1 and A1

Or for 0.163×419.1 oe

= £68.30 A1 Allow £68.29 to £68.32 Allow 68.3

Do not penalise lack of units in mean or sd

[3]

Deduct at most 1 mark overall in whole question for over-specification of either mean or SD or both

8


Question Answer Marks Guidance 2 (i) (A)

P(Watched cyc but not fb) = 15250

=3

50 = 0.06 B1 CAO (aef)

[1] 2 (i) (B)

P(Watched one or two) = 33 12 21 14 3 65

250+ + + + +

M1

250 (64 38)OR: 250− +

=

For M1 terms must be added with no extra terms (added or subtracted)

148250

= =74

125= 0.592 A1 CAO (aef)

[2] 2 (ii)

P(Not watched fb|watched cyc) = 1567

= 0.224 (0.223880597…) M1 For denominator of either 67 or 67/250 or 0.268

A1 CAO (aef) Allow 0.22 with working [2] 3 (i)

P(Neither is an ace) = 241

52 −

M1 For 48/52 oe seen

2304 144 0.8522704 169

= = = (0.8572071…) A1 CAO Allow 0.85 with working

[2] 3 (ii) Expected number = 10 × 0.852 = 8.52 B1 FT their (i) if seen

Do not allow whole number final answer even if 8.52 seen first. Allow fractional answer

[1] 4 (i) 25

15

M1 Accept 25C15 or 25!/(15!10!) or equivalent

for M1 No marks for permutations

= 3268760 A1 Exact answer required [2] 4 (ii) 13

8

×127

= 1287 × 792 M1 For product of both

correct combinations No marks for permutations

= 1019304 A1 CAO Exact answer required [2]

9


Question Answer Marks Guidance 4 (iii) 1019304/3268760 M1 For their (ii) divided by

their (i)

= 0.312 (0.311832) Allow fully simplified fraction11583/37145

A1 FT Allow 0.31 with working

[2] OR 15 13 12 11 10 9 8 7 6 12 11 10 9 8 7 6

8 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11

× × × × × × × × × × × × × × ×

(M1) For product of fractions with coefficient

SC1 for 8 715 13 12

8 25 25 × ×

Allow 15 15 or

8 7

= 0.312 (A1) 5 (i) 0 6

1 5 8 2 1 5 8 3 1 1 3 5 8 9 Key 1 8 represents 18 people

G1

Stem (in either order) and leaves

Do not allow leaves 21 ,25, 28 etc Ignore commas between leaves Allow stem 0, 10, 20, 30

G1

Sorted and aligned

Allow errors in leaves if sorted and aligned. Use paper test if unsure about alignment – hold a piece of paper vertically and the columns of leaves should all be separate. Alternatively place a pencil vertically over each column. If any figures protrude then deem this as non-alignment. Highlight this error

G1 Key

[3] 5 (ii) Negative B1 Allow -ve but NOT skewed to the left

Do not allow ‘negative correlation’ [1] 5 (iii) Median = 29.5 B1 CAO Mean = 26.7 (26.6666) or 262/3 or 80/3 or 26.6 B1 CAO Do not allow 27

but condone 26.6 www Mode = 31 B1 CAO The mode is not at all useful as it is just by chance that it is 31. E1 Allow any reasonable

10


Question Answer Marks Guidance Mark awarded for stating not useful and -not representative of data -does not represent Central Tendency -happened by chance (or similar) -comment about not appearing significantly more (only one repetition/only twice/ etc) No mark for stating it would be useful OR NOT USEFUL because of -spread/range -sample size -negatively skewed -unaffected by outliers -isn't close to mean and median

comment

[4] 6 (i)

P(X = 6) = 1 - P(X < 6) = 1 - 35

6

= 1 - 125216

M1 For 35

6

M1 For

3516

−

=

91216

A1 NB ANSWER GIVEN

[3]

OR: 3 2 21 5 1 5 13 3

6 6 6 6 6 = + × × + × ×

M1 For second or third product term

Correct, including ×3 or probabilities seen on correct tree diagram

M1 For attempt at three terms

With no extras, but allow omission of ×3

91216

= A1 NB ANSWER GIVEN

NB Zero for 1 – (sum of probs given in part (ii))

OR: 1 + 15 + 75 M1 for 15 or 75 seen 1 15 75

216+ +

= M1

11


Question Answer Marks Guidance 91

216= A1 NB ANSWER

GIVEN

6 (ii) E(X) =

1 7 19 37 61 911 2 3 4 5 6216 216 216 216 216 216

× + × + × + × + × + ×

M1

For Σrp (at least 3 terms correct)

=

1071216

= 11924

= 4.96 (exact answer 4.9583333) A1 CAO Accept fractional answers Do not allow answer of 5 unless more accurate answer given first

E(X2) = 1 7 19 37 61 91

1 4 9 16 25 36216 216 216 216 216 216

× + × + × + × + × + ×

M1* For Σr2p (at least 3 terms correct)

Use of E(X-µ)2 gets M1 for attempt at (x-µ)2 should see (-3.96)2, (-2.96)2, (-1.96)2, (-0.96)2 0.042, 1.042, (if E(X) wrong FT their E(X)) (all 6 correct for M1), then M1 for Σp(x-µ)2 (at least 3 terms correct) Division by 6 or other spurious value at end and/or rooting final answer gives max M1A1M1M1A0, or M1A0M1M1A0 if E(X) also divided by 6.

=

5593216

= 25.89

Var(X) = 25.89.. – 4.958…2

M1* dep for – their (E( X ))²

= 1.31 Accept answers in range 1.28 to 1.31 with correct working or 2261/1728 (Exact answer 1.308449…)

A1 FT their E(X) provided Var( X ) > 0

Do not FT E(X) = 5 if full marks given for E(X) Deduct at most 1 mark for over-specification of either mean or variance or both Unsupported correct answers get 5 marks (Probably from calculator)

[5] 7 (i) (A) X ~ B(20, 0.78)

12



P(Exactly 19 cured) = 19 1200.78 0.22

19

× ×

M1 For 19 10.78 0.22× Allow M2A0 for linear interpolation

from tables leading to 0.9918 – 0.9488 = 0.0430 But zero for use of tables with 0.8 leading to 0.9885 - 0.9308 = 0.0577

M1 For 19 120

19p q

× ×

With p + q = 1 Also for 20 × 0.00196

= 0.0392 (0.039197)

A1 CAO Allow 0.039 or better Condone 0.03919 but not 0.0391

[3] 7 (i) (B)

P(Exactly 20 cured) = 20 0200.78 0.22

20

× ×

= 0.0069 M1

For 0.7820 oe

Allow M2 for 0.9488 for linear interpolation from tables or M1 for 1 – 0.9918 = 0.0082 and second M1 for correct FT using answer to (i)(A) Zero for use of p = 0.8 here

P(At most 18 cured) = 1 – (0.0069 + 0.0392) M1 For P(19) + P(20) Not necessarily correct, but both attempts at binomial, including coefficient in (i) and no extra terms (such as P(X=18)) Condone use of p = 0.8

= 0.954 (0.95385) A1 CAO Allow 0.95 with working [3] 7 (i) (C) E(X) = np = 20 × 0.78 = 15.6 B1 CAO Do not allow final answer of 15 or 16

even if correct 15.6 given earlier [1] 7 (ii) Let X ~ B(20, 0.78)

Let p = probability of a patient being cured (for population) B1 For definition of p In context See below for additional notes

H0: p = 0.78 B1 For H0 H1: p > 0.78

B1 For H1

No further marks if point probabilities

13


Question Answer Marks Guidance used

P(X ≥ 19) = 0.0392 + 0. 0069 B1 For NOTATION P(X ≥ 19) or P(X > 18) or 1– P(X ≤ 18) or 1– P(X < 19)

Notation P(X = 19) scores B0. If they have the correct P(X ≥ 19) then give B1 and ignore any further incorrect notation.

= 0.0461

B1* CAO For 0.0461 allow 0.0462

FT answer to (i)B for following three marks provided based on 1 – (P(19) + P(20))

0.0461 > 1% M1* dep

For comparison with 1%

Dep on sensible attempt at P(X ≥ 19)

So not significant. A1 Allow ‘accept H0’ or ‘reject H1’ Conclude that there is not enough evidence to suggest that the new drug

is more effective than the old one. E1 Must include ‘insufficient evidence’

or something similar such as ‘to suggest that’ ie an element of doubt either in the A or E mark. Must be in context to gain E1 mark. Do NOT allow ‘sufficient evidence to suggest proportion cured is 0.78’ or similar 99% method: P(X≤18) = 0.9539 B1B1* CAO 0.9539 < 99% M1* then as per scheme

[8] ALTERNATIVE METHOD FOR FINAL 5 MARKS If combination of

methods used, mark both and give higher mark.

No further marks if point probabilities used

P(X ≥ 19) = 0.0461 > 1% B1 For either probability Do not insist on correct notation as candidates have to work out two probabilities for full marks.

14


Question Answer Marks Guidance P(X ≥ 20) = 0.0069 < 1%

M1 For at least one

comparison with 1% Allow comparison in form of statement ‘critical region at 1% level is …’

So critical region is {20} B1* CAO dep on the two correct probabilities

No marks if CR not justified Condone X ≥ 20, X = 20, oe but not P(X ≥ 20,) etc

(19 not in CR so) not significant. A1* dep

Dep on correct CR Allow ‘accept H0’ or ‘reject H1’

Conclude that there is not enough evidence to suggest that the new drug is more effective than the old one.

E1* dep

Ignore any work on lower critical region

7 (iii) With a 5% significance level rather than a 1% level, the null hypothesis would have been rejected. OR: ‘there would be enough evidence to suggest that the new drug is more effective than the old one.’

B1* oe FT their probability from (ii) but NO marks if point probabilities used There must be a sensible attempt to use P(X = 19) + P(X =20) or must have correct CR.

This is because 0.0461 < 5% B1* dep

oe Dep on correct answer of 0.0461 compared with 5% or 0.9539 compared with 95% or correct CR.

[2]

8 (i) Inter-quartile range = 18.1 – 17.8 = 0.3 B1

Lower limit 17.8 – (1.5 × 0.3) (= 17.35) M1 FT their IQR for M marks only No outliers at lower end. A1 dep on 17.35

Allow ‘No values below 17.35 for first A1 Allow ‘Lower limit = 17.35 so no outliers (at lower end)’ Watch for use of median giving 17.45 which gets M0A0 You must be convinced that comments about no outliers refer to lower tail only.

Upper limit 18.1 + (1.5 × 0.3) ( = 18.55) M1 (Max is 18.6) so at least one outlier at upper end. A1 dep on 18.55 Allow ‘At least one value above

18.55’ for second A1 Allow ‘any value above 18.55 is an outlier’ so at least one outlier.

15


Question Answer Marks Guidance Do not allow ‘There MAY be one outlier’ oe Condone ‘one outlier’ Condone ‘there are outliers’ Watch for use of median giving 18.35 which gets M0A0 You must be convinced that comments about some outliers refer to upper tail only.

[5] 8 (ii)

P(A) = P(All 3 have orange centres) = 7 6 5 720 19 18 228

× × = M1

For 7/20×

Allow final answer of 0.031 with working ALTERNATIVE SCHEME 7C3/20C3 = 35/1140 = 7/228 = 0.0307 M1 for either term in correct position in a fraction M1 for correct fraction A1 CAO

M1 For product of correct three fractions Without extra terms

= 0.0307 (0.030702) A1 CAO Allow full marks for fully simplified fractional answers

P(B) = P(All 3 have same centres) = 7 6 5 6 5 4 4 3 2 3 2 120 19 18 20 19 18 20 19 18 20 19 180.0307 0.0175 0.0035 0.0009

× × + × × + × × + × × = = + + +

M1

For at least two correct triple products or fractions or decimals

ALTERNATIVE SCHEME 7C3/20C3 + 6C3/20C3 +4C3/20C3 +3C3/20C3 M1 for at least two correct terms

M1 For sum of all four correct

M1 for sum of all four (all correct) either as combinations or decimals

0.0526=

119

= (0.052632) A1 CAO A1 CAO

7 1 1 1228 57 285 1140

= + + +

[6] Allow 0.053 or

anything which rounds up to 0.053 with working

Please check all of the answer space for this part

8 (iii) P( | )A B =

0.0307..0.0526..

M1 For their ‘A’ divided

by their ‘B’

Allow 0.584 from 0.03070.0526

16


Question Answer Marks Guidance = 0.583 (= 0.58333) A1 FT their answers to (ii)

provided answer < 1 Allow 7

12

P( | )B A = 1 B1 CAO [3] 8 (iv)

P(All have orange centres) = 20.0307 0.00094= or = 49

51984

= (0.00094260)

M1 For their 0.03072 Allow 9.4 × 10-4

condone 0.0009 or 9 × 10-4

A1 FT [2] 8 (v) P(Has to select > 2) = 1 - P(Has to select ≤ 2)

= ( )14 6 141 1 0.7 0.221 1 0.92120 20 19

− + × = − + = −

M1 For

6 1420 19

×

For any of the methods below allow SC2 for 1 – 0.079 = 0.921 or 1 – 3/38 = 35/38 o.e. as final answer

M1 For 1 – sum of both This is 1 – P(C’ + CC’) = 0.079 (=0.078947) A1 CAO [3] OR

P(Has to select > 2) = P(First 2 both cherry) = 6 520 19

×

M2 For whole product Without extra terms added

M1 if multiplied by k/18 only where 0<k<18 (seen as a triple product only)

= 0.079

338

= A1 CAO This is P(CC).

OR 1 – (P(0 cherries) + P(1 cherry))=

( )

14 13 6 14 14 6120 19 20 19 20 19

1 0.4789 0.2211 0.2211 1 0.9209

− × + × + × = − + + = −

M1 For any term M1 For 1 – sum of all three This is 1 – P(C’C’ + CC’ + C’C)

= 0.079 A1 CAO OR 6 5 14 6 5 4 14 6 5 4 3 14 6 5 4 3 2 14 6 5 4 3 2 1 14

20 19 18 20 19 18 17 20 19 18 17 16 20 19 18 17 16 15 20 19 18 17 16 15 14

× × + × × × + × × × × + × × × × × + × × × × × ×

M1 For any term This is P(CCC’ + CCCC’ + CCCCC’ + CCCCCC’ + CCCCCCC’)

17



=7 14 7 7 1

114 969 2584 19380 38760+ + + +

M1 For sum of all five terms (all correct)

= 0.079

A1 CAO

NOTE RE OVER-SPECIFICATION OF ANSWERS If answers are grossly over-specified, deduct the final answer mark (max once per question). Probabilities should also be rounded to a sensible degree of accuracy. In general final non probability answers should not be given to more than 4 significant figures. Allow probabilities given to 5 sig fig. PLEASE HIGHLIGHT ANY OVER-SPECIFICATION and also non alignment in question 5(i) Please note that there are no G or E marks in scoris, so use B instead

NB Please annotate every additional answer sheet even if full marks awarded or the page is blank – Use BP symbol Other rules for annotation: In the 10 standardisation scripts annotate everything that gets anything other than zero or full marks. After this, annotate down the right hand side in Q7(ii), up to where marks stop being scored, in the order given in the mark scheme. Annotate anywhere else where necessary to clarify how you have awarded the marks. Additional notes re Q7 part ii (first three marks) Minimum needed for B1 is p = probability of being cured. Allow p = P(patient being cured) Definition of p must include word probability (or chance or proportion or percentage or likelihood but NOT possibility, number or amount). Preferably given as a separate comment. However can be at end of H0 as long as it is a clear definition ‘p = the probability of patient being cured.’ Do NOT allow ‘p = probability of patient in the sample being cured’

18


Do NOT allow ‘p = the probability of patient being cured is different’ Allow p=78%, allow only p or θ or π or ρ. However allow any single symbol if defined (including x) Allow H0 = p=0.78, Allow H0 : p=39/50 or p=78/100 Allow NH and AH in place of H0 and H1 Do not allow H0 : P(X=x) = 0.78 Do not allow H0: =0.78, =78%, P(0.78), p(x)=0.78, x=0.78 (unless x correctly defined as a probability) Do not allow H0 and H1 reversed For hypotheses given in words allow Maximum B0B1B1 Hypotheses in words must include probability (or chance or proportion or percentage) and the figure 0.78 oe Thus eg H0 : P(patient being cured) = 0.78, H1 : P(patient being cured) > 0.78 gets B0B1B1 Do not allow if H1 wrong Additional notes re Q7 part ii Smallest critical region method: Smallest critical region that 19 could fall into is {19, 20} and has size 0.0461 gets B1, This is > 1% gets M1, B1, A1, E1 as per scheme NB These marks only awarded if 19 used, not other values. Use of k method with no probabilities quoted: This gets zero marks. Use of k method with one probability quoted: Mark as per scheme Line diagram method and Bar chart method No marks unless correct probabilities shown on diagram, then mark as per scheme. Lower tailed test done with H1: p<0.78 Hyp gets max B1B1B0 If compare with 1% give SC2 for P(X ≤ 19) = 1 – 0.0069 = 0.9931 > 1% and SC1 for final conclusion, otherwise give zero. Two-tailed test done with H1: p ≠ 0.78 Provided compare with 0.5% give SC2 for P(X ≥ 19) = 0.0461 > 0.5% and SC1 for final conclusion

19

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GCE

Mathematics (MEI)

Unit 4766: Statistics 1 Advanced Subsidiary GCE


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 2016


3

Annotations and abbreviations

Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in mark scheme

Meaning



4

Subject-specific Marking Instructions for GCE Mathematics (MEI) Statistics strand a Annotations should be used whenever appropriate during your marking.

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


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


5

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



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. 3 significant figures may often be the norm for this, but this always needs to be considered in the context of the problem in hand. For example, in quoting probabilities from Normal tables, we generally expect some evidence of interpolation and so quotation to 4 decimal places will often be appropriate. But even this does not always apply – quotations of the standard critical points for significance tests such as 1.96, 1.645, 2.576 (maybe even 2.58 – but not 2.57) will commonly suffice, especially if the calculated value of a test statistic is nowhere near any of these values. Sensible discretion must be exercised in such cases.


6

Discretion must also be exercised in the case of small variations in the degree of accuracy to which an answer is given. For example, if 3 significant figures are expected (either because of an explicit instruction or because the general context of a problem demands it) but only 2 are given, loss of an accuracy ("A") mark is likely to be appropriate; but if 4 significant figures are given, this should not normally be penalised. Likewise, answers which are slightly deviant from what is expected in a very minor manner (for example a Normal probability given, after an attempt at interpolation, as 0.6418 whereas 0.6417 was expected) should not be penalised. However, answers which are grossly over- or under-specified should normally result in the loss of a mark. This includes cases such as, for example, insistence that the value of a test statistic is (say) 2.128888446667 merely because that is the value that happened to come off the candidate's calculator. Note that this applies to answers that are given as final stages of calculations; intermediate working should usually be carried out, and quoted, to a greater degree of accuracy to avoid the danger of premature approximation. NB See note below scheme 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.





7



8


1 (i) Median = 29.0 B1 Condone wrong method

IQR = 31.8 – 27.8

= 4.0

M1

A1

[3]

For either quartile – allow

alternative definitions of

quartiles

Allow 27.75 and 31.9 leading to

4.15

Do not allow 27.7, 27.9, 31.6,32.0

1 (ii) Lower limit = 27.8 - 1.5 × 4.0 = 21.8

27.75, 31.9 lead to 21.525 and 38.125

27.7, 31.6 lead to 21.85 and 37.45

M1

A1

Method for either

FT sensible quartiles and

IQR

For use of mean (29.44) and SD

(2.516765…)

29.44 ± 2×2.516765 M1

Lower Limit = 24.4 A1

Upper limit = 34.5 A1

So no outliers B1 Upper limit = 31.8 + 1.5 × 4.0 = 37.8 A1 FT sensible quartiles and

IQR

So there are no outliers (at either end of the distribution) B1

[4]

Dep on at least one A1

Use of median scores 0/4

2 (i) P(Does not lose any match) = 0.83 = 0.512 =

64/125 B1

[1]

2 (ii) P(Wins all 3 or draws all 3 or loses all 3) = 0.53 + 0.3

3 + 0.2

3

M1

Including addition

= 0.16 = 4/25 A1

[2]

2 (iii) P(all three outcomes occur) = 3! × 0.5 × 0.3 × 0.2 M1* Allow M1 for k × 0.5 × 0.3 × 0.2

even if k = 1 Even if cubed

= 0.18 A1

Required probability = 1 – ‘0.18’ – 0.16

*M1

dep

Not if cubed

= 0.66 = 33

/50 A1

[4]

CAO

OR:

P(WWW’) + P(DDD’) + P(LLL’) M1 For any one product (no

need for ‘3 ×’)

Even if cubed

3 × 0.52 × 0.5 + 3 × 0.3

2 × 0.7 + 3 × 0.2

2 × 0.8 M1 For ‘3 ×’ Dep on at least 1 correct term

0.375 + 0.189 + 0.096 M1 For sum of three correct

terms (no need for ‘3 ×’)

And no incorrect terms

NB common wrong answer of 0.22

from omitting ‘3 ×’ or 0.44 from

‘2×’ scores M1M0M1A0

Not if cubed

0.66 A1 CAO


9


OR:

P(WWD) + P(WWL) + P(DDW) + P(DDL) + P(LLW) + P(LLD) M1 For any one product (no

need for ‘3 ×’)

Even if cubed

3×0.52×0.3+3×0.5

2×0.2+3×0.3

2×0.5+3×0.3

2×0.2+3×0.2

2×0.5+3×0.2

2×0.3 M1 For ‘3 ×’ Dep on at least 1 correct term

0.225 + 0.15 + 0.135 + 0.054 + 0.06 + 0.036 M1 For sum of six correct

terms (no need for ‘3 ×’)

And no incorrect terms

Not if cubed

0.66 A1 CAO

3 (i) Number of ways = 5! = 120 B1

[1]

3 (ii) Probability = 2/120 M1

For division by their 120 M1 for k/120

= 1/60 or 0.0167 or 0.01 6 A1

[2]

CAO Condone final answer of 2/120

Do not allow 0.016

3 (iii) 1 1 1 1

5 4 3 60 or 0.0167 or 0.01 6

B1

[1]

Condone 2/120 for B1 Do not allow 0.016

3 (iv) 3 2 1 1

5 4 3 10 M1

A1

For 3/5 ×

CAO

Listing options gives 12/120

Or (3! / 120) × 2

Or (3P3×

2P2)/120)

or 1/5C3 = 1/10 M1

A1

[2]

For division by 5C3 SC2 for 3! × their part (iii) or 6 ×

their part (iii)


10


4 (i) k /2 + k /6 + k /12 + k /20 + k /30 = 1

(30 + 10 + 5 + 3 + 2)k/60 = 1

50k = 60

k = 1.2

M1

A1

For correct equation

including = 1

Need one further

intermediate step after

equation

NB Answer Given

Allow substitution of k = 1.2 to

show probabilities add to 1 with

convincing working which must be

more than just

1.2/2+1.2/6+1.2/12+1.2/20+1.2/30

= 1 This latter gets M1A0

r 2 3 4 5 6

P(X = r)

0.6

= 3/5

0.2

= 1/5

0.1

= 1/10

0.06

= 3/50

0.04

= 1/25

B1

[3]

Complete correct table in

fraction or decimal form

NOT in terms of k

Must tabulate probabilities, though

may be seen in part(ii)

If fractions any denominator is ok

provided numerators are integers

4 (ii) E(X) = (2 0.6) (3 0.2) (4 0.1) (5 0.06) (6 0.04)

E(X) = 2.74 or 137

/50

M1

A1

For Σrp (at least 3 terms

correct) Provided 5

reasonable probabilities

seen.

CAO

If probs wrong but sum =

1 allow max

M1A0M1M1A1. If sum ≠

1 allow max

M1A0M1M1A0 (provided

all probabilities ≥ 0 and

<1)

Use of E(X-µ)2 gets M1 for

attempt at (x-µ)2 should see (-

0.74)2, 0.26

2, 1.26

2, 2.26

2, 3.26

2,

(if E(X) wrong FT their E(X)) (all

5 correct for M1), then M1 for

Σp(x-µ)2 (at least 3 terms correct

with their probabilities)

E(X2) = (4 0.6) (9 0.2) (16 0.1) (25 0.06) (36 0.04)

= 8.74 or 437

/50

M1*

For Σr2p (at least 3 terms

correct)

Division by 5 or other spurious

value at end and/or rooting final

answer gives max

M1A1M1M1A0, or

M1A0M1M1A0 if E(X) also

divided by 5.

Var(X) = 8.74 – 2.742 = 1.23 or 1.232 or

3081/2500

M1*

dep

A1

[5]

for – their E( X )²

FT their E(X) provided

Var(X ) > 0

Unsupported correct answers get 5

marks (Probably from calculator)

Condone 1.2324 despite the fact

that this is over-specified since it is

the exact answer


11

0.286

0.21

0.06

R

S

0.66 0.07


5 (i) P(R ∩ S) = P(R) + P(S) – P(R ∪ S)

= 0.28 + 0.87 – 0.94

M1 For correct use of formula Or 0.28 – x + 0.87 – x + x = 0.94

= 0.21 A1

[2]

5 (ii)

G1

G1

G1

[3]

For two labelled

intersecting circles

For at least 2 correct

probabilities.

FT their P(R ∩ S)

For remaining

probabilities. FT their P(R

∩ S)

Allow labels such as P(R) and P(S)

Allow other sensible shapes in

place of circles

Allow their P(R ∩ S) rounded to

2dp

For both G1 marks FT their 0.21

provided < 0.28

For FT if P(R ∩ S) = x then others

are 0.28 – x, 0.87 – x , x – 0.15

0.2436 leads to 0.0364, 0.6264,

0.0936

5 (iii)

( ) 0.21 21( | ) 0.241

( ) 0.87 87

P R SP R S

P S

Exact answer 0.241379…

M1

A1

for fraction

CAO

FT their part (i) (for M1

only) but M0 if their

answer to part (i) is P(R) ×

P(S)

Allow 7 21

or 29 87

as final answer

Allow 0.24 with working

Condone ‘if’ or ‘when’ for ‘given

that’ but not the words ‘and’ or

‘because’ or ‘due to’ for E1.

E1 (independent of M1): the

order/structure must be correct i.e.

no reverse statement

Allow ‘The probability that on a

randomly selected day when there

is at least 1 hour of sun there is at

least 1 mm of rain.’ oe

This is the probability that (on a randomly selected day) there is at

least 1 mm of rain, given that there is at least 1 hour of sun.

E1

[3]

Need more than just

probability of rain given

sun

Must include ‘probability’

or ‘chance’ oe

Do not allow just P(at least

1 mm of rain, given that

there is at least 1 hour of

sun)


12


6 (i) Mean =

(25 147) (45 109) (55 182) (70 317) (140 175)

930

= (£)70.19 or

(£)70.2

M1

A1

For midpoints (at least 3

correct)

(allow 25.005, 45.005 etc

leading to answer 70.20)

CAO


Correct answers obtained

from use of calculator

statistical functions gain

full marks

Condone answer of

(£)70.20

M0A0M0A0 unless using

midpoints

Answer must NOT be left as

improper fraction as this is an

estimate

Accept correct answers for mean

and sd from calculator even if eg

wrong Sxx given

x2f =

2 2 2 2 2(25 147) (45 109) (55 182) (70 317) (140 175)

= 91875 + 220725 + 550550 + 1553300 + 3430000

= 5846450

2652805846450 1264215.161

930xxS or 5846450 – 930×70.19

2

12642151360.83 36.89

929s or (£)36.9

Allow any answer between 36.87 and 36.90 without checking working

M1

A1

[4]

For attempt at xxS

Should include sum of at

least 3 correct multiples fx2

– Ʃx2/n

Do not FT their incorrect

mean for A1


Condone answer of

(£)36.90

If both mean and sd over-

specified, just deduct one

mark

For use of midpoints 25.5, 45.5,

55.5, 70.5, 140.5 allow SC1 for

£70.69 and SC1 for 36.89

If using (x – x )2 method, B2 if

36.9 or better, otherwise B0

Allow use of 70.2 in calculation of

Sxx = 1263372.8 leading to

36.87719…

Condone RMSD of 36.87

(36.86985…) since n is so large

6 (ii) 100/120 × 175 = 145.83

145.83/930 = 0.1568

So 15.7%

M1*

*M1dep

A1

[3]

For 175/120 Or 20/120 × 175 = 29.166 oe

(175 – 29.166)/930

Accept 16% with working


13


6 (iii)

Price Frequency

Group

width

Frequency

density

10 ≤ x ≤ 40 147 30 4.90

40 < x ≤ 50 109 10 10.90

50 < x ≤ 60 182 10 18.20

60 < x ≤ 80 317 20 15.85

80 < x ≤ 200 175 120 1.46

M1

A1

For fds - at least 3 correct

Accept any suitable unit

for fd such as eg freq per

cm.

Allow 15.9 and 1.5 and

condone 1.45

M1 can be also be gained from

freq per 10 – 4.9, 10.9, 18.2,

15.35, 0.146 (at least 3 correct) or

similar.

If fd not explicitly given, M1 A1

can be gained from all heights

correct (within ≤ one square) on

histogram (and M1A0 if at least 3

correct)

NB If not using fd’s only mark available is B1 for width of bars

Heights must be within ≤ 1 square of overlay (only for scales 2cm = 4

units (blue) or 5 units (red)) – otherwise check heights.

Note that you must make sure that the overlay is aligned correctly

with the vertical axis.

B1

B1

B1

[5]

linear scales on both axes

and label on both axes

(Allow horizontal axis

labelled x)

Vertical scale starting

from zero (not broken -

but can get final mark for

heights if broken)

width of bars (within half

a square)

(NO GAPS ALLOWED)

height of bars

Linear scale and label on vertical

axis IN RELATION to first M1

mark ie fd or frequency density or

if relevant freq/10, etc (NOT eg

fd/10).

However allow scale given as

fd×10, or similar

Accept f/w or f/cw (freq/width or

freq/class width)

Can also be gained from an

accurate key

G0 if correct label but not fd’s.

Must have linear scale. Condone

starting at 10 rather than 0.

For inequality labels see additional

notes below.

Height of bars – must be linear

vertical scale.

FT of heights dep on at least 3

heights correct and all must agree

with their fds

If fds not given and 3 or 4 heights

correct then max M1A0G1G1G0


14


6 (iv) Positive skewness B1

[1]

Allow +ve

6 (v) Area for men from 100 to 200 = 100 × 2 = 200 M1 Or 100

/120 × 240

200/990 = 0.202

So 20.2%

A1

20% with working

Cannot be certain as both figures are estimates E1

[3]

Independent Allow comments such as ‘grouped

data so cannot be certain’

or ‘Values are not exact so cannot

be certain’ oe

or ‘midpoints have been used so

cannot be certain’ oe

6 (vi) Men’s running shoes have a lower average price than women’s (as

their mean is only £68.83 compared to £70.19).

Or equivalent for women

Men’s running shoes have a more variation in price than women’s (as

their sd is £42.93 compared to £36.89).

Or equivalent for women

E1

E1

[2]

FT their mean

Do NOT condone lower

central tendency or lower

mean

FT their SD

Allow ‘on the whole’ or similar in

place of ‘average’.

Allow ‘more spread’ or similar but

not ‘higher range’ or ‘higher

variance’ or ‘less distributed’

Condone less consistent

7 (i) (A) X ~ B(16, 0.1)

P(X = 3) = = 0.1423

M1

M1

A1

For 3 130.1 0.9

For 3 13

16

3p q

CAO

With p + q = 1

Also for 560 × 0.000254..

Allow 0.14 or better

Or: From tables

P(X ≤ 3) – P(X ≤ 2) = 0.9316 – 0.7892 = 0.1424

M2

A1

[3]

For 0.9316 – 0.7892

CAO

7 (i) (B)

P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – 0.7892 = 0.2108

M1

A1

[2]

For 0.7892

CAO

If calculating P(X = 0)+ P(X = 1)+

P(X = 2) allow M1 for 0.79 or

better and A1 for 0.21 or better.


15


7 (i) (C) Expected number = 16 × 0.1 = 1.6 B1

[1]

Do not allow final answer of 1 or 2

even if correct 1.6 given earlier

7 (ii) Let p = probability of a randomly chosen person using 1234 as their

PIN (in the population)

H0: p = 0.1

H1: p < 0.1

B1

B1

B1

For definition of p (in

context)

For H0

For H1

Do NOT allow number in place of

probability.

See below for additional notes

The alternative hypothesis has this form as the advertising campaign

aims to reduce the proportion of the population who use 1234 as their

PIN.

B1

[4]

Dep on < 0.1 used in H1

Do Not allow just

‘proportion will be lower’

or similar.

For use of 0.9 as P(do not use

1234), contact team leader.

E0 for simply stating H1 in words

Condone number instead of

proportion.

7 (iii) (A) For n = 20 , P(X 0) = 0.1216

0.1216 > 0.10

M1*

*M1dep

For sight of 0.1216

For > 0.10 or > 10% Do

NOT FT wrong H1

Condone P(X = 0) in place of P(X

≤ 0)

Need to see a comparison with 0.1

or 10% explicitly, not just

mentioning significance level.

So no point in carrying out the test as H0 could not be rejected (even if

nobody in the sample uses 1234 as their PIN). oe

A1

[3]

or state ‘There is no

critical region’ oe

For A1 need P(X 0) or

P(X = 0) somewhere oe

Allow SC2 for clearly indicating

use of B(20, 0.1) but with no

mention of 0.1216 with convincing

reasoning and final answer correct

Allow CR is empty but NOT CR is

zero

(B) Lowest value of k is 13 B1

[1]

Or 13%

7 (iv) P(X ≤ 2) = 0.0530 B1 For use of P(X ≤ 2) only No marks unless H1 correct

If B0 then no further marks


Must include ‘insufficient

evidence to suggest that’ or

something similar i.e. an element

of doubt either in the A or E mark.

0.0530 > 0.05 M1 For comparison of 0.0530

with 5%

So not significant. Do not reject H0

Conclude that there is not enough evidence to support the suggestion

that the advertising campaign has been successful.

Reminder: When you mark this question part, if you ‘fit to height’

you can check the last page for working or mark it BP if there none

A1*

*E1

dep

[4]

Also allow P(X ≤ 2) >

0.05, (P(X ≤ 1) < 0.05) so

CR is {0, 1} for first two

marks then A1E1 as usual

Condone ‘number of

people’ in conclusion


16

NOTE RE OVER-SPECIFICATION OF ANSWERS

If answers are grossly over-specified, deduct the final answer mark. Note in Q6i only deduct 1 mark even if both mean and sd over-specified. Probabilities should

also be rounded to a sensible degree of accuracy. In general final non probability answers should not be given to more than 4 significant figures. Allow probabilities

given to 5 sig fig.

ANNOTATION RULES

See note 12 above and particularly 12a. Remember to put full annotation on all practice and standardisation scripts unless the candidate has scored full marks or

zero. In addition for all marking in Q6(iii), if the candidate has not scored full marks then show which B marks have been awarded in the right hand margin, in the

same order as they are given in the mark scheme. You should indicate any errors made. Additional notes re Q6 part iii G mark

10 ≤ x < 40 ≤ x < 50 ≤ x < 60 SCORES G1

10 ≤ x < 40 40 ≤ x <50 50 ≤ x < 60 SCORES G0

BUT

30 10 ≤ x < 40 40 40≤ x<50 50 50≤x<60 60 SCORES G1


17

Additional notes re Q7 part ii Minimum needed for B1 is p = probability of using 1234.

Allow p = P(using 1234)

Definition of p must include word probability (or chance or proportion or percentage or likelihood but NOT possibility, number or amount).

Preferably given as a separate comment. However can be at end of H0 as long as it is a clear definition ‘p = the probability of using 1234.’

Do NOT allow ‘p = the probability of using 1234 is different’

Allow p=10%, allow only p or θ or π or ρ. However allow any single symbol if defined (including x)

Allow H0 = p=0.1, Allow H0 : p=1/10

Allow NH and AH in place of H0 and H1

Do not allow H0 : P(X=x) = 0.1

Do not allow H0: =0.1, =10%, P(0.1), p(x)=0.1, x=0.1 (unless x correctly defined as a probability)

Do not allow H0 and H1 reversed

For hypotheses given in words allow Maximum B0B1B1

Hypotheses in words must include probability (or chance or proportion or percentage) and the figure 0.1 oe

Thus eg H0 : P(using 1234) = 0.1, H1 : P(using 1234) < 0.1 gets B0B1B1

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Documents

Mark Scheme 4766 June 2005 - Mathematics in …mei.org.uk/files/papers/All packs/MS_MEI_S1.pdfMark Scheme 4766 January 2006. Q 1 (i) The range = 55 – 15 = 40 The interquartile range