A Level Mathematics - Weebly · 2020-02-22 · A Level Mathematics Sample Assessment Materials DRAFT Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) First teaching from

A Level Mathematics

Sample Assessment Materials DRAFTPearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0)First teaching from September 2017First certification from 2018

This draft qualification has not yet been accredited by Ofqual. It is published to enable teachers to have early sight of our proposed approach to Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0). Further changes may be required and no assurance can be given at this time that the proposed qualification will be made available in its current form, or that it will be accredited in time for first teaching in September 2017 and first award in 2018.

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Edexcel, BTEC and LCCI qualifications Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding body offering academic and vocational qualifications that are globally recognised and benchmarked. For further information, please visit our qualifications website at qualifications.pearson.com. Alternatively, you can get in touch with us using the details on our contact us page at qualifications.pearson.com/contactus

About Pearson

Pearson is the world's leading learning company, with 35,000 employees in more than 70 countries working to help people of all ages to make measurable progress in their lives through learning. We put the learner at the centre of everything we do, because wherever learning flourishes, so do people. Find out more about how we can help you and your learners at qualifications.pearson.com

References to third party material made in this sample assessment materials are made in good faith. Pearson does not endorse, approve or accept responsibility for the content of materials, which may be subject to change, or any opinions expressed therein. (Material may include textbooks, journals, magazines and other publications and websites.)

All information in this document is correct at time of publication.

Original origami artwork: Mark Bolitho Origami photography: Pearson Education Ltd/Naki Kouyioumtzis

ISBN 978 1 4469 3344 2

All the material in this publication is copyright

© Pearson Education Limited 2017

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Contents

Introduction 1

General marking guidance 3

Paper 1 – sample question paper and mark scheme 5



Mathematical formulae and statistical tables

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Introduction

The Pearson Edexcel Level 3 Advanced GCE in Mathematics is designed for use in schools and colleges. It is part of a suite of AS/A Level qualifications offered by Pearson.

These sample assessment materials have been developed to support this qualification and will be used as the benchmark to develop the assessment students will take.

Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample assessment materials (SAMs) –Draft 1.2 – January 2017 © Pearson Education Limited 2017

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General marking guidance

All candidates must receive the same treatment. Examiners must mark the last candidate in exactly the same way as they mark the first.

Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than be penalised for omissions.

Examiners should mark according to the mark scheme – not according to their perception of where the grade boundaries may lie.

All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme.

Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification/indicative content will not be exhaustive. However different examples of responses will be provided at standardisation.

When examiners are in doubt regarding the application of the mark scheme to a candidate’s response, a senior examiner must be consulted before a mark is given.

Crossed-out work should be marked unless the candidate has replaced it with an alternative response.

Specific guidance for mathematics

1. These mark schemes use the following types of marks:

M marks: Method marks are awarded for ‘knowing a method and attempting to apply it’, unless otherwise indicated.

A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned.

B marks are unconditional accuracy marks (independent of M marks)

Marks should not be subdivided.

2. Abbreviations

These are some of the traditional marking abbreviations that may appear in the mark schemes.

bod benefit of doubt

ft follow through

this symbol is used for correct ft

cao correct answer only

cso correct solution only. There must be no errors in this part of the question to obtain this mark

isw ignore subsequent working

awrt answers which round to

SC: special case

o.e. or equivalent (and appropriate)

d… dependent or dep

indep independent

dp decimal places

sf significant figures

The answer is printed on the paper or ag- answer given


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or d… The second mark is dependent on gaining the first mark

3. All M marks are follow through.

All A marks are ‘correct answer only’ (cao.), unless shown, for example, as A1 ft toindicate that previous wrong working is to be followed through. After a misreadhowever, the subsequent A marks affected are treated as A ft, but answers that don’tlogically make sense e.g. if an answer given for a probability is >1 or <0, shouldnever be awarded A marks.

4. For misreading which does not alter the character of a question or materially simplifyit, deduct two from any A or B marks gained, in that part of the question affected.

5. Where a candidate has made multiple responses and indicates which response theywish to submit, examiners should mark this response. If there are several attempts ata question which have not been crossed out, examiners should mark the final answerwhich is the answer that is the most complete.

6. Ignore wrong working or incorrect statements following a correct answer.

7. Mark schemes will firstly show the solution judged to be the most common responseexpected from candidates. Where appropriate, alternative answers are provided in thenotes. If examiners are not sure if an answer is acceptable, they will check the markscheme to see if an alternative answer is given for the method used. If no suchalternative answer is provided but deemed to be valid, examiners must escalate theresponse to a senior examiner to review.


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Pearson Edexcel Level 3 GCE

Mathematics Advanced Paper 1: Pure Mathematics 1 Sample assessment material for first teaching September 2017 Time: 2 hours

Paper Reference(s)

9MA0/01

You must have: Mathematical Formulae and Statistical Tables Calculator

Candidates may use any calculator permitted by Pearson regulations. Calculators must not have the facility for algebraic manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them.

Instructions Use black ink or ball-point pen. If pencil is used for diagrams/sketches/graphs it must be dark (HB or B). Fill in the boxes at the top of this page with your name, centre

number and candidate number. Answer all questions and ensure that your answers to parts of

questions are clearly labelled. Answer the questions in the spaces provided

– there may be more space than you need. You should show sufficient working to make your methods clear.

Answers without working may not gain full credit. Inexact answers should be given to three significant figures unless

otherwise stated. Information A booklet ‘Mathematical Formulae and Statistical Tables’ is provided. There are 15 questions in this question paper. The total mark for this

paper is 100. The marks for each question are shown in brackets

– use this as a guide as to how much time to spend on each question. Advice Read each question carefully before you start to answer it. Try to answer every question. Check your answers if you have time at the end. If you change your mind about an answer cross it out and put your

new answer and any working out underneath.


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Answer ALL questions. Write your answers in the spaces provided.

1. The curve C has equation

4 33 8 3y x x

(a) Find (i) d

d

y

x

(ii) 2

2

d

d

y

x

(3)

(b) Verify that C has a stationary point when 2x

(2)

(c) Determine the nature of this stationary point, giving a reason for your answer.

(2)

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Question 1 continued

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(Total for Question 1 is 7 marks)


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2.

The shape ABCDOA, as shown in Figure 1, consists of a sector COD of a circle centre O joined to a sector AOB of a different circle, also centre O.

Given that arc length CD = 3 cm, 0.4COD radians and AOD is a straight line of length 12 cm,

(a) find the length of OD, (2)

(b) find the area of the shaded sector AOB. (3)

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3. A circle C has equation

2 2 4 10x y x y k

where k is a constant.

(a) Find the coordinates of the centre of C. (2)

(b) State the range of possible values for k. (2)

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4. Given that a is a positive constant and

2

1d ln 7

a

a

tt

t

show that ln ,a k where k is a constant to be found. (4)

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5. A curve C has parametric equations

3 2 1, 4 7 + , 0x t y t t

t

Show that the Cartesian equation of the curve C can be written in the form

y = 22

1

x ax b

x

, 1x

where a and b are integers to be found. (3)

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6. Jess starts working for a company.

In year 1 her annual salary will be £17 000

In year 11 her annual salary is predicted to be £32 000

Model A assumes that her salary each year will form the terms of an arithmetic sequence.

(a) According to model A, determine Jess's salary in year 2.

(2)

Model B assumes that her salary each year will form the terms of a geometric sequence.

(b) According to model B, determine Jess's salary in year 2.

(3)

(c) If Jess is to remain with the company for more than 20 years, which model would predict the greater value for the sum of all of her annual salaries. Give reasons for your answer.

(2)

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7.

Figure 2

Figure 2 shows a sketch of a triangle ABC.

Given =2 +3 +AB i j k

and = 9 +3 ,BC i j k

show that 105.9BAC to one decimal place. (5)

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8.

2f ( ) ln(2 5) 2 30, 2.5x x x x

(a) Show that f ( ) 0x has a root in the interval 3.5,4

(2)

A student takes 4 as the first approximation to .

Given f (4) 3.099 and f (4) 16.67 to 4 significant figures,

(b) apply the Newton-Raphson procedure once to obtain a second approximation for α, giving your answer to 3 significant figures.

(2)

(c) Show that is the only root of f ( ) 0x (2)

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9. (a) Prove that

tan cot 2cosec2 , ,2

nn

(4)

(b) Hence explain why the equation

tan cot 1

does not have any real solutions. (1)

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10. Given that is measured in radians, prove, from first principles, that the derivative of sin is cos

You may assume the formula for sin( )A B and that as 0,h sin

1h

h and

cos 10

h

h

(5)

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11. An archer shoots an arrow.

The height, H metres, of the arrow above the ground is modelled by the formula

21.8 0.4 0.002 , 0H d d d

where d is the horizontal distance of the arrow from the archer, measured in metres.

Given that the arrow travels in a vertical plane until it hits the ground,

(a) find the horizontal distance travelled by the arrow, as given by this model. (3)

(b) With reference to the model, interpret the significance of the constant 1.8 in the formula.

(1)

(c) Write 21.8 0.4 0.002d d in the form

2( )A B d C

where A, B and C are constants to be found. (3)

It is decided that the model should be adapted for a different archer.

The adapted formula for this archer is 22.1 0.4 0.002 , 0H d d d

Hence or otherwise, find, for the adapted model

(d) (i) the maximum height of the arrow above the ground.

(ii) the horizontal distance, from the archer, of the arrow when it is at its maximum height.

(2)

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Total for Question 11 is 9 marks)


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12. In a controlled experiment, the number of microbes, N, present in a culture T days after the start of the experiment were counted.

N and T are expected to satisfy a relationship of the form

,bN aT where a and b are constants

(a) Show that this relationship can be expressed in the form

10 10log logN m T c

giving m and c in terms of the constants a and/or b. (2)

Figure 3

Figure 3 shows the line of best fit for values of 10

log N plotted against values of 10

log T

(b) Use the information provided to estimate the number of microbes present in the culture 3 days after the start of the experiment.

(4)

(c) Explain why the information provided could not reliably be used to estimate the day when the number of microbes in the culture first exceeds 1 000 000.

(2)

(d) With reference to the model, interpret the value of the constant a. (1)


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13. The curve C has parametric equations

x = 2cos t, y = 3 cos 2 ,t 0 t π

(a) Find an expression for d

d

y

x in terms of t.

(2)

The point P lies on C where 2

3t

.

The line l is the normal to C at P.

(b) Show that an equation for l is

2 2 3 1 0x y

(5)

The line l intersects the curve C again at the point Q.

(c) Find the exact coordinates of Q.

You must show clearly how you obtained your answers. (6)

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29

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14.

Figure 4

Figure 4 shows a sketch of part of the curve C with equation

2 ln2 5, 0

3

x xy x x

The finite region S, shown shaded in Figure 4, is bounded by the curve C, the line with equation x = 1, the x-axis and the line with equation x = 3

The table below shows corresponding values of x and y with the values of y given to 4 decimal places as appropriate.

x 1 1.5 2 2.5 3

y 3 2.3041 1.9242 1.9089 2.2958

(a) Use the trapezium rule, with all the values of y in the table, to obtain an estimate for the area of S, giving your answer to 3 decimal places.

(3)

(b) Explain how the trapezium rule could be used to obtain a more accurate estimate for the area of S.

(1)

(c) Show that the exact area of S can be written in the form ln ,a

cb where a, b and c are

integers to be found.

(In part (c) solutions based entirely on graphical or numerical methods are not

acceptable.) (6)


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33

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15.

Figure 5

Figure 5 shows a sketch of the curve with equation f ( ),y x where

2 1

4sin 2f ( ) , 0

e x

xx x

The curve has a maximum turning point at P and a minimum turning point at Q as shown in Figure 5.

(a) Show that the x coordinates of point P and point Q are solutions of the equation

tan 2 2x (4)

(b) Using your answer to part (a), find the x-coordinate of the minimum turning point on the curve with equation

(i) f (2 ),y x

(ii) 3 2f ( ).y x (4)

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TOTAL FOR PAPER IS 100 MARKS


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Paper 1: Pure Mathematics 1 Mark Scheme

Question Scheme Marks AOs

1(a) (i) 3 2d

12 24d

yx x

x

M1

A1

1.1b

1.1b

(ii) 2

22

d36 48

d

yx x

x A1ft 1.1b

(3)

(b) Substitutes 2x into their 3 2d

12 2 24 2d

y

x M1 1.1b

Shows d

0d

y

x and states ''hence there is a stationary point'' A1 2.1

(2)

(c)

Substitutes 2x into their2

22

d36 2 48 2

d

y

x M1 1.1b

2

2

d48 0

d

y

x and states ''hence the stationary point is a minimum'' A1ft 2.2a

(2)

(7 marks)

Notes: (a)(i) M1: Differentiates to a cubic form

A1: 3 2d12 24

d

yx x

x

(a)(ii)

A1ft: Achieves a correct 2

2

d

d

y

xfor their 2d

36 48d

yx x

x

(b)

M1: Substitutes 2x into their d

d

y

x

A1: Shows d

0d

y

x and states ''hence there is a stationary point'' All aspects of the proof must be correct

(c)

M1: Substitutes 2x into their2

2

d

d

y

x

Alternatively calculates the gradient of C either side of 2x A1ft: For a correct calculation, a valid reason and a correct conclusion.

Follow through on an incorrect 2

2

d

d

y

x


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2(a) Uses 3 0.4s r r M1 1.2

7.5 cmOD A1 1.1b

(2)

(b) Uses angle 0.4AOB or uses radius is 12 '7.5' cm M1 3.1a

Uses area of sector 2 21 1(12 7.5) ( 0.4)

2 2r M1 1.1b

227.8cm A1ft 1.1b

(3)

(5 marks)

Notes: (a) M1: Attempts to use the correct formula s r with 3s and 0.4 A1: 7.5 cmOD (An answer of 7.5cm implies the use of a correct formula and scores both marks)

(b)

M1: 0.4AOB may be implied by the use of AOB = awrt 2.74 or uses radius is 12 their '7.5'

M1: Follow through on their radius (12 their OD ) and their angle

A1ft: Allow awrt 27.8 cm2. (Answer 27.75862562). Follow through on their 12 their '7.5'

Note: Do not follow through on a radius that is negative.


3.(a)

Attempts 2 22 5 ....x y M1 1.1b

Centre 2, 5 A1 1.1b

[2]

(b) Sets 2 22 5 0k M1 2.2a

29k A1ft 1.1b

[2]

(4 marks)

Notes: (a)

M1: Attempts to complete the square so allow 2 22 5 ....x y

A1: States the centre is at 2, 5 . Also allow written separately 2, 5x y 2, 5 implies both

marks (b)

M1: Deduces that the right hand side of their 2 2... ... ...x y is > 0 or 0

A1ft: 29k Also allow 29k Follow through on their rhs of 2 2... ... ...x y


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4

Writes 1 1

d 1 dt

t tt t

and attempts to integrate M1 2.1

lnt t c M1 1.1b

2 ln 2 ln ln 7a a a a M1 1.1b

7

ln2

a with 7

2k A1 1.1b

(4 marks)

Notes: M1: Attempts to divide each term by t or alternatively multiply each term by t -1

M1: Integrates each term and knows 1

d ln .t tt

The + c is not required for this mark

M1: Substitutes in both limits, subtracts and sets equal to ln7

A1: Proceeds to 7

ln2

a and states 7

2k or exact equivalent such as 3.5


5 Attempts to substitute

1

2

xt

into

1 6 4 7

2 ( 1)

xy y

x

M1 2.1

Attempts to write as a single fraction (2 5)( 1) 6

( 1)

x xy

x

M1 2.1

22 3 1

3, 11

x xy a b

x

A1 1.1b

(3 marks)

Notes:

M1: Score for an attempt at substituting 1

2

xt

or equivalent into

34 7 +y t

t

M1: Award this for an attempt at a single fraction with a correct common denominator.

Their 1

4 7 2

x

term may be simplified first

A1: Correct answer only 22 3 1

3, 11

x xy a b

x


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6 (a) Attempts 32000 17000 10 (1500)d d M1 3.3

Salary in Year 2 = £18 500 A1 1.1b

(2)

(b) Attempts 1032000 17000 (1.065)r r M1 3.3

Attempts 17000 ' 'r M1 1.1b

Salary in Year 2 = £18 110 A1 1.1b

(3)

(c) Model A predicts that Jess's salary rises by the same amount each year Model B predicts that Jess's salary rises by a greater amount each year

B1 ft 3.4

Long term, the annual rise under model B will become much greater than £1500 and so will predict the greater total salary for model B.

B1 2.4

(2)

(7 marks)

Notes: (a) M1: Attempts 32000 17000 10d and proceeds as far as a value for d A1: £18 500. An answer of £18 500 without any working would score both marks (b)

M1: Attempts 1032000 17000 r and proceeds as far as a value for r

M1: Attempts 17000 their ' 'r but do not accept a value of 1r as this would not represent growth A1: awrt £18 110 (c) B1 ft: Makes a valid statement that compares the annual growth in salary of the two models. Allow the candidates to compare the salaries for years say 20, 21 and 22 following through on reasonable values for r ( 0 1.1r ). Allow the candidates to compare the sum of salaries for N years ( 20N ) under both models. Allow candidates to graph the annual salaries under each model

B1: Explains why model B would give a greater total for the sum of the annual salaries. For example states that the salary increase for model A stays the same whereas for model B it increases each year. Hence the total salary under model B will eventually get larger and larger compared to the total salary under model A.

Annual

Salary

Years

Model B

Model A


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7 Attempts

2 3 9 3 3 6 4A C A B B C i j k i j k i j k

M1 3.1a

Attempts to find any one length using 3-d Pythagoras M1 2.1

Finds all of 14, 61, 91AB AC BC A1ft 1.1b

14 61 91cos

2 14 61BAC

M1 2.1

angle 105.9BAC * A1* 1.1b

(5)

(5 marks)

Notes:

M1: Attempts to find A C

by using AC AB BC

M1: Attempts to find any one length by use of Pythagoras' Theorem

A1ft: Finds all three lengths in the triangle. Follow through on their AC

M1: Attempts to find BAC using 2 2 2

cos2

AB AC BCBAC

AB AC

Allow this to be scored for other methods such as .cos

AB ACBAC

AB AC

A1*: This is a show that and all aspects must be correct. Angle 105.9BAC


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8 (a) f (3.5) 4.8 , f (4) ( )3.1 M1 1.1b

Change of sign and function continuous in interval 3.5,4 Root * A1* 2.4

(2)

(b) Attempts 0

1 00

f ( )

f ( )

xx x

x

1

3.0994

16.67x M1 1.1b

1 3.81x A1 1.1b

(2)

(c)

Attempts to sketch both ln(2 5)y x and 230 2y x

M1 3.1

States that ln(2 5)y x meets 230 2y x in just one place ,

therefore 2ln(2 5) 30 2x x has just one root f ( ) 0x has just

one root

A1 2.4

(2)

(6 marks)

Notes: (a) M1: Attempts f ( )x at both 3.5x and 4x with at least one correct to 1 significant figure

A1*: f (3.5) and f (4) correct to 1 sig figure (rounded or truncated) with a correct reason and

conclusion. A reason could be change of sign, or f (3.5) f (4) 0 or similar with f ( )x being continuous

in this interval. A conclusion could be 'Hence root' or 'Therefore root in interval' (b)

M1: Attempts 01 0

0

f ( )

f ( )

xx x

x

evidenced by 1

3.0994

16.67x

A1: Correct answer only 1 3.81x

(c) M1: For a valid attempt at showing that there is only one root. This can be achieved by

Sketching graphs of ln(2 5)y x and 230 2y x on the same axes

Showing that 2f ( ) ln(2 5) 2 30x x x has no turning points

Sketching a graph of 2f ( ) ln(2 5) 2 30x x x

A1: Scored for correct conclusion

ln(2 5)y x

230 2y x


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9(a)

sin costan cot

cos sin

M1 2.1

2 2sin cos

sin cos

A1 1.1b

1

1 sin 22 M1 2.1

2cosec2 * A1* 1.1b

(4)

(b) States tan cot 1 sin 2 2

AND no real solutions as 1 sin 2 1 B1 2.4

(1)

(5 marks)

Notes: (a)

M1: Writes sin

tancos

and cos

cotsin

A1: Achieves a correct intermediate answer of 2 2sin cos

sin cos

M1: Uses the double angle formula sin 2 2sin cos A1*: Completes proof with no errors. This is a given answer. Note: There are many alternative methods. For example

1tan cot tan

tan

2 2

2

tan 1 sec 1 1sintan tan cos sincos cos

then as the

main scheme. (b) B1: Scored for sight of sin 2 2 and a reason as to why this equation has no real solutions.

Possible reasons could be 1 sin 2 1 ........and therefore sin 2 2

Or sin 2 2 2 arcsin 2 which has no answers as 1 sin 2 1


43

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10 Use of

sin( ) sin

( )

h

h

B1 2.1

Uses the compound angle identity for sin( )A B with

,A B h sin( ) sin cos cos sinh h+ h M1 1.1b

Achieves sin( ) sin sin cos cos sin sinh h+ h

h h

A1 1.1b

sin cos 1

cos sinh h

h h

M1 2.1

Uses 0,h sin

1h

h and

cos 10

h

h

Hence the 0

sin( ) sinlimit cos

( )h

h

h

and the gradient

of the chord gradient of the curve d

cosd

y

*

A1* 2.5

(5 marks)

Notes:

B1: States or implies that the gradient of the chord is sin( ) sinh

h

or similar such as

sin( ) sin

for a small h or

M1: Uses the compound angle identity for sin( )A B with ,A B h or

A1: Obtains sin cos cos sin sinh+ h

h

or equivalent

M1: Writes their expression in terms of sin h

hand

cos 1h

h

A1*: Uses correct language to explain that d

cosd

y

For this method they should use all of the given statements 0,h sin

1h

h

cos 10

h

h

meaning that the 0

sin( ) sinlimit cos

( )h

h

h

and therefore the gradient of the chord gradient of the curve d

cosd

y


44

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10alt Use of

sin( ) sin

( )

h

h

B1 2.1

Sets

sin sinsin( ) sin 2 2 2 2

( )

h h h hh

h h

and uses the compound angle identity for sin( )A B and sin( )A B

with ,2 2

h hA B

M1 1.1b

Achieves sin( ) sinh

h

sin cos cos sin sin cos cos sin2 2 2 2 2 2 2 2h h h h h h h h

+

h

A1 1.1b

sin2

cos2

2

hh

h

M1 2.1

Uses 0, 02

hh hence

sin2

1

2

h

h

and cos cos

2

h

Therefore the 0

sin( ) sinlimit cos

( )h

h

h

and the gradient of

the chord gradient of the curve d

cosd

y

*

A1* 2.5

(5 marks)

Additional notes:

A1*: Uses correct language to explain that d

cosd

y

. For this method they should use the (adapted)

given statement 0, 02

hh hence

sin2

1

2

h

h

with cos cos

2

h

meaning that the

0

sin( ) sinlimit cos

( )h

h

h

and therefore the gradient of the chord gradient of the curve

dcos

d

y


45

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11(a) Sets 20 1.8 0.4 0.002 0H d d M1 3.4

Solves using an appropriate method, for example

20.4 0.4 4 0.002 1.8

2 0.002d

dM1 1.1b

Distance awrt 204 m only A1 2.2a

(3)

(b) States the initial height of the arrow above the ground. B1 3.4

(1)

(c) 2 21.8 0.4 0.002 0.002 200 1.8d d d d M1 1.1b

20.002 ( 100) 10000 1.8d M1 1.1b

221.8 0.002( 100)d A1 1.1b

(3)

(d) (i) 22.1 metres B1ft 3.4

(ii) 100 metres B1ft 3.4

(2)

(9 marks)

Notes: (a) M1: Sets 20 1.8 0.4 0.002 0H d d M1: Solves using formula, which if stated must be correct, by completing square (look for

2100 10900 ..d d ) or even allow answers coming from a graphical calculator

A1: Awrt 204 m only (b) B1: States it is the initial height of the arrow above the ground. Do not allow '' it is the height of the archer'' (c)

M1: Score for taking out a common factor of 0.002 from at least the 2d and d terms

M1: For completing the square for their 2 200d d term

A1: 221.8 0.002( 100)d or exact equivalent

(d)

B1ft: For their '21.8+0.3' =22.1m

B1ft: For their 100m


46

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12 (a) 10 10 10log log logb bN aT N a T M1 2.1

10 10 10log log logN a b T so 10logandm b c a A1 1.1b

(2)

(b) Uses the graph to find either a or b intercept10a or b = gradient M1 3.1b

Uses the graph to find both a and b intercept10a and b = gradient M1 1.1b

Uses 3T in bN aT with their a and b M1 3.1b

Number of microbes 800 A1 1.1b

(4)

(c) 101000000 log 6N N M1 3.4

We cannot ‘extrapolate’ the graph and assume that the model still holds

A1 3.5b

(2)

(d) States that 'a' is the number of microbes 1 day after the start of the experiment

B1 3.2a

(1)

(9 marks)


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Notes: (a) M1: Takes logs of both sides and shows the addition law

M1: Uses the power law, writes 10 10 10log log logN a b T and states 10logandm b c a

(b)

M1: Uses the graph to find either a or b intercept10a or b = gradient. This would be implied by the sight

of 1.82.3 10 63b a or

M1: Uses the graph to find both a and b intercept10a and b = gradient. This would be implied by the sight

of 1.82.3 10 63b a and

M1: Uses 3 bT N aT with their a and b. This is implied by an attempt at 2.363 3

A1: Accept a number of microbes that are approximately 800. Allow 800 150 following correct work. There is an alternative to this using a graphical approach. M1: Finds the value of

10log T from T =3. Accept as 103 log 0.48T T .

M1: Then using the line of best fit finds the value of 10

log N from their ''0.48'' Accept 10log 2.9N .

M1: Finds the value of N from their value of 10

log N 10log 2.9N '2.9 '10N

A1: Accept a number of microbes that are approximately 800. Allow 800 150 following correct work. (c)

M1 For using 1000000N and stating that 10log 6N

A1: Statement to the effect that ''we only have information for values of log N between 1.8 and 4.5 so we

cannot be certain that the relationship still holds''. ''We cannot extrapolate with any certainty, we could only interpolate'' There is an alternative approach that uses the formula.

M1: Use 1000000N in their 2.363N T 10

10

1000000log

63log 1.83

2.3T

.

A1: The reason would be similar to the main scheme as we only have 10log T values from 0 to 1.2. We

cannot ‘extrapolate’ the graph and assume that the model still holds (d)

B1: Allow a numerical explanation 1 1bT N a N a giving a is the value of N at T =1


48

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13(a)

Attempts

dd d

ddd

yy t

xxt

M1 1.1b

d 3 sin 22 3 cos

d sin

y tt

x t A1 1.1b

(2)

(b) Substitutes

2

3t

in d 3 sin 2

3d sin

y t

x t M1 2.1

Uses gradient of normal = 1 1

d 3d

yx

M1 2.1

Coordinates of P = 3

1,2

B1 1.1b

Correct form of normal 3 11

2 3y x M1 2.1

Completes proof 2 2 3 1 0x y * A1* 1.1b

(5)

(c) Substitutes x = 2cos t and y = 3 cos 2t into 2 2 3 1 0x y M1 3.1a

Uses the identity 2cos 2 2cos 1t t to produce a quadratic in cos t

M1 3.1a

212cos 4cos 5 0t t A1 1.1b

Finds 5 1

cos ,6 2

t M1 2.4

Substitutes their 5

cos6

t into x = 2cos t, y = 3 cos 2t , M1 1.1b

5 7

, 33 18

Q

A1 1.1b

(6)

(13 marks)


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Notes: (a)

M1: Attempts

dd d

ddd

yy t

xxt

and achieves a form sin 2

sin

tk

t Alternatively candidates may apply the double

angle identity for cos2t and achieve a form sin cos

sin

t tk

t

A1: Scored for a correct answer, either 3 sin 2

sin

t

t or 2 3 cos t

(b)

M1: For substituting 2

3t

in their

d

d

y

xwhich must be in terms of t

M1: Uses the gradient of the normal is the negative reciprocal of the value of d

d

y

x. This may be seen in the

equation of l.

B1: States or uses (in their tangent or normal) that P = 3

1,2

M1: Uses their numerical value of d

1dyx

with their 3

1,2

to form an equation of the normal at P

A1*: This is a proof and all aspects need to be correct. Correct answer only 2 2 3 1 0x y

(c)

M1: For substituting x = 2cos t and y = 3 cos 2t into 2 2 3 1 0x y to produce an equation in t.

Alternatively candidates could use 2cos 2 2cos 1t t to set up an equation of the form 2y Ax B .

M1: Uses the identity 2cos 2 2cos 1t t to produce a quadratic equation in cos t

In the alternative method it is for combining their 2y Ax B with 2 2 3 1 0x y to get an equation in

just one variable

A1: For the correct quadratic equation 212cos 4cos 5 0t t

Alternatively the equations in x and y are 23 2 5 0x x 212 3 4 7 3 0y y

M1: Solves the quadratic equation in cos t (or x or y) and rejects the value corresponding to P.

M1: Substitutes their 5

cos6

t or their 5

arccos6

t

in x = 2cos t and y = 3 cos 2t

If a value of x or y has been found it is for finding the other coordinate.

A1: 5 7

, 33 18

Q

. Allow 5 7

, 33 18

x y but do not allow decimal equivalents.


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14(a) Uses or implies h = 0.5 B1 1.1b

For correct form of the trapezium rule = M1 1.1b

0.53 2.2958 2 2.304 1.92421 1.9089 4.393

2 A1 1.1b

(3)

(b) Any valid statement reason, for example

Increase the number of strips Decrease the width of the strips Use more trapezia

B1 2.4

(1)

(c) For integration by parts on 2 ln dx x x M1 2.1

3 2

ln d3 3

x xx x A1 1.1b

22 5 d 5x x x x c B1 1.1b

All integration attempted and limits used

Area of S =

3 32 3 32

11

ln2 5 d ln 5

3 9 27

x

x

x x x xx x x x x

M1 2.1

Uses correct ln laws, simplifies and writes in required form M1 2.1

Area of S = 28

ln 2727

( 28, 27, 27)a b c A1 1.1b

(6)

(10 marks)

Notes:


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Notes: (a)

B1: States or uses the strip width h = 0.5. This can be implied by the sight of 0.5...

2 in the trapezium rule

M1: For the correct form of the bracket in the trapezium rule. Must be y values rather than x values

first value last value 2 sum of other valuesy y y

A1: 4.393 (b) B1: See scheme (c)

M1: Uses integration by parts the right way around. Look for 2 ln dx x x 3 2ln dAx x Bx x

A1: 3 2

ln d3 3

x xx x

B1: Integrates the 2 5x term correctly 2 5x x M1: All integration completed and limits used

M1: Simplifies using ln law(s) to a form lna

cb

A1: Correct answer only 28

ln 2727


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15(a) Attempts to differentiate using the quotient rule or otherwise M1 2.1

2 1 2 1

22 1

e 8cos 2 4sin 2 2ef ( )

e

x x

x

x xx

A1 1.1b

Sets f ( ) 0x and divides/ factorises out the 2 1e x terms M1 2.1

Proceeds via sin 2 8

cos2 4 2

x

x to tan 2 2x * A1* 1.1b

(4)

(b) (i) Solves tan 4 2x and attempts to find the 2nd solution M1 3.1a

1.02x A1 1.1b

(ii) Solves tan 2 2x and attempts to find the 1st solution M1 3.1a

0.478x A1 1.1b

(4)

(8 marks)

Notes: (a)

M1: Attempts to differentiate by using the quotient rule with 4sin 2u x and 2 1e xv or alternatively

uses the product rule with 4sin 2u x and 1 2e xv

A1: For achieving a correct f ( )x . For the product rule 1 2 1 2f ( ) e 8cos 2 4sin 2 2ex xx x x

M1: This is scored for cancelling/ factorising out the exponential term. Look for an equation in just cos 2x

and sin 2x

A1*: Proceeds to tan 2 2x . This is a given answer. (b) (i)

M1: Solves tan 4 2x attempts to find the 2nd solution. Look for arctan 2

4x

Alternatively finds the 2nd solution of tan 2 2x and attempts to divide by 2

A1: Allow awrt 1.02x . The correct answer, with no incorrect working scores both marks (b)(ii)

M1: Solves tan 2 2x attempts to find the 1st solution. Look for arctan 2

2x

A1: Allow awrt 0.478x . The correct answer, with no incorrect working scores both marks


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Mathematics Advanced Paper 2: Pure Mathematics 2 Sample assessment material for first teaching September 2017 Time: 2 hours

Paper Reference(s)

9MA0/02

You must have: Mathematical Formulae and Statistical Tables Calculator


Instructions Use black ink or ball-point pen. If pencil is used for diagrams/sketches/graphs it must be dark (HB or B). Fill in the boxes at the top of this page with your name, centre

number and candidate number. Answer all questions and ensure that your answers to parts of

questions are clearly labelled. Answer the questions in the spaces provided


Answers without working may not gain full credit. Inexact answers should be given to three significant figures unless

otherwise stated. Information A booklet ‘Mathematical Formulae and Statistical Tables’ is provided. There are 16 questions in this question paper. The total mark for this





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1. 3 2f ( ) 2 5x x x ax a

Given that 2x is a factor of f ( ),x find the value of the constant a.

(3)

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2. Some A level students were given the following question.

Solve, for 90 90 , the equation

cos 2sin

The attempts of two of the students are shown below.

(a) Identify the error made by student A. (1)

Student B gives 26.6 as one of the answers to cos 2sin .

(b) (i) Explain why this answer is incorrect.

(ii) Explain how this incorrect answer arose. (2)

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Student

cos 2sin

cos2

sintan 2

63.4

A

2 2

2 2

2

Student

cos 2sin

cos 4sin

1 sin 4sin

1sin

51

sin5

26.6

B


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3. Given 4(2 1) ,y x x show that

d2 1

d

nyx Ax B

x

where n, A and B are constants to be found. (4)

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4. Given f ( ) e ,xx x

g( ) 3ln , 0,x x x x

(a) find an expression for gf ( ),x simplifying your answer.

(2) (b) Show that there is only one real value of x for which gf ( ) fg( )x x

(3)

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5. (a) Use the binomial expansion, in ascending powers of x, to show that

214 2 ...

4x x kx

where k is a rational constant to be found. (4)

A student attempts to substitute 1x into both sides of this equation to find an approximate value for 3.

(b) State, giving a reason, if the expansion is valid for this value of x. (1)

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6. Complete the table below. The first one has been done for you.

For each statement you must state if it is always true, sometimes true or never true, giving a reason in each case.

Statement Always True

Sometimes True

Never True

Reason

The quadratic equation 2 0, ( 0)ax bx c a

has 2 real roots.

It only has 2 real roots when 2 4 0.b ac

When 2 4 0b ac it has 1 real

root and when 2 4 0b ac it has 0 real roots.

(i)

When a real value of x is substituted into

2 6 10x x the result is positive.

(2)

(ii)

If ax b then b

xa

(2)

(iii)

The difference between consecutive square numbers is odd.

(2)



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7.

Figure 1

Figure 1 shows a rectangle ABCD.

The point A lies on the y-axis and the points B and D lie on the x-axis as shown in Figure 1.

Given that the straight line through the points A and B has equation 5y + 2x = 10

(a) show that the straight line through the points A and D has equation 2y − 5x = 4 (4)

(b) find the area of the rectangle ABCD. (3)

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n


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63

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8. Given that A is constant and

4

2

1

3 d 2x A x A

show that there are exactly two possible values for A. (5)

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Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample assessment materials (SAMs) – Draft 1.2 – January 2017 © Pearson Education Limited 2017

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9. In a geometric series the common ratio is r and sum to n terms is nS

Given

6

8

7S S

show that 1

,rk

where k is an integer to be found.

(4)

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10.

Figure 2 shows a sketch of part of the graph y = f(x) where

f ( ) 2 3 5, 0x x x

(a) State the range of f (1)

(b) Solve the equation 1

f ( ) 302

x x

(3)

Given that the equation f(x) = k, where k is a constant, has two distinct roots,

(c) state the set of possible values for k. (2)

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66

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67

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11. The mass, A

m grams, of a radioactive substance A, t years after first being observed, is

modelled by the equation 0.0525e t

Am

According to the model,

(a) find the mass of the radioactive substance A six months after it was first observed, (2)

(b) show that d,

dA

A

mkm

t where k is a constant to be found.

(2)

100 grams of a radioactive substance B is being studied over the same time period as radioactive substance A.

The rate of decay of substance B is twice that of radioactive substance A.

(c) State, giving reasons for your answer, an expression for the mass of the radioactive substance B in terms of t.

(2)

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12. (a) Solve, for –180° x < 180°, the equation

2 23sin sin 8 9 cosx x x

giving your answers to 2 decimal places. (6)

(b) Hence find the smallest positive solution of the equation

2 23sin 2 30 sin 2 30 8 9cos 2 30

giving your answer to 2 decimal places. (2)

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13. (a) Express 10 cos – 3 sin in the form R cos ( + ), where R > 0 and 0 < < 90°

Give the exact value of R and give the value of , in degrees, to 2 decimal places. (3)

Figure 3

The height above the ground of a passenger on a Ferris wheel is modelled by the equation

H = 11 – 10 cos (80t)° + 3 sin (80t)°

where the height of the passenger above the ground is H metres, t minutes after the wheel starts turning.

Figure 3 shows the graph of H against t for two complete cycles of the wheel.

Use the model to find

(b) the maximum height above the ground reached by the passenger, (1)

(c) the time taken, to the nearest second, for the passenger to reach the maximum height on the second cycle.

(Solutions based entirely on graphical or numerical methods are not acceptable.) (3)

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14. A company decides to manufacture a soft drinks can with a capacity of 500 ml.

The company models the can in the shape of a right circular cylinder with radius r cm and height h cm.

In the model they assume that the can is made from a metal of negligible thickness.

(a) Prove that the total surface area, S cm2, of the can is given by

2 10002S r

r

(3)

Given that r can vary,

(b) find the dimensions of a can that has minimum surface area. (5)

(c) With reference to the shape of the can, suggest a reason why the company may choose not to manufacture a can with minimum surface area.

(1)

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15.

Figure 4

Figure 4 shows a sketch of the curve C with equation

3

25 9 11, 0y x x x

The point P with coordinates (4, 15) lies on C.

The line l is the tangent to C at the point P.

The region R, shown shaded in Figure 4, is bounded by the curve C, the line l and the y-axis.

Show that the area of R is 24, making your method clear.

(Solutions based entirely on graphical or numerical methods are not acceptable.) (10)

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n


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16. (a) Express 1

(11 2 )P P in partial fractions.

(3)

A population of meerkats is being studied.

The population is modelled by the differential equation

t

P

d

d =

1

22P (11 – 2P), t 0, 0 5.5P

where P, in thousands, is the population of meerkats and t is the time measured in years since the study began.

Given that there are 1000 meerkats in the population when the study began,

(b) determine the time taken, in years, for this population of meerkats to double, (6)

(c) show that

1

2et

AP

B C

where A, B and C are integers to be found. (3)

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n


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Paper 2: Pure Mathematics 2 Mark Scheme


1 Sets 3 2f 2 0 2 2 5 2 2 0a a M1 3.1a

Solves linear equation 2 36a a a dM1 1.1b

36a A1 1.1b

(3 marks)

Notes: M1: Selects a suitable method given that ( 2)x is a factor of f ( ).x

Accept either setting f 2 0 or attempted division of f ( )x by ( 2)x

dM1: Solves linear equation in a. Minimum requirement is that there are two terms in 'a' which must be collected to get .. ..a a A1: 36a


2(a) Error for student A: They use

costan

sin

It should be sin

tancos

B1 2.3

(1) (b) (i) Shows 26.6 26.6cos 2sin , so cannot be a solution

B1 2.4

(ii) Explains that the incorrect answer was introduced by squaring

B1 2.4

(2)

(3 marks)

Notes: (a)

B1: A response of the type 'They use cos

tansin

. This is incorrect as

sintan

cos

'

It can be implied by a response such as 'They should get 1

tan2

not tan 2 '

(b) B1: Accept a response where the candidate shows that 26.6 is not a solution of cos 2sin . This can be shown by, for example, finding both 26.6cos and

26.62sin and stating that they are not equal. An acceptable alternative is to state that

26.6cos ve and 26.62sin ve and stating that they therefore cannot be equal.

B1: Explains that the incorrect answer was introduced by squaring Accept an example showing this. For example 5x squared gives 2 25x which has answers 5


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3 Attempts the product and chain rule on 4(2 1)y x x M1 2.1

4 3d(2 1) 8 (2 1)

d

yx x x

x A1 1.1b

Takes out a common factor 3d(2 1) (2 1) 8

d

yx x x

x M1 1.1b

3d

(2 1) (10 1) 3, 10, 1d

yx x n A B

x A1 1.1b

(4 marks)

Notes:

M1: Applies the product rule to reach 4 3d(2 1) (2 1)

d

yx Bx x

x

A1: 4 3d(2 1) 8 (2 1)

d

yx x x

x

M1: Takes out a common factor of 3(2 1)x

A1: The form of this answer is given. Look for 3d(2 1) (10 1) 3, 10, 1

d

yx x n A B

x


4 (a) gf ( ) 3ln exx M1 1.1b

3 ,x x A1 1.1b

(2)

(b) 3gf ( ) fg( ) 3x x x x M1 1.1b

3 3 0x x x M1 1.1b

3x only as ln x is not defined at 0x and 3 A1 2.3

(3)

(5 marks)Notes: (a) M1: For applying the functions in the correct order A1: The simplest form is required so it must be 3x and not left in the form 3ln ex An answer of 3x with no working would score both marks (b) M1: Allow the candidates to score this mark if they have 3lne their 3x x M1: For solving their cubic in x and obtaining at least one solution. A1: For either stating that 3x only as ln x (or3ln x ) is not defined at 0x and 3

or stating that 33x x would have three answers, one positive one negative and one zero but ln x (or3ln x ) is not defined for 0x so therefore there is only one (real) answer.

Note: Student who mix up fg and gf can score full marks in part (b) as they have already been penalised in part (a)


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5 (a)

1

214 2 1

4x x

M1 2.1

12

2

1 1

1 1 1 12 21 1 ...4 2 4 2! 4

x x x

M1 1.1b

21 14 2 1 ..

8 128x x x

A1 1.1b

4 x 21 12 ...

4 64x x and

1

64k A1 1.1b

(4) (b) The expansion is valid for 4,x so 1x can be used B1 2.4

(1)

(5 marks)

Notes: (a)

M1: Takes out a factor of 4 and writes 1

24 2 1 ...x

M1: For an attempt at the binomial expansion with 1

2n Eg.

1

22

1 1

1 2 21 1 ...2 2!

ax ax ax

A1: Correct expression inside the bracket 21 11

8 128x x which may be left unsimplified

A1: 4 x 21 12 ...

4 64x x and

1

64k

(b) B1: The expansion is valid for 4,x so 1x can be used


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Question Scheme

Marks AOs

6(i) 22 6 10 3 1x x x M1 2.1

Deduces ''always true''

as 23 0x 2

3 1 1x and so is always positive A1 2.2a

(2) (ii) For an explanation that it need not (always) be true

This could be if 0 thenb

a ax b xa

M1 2.4

States 'sometimes' and explains if 0 thenb

a ax b xa

if 0 thenb

a ax b xa

A1 2.4

(2) (iii) Difference 2 21 2 1n n n M1 3.1a

Deduces ''Always true'' as 2 1n = (even +1) = odd A1 2.2a

(2)

(6 marks)

Notes: (i) M1: Attempts to complete the square or any other valid reason. Allow for a graph of

2 6 10y x x or an attempt to find the minimum by differentiation A1: States always true with a valid reason for their method (ii) M1: For an explanation that it need not be true (sometimes). This could be if

0 thenb

a ax b xa

or simply 3 6 2x x

A1: Correct statement (sometimes true) and explanation (iii) M1: Sets up the proof algebraically.

For example by attempting 2 21 2 1n n n or 2 2 ( )( )m n m n m n with 1m n

A1: States always true with reason and proof


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7(a) Gradient AB = 2

5 B1 2.1

y coordinate of A is 2 B1 2.1

Uses perpendicular gradients 5

2y x c M1 2.2a

2 5 4 *y x A1* 1.1b

(4) (b) Uses Pythagoras' theorem to find AB or AD

Either 2 25 2 or

224

25

M1 3.1a

Uses area ABCD = 11629

25AD AB M1 1.1b

area ABCD = 11.6 A1 1.1b

(3)

(7 marks)

Notes: (a) It is important that the student communicates each of these steps clearly

B1: States the gradient of AB is 2

5

B1: States that y coordinate of A = 2

M1: Uses the form y mx c with 2

their adapted5

m and their 2c

Alternatively uses the form 1 1y y m x x with

2their adapted

5m and 1 1

, 0, 2x y

A1*: Proceeds to given answer (b)

M1: Finds the lengths of AB or AD using Pythagoras' Theorem. Look for 2 25 2 or

224

25

Alternatively finds the lengths BD and AO using coordinates. Look for 4

55

and 2

M1: For a full method of finding the area of the rectangle ABCD. Allow for AD AB

Alternatively attempts area ABCD = 1

22

BD AO 1

2 '5.8' '2'2

A1: Area ABCD = 11.6 or other exact equivalent such as 58

5


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8

0.5 1.53 d 2x A x x Ax c M1 A1

3.1a 1.1b

Uses limits and sets = 2 22 2 8 4 2 1 2A A A A M1 1.1b

Sets up quadratic and attempts to solve

Sets up quadratic and attempts 2 4b ac

M1 1.1b

72,

2A and states that

there are two roots

States 2 4 121 0b ac and hence there are two roots

A1 2.4

(5 marks)

Notes: M1: Integrates the given function and achieves an answer of the form 1.5kx Ax c where k is a

non- zero constant A1: Correct answer but may not be simplified

M1: Substitutes in limits and subtracts. This can only be scored if dA x Ax and not 2

2

A

M1: Sets up quadratic equation in A and either attempts to solve or attempts 2 4b ac

A1: Either 72,

2A and states that there are two roots

Or states 2 4 121 0b ac and hence there are two roots

Question Scheme

Marks AOs

9

Attempts 6

6

8 8 (1 )

7 1 7 1

a a rS S

r r

M1 2.1

681 (1 )

7r M1 2.1

6 1..

8r r M1 1.1b

1

2r (so 2)k A1 1.1b

(4 marks)Notes:

M1: Substitutes the correct formulae for S and 6S into the given equation 6

8

7S S

M1: Proceeds to an equation just in r M1: Solves using a correct method

A1: Proceeds to 1

2r giving 2k


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Question Scheme

Marks AOs

10 (a) f ( ) 5x B1 1.1b

(1) (b)

Uses 1

2(3 ) 5 302

x x M1 3.1a

Attempts to solve by multiplying out bracket,

collect terms etc 3

312

x M1 1.1b

62

3x only A1 1.1b

(3) (c) Makes the connection that there must be two intersections.

Implied by either end point 5k or 11k M1 2.2a

: , 5 11k k k A1 2.5

(2)

(6 marks)

Notes: (a) B1: f ( ) 5x Also allow f ( ) 5,x

(b)

M1: Deduces that the solution to1

f ( ) 302

x x can be found by solving 1

2(3 ) 5 302

x x

M1: Correct method used to solve their equation. Multiplies out bracket/ collects like terms

A1: 62

3x only. Do not allow 20.6

(c) M1: Deduces that two distinct roots occurs when y k intersects f ( )y x in two places. This may be implied by the sight of either end point. Score for sight of either 5k or 11k

A1: Correct solution only : , 5 11k k k


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Question Scheme

Marks AOs

11(a) Substitutes 0.05 0.05 0.50.5 into 25e 25etA At m m M1 3.4

24.4gAm A1 1.1b

(2) (b) States or uses 0.05 0.05d

e ed

t tCt

M1 2.1

0.05d

0.05 25e 0.05 0.05d

tAA

mm k

t A1 1.1b

(2) (c) States ...100eBm AND explains that the '100' in the

equation represents the initial value. B1 3.3

0.1...e tBm AND explains that the value of k must

be doubled B1ft 3.3

(2)

(6 marks)

Notes: (a) M1: Substitutes 0.05 0.05 0.50.5 into 25e 25ett m m A1: 24.4gm An answer of 24.4gm with no working would score both marks (b)

M1: Applies the rule de e

dkx kxk

t in this context by stating or using 0.05 0.05d

e ed

t tCt

A1: 0.05d0.05 25e 0.05 0.05

dtm

m kt

(c) B1: States ...100eBm AND explains that the '100' in the equation represents the initial value.

B1ft: 0.1...e tBm AND explains that the value of k must be doubled. Follow through on their

value of k in part (b)


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12(a) Uses 2 2 2 2cos 1 sin 3sin sin 8 9 1 sinx x x x x M1 3.1a

212sin sin 1 0x x A1 1.1b

4sin 1 3sin 1 0x x M1 1.1b

1 1

sin ,4 3

x A1 1.1b

Uses arcsin to obtain two correct values M1 1.1b

All four of 14.48 ,165.52 , 19.47 , 160.53x A1 1.1b

(6) (b) Attempts 2 30 19.47 M1 3.1a

5.26 A1ft 1.1b

(2)

(8 marks)

Notes: (a) M1: Substitutes 2 2cos 1 sinx x into 2 23sin sin 8 9cosx x x to create a quadratic equation in just sin x A1: 212sin sin 1 0x x or exact equivalent M1: Attempts to solve their quadratic equation in sin x by a suitable method. These could include factorisation, formula or completing the square.

A1: 1 1

sin ,4 3

x

M1: Obtains two correct values for their sin x k A1: All four of 14.48 ,165.52 , 19.47 , 160.53x (b) M1: For setting 2 30 their ' 19.47 ' A1ft: 5.26 but allow a follow through on their ' 19.47 '


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Question Scheme

Marks AOs

13(a) 109R B1 1.1b

3

tan10

M1 1.1b

16.70 so 109 cos 16.70 A1 1.1b

(3) (b) 11 109 or 21.44 m B1ft 3.4

(1)

(c) Sets 80 ''16.70 '' 540t M1 3.4

540 ''16.70 ''6.54

80t

M1 1.1b

6 mins 32secondst A1 1.1b

(3)

(7 marks)

Notes: (a) B1: 109R Do not allow decimal equivalents

M1: Allow for 3

tan10

A1: 16.70 (b) B1ft: 11 their 109 Allow decimals here. (c) M1: Sets 80 ''16.70 '' 540t . Follow through on their 16.70 M1: Solves their 80 ''16.70 '' 540t correctly to find t A1: 6 mins 32secondst


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Question Scheme

Marks AOs

14(a) Sets 2500 r h B1 2.1

Substitute 2

500h

r into 2 2

2

5002 2 2 2S r rh r r

r

M1 2.1

Simplifies to reach given answer 2 10002 *S r

r A1* 1.1b

(3) (b) Differentiates S with both indices correct in d

d

S

r M1 3.4

2

d 10004

d

Sr

r r A1 1.1b

Sets d0

d

S

r and proceeds to 3 ,r k k is a constant M1 2.1

Radius 4.30cm A1 1.1b

Substitutes their 4.30r into 2

500h

r Height 8.60cm A1 1.1b

(5) (c) States a valid reason such as

The radius is too big for the size of our hands If 4.3cmr and 8.6cmh the can is square in profile.

All drinks cans are taller than they are wide The radius is too big for us to drink from They have different dimensions to other drinks cans

and would be difficult to stack on shelves with other drinks cans

B1 3.5a

(1)

9 marks Notes: (a) B1: Uses the correct volume formula with V =500. Accept 2500 r h

M1: Substitutes 2

500h

r or 500

rhr

into 22 2S r rh to get S as a function of r

A1*: 2 10002S r

r Note that this is a given answer.

(b)

M1: Differentiates the given S to reach 2d

d

SAr Br

r

A1: 2

d 10004

d

Sr

r r or exact equivalent

M1: Sets d0

d

S

r and proceeds to 3 ,r k k is a constant

A1: R = awrt 4.30cm A1: H = awrt 8.60 cm (c) B1: Any valid reason. See scheme for alternatives


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15

1

2d 159

d 2

yx

x

M1 A1

3.1a 1.1b

Substitutes d

4 6d

yx

x M1 2.1

Uses 4,15 and gradient 15 6( 4)y x M1 2.1

Equation of l is 6 9y x A1 1.1b

Area R = 4 3

2

0

5 9 11 6 9 dx x x x

M1 3.1a

45

22

0

152 20

2x x x c

A1 1.1b

Uses both limits of 4 and 0 45 5

2 22 2

0

15 152 20 2 4 4 20 4 0

2 2x x x

M1 2.1

Area of R = 24 *

A1* 1.1b

Correct notation with good explanations A1 2.5

(10)

(10 marks)


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Notes:

M1: Differentiates 3

25 9 11x x to a form 1

2Ax B

A1: 1

2d 159

d 2

yx

x but may not be simplified

M1: Substitutes 4x in their d

d

y

x to find the gradient of the tangent

M1: Uses their gradient and the point 4,15 to find the equation of the tangent

A1: Equation of l is 6 9y x

M1: Uses Area R = 4 3

2

0

5 9 11 6 9 dx x x x

following through on their 6 9y x

Look for a form 5

22Ax Bx Cx

A1: 45

22

0

152 20

2x x x c

This must be correct but may not be simplified

M1: Substitutes in both limits and subtracts A1*: Correct area for R = 24 A1: Uses correct notation and produces a well explained and accurate solution. Look for

Correct notation used consistently and accurately for both differentiation and integration Correct explanations in producing the equation of l. See scheme. Correct explanation in finding the area of R. In way 2 a diagram may be used.

Alternative method for the area using area under curve and triangles. (Way 2)

M1: Area under curve =4 3

2

0

5 9 11x x

4522

0

Ax Bx Cx

A1: 45

22

0

92 11 36

2x x x

M1: This requires a full method with all triangles found using a correct method

Look for 1 3 1 3

Area = their 36 15 4 their their 9 their2 2 2 2

R


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Question Scheme

Marks AOs

16(a) Sets

1

(11 2 ) (11 2 )

A B

P P P P

B1 1.1a

Substitutes either 0P or 11

2P into

1 (11 2 )A P BP A or B M1 1.1b

1 21 11 11

(11 2 ) (11 2 )P P P P

A1 1.1b

(3) (b) Separates the variables

22d 1d

(11 2 )P t

P P

B1 3.1b

Uses (a) and attempts to integrate 2 4

d(11 2 )

P t cP P

M1 1.1b

2 ln 2 ln 11 2P P t c A1 1.1b

Substitutes 0, 1t P 2ln 9c M1 3.1b

Substitutes 2 2 ln 2 2 ln 9 2 ln 7P t M1 3.1b

Time 1.89 years A1 3.2a

(6) (c) Uses ln laws 2 ln 2 ln 11 2 2 ln 9P P t

9 1

ln11 2 2

Pt

P

M1 2.1

Makes 'P' the subject 1

29

e11 2

tP

P

1

29 11 2 et

P P

1

2f et

P

or 1

2f et

P

M1 2.1

1

2

1111, 2, 9

2 9et

P A B C

A1 1.1b

(3)

(12 marks)


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16 continued

Notes: (a)

B1: Sets 1

(11 2 ) (11 2 )

A B

P P P P

M1: Substitutes 0P or 11

2P into 1 (11 2 )A P BP A or B

Alternatively compares terms to set up and solve two simultaneous equations in A and B

A1: 1 21 11 11

(11 2 ) (11 2 )P P P P

or equivalent

1 1 2

(11 2 ) 11 11(11 2 )P P P P

Note: The correct answer with no working scores all three marks. (b)

B1: Separates the variables to reach 22

d 1d(11 2 )

P tP P

or equivalent

M1: Uses part (a) and d ln ln(11 2 )(11 2 )

A BP A P C P

P P

A1: Integrates both sides to form a correct equation including a 'c' Eg 2 ln 2 ln 11 2P P t c

M1: Substitutes t= 0 and P =1 to find c M1: Substitutes P = 2 to find t. This is dependent upon having scored both previous M's A1: Time 1.89 years (c)

M1: Uses correct log laws to move from 2 ln 2 ln 11 2P P t c to 1

ln11 2 2

Pt d

P

for

their numerical 'c'

M1: Uses a correct method to get P in terms of 1

2et

This can be achieved from 1

21

ln e11 2 2 11 2

t dP Pt d

P P

followed by cross multiplication

and collection of terms in P (See scheme)

Alternatively uses a correct method to get P in terms of 1

2et For example

1 1 112 2 22

11 2 11 11e e 2 e 2 e

11 2

t d t d t dt dP P

P P P P

followed by division

A1: Achieves the correct answer in the form required. 1

2

1111, 2, 9

2 9et

P A B C

oe


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Mathematics Advanced Paper 3: Statistics and Mechanics Sample assessment material for first teaching September 2017 Time: 2 hours

Paper Reference(s)

9MA0/03

You must have: Mathematical Formulae and Statistical Tables, calculator


Instructions Use black ink or ball-point pen If pencil is used for diagrams/sketches/graphs it must be dark (HB or B). Fill in the boxes at the top of this page with your name, centre

number and candidate number. There are two sections in this question paper. Answer all the

questions in Section A and all the questions in Section B. Answer the questions in the spaces provided


Answers without working may not gain full credit. Answers should be given to three significant figures unless otherwise stated. Information A booklet ‘Mathematical Formulae and Statistical Tables’ is provided. There are 10 questions in this question paper. The total mark for this





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SECTION A: STATISTICS


1. The number of hours of sunshine each day, y, for the month of July at Heathrow are summarised in the table below.

Hours 0 5y 5 8y 8 11y 11 12y 12 14y

Frequency 12 6 8 3 2

A histogram was drawn to represent these data. The 8 11y group was represented by a bar of width 1.5 cm and height 8 cm.

(a) Find the width and the height of the 0 5y group. (3)

(b) Use your calculator to estimate the mean and the standard deviation of the number of hours of sunshine each day, for the month of July at Heathrow.

(3)

(c) Use linear interpolation to estimate the 60th percentile of the number of hours of sunshine each day, for the month of July at Heathrow.

(2)

(d) Estimate the number of days in July at Heathrow where the number of hours of sunshine is more than 1 standard deviation above the mean.

(2)

The mean and standard deviation for the number of hours of daily sunshine for the same

month in Hurn are 5.98 hours and 4.12 hours respectably. Thomas believes that the further south you are the more consistent should be the number

of hours of daily sunshine.

(e) State, giving a reason, whether or not the calculations in part (b) support Thomas’ belief.

(2)


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Question 1 Continued

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2. A meteorologist believes that there is a relationship between the daily mean windspeed, w kn, and the daily mean temperature, t °C. A random sample of 9 consecutive days is taken from past records from a town in the UK in July and the relevant data is given in the table below.

The meteorologist calculated the product moment correlation coefficient for the 9 days and obtained r = 0.609

(a) Explain why a linear regression model based on these data is unreliable on a day when the mean temperature is 24oC

(1)

(b) State what is measured by the product moment correlation coefficient. (1)

(c) Stating your hypotheses clearly test, at the 5% significance level, whether or not the product moment correlation coefficient for the population is greater than zero.

(3)

Using the same 9 days a location from the large data set gave 27.2 and 3.5t w

(d) Using your knowledge of the large data set, suggest, giving your reason, the location

that gave rise to these statistics. (1)

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t 13.3 16.2 15.7 16.6 16.3 16.4 19.3 17.1 13.2

w 7 11 8 11 13 8 15 10 11


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3. A machine cuts strips of metal to length L cm, where L is normally distributed with standard deviation 0.5 cm.

Strips with length either less than 49 cm or greater than 50.75 cm cannot be used.

Given that 2.5% of the cut lengths exceed 50.98 cm,

(a) find the probability that a randomly chosen strip of metal can be used. (5)

Ten strips of metal are selected at random.

(b) Find the probability fewer than 4 of these strips cannot be used. (2)

A second machine cuts strips of metal of length X cm, where X is normally distributed with standard deviation 0.6 cm

A random sample of 15 strips cut by this second machine was found to have a mean length of 50.4 cm

(c) Stating your hypotheses clearly and using a 1% level of significance, test whether or not the mean length of all the strips, cut by the second machine, is greater than 50.1 cm

(4)

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102

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103

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4. Given that

P(A) = 0.35 P(B) = 0.45 and P(A B) = 0.13

find

(a) P(A B) (2)

(b) P |A B

(2)

(c) Explain why the events A and B are not independent. (1)

The event C has P( C ) = 0.20

The events A and C are mutually exclusive and the events B and C are statistically independent.

(d) Draw a Venn diagram to illustrate the events A, B and C, giving the probabilities for each region.

(5)

(e) Find P( [B C] ) (2)

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104

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n


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(Total for Question 4 is 12marks)


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5. A company sells seeds and claims that 55% of its pea seeds germinate.

In a random sample of 35 pea seeds, X germinate.

(a) Calculate the value of P(X = 14). (1)

(b) Write down a reason why the company should not justify their claim by testing all the pea seeds they produce.

(1)

(c) Write down two conditions under which the normal distribution may be used as an approximation to the binomial distribution.

(2)

A random sample of 220 pea seeds was planted and 138 of these seeds germinated.

(d) Assuming that the company’s claim is correct, use a normal approximation to find the probability that at least 138 pea seeds germinate.

(3)

(e) Using your answer to part (d)

(i) comment on the company’s claim that 55% of its pea seeds germinate,

(ii) suggest a refinement to the company’s claim. (2)

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106

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TOTAL FOR SECTION A IS 50 MARKS


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SECTION B: MECHANICS


Unless otherwise indicated, whenever a numerical value of g is required, take g = 9.8 m s-2 and give your answer to either 2 significant figures or 3 significant figures.

6. At time t seconds, where t 0, a particle P moves so that its acceleration a 2ms is given

by

125 15t t a i j

When t = 0, the velocity of P is 20i 1ms

Find the speed of P when t = 4 (6)

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108

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109

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7. A rough plane is inclined to the horizontal at an angle , where3

tan4

.

A particle of mass m is placed on the plane and then projected up a line of greatest slope of the plane.

The coefficient of friction between the particle and the plane is µ.

The particle moves up the plane with a constant deceleration of 4

5g .

(a) Find the value of µ. (6)

The particle comes to rest at the point A on the plane.

(b) Determine whether the particle will remain at A, carefully justifying your answer.

(2)

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110

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111

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8. [In this question i and j are horizontal unit vectors due east and due north respectively]

A radio controlled model boat is placed on the surface of a large pond.

The boat is modelled as a particle.

At time t = 0, the boat is at the fixed point O and is moving due north with

speed 0.6 1m s .

Relative to O, the position vector of the boat at time t seconds is r metres.

At time t = 15, the velocity of the boat is (10.5i – 0.9j) 1ms .

The acceleration of the boat is constant.

(a) Show that the acceleration of the boat is (0.7i – 0.1j) 2ms . (2)

(b) Find r in terms of t. (2)

(c) Find the value of t when the boat is north-east of O. (3)

(d) Find the value of t when the boat is moving in a north-east direction. (3)

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112

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113

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9.

A uniform ladder AB, of length 2a and weight W, has its end A on rough horizontal ground.

The coefficient of friction between the ladder and the ground is 1

4.

The end B of the ladder is resting against a smooth vertical wall, as shown in Figure 1.

A builder of weight 7W stands at the top of the ladder. To stop the ladder from slipping, the builder’s assistant applies a horizontal force of magnitude P to the ladder at A, towards the wall. The force acts in a direction which is perpendicular to the wall. The ladder rests in equilibrium in a vertical plane perpendicular to the wall and makes

an angle with the horizontal ground, where5

tan2

.

The builder is modelled as a particle and the ladder is modelled as a uniform rod.

(a) Show that the reaction of the wall on the ladder at B has magnitude 3W. (5)

(b) Find, in terms of W, the range of possible values of P for which the ladder remains in equilibrium.

(5)

Often in practice, the builder’s assistant will simply stand on the bottom of the ladder.

(c) Explain briefly how this helps to stop the ladder from slipping. (3)

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114

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115

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116

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117

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10.

[In this question use g = 10 m s-2]

A boy throws a stone with speed U m s-1 from a point O at the top of a vertical cliff. The point O is 18 m above sea level.

The stone is thrown at an angle above the horizontal, where tan3

4 .

The stone hits the sea at the point S which is at a horizontal distance of 36 m from the foot of the cliff, as shown in Figure 2. The stone is modelled as a particle moving freely under gravity. Find (a) the value of U,

(6) (b) the time taken by the stone to travel from O to S,

(2)

(c) the speed of the stone when it is 10.8 m above sea level, giving your answer to 2 significant figures.

(5)

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118

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119

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120

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TOTAL FOR SECTION B IS 50 MARKS



121

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122

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Paper 3: Statistics and Mechanics Mark Scheme

Question Scheme

Marks AOs

1(a) Area = 28 1.5 12 cm Frequency = 8 so 21 cm = 2

3 hour

(o.e.) M1 3.1a

Frequency of 12 corresponds to area of 18 so height = 18 2.5 = 7.2 (cm)

A1 1.1b

Width = 5 0.5 = 2.5 (cm) B1cao 1.1b (3)

(b) 205.5 awrt 6.63

31y B1cao 1.1b

21785.25

31y y = 13.644641 = awrt 3.69

allow 21785.25 31

30

ys

= awrt 3.69

M1 A1

1.1a 1.1b

(3) (c) 60

0.6 1.2 8 3 (use of ( 1)) (8 ) 3

8 8P n or

M1 1.1b

= 8.225 or 8.45 A1 1.1b (2)

(d) 10.3x so number of days is e.g. (11 "10.3")8 5

3

M1 1.1b

= 6.86 so 7 days A1 1.1b (2)

(e) Mean of Heathrow is higher than Hurn and standard deviation smaller suggesting Heathrow is more reliable

M1 2.4

Hurn is South of Heathrow so does not support his belief A1 2.2b

(2) (12 marks)

Notes (a) M1 for clear attempt to relate the area to frequency. Can also award if their

height their width = 18 A1 for height = 7.2 (cm)

(b) M1 for a correct expression for or s, can ft their value for mean A1 awrt 3.69

(c) M1 for an attempt to interpolate…condone error in end point but the rest must be correct. May “work down” so use end point 11. Allow use of n + 1 A1ft their use of n or (n+1)

(d) M1 for a correct expression – ft their A1 for 7 days but accept 6 (rounding down) following a correct expression

(e) M1 for a suitable comparison of standard deviations to comment on reliability. A1 for stating Hurn is south of Heathrow and a correct conclusion

10.3x


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Question Scheme

Marks AOs

2(a) e.g. It requires extrapolation so will be unreliable (o.e.)

B1 1.2

(1) (b) e.g. Linear association between w and t B1 1.2

(1)

(c) H0: 0 H1: 0 B1 2.5

Critical value 0.5822 M1 1.1a

Reject H0

There is evidence that the product moment correlation coefficient is greater than 0

A1 2.2b

(3)

(d) Higher t suggests overseas and not Perth…lower wind speed so perhaps not close to the sea so suggest Beijing

B1 2.4 (1)

(6 marks)Part Notes

(a) B1 for a correct statement (unreliable) with a suitable reason (b) B1 for a correct statement (c) B1 for both hypotheses in terms of

M1 for selecting a suitable 5% critical value compatible with their H1 A1 for a correct conclusion stated

(d) B1 for suggesting Beijing with some supporting reason based on t or w Allow Jacksonville with a reason based just on higher t


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Question Scheme

Marks AOs

Q3(a)

P (L > 50.98) = 0.025 B1cao 3.1b

P

5.0

98.50 Z = 0.025 M1 3.4

5.0

98.50 = 1.96 M1 3.4

= 50 A1cao 1.1b

P(49 < L < 50.75) = 0.9104 = awrt 0.910

B1 1.1b

(5)

(b) S = number of strips that cannot be used so S~B(10, 0.090) M1 3.3

= P( S 3) = 0.991166… awrt 0.991 A1 1.1b

(2)

(c) H0 : μ = 50.1 H1 : μ > 50.1 B1 2.5

z =

2

50.4 50.1

0.615

= 1.936…

M1 3.4

z = 2.3263 (p = 0.0264) B1 1.1b

Not significant; There is insufficient evidence that the mean length of strips is not greater than 50.1

A1 2.2b

(4) (11 marks)

Part Notes (a) 1st M1 for use of normal and standardizing with and 0.5 (allow +)

2nd M1 for forming an equation for with a suitable z value (|z| > 1) 2nd B1 awrt 0.910

(b) M1 for identifying a suitable binomial distribution A1 awrt 0.991 (from calculator)

(c) B1 hypotheses stated correctly M1 for a correct expression for z B1cao for z or p A1 for a correct conclusion mentioning “mean length” and 50.1

49 50.75


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4(a) P( ) 0.35 0.45 0.13 or 0.22 0.13 .32 0 A B M1 1.1b

= 0.67 A1 1.1b

(2)

(b)

P( ) 0.33P( | ) or

P( ) 0.55

A BA B

B

M1 3.1a

=

3 or 0.6

5 A1 1.1b

(2)

(c) e.g. 7 9 63 5220 20 400 400P( ) P( ) P( ) 0.13A B A B

or P( | ) 0.6 P( ) 0.65A B A B1 2.4

(1) (d)

B1 2.5

M1 3.1a

A1 1.1b

M1 1.1b

A1 1.1b

(5)

(e) P( ) 1 [0.56]

or 1

0.22 0.

[0.13 0.23 0.09 0.1

22

]

or

1

B C

o.e. M1 1.1b

= 0.44 A1 1.1b

(2) (12 marks)

Part Notes (a) M1 for a correct expression

A1 cao (b) M1 for a correct ratio of probabilities formula and at least one correct value.

A1 a correct answer (c) B1 for a fully correct explanation: correct probabilities and correct comparisons.

(d) B1 for box with B intersecting A and C but C not intersecting A.( Or accept three intersecting circles, but with zeros entered for A C and A B C )No box is B0

M1 for method for finding P( B C ) A1 for 0.09

M1 for 0.13 and their 0.09 in correct places and method for their 0.23 A1 fully correct

(e) M1 for a correct expression – ft their probabilities from their Venn diagram. A1 cao


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Q5(a) P(X = 14) = 0.0280556… awrt 0.0281 B1 1.1b

(1)

(b) The seeds would be destroyed in the process so they would have none to sell

B1 2.4

(1)

(c) n is large B1 1.2

and p close to 0.5 B1 1.2

(2)

(d) X~N(121, 54.45) B1 1.1b

P(X 137.5) = 137.5 121

P 54.45

Z M1 3.4

= 0.0127 A1cso 1.1b

(3)

(e)(i) e.g The probability is very small therefore there is evidence that the company’s claim is incorrect.

B1 2.2b

(e)(ii) e.g Use a binomial distribution with

138ˆ or 0.63

220p p B1 3.5c

(2)

(9 marks)Part Notes

(a) B1 cao (b) B1 a correct reason (c) B1 one correct condition

B1 two correct conditions (d) B1 for correct normal distribution

M1 for correct use of continuity correction A1 cso

(e)(i) B1 correct statement (ii) B1 a correct refinement


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6 Integrate a w.r.t. time M1 1.1a

v 5t 2

2i –10t

32 j C (allow omission of C) A1 1.1b

v 5t 2

2i –10t

32 j 20i A1 1.1b

When t = 4, v = 60i – 80 j M1 1.1b

Attempt to find magnitude: √ (602 + 802 ) M1 3.1a

Speed = 100 m s-1 A1ft 1.1b

(6 marks)

Notes 1st M1 for integrating a w.r.t. time (powers of t increasing by 1) 1stA1 for a correct v expression without C 2nd A1 for a correct v expression including C 2nd M1 for putting t = 4 into their v expression 3rd M1 for finding magnitude of their v 3rd A1 ft for 100 m s-1, follow through on an incorrect v


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7(a)

R = mgcos B1 3.1b

Resolve parallel to the plane M1 3.1b

- F - mgsin 0.8mg A1 1.1b

F =R M1 1.2

Produce an equation in only and solve for M1 2.2a

1

4 A1 1.1b

(b)

(6)

Compare mgcoswith mgsin M1 3.1b

Deduce an appropriate conclusion A1 ft 2.2a

(2)

(8 marks)

Notes (a) B1 for R = mgcos 1st M1 for resolving parallel to the plane 1st A1 for a correct equation 2nd M1 for use of F =R 3rd M1 for eliminating F and R to give a value for

2nd A1 for 1

4

(b) M1 comparing size of limiting friction with weight component down the plane A1ft for an appropriate conclusion from their values


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8(a)

Use of v = u + at : (10.5i – 0.9j) = 0.6j + 15a M1 3.1b

a = (0.7i – 0.1j) m s-2 Given answer A1 1.1b

(2)

(b) Use of r = ut +

1

2 at2 M1 3.1b

r = 0.6j t + 1

2(0.7i – 0.1j) t2 A1 1.1b

(2)

(c)

Equating the i and j components of r M1 3.1b

1

2 0.7 t2 = 0.6 t –

1

2 0.1 t2 A1ft 1.1b

t = 1.5 A1 1.1b

(3)

(d) Use of v = u + at: v = 0.6j + (0.7i – 0.1j) t M1 3.1b

Equating the i and j components of v M1 3.1b

t = 0.75 A1 ft 1.1b

(3)

(10 marks)

Notes (a) M1 for use of v = u + at A1 for given answer correctly obtained

(b) M1 for use of r = ut + 1

2 at2

A1 for a correct expression for r in terms of t (c) M1 for equating the i and j components of their r A1ft for a correct equation following their r A1 for t = 1.5 (d) M1 for use of v = u + at for a general t M1 for equating the i and j components of their v A1ft for t = 0.75, or a correct follow through answer from an incorrect equation


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9(a)

Take moments about A (or any other complete method to produce an equation in S , W and only)

M1 3.3

Wacos7W2acos= S 2asin A1 A1

1.1b 1.1b

Use of tan5

2 to obtain S M1 2.1

S = 3W Given answer A1 2.2a

(5)

(b)

R = 8W B1 3.4

F = 1

4 R (= 2W) M1 3.4

PMAX = 3W + F or PMIN = 3W - F M1 3.4

PMAX = 5W or PMIN = W A1 1.1b

W ≤ P ≤ 5W A1 2.5

(5)

(c) M(A) shows that the reaction on the ladder at B is unchanged

M1 2.4

also R increases (resolving vertically) M1 2.4

which increases max F available M1 2.4

(3)

(13 marks)

Notes (a) 1st M1 for producing an equation in S, W and only 1st A1 for an equation that is correct, or which has one error or omission

2nd A1 for a fully correct equation

2nd M1 for use of tan5

2 to obtain S in terms of W only

3rd A1 for given answer S = 3W correctly obtained (b) B1 for R = 8W

1st M1 for use of F = 1

4 R

2nd M1 for either P = (3W + their F) or P = (3W - their F) 1st A1 for a correct max or min value for a correct range for P 2nd A1 for a correct range for P

(c) 1st M1 for showing, by taking moments about A, that the reaction at B is unchanged by the builder’s assistant standing on the bottom of the ladder

2nd M1 for showing, by resolving vertically, that R increases as a result of the builder’s


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assistant standing on the bottom of the ladder 3rd M1 for concluding that this increases the limiting friction at A


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10(a)

Using the model and horizontal motion: s = ut M1 3.4

36 = Utcos A1 1.1b

Using the model and vertical motion: s = ut + 1

2at2 M1 3.4

-18 = Utsin - 1

2gt2 A1 1.1b

Correct strategy for solving the problem by setting up two equations in t and U and solving for U

M1 3.1b

U = 15 A1 1.1b

(6) (b)

36 = (their U) t 4

5 M1 1.1b

t = 3 (follow through on their U) A1 ft 1.1b

(2)

(c) Using the model and horizontal motion: Ucos B1 3.4

Using the model and vertical motion: v2 = (Usin2 + 2(-10)(-7.2)

M1 3.4

v = 15 A1 1.1b

Correct strategy for solving the problem by finding the horizontal and vertical components of velocity and combing using Pythagoras: Speed = √(122 + 152 )

M1 3.1b

√369 = 19 m s -1 (2sf) A1 ft 1.1b

(5)

(13 marks)

Notes (a) 1st M1 for use of s = ut horizontally 1st A1 for a correct equation

2nd M1 for use of s = ut + 1

2at2 vertically

2nd A1 for a correct equation 3rd M1 for correct strategy (need both equations) 2nd A1 for U = 15 (b) M1 for use of their horizontal motion equation with 36, their U and cos

A1ft for their t following their U

(c) B1 for Ucos used as horizontal velocity component 1st M1 for attempt to find vertical component 1st A1 for 15 2nd M1 for correct strategy (need both components) 2nd A1 ft for 19 m s -1 (2sf) following through on in correct component(s)


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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE Mathematics and Further Mathematics Mathematical formulae and statistical tables

For first certification from June 2018 for:

Advanced Subsidiary GCE in Mathematics (8MA0)

Advanced GCE in Mathematics (9MA0)

Advanced Subsidiary GCE in Further Mathematics (8FM0)

For first certification from June 2018 for:

Advanced GCE in Further Mathematics (9FM0)

This copy is the property of Pearson. It is not to be removed from the examination room or marked in any way.

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Contents

1 Introduction 1

Pure Mathematics 2 Statistics 2 Mechanics 3

3 A Level in Mathematics 4


4 AS Level in Further Mathematics 8


5 A Level in Further Mathematics 15


6 Statistical Tables 26

Binomial Cumulative Distribution Function 26 Percentage Points Of The Normal Distribution 31 Poisson Cumulative Distribution Function 32 Percentage Points of the 2 Distribution 33 Critical Values for Correlation Coefficients 34 Random Numbers 35 Percentage Points of Student’s t Distribution 36 Percentage Points of the F Distribution 37

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 1 Mathematical Formulae and Statistical Tables – Draft 1.2 – January 2017 - © Pearson Education Limited 2017

1 Introduction The formulae in this booklet have been arranged by qualification. Students sitting AS or A Level Further Mathematics papers may be required to use the formulae that were introduced in AS or A Level Mathematics papers.

It may also be the case that students sitting Mechanics and Statistics papers will need to use formulae introduced in the appropriate Pure Mathematics papers for the qualification they are sitting.

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2 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.2 – January 2017 - © Pearson Education Limited 2017

2 AS Level in Mathematics

Pure Mathematics

Mensuration

Surface area of sphere = 4 r 2

Area of curved surface of cone = r slant height

Binomial series

1 2 2 2

( )1

n n n r r nn n n na b a b a b b

r

na b a

(n )

where C( )

nr

n n!

r r ! n r !

Logarithms and exponentials

loglog

logb

ab

xx

a

lnex a xa

Differentiation

First Principles

0

f ( ) f ( )f ( ) = lim

h

x h xx

h

Statistics

Probability

P(A) = 1 – P(A)

Standard deviation

Standard deviation = (Variance)

Interquartile range = IQR = Q3 – Q1

For a set of n values 1 2, ,... ,...i nx x x x

22 2 ( )

S ( ) ixx i i

xx x x

n

Standard deviation = 2

2or S

xxx

xn n

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Statistical tables

The following statistical tables are required for A Level Mathematics:

Binomial Cumulative Distribution Function (see page 25)

Random Numbers (see page 34)

Mechanics

Kinematics

For motion in a straight line with constant acceleration:

v = u + at

s = ut + ½ at2

s = vt - ½ at2

v2 = u2 + 2as

s = ½ (u + v)t

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3 A Level in Mathematics

Pure Mathematics

Mensuration

Surface area of sphere = 4 r 2

Area of curved surface of cone = r slant height

Arithmetic series

Sn = 1

2n(a + l) =

1

2n[2a + (n 1)d]

Binomial series

1 2 2 21

n n n r r nn n n na b a b a b b

r

n( a b ) a

(n )

where C( )

nr

n n!

r r ! n r !

2( 1) ( 1) ( 1)(1 ) 1 ( 1 )

1 2 1 2n rn n n n n r

x nx x x x , nr

Logarithms and exponentials

loglog

logb

ab

xx

a

lnx a xae

Geometric series

Sn = (1 )

1

na r

r

S =1

a

r for r < 1

Numerical integration

The trapezium rule:

b

a

xy d 21 h{(y0 + yn) + 2(y1 + y2 + ... + yn – 1)}, where

b ah

n

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Trigonometric identities

sin ( ) sin cos cos sinA B A B A B

cos ( ) cos cos sin sinA B A B A B

12

tan tantan ( ) ( ( ) )

1 tan tan

A BA B A B k

A B

sin sin 2sin cos2 2

A B A BA B

sin sin 2cos sin2 2

A B A BA B

cos cos 2cos cos2 2

A B A BA B

cos cos 2sin sin2 2

A B A BA B

Differentiation

First Principles

0

f ( ) f ( )f ( ) = lim

h

x h xx

h

f(x) f (x)

tan kx k sec2 kx

seckx kseckx tankx

cotkx – kcosec2kx

cosec kx – kcosec kx cot kx

f( )

g( )

x

x

2

f ( ) g( ) f( ) g ( )

(g( ))

x x x x

x

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Integration (+ constant)

f(x) f( ) d

x x

sec2 kx k

1 tan kx

tankx k

1ln sec kx

cot kx k

1ln sin kx

coseckx 12

1 1ln cosec cot , ln tan( )kx kx kx

k k

seckx 1 12 4

1 1ln sec tan , ln tan( )kx kx kx

k k

xx

uvuvx

x

vu d

d

dd

d

d

Numerical solution of equations

The Newton-Raphson iteration for solving 0)f( x : )(f

)f(1

n

nnn x

xxx

Statistics

Probability

P(A) = 1 – P(A)

P( ) P( ) P( ) P( )A B A B A B

P( ) P( )P( )A B A B | A

)P()|P()P()|P(

)P()|P()|P(

AABAAB

AABBA

For independent events A and B,

P(BA) = P(B), P(AB) = P(A),

P(A B) = P(A) P(B)

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Standard deviation

Standard deviation = (Variance)

Interquartile range = IQR = Q3 – Q1

For a set of n values 1 2, ,... ,...i nx x x x

n

xxxxS i

iixx

222 )(

)(

Standard deviation = 2

2xxxS

xn n

or Discrete distributions

Distribution of X P(X = x) Mean Variance

Binomial ),B( pn (1 )x n xnp p

x

np )1( pnp

Sampling distributions

For a random sample of n observations from 2N( , )

~ N(0, 1)/

X

n

Statistical tables

The following statistical tables are required for A Level Mathematics:


Percentage Points Of The Normal Distribution (30)

Critical Values for Correlation Coefficients: Product Moment Coefficient (see page 33)


Mechanics

Kinematics

For motion in a straight line with constant acceleration:

v = u + at

s = ut + ½ at2

s = vt - ½ at2

v2 = u2 + 2as

s = ½ (u + v)t

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4 AS Level in Further Mathematics Students sitting a AS Level Further Mathematics paper may also require those formulae listed for A Level Mathematics in Section 3.

Pure Mathematics

Summations

)12)(1(61

1

2

nnnrn

r

2241

1

3 )1(

nnrn

r

Matrix transformations

Anticlockwise rotation through about O:

cos sin

sincos

Reflection in the line xy )(tan :

2cos2sin

2sin 2cos

Area of a sector

A = d

2

1 2r (polar coordinates)

Complex numbers

{ (cos i sin )} (cos i sin )n nr r n n

The roots of 1nz are given by 2 i

ek

nz

, for 1 , ,2 ,1 ,0 nk

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Maclaurin’s and Taylor’s Series

)0(f!

)0(f!2

)0(f)0f()f( )(2

rr

r

xxxx

xr

xxxx

rx allfor

!

!21)exp(e

2

2 31ln (1 ) ( 1) ( 1 1)

2 3

rrx x x

x x xr

xr

xxxxx

rr allfor

)!12()1(

!5!3sin

1253

xr

xxxx

rr allfor

)!2()1(

!4!21cos

242

3 5 2 1

arctan ( 1) ( 1 1)3 5 2 1

rrx x x

x x xr

Vectors

Vector product:

1221

3113

2332

321

321ˆ sin

baba

baba

baba

bbb

aaa

kji

nbaba

)()()(

321

321

321

bac.acb.cba. ccc

bbb

aaa

If A is the point with position vector kjia 321 aaa and the direction vector b is given by

kjib 321 bbb , then the straight line through A with direction vector b has cartesian equation

)( 3

3

2

2

1

1

b

az

b

ay

b

ax

The plane through A with normal vector kjin 321 nnn has cartesian equation

1 2 3 0 where n x n y n z d d a.n

The plane through non-collinear points A, B and C has vector equation cbaacabar )1()()(

The plane through the point with position vector a and parallel to b and c has equation

cbar ts

The perpendicular distance of ) , ,( from 1 2 3 0n x n y n z d is 1 2 3

2 2 21 2 3

n n n d

n n n

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Hyperbolic functions

1sinhcosh 22 xx

xxx coshsinh22sinh

xxx 22 sinhcosh2cosh

)1( 1lnarcosh }{ 2 xxxx

}{ 1lnarsinh 2 xxx

12

1artanh ln ( 1)

1

xx x

x

Differentiation

f(x) f(x)

xarcsin 21

1

x

xarccos 21

1

x

xarctan 21

1

x

xsinh xcosh

xcosh xsinh

xtanh x2sech

xarsinh 21

1

x

xarcosh 1

12 x

artanh x 21

1

x

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Integration (+ constant; 0a where relevant)

f(x)

xx d)f(

xsinh xcosh

xcosh xsinh

xtanh xcoshln

22

1

xa )( arcsin ax

a

x

22

1

xa

a

x

aarctan

1

22

1

ax )( ln,arcosh }{ 22 axaxx

a

x

22

1

xa }{ 22ln,arsinh axx

a

x

22

1

xa )( artanh

1ln

2

1ax

a

x

axa

xa

a

22

1

ax

ax

ax

a

ln2

1

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Statistics Discrete distributions

For a discrete random variable X taking values ix with probabilities P(X = xi)

Expectation (mean): E(X) = = ix P(X = ix )

Variance: Var(X) = 2 = ( ix – )2 P(X = ix ) = 2i

x P(X = ix ) – 2

Discrete distributions

Standard discrete distributions:

Distribution of X P(X = x) Mean Variance

Binomial B(n, p) 1n xxn

p px

np np(1 – p)

Poisson Po ( ) e!

x

x

Continuous distributions

For a continuous random variable X having probability density function f

Expectation (mean): xxxX d)f()E(

Variance: 2222 d)f(d)f()()Var( xxxxxxX

For a function )g( X : xxxX d)f()g())E(g(

Cumulative distribution function: 0

0 0F( ) P( ) f ( ) dx

x X x t t

Standard continuous distribution:

Distribution of X P.D.F. Mean Variance

Normal ) ,N( 2

2

21

e2

1

x

2

Uniform (Rectangular) on [a, b] ab

1 1

2( )a b 2

121 )( ab

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Correlation and regression

For a set of n pairs of values ) ,(ii

yx

22 2 ( )

S ( ) ixx i i

xx x x

n

22 2 ( )

S ( ) iyy i i

yy y y

n

( )( )S ( )( ) i i

xy i i i ix y

x x y y x yn

The product moment correlation coefficient is:

2 2 2 22 2

( )( )S ( )( )

S S ( ) ( ) ( ) ( )

{ }{ }

i ii ixy i i

xx yy i i i ii i

x yx yx x y y nr

x x y y x yx y

n n

The regression coefficient of y on x is 2

S ( )( )

S ( )

xy i i

xx i

x x y yb

x x

Least squares regression line of y on x is bxay where xbya

Residual Sum of Squares (RSS) =

2

2S

S S 1S

xy

yy yyxx

r

Spearman’s rank correlation coefficient is 2

2

61

( 1)s

d

n nr

Non-parametric tests

Goodness-of-fit test and contingency tables: 22

~)(

i

ii

E

EO

Statistical tables

The following statistical tables are required for AS Level Further Mathematics:


Poisson Cumulative Distribution Function (see page 31)

Percentage Points of the 2 Distribution (see page 32)

Critical Values for Correlation Coefficients: Product Moment Coefficient and Spearman’s Coefficient (see page 33)


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Mechanics

Centres of mass

For uniform bodies:

Triangular lamina: 23 along median from vertex

Circular arc, radius r, angle at centre 2 : sinr

from centre

Sector of circle, radius r, angle at centre 2 : 2 sin

3

r

from centre

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5 A Level in Further Mathematics Students sitting a A Level Further Mathematics paper may also require those formulae listed for A Level Mathematics in Section 3.

Pure Mathematics

Summations

)12)(1(61

1

2

nnnrn

r

2241

1

3 )1(

nnrn

r

Matrix transformations

Anticlockwise rotation through about O:

cos sin

sincos

Reflection in the line xy )(tan :

2cos2sin

2sin 2cos

Area of a sector

A = d

2

1 2r (polar coordinates)

Complex numbers

{ (cos i sin )} (cos i sin )n nr r n n

The roots of 1nz are given by 2 i

ek

nz

, for 1 , ,2 ,1 ,0 nk

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Maclaurin’s and Taylor’s Series

)0(f!

)0(f!2

)0(f)0f()f( )(2

rr

r

xxxx

xr

xxxx

rx allfor

!

!21)exp(e

2

2 31ln (1 ) ( 1) ( 1 1)

2 3

rrx x x

x x xr

xr

xxxxx

rr allfor

)!12()1(

!5!3sin

1253

xr

xxxx

rr allfor

)!2()1(

!4!21cos

242

3 5 2 1

arctan ( 1) ( 1 1)3 5 2 1

rrx x x

x x xr

Vectors

Vector product:

1221

3113

2332

321

321ˆ sin

baba

baba

baba

bbb

aaa

kji

nbaba

)()()(

321

321

321

bac.acb.cba. ccc

bbb

aaa

If A is the point with position vector kjia 321 aaa and the direction vector b is given by

kjib 321 bbb , then the straight line through A with direction vector b has cartesian equation

)( 3

3

2

2

1

1

b

az

b

ay

b

ax

The plane through A with normal vector kjin 321 nnn has cartesian equation

1 2 3 0 where n x n y n z d d a.n

The plane through non-collinear points A, B and C has vector equation cbaacabar )1()()(

The plane through the point with position vector a and parallel to b and c has equation

cbar ts The perpendicular distance of ) , ,( from 1 2 3 0n x n y n z d is

1 2 3

2 2 21 2 3

n n n d

n n n

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Hyperbolic functions 1sinhcosh 22 xx

xxx coshsinh22sinh

xxx 22 sinhcosh2cosh

)1( 1lnarcosh }{ 2 xxxx

}{ 1lnarsinh 2 xxx

12

1artanh ln ( 1)

1

xx x

x

Conics

Ellipse Parabola Hyperbola Rectangular Hyperbola

Standard Form 1

2

2

2

2

b

y

a

x axy 42 1

2

2

2

2

b

y

a

x 2cxy

Parametric Form

)sin ,cos( ba )2 ,( 2 atat (a sec , b tan ) (a cosh , b sinh )

t

cct,

Eccentricity 1e

)1( 222 eab 1e

1e

)1( 222 eab e = 2

Foci )0 ,( ae )0 ,(a )0 ,( ae ( 2c , 2c )

Directrices e

ax ax

e

ax x + y = 2c

Asymptotes none none b

y

a

x 0 ,0 yx

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Differentiation

f(x) f(x)

xarcsin 21

1

x

xarccos 21

1

x

xarctan 21

1

x

xsinh xcosh

xcosh xsinh

xtanh x2sech

xarsinh 21

1

x

xarcosh 1

12 x

artanh x 21

1

x

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Integration (+ constant; 0a where relevant)

f(x)

xx d)f(

xsinh xcosh

xcosh xsinh

xtanh xcoshln

22

1

xa )( arcsin ax

a

x

22

1

xa

a

x

aarctan

1

22

1

ax )( ln,arcosh }{ 22 axaxx

a

x

22

1

xa }{ 22ln,arsinh axx

a

x

22

1

xa )( artanh

1ln

2

1ax

a

x

axa

xa

a

22

1

ax

ax

ax

a

ln2

1

Arc length

xx

ys d

d

d1

2

(cartesian coordinates)

tt

y

t

xs d

d

d

d

d22

(parametric form)

22 dr

dd

s r

(polar form)

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Surface area of revolution

Sx =

2d

2 1 dd

yy x

x

(cartesian coordinates)

Sx = 2 2

d d2 d

d d

x yy t

t t

(parametric form)

Sx = 2

2 dr2 sin d

dr r

(polar form)

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Statistics


For a discrete random variable X taking values ix with probabilities P(X = xi)

Expectation (mean): E(X) = = xi P(X = xi)

Variance: Var(X) = 2 = (xi – )2 P(X = xi) = 2i

x P(X = xi) – 2

For a function )g( X : E(g(X)) = g(xi) P(X = xi)

The probability generating function of X is G ( ) E X

X t t and

2E( ) G (1) and Var( ) G (1) G (1) G (1)X X X XX X

For Z = X + Y, where X and Y are independent: G ( ) G ( ) G ( )Z X Yt t t


Standard discrete distributions:

Distribution of X P(X = x) Mean Variance P.G.F.

Binomial B(n, p) 1n xxn

p px

np np(1 – p) 1

np pt

Poisson Po ( ) e!

x

x 1e t

Geometric Geo(p) on 1, 2, … 1

1x

p p

1

p

2

1 p

p

1 (1 )

pt

p t

Negative binomial on r, r + 1, …

1(1 )

1r x rx

p pr

r

p

2

(1 )r p

p

1 (1 )

rpt

p t

Continuous distributions

For a continuous random variable X having probability density function f

Expectation (mean): xxxX d)f()E(

Variance: 2222 d)f(d)f()()Var( xxxxxxX

For a function )g( X : xxxX d)f()g())E(g(

Cumulative distribution function: 0

0 0F( ) P( ) f ( ) dx

x X x t t

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Standard continuous distribution:

Distribution of X P.D.F. Mean Variance

Normal ) ,N( 2

2

21

e2

1

x

2

Uniform (Rectangular) on [a, b] ab

1 1

2( )a b 2

121 )( ab

Correlation and regression

For a set of n pairs of values ) ,(ii

yx

22 2 ( )

S ( ) ixx i i

xx x x

n

22 2 ( )

S ( ) iyy i i

yy y y

n

( )( )S ( )( ) i i

xy i i i ix y

x x y y x yn

The product moment correlation coefficient is

2 2 2 22 2

( )( )S ( )( )

S S ( ) ( ) ( ) ( )

{ }{ }

i ii ixy i i

xx yy i i i ii i

x yx yx x y y nr

x x y y x yx y

n n

The regression coefficient of y on x is 2

S ( )( )

S ( )

xy i i

xx i

x x y yb

x x

Least squares regression line of y on x is bxay where xbya

Residual Sum of Squares (RSS) =

2

2S

S S 1S

xy

yy yyxx

r

Spearman’s rank correlation coefficient is 2

2

61

( 1)s

d

n nr

Expectation algebra

For independent random variables X and Y

E( ) E( ) E( )XY X Y , 2 2Var( ) Var( ) Var( )aX bY a X b Y

DRAFT

Subject toOfq


tio

n


Sampling distributions

(i) Tests for mean when is known

For a random sample n

XXX , , ,21 of n independent observations from a distribution having

mean and variance 2 :

X is an unbiased estimator of , with n

X2

)Var(

2S is an unbiased estimator of 2 , where 1

)( 22

n

XXS i

For a random sample of n observations from 2N( , ) , ~ N(0, 1)/

X

n

For a random sample of xn observations from ) ,N( 2xx

and, independently, a random

sample of yn observations from ) ,N( 2yy

, )1 ,0N(~)()(

22

y

y

x

x

yx

nn

YX

(ii) Tests for variance and mean when is not known

For a random sample of n observations from ) ,N( 2 :

212

2

~)1(

n

Sn

1~

/

n

tnS

X (also valid in matched‐pairs situations)

For a random sample of xn observations from ) ,N( 2xx

and, independently, a random

sample of y

n observations from ) ,N( 2yy

1 ,122

22

~/

/ ynxn

yy

xx FS

S

If 222

yx (unknown) then

2

2

( ) ( )~

1 1

x yn nx y

px y

X Yt

Sn n

where

2 22 ( 1) ( 1)

2

x x y yp

x y

n S n SS

n n

DRAFT

Subject toOfq


tio

n


Non-parametric tests

Goodness-of-fit test and contingency tables: 22

~)(

i

ii

E

EOStatistical tables

The following statistical tables are required for A Level Further Mathematics:


Percentage Points Of The Normal Distribution (see page 30)

Poisson Cumulative Distribution Function (see page 31)

Percentage Points of the 2 Distribution (see page 32)

Critical Values for Correlation Coefficients: Product Moment Coefficient and Spearman’s Coefficient (see page 33)


Percentage Points of Student’s t Distribution (see page 35)

Percentage Points of the F Distribution (see page 36

DRAFT

Subject toOfq


tio

n


Mechanics

Centres of mass

For uniform bodies:

Triangular lamina: 32 along median from vertex

Circular arc, radius r, angle at centre 2 : sinr

from centre

Sector of circle, radius r, angle at centre 2 : 2 sin

3

r

from centre

Solid hemisphere, radius r: r83 from centre

Hemispherical shell, radius r: r21 from centre

Solid cone or pyramid of height h: 14 h above the base on the line from centre of base to vertex

Conical shell of height h: 13 h above the base on the line from centre of base to vertex

Motion in a circle

Transverse velocity: rv

Transverse acceleration: rv

Radial acceleration: r

vr

22

DRAFT

Subject toOfq


tio

n


6 Statistical Tables

Binomial Cumulative Distribution Function

The tabulated value is P(X x), where X has a binomial distribution with index n and parameter p.

p = 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50n = 5, x = 0 0.7738 0.5905 0.4437 0.3277 0.2373 0.1681 0.1160 0.0778 0.0503 0.0312

1 0.9774 0.9185 0.8352 0.7373 0.6328 0.5282 0.4284 0.3370 0.2562 0.18752 0.9988 0.9914 0.9734 0.9421 0.8965 0.8369 0.7648 0.6826 0.5931 0.50003 1.0000 0.9995 0.9978 0.9933 0.9844 0.9692 0.9460 0.9130 0.8688 0.81254 1.0000 1.0000 0.9999 0.9997 0.9990 0.9976 0.9947 0.9898 0.9815 0.9688

n = 6, x = 0 0.7351 0.5314 0.3771 0.2621 0.1780 0.1176 0.0754 0.0467 0.0277 0.01561 0.9672 0.8857 0.7765 0.6554 0.5339 0.4202 0.3191 0.2333 0.1636 0.10942 0.9978 0.9842 0.9527 0.9011 0.8306 0.7443 0.6471 0.5443 0.4415 0.34383 0.9999 0.9987 0.9941 0.9830 0.9624 0.9295 0.8826 0.8208 0.7447 0.65634 1.0000 0.9999 0.9996 0.9984 0.9954 0.9891 0.9777 0.9590 0.9308 0.89065 1.0000 1.0000 1.0000 0.9999 0.9998 0.9993 0.9982 0.9959 0.9917 0.9844

n = 7, x = 0 0.6983 0.4783 0.3206 0.2097 0.1335 0.0824 0.0490 0.0280 0.0152 0.00781 0.9556 0.8503 0.7166 0.5767 0.4449 0.3294 0.2338 0.1586 0.1024 0.06252 0.9962 0.9743 0.9262 0.8520 0.7564 0.6471 0.5323 0.4199 0.3164 0.22663 0.9998 0.9973 0.9879 0.9667 0.9294 0.8740 0.8002 0.7102 0.6083 0.50004 1.0000 0.9998 0.9988 0.9953 0.9871 0.9712 0.9444 0.9037 0.8471 0.77345 1.0000 1.0000 0.9999 0.9996 0.9987 0.9962 0.9910 0.9812 0.9643 0.93756 1.0000 1.0000 1.0000 1.0000 0.9999 0.9998 0.9994 0.9984 0.9963 0.9922

n = 8, x = 0 0.6634 0.4305 0.2725 0.1678 0.1001 0.0576 0.0319 0.0168 0.0084 0.00391 0.9428 0.8131 0.6572 0.5033 0.3671 0.2553 0.1691 0.1064 0.0632 0.03522 0.9942 0.9619 0.8948 0.7969 0.6785 0.5518 0.4278 0.3154 0.2201 0.14453 0.9996 0.9950 0.9786 0.9437 0.8862 0.8059 0.7064 0.5941 0.4770 0.36334 1.0000 0.9996 0.9971 0.9896 0.9727 0.9420 0.8939 0.8263 0.7396 0.63675 1.0000 1.0000 0.9998 0.9988 0.9958 0.9887 0.9747 0.9502 0.9115 0.85556 1.0000 1.0000 1.0000 0.9999 0.9996 0.9987 0.9964 0.9915 0.9819 0.96487 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9998 0.9993 0.9983 0.9961

n = 9, x = 0 0.6302 0.3874 0.2316 0.1342 0.0751 0.0404 0.0207 0.0101 0.0046 0.00201 0.9288 0.7748 0.5995 0.4362 0.3003 0.1960 0.1211 0.0705 0.0385 0.01952 0.9916 0.9470 0.8591 0.7382 0.6007 0.4628 0.3373 0.2318 0.1495 0.08983 0.9994 0.9917 0.9661 0.9144 0.8343 0.7297 0.6089 0.4826 0.3614 0.25394 1.0000 0.9991 0.9944 0.9804 0.9511 0.9012 0.8283 0.7334 0.6214 0.50005 1.0000 0.9999 0.9994 0.9969 0.9900 0.9747 0.9464 0.9006 0.8342 0.74616 1.0000 1.0000 1.0000 0.9997 0.9987 0.9957 0.9888 0.9750 0.9502 0.91027 1.0000 1.0000 1.0000 1.0000 0.9999 0.9996 0.9986 0.9962 0.9909 0.98058 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9992 0.9980

n = 10, x = 0 0.5987 0.3487 0.1969 0.1074 0.0563 0.0282 0.0135 0.0060 0.0025 0.00101 0.9139 0.7361 0.5443 0.3758 0.2440 0.1493 0.0860 0.0464 0.0233 0.01072 0.9885 0.9298 0.8202 0.6778 0.5256 0.3828 0.2616 0.1673 0.0996 0.05473 0.9990 0.9872 0.9500 0.8791 0.7759 0.6496 0.5138 0.3823 0.2660 0.17194 0.9999 0.9984 0.9901 0.9672 0.9219 0.8497 0.7515 0.6331 0.5044 0.37705 1.0000 0.9999 0.9986 0.9936 0.9803 0.9527 0.9051 0.8338 0.7384 0.62306 1.0000 1.0000 0.9999 0.9991 0.9965 0.9894 0.9740 0.9452 0.8980 0.82817 1.0000 1.0000 1.0000 0.9999 0.9996 0.9984 0.9952 0.9877 0.9726 0.94538 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9995 0.9983 0.9955 0.98939 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9990

DRAFT

Subject toOfq


tio

n


p = 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50n = 12, x = 0 0.5404 0.2824 0.1422 0.0687 0.0317 0.0138 0.0057 0.0022 0.0008 0.0002

1 0.8816 0.6590 0.4435 0.2749 0.1584 0.0850 0.0424 0.0196 0.0083 0.00322 0.9804 0.8891 0.7358 0.5583 0.3907 0.2528 0.1513 0.0834 0.0421 0.01933 0.9978 0.9744 0.9078 0.7946 0.6488 0.4925 0.3467 0.2253 0.1345 0.07304 0.9998 0.9957 0.9761 0.9274 0.8424 0.7237 0.5833 0.4382 0.3044 0.19385 1.0000 0.9995 0.9954 0.9806 0.9456 0.8822 0.7873 0.6652 0.5269 0.38726 1.0000 0.9999 0.9993 0.9961 0.9857 0.9614 0.9154 0.8418 0.7393 0.61287 1.0000 1.0000 0.9999 0.9994 0.9972 0.9905 0.9745 0.9427 0.8883 0.80628 1.0000 1.0000 1.0000 0.9999 0.9996 0.9983 0.9944 0.9847 0.9644 0.92709 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9992 0.9972 0.9921 0.9807

10 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9989 0.996811 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9998

n = 15, x = 0 0.4633 0.2059 0.0874 0.0352 0.0134 0.0047 0.0016 0.0005 0.0001 0.00001 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353 0.0142 0.0052 0.0017 0.00052 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268 0.0617 0.0271 0.0107 0.00373 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969 0.1727 0.0905 0.0424 0.01764 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155 0.3519 0.2173 0.1204 0.05925 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216 0.5643 0.4032 0.2608 0.15096 1.0000 0.9997 0.9964 0.9819 0.9434 0.8689 0.7548 0.6098 0.4522 0.30367 1.0000 1.0000 0.9994 0.9958 0.9827 0.9500 0.8868 0.7869 0.6535 0.50008 1.0000 1.0000 0.9999 0.9992 0.9958 0.9848 0.9578 0.9050 0.8182 0.69649 1.0000 1.0000 1.0000 0.9999 0.9992 0.9963 0.9876 0.9662 0.9231 0.8491

10 1.0000 1.0000 1.0000 1.0000 0.9999 0.9993 0.9972 0.9907 0.9745 0.940811 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9995 0.9981 0.9937 0.982412 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9989 0.996313 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.999514 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

n = 20, x = 0 0.3585 0.1216 0.0388 0.0115 0.0032 0.0008 0.0002 0.0000 0.0000 0.00001 0.7358 0.3917 0.1756 0.0692 0.0243 0.0076 0.0021 0.0005 0.0001 0.00002 0.9245 0.6769 0.4049 0.2061 0.0913 0.0355 0.0121 0.0036 0.0009 0.00023 0.9841 0.8670 0.6477 0.4114 0.2252 0.1071 0.0444 0.0160 0.0049 0.00134 0.9974 0.9568 0.8298 0.6296 0.4148 0.2375 0.1182 0.0510 0.0189 0.00595 0.9997 0.9887 0.9327 0.8042 0.6172 0.4164 0.2454 0.1256 0.0553 0.02076 1.0000 0.9976 0.9781 0.9133 0.7858 0.6080 0.4166 0.2500 0.1299 0.05777 1.0000 0.9996 0.9941 0.9679 0.8982 0.7723 0.6010 0.4159 0.2520 0.13168 1.0000 0.9999 0.9987 0.9900 0.9591 0.8867 0.7624 0.5956 0.4143 0.25179 1.0000 1.0000 0.9998 0.9974 0.9861 0.9520 0.8782 0.7553 0.5914 0.4119

10 1.0000 1.0000 1.0000 0.9994 0.9961 0.9829 0.9468 0.8725 0.7507 0.588111 1.0000 1.0000 1.0000 0.9999 0.9991 0.9949 0.9804 0.9435 0.8692 0.748312 1.0000 1.0000 1.0000 1.0000 0.9998 0.9987 0.9940 0.9790 0.9420 0.868413 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9985 0.9935 0.9786 0.942314 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9984 0.9936 0.979315 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9985 0.994116 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.998717 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999818 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

DRAFT

Subject toOfq


tio

n


p = 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50n = 25, x = 0 0.2774 0.0718 0.0172 0.0038 0.0008 0.0001 0.0000 0.0000 0.0000 0.0000

1 0.6424 0.2712 0.0931 0.0274 0.0070 0.0016 0.0003 0.0001 0.0000 0.00002 0.8729 0.5371 0.2537 0.0982 0.0321 0.0090 0.0021 0.0004 0.0001 0.00003 0.9659 0.7636 0.4711 0.2340 0.0962 0.0332 0.0097 0.0024 0.0005 0.00014 0.9928 0.9020 0.6821 0.4207 0.2137 0.0905 0.0320 0.0095 0.0023 0.00055 0.9988 0.9666 0.8385 0.6167 0.3783 0.1935 0.0826 0.0294 0.0086 0.00206 0.9998 0.9905 0.9305 0.7800 0.5611 0.3407 0.1734 0.0736 0.0258 0.00737 1.0000 0.9977 0.9745 0.8909 0.7265 0.5118 0.3061 0.1536 0.0639 0.02168 1.0000 0.9995 0.9920 0.9532 0.8506 0.6769 0.4668 0.2735 0.1340 0.05399 1.0000 0.9999 0.9979 0.9827 0.9287 0.8106 0.6303 0.4246 0.2424 0.1148

10 1.0000 1.0000 0.9995 0.9944 0.9703 0.9022 0.7712 0.5858 0.3843 0.212211 1.0000 1.0000 0.9999 0.9985 0.9893 0.9558 0.8746 0.7323 0.5426 0.345012 1.0000 1.0000 1.0000 0.9996 0.9966 0.9825 0.9396 0.8462 0.6937 0.500013 1.0000 1.0000 1.0000 0.9999 0.9991 0.9940 0.9745 0.9222 0.8173 0.655014 1.0000 1.0000 1.0000 1.0000 0.9998 0.9982 0.9907 0.9656 0.9040 0.787815 1.0000 1.0000 1.0000 1.0000 1.0000 0.9995 0.9971 0.9868 0.9560 0.885216 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9992 0.9957 0.9826 0.946117 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9988 0.9942 0.978418 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9984 0.992719 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9996 0.998020 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.999521 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999922 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

n = 30, x = 0 0.2146 0.0424 0.0076 0.0012 0.0002 0.0000 0.0000 0.0000 0.0000 0.00001 0.5535 0.1837 0.0480 0.0105 0.0020 0.0003 0.0000 0.0000 0.0000 0.00002 0.8122 0.4114 0.1514 0.0442 0.0106 0.0021 0.0003 0.0000 0.0000 0.00003 0.9392 0.6474 0.3217 0.1227 0.0374 0.0093 0.0019 0.0003 0.0000 0.00004 0.9844 0.8245 0.5245 0.2552 0.0979 0.0302 0.0075 0.0015 0.0002 0.00005 0.9967 0.9268 0.7106 0.4275 0.2026 0.0766 0.0233 0.0057 0.0011 0.00026 0.9994 0.9742 0.8474 0.6070 0.3481 0.1595 0.0586 0.0172 0.0040 0.00077 0.9999 0.9922 0.9302 0.7608 0.5143 0.2814 0.1238 0.0435 0.0121 0.00268 1.0000 0.9980 0.9722 0.8713 0.6736 0.4315 0.2247 0.0940 0.0312 0.00819 1.0000 0.9995 0.9903 0.9389 0.8034 0.5888 0.3575 0.1763 0.0694 0.0214

10 1.0000 0.9999 0.9971 0.9744 0.8943 0.7304 0.5078 0.2915 0.1350 0.049411 1.0000 1.0000 0.9992 0.9905 0.9493 0.8407 0.6548 0.4311 0.2327 0.100212 1.0000 1.0000 0.9998 0.9969 0.9784 0.9155 0.7802 0.5785 0.3592 0.180813 1.0000 1.0000 1.0000 0.9991 0.9918 0.9599 0.8737 0.7145 0.5025 0.292314 1.0000 1.0000 1.0000 0.9998 0.9973 0.9831 0.9348 0.8246 0.6448 0.427815 1.0000 1.0000 1.0000 0.9999 0.9992 0.9936 0.9699 0.9029 0.7691 0.572216 1.0000 1.0000 1.0000 1.0000 0.9998 0.9979 0.9876 0.9519 0.8644 0.707717 1.0000 1.0000 1.0000 1.0000 0.9999 0.9994 0.9955 0.9788 0.9286 0.819218 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9986 0.9917 0.9666 0.899819 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.9971 0.9862 0.950620 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9991 0.9950 0.978621 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9984 0.991922 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.997423 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.999324 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999825 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

DRAFT

Subject toOfq


tio

n


p = 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 n = 40, x = 0 0.1285 0.0148 0.0015 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

1 0.3991 0.0805 0.0121 0.0015 0.0001 0.0000 0.0000 0.0000 0.0000 0.00002 0.6767 0.2228 0.0486 0.0079 0.0010 0.0001 0.0000 0.0000 0.0000 0.00003 0.8619 0.4231 0.1302 0.0285 0.0047 0.0006 0.0001 0.0000 0.0000 0.00004 0.9520 0.6290 0.2633 0.0759 0.0160 0.0026 0.0003 0.0000 0.0000 0.00005 0.9861 0.7937 0.4325 0.1613 0.0433 0.0086 0.0013 0.0001 0.0000 0.00006 0.9966 0.9005 0.6067 0.2859 0.0962 0.0238 0.0044 0.0006 0.0001 0.00007 0.9993 0.9581 0.7559 0.4371 0.1820 0.0553 0.0124 0.0021 0.0002 0.00008 0.9999 0.9845 0.8646 0.5931 0.2998 0.1110 0.0303 0.0061 0.0009 0.00019 1.0000 0.9949 0.9328 0.7318 0.4395 0.1959 0.0644 0.0156 0.0027 0.0003

10 1.0000 0.9985 0.9701 0.8392 0.5839 0.3087 0.1215 0.0352 0.0074 0.001111 1.0000 0.9996 0.9880 0.9125 0.7151 0.4406 0.2053 0.0709 0.0179 0.003212 1.0000 0.9999 0.9957 0.9568 0.8209 0.5772 0.3143 0.1285 0.0386 0.008313 1.0000 1.0000 0.9986 0.9806 0.8968 0.7032 0.4408 0.2112 0.0751 0.019214 1.0000 1.0000 0.9996 0.9921 0.9456 0.8074 0.5721 0.3174 0.1326 0.040315 1.0000 1.0000 0.9999 0.9971 0.9738 0.8849 0.6946 0.4402 0.2142 0.076916 1.0000 1.0000 1.0000 0.9990 0.9884 0.9367 0.7978 0.5681 0.3185 0.134117 1.0000 1.0000 1.0000 0.9997 0.9953 0.9680 0.8761 0.6885 0.4391 0.214818 1.0000 1.0000 1.0000 0.9999 0.9983 0.9852 0.9301 0.7911 0.5651 0.317919 1.0000 1.0000 1.0000 1.0000 0.9994 0.9937 0.9637 0.8702 0.6844 0.437320 1.0000 1.0000 1.0000 1.0000 0.9998 0.9976 0.9827 0.9256 0.7870 0.562721 1.0000 1.0000 1.0000 1.0000 1.0000 0.9991 0.9925 0.9608 0.8669 0.682122 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9970 0.9811 0.9233 0.785223 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9989 0.9917 0.9595 0.865924 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.9966 0.9804 0.923125 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9988 0.9914 0.959726 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.9966 0.980827 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9988 0.991728 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.996829 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.998930 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999731 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999932 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

DRAFT

Subject toOfq


tio

n


p = 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 n = 50, x = 0 0.0769 0.0052 0.0003 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

1 0.2794 0.0338 0.0029 0.0002 0.0000 0.0000 0.0000 0.0000 0.0000 0.00002 0.5405 0.1117 0.0142 0.0013 0.0001 0.0000 0.0000 0.0000 0.0000 0.00003 0.7604 0.2503 0.0460 0.0057 0.0005 0.0000 0.0000 0.0000 0.0000 0.00004 0.8964 0.4312 0.1121 0.0185 0.0021 0.0002 0.0000 0.0000 0.0000 0.00005 0.9622 0.6161 0.2194 0.0480 0.0070 0.0007 0.0001 0.0000 0.0000 0.0000 6 0.9882 0.7702 0.3613 0.1034 0.0194 0.0025 0.0002 0.0000 0.0000 0.00007 0.9968 0.8779 0.5188 0.1904 0.0453 0.0073 0.0008 0.0001 0.0000 0.00008 0.9992 0.9421 0.6681 0.3073 0.0916 0.0183 0.0025 0.0002 0.0000 0.00009 0.9998 0.9755 0.7911 0.4437 0.1637 0.0402 0.0067 0.0008 0.0001 0.0000

10 1.0000 0.9906 0.8801 0.5836 0.2622 0.0789 0.0160 0.0022 0.0002 0.0000 11 1.0000 0.9968 0.9372 0.7107 0.3816 0.1390 0.0342 0.0057 0.0006 0.000012 1.0000 0.9990 0.9699 0.8139 0.5110 0.2229 0.0661 0.0133 0.0018 0.000213 1.0000 0.9997 0.9868 0.8894 0.6370 0.3279 0.1163 0.0280 0.0045 0.000514 1.0000 0.9999 0.9947 0.9393 0.7481 0.4468 0.1878 0.0540 0.0104 0.001315 1.0000 1.0000 0.9981 0.9692 0.8369 0.5692 0.2801 0.0955 0.0220 0.0033 16 1.0000 1.0000 0.9993 0.9856 0.9017 0.6839 0.3889 0.1561 0.0427 0.007717 1.0000 1.0000 0.9998 0.9937 0.9449 0.7822 0.5060 0.2369 0.0765 0.016418 1.0000 1.0000 0.9999 0.9975 0.9713 0.8594 0.6216 0.3356 0.1273 0.032519 1.0000 1.0000 1.0000 0.9991 0.9861 0.9152 0.7264 0.4465 0.1974 0.059520 1.0000 1.0000 1.0000 0.9997 0.9937 0.9522 0.8139 0.5610 0.2862 0.1013 21 1.0000 1.0000 1.0000 0.9999 0.9974 0.9749 0.8813 0.6701 0.3900 0.161122 1.0000 1.0000 1.0000 1.0000 0.9990 0.9877 0.9290 0.7660 0.5019 0.239923 1.0000 1.0000 1.0000 1.0000 0.9996 0.9944 0.9604 0.8438 0.6134 0.335924 1.0000 1.0000 1.0000 1.0000 0.9999 0.9976 0.9793 0.9022 0.7160 0.443925 1.0000 1.0000 1.0000 1.0000 1.0000 0.9991 0.9900 0.9427 0.8034 0.5561 26 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9955 0.9686 0.8721 0.664127 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9981 0.9840 0.9220 0.760128 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9993 0.9924 0.9556 0.838929 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9966 0.9765 0.898730 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9986 0.9884 0.9405 31 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9995 0.9947 0.967532 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9978 0.983633 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9991 0.992334 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.996735 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9987 36 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999537 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999838 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

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Percentage Points Of The Normal Distribution The values z in the table are those which a random variable Z N(0, 1) exceeds with probability p; that is, P(Z > z) = 1 (z) = p.

p z p z

0.5000 0.0000 0.0500 1.6449

0.4000 0.2533 0.0250 1.9600

0.3000 0.5244 0.0100 2.3263 0.2000 0.8416 0.0050 2.5758

0.1500 1.0364 0.0010 3.0902

0.1000 1.2816 0.0005 3.2905

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Poisson Cumulative Distribution Function

The tabulated value is P(X x), where X has a Poisson distribution with parameter .

= 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 x = 0 0.6065 0.3679 0.2231 0.1353 0.0821 0.0498 0.0302 0.0183 0.0111 0.0067

1 0.9098 0.7358 0.5578 0.4060 0.2873 0.1991 0.1359 0.0916 0.0611 0.0404 2 0.9856 0.9197 0.8088 0.6767 0.5438 0.4232 0.3208 0.2381 0.1736 0.1247 3 0.9982 0.9810 0.9344 0.8571 0.7576 0.6472 0.5366 0.4335 0.3423 0.2650 4 0.9998 0.9963 0.9814 0.9473 0.8912 0.8153 0.7254 0.6288 0.5321 0.4405 5 1.0000 0.9994 0.9955 0.9834 0.9580 0.9161 0.8576 0.7851 0.7029 0.6160 6 1.0000 0.9999 0.9991 0.9955 0.9858 0.9665 0.9347 0.8893 0.8311 0.7622 7 1.0000 1.0000 0.9998 0.9989 0.9958 0.9881 0.9733 0.9489 0.9134 0.8666 8 1.0000 1.0000 1.0000 0.9998 0.9989 0.9962 0.9901 0.9786 0.9597 0.9319 9 1.0000 1.0000 1.0000 1.0000 0.9997 0.9989 0.9967 0.9919 0.9829 0.9682

10 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9990 0.9972 0.9933 0.9863 11 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9991 0.9976 0.9945 12 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9992 0.9980 13 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9993 14 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9998 15 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 16 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 17 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 18 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 19 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

= 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 x = 0 0.0041 0.0025 0.0015 0.0009 0.0006 0.0003 0.0002 0.0001 0.0001 0.0000

1 0.0266 0.0174 0.0113 0.0073 0.0047 0.0030 0.0019 0.0012 0.0008 0.0005 2 0.0884 0.0620 0.0430 0.0296 0.0203 0.0138 0.0093 0.0062 0.0042 0.0028 3 0.2017 0.1512 0.1118 0.0818 0.0591 0.0424 0.0301 0.0212 0.0149 0.0103 4 0.3575 0.2851 0.2237 0.1730 0.1321 0.0996 0.0744 0.0550 0.0403 0.0293 5 0.5289 0.4457 0.3690 0.3007 0.2414 0.1912 0.1496 0.1157 0.0885 0.0671 6 0.6860 0.6063 0.5265 0.4497 0.3782 0.3134 0.2562 0.2068 0.1649 0.1301 7 0.8095 0.7440 0.6728 0.5987 0.5246 0.4530 0.3856 0.3239 0.2687 0.2202 8 0.8944 0.8472 0.7916 0.7291 0.6620 0.5925 0.5231 0.4557 0.3918 0.3328 9 0.9462 0.9161 0.8774 0.8305 0.7764 0.7166 0.6530 0.5874 0.5218 0.4579

10 0.9747 0.9574 0.9332 0.9015 0.8622 0.8159 0.7634 0.7060 0.6453 0.5830 11 0.9890 0.9799 0.9661 0.9467 0.9208 0.8881 0.8487 0.8030 0.7520 0.6968 12 0.9955 0.9912 0.9840 0.9730 0.9573 0.9362 0.9091 0.8758 0.8364 0.7916 13 0.9983 0.9964 0.9929 0.9872 0.9784 0.9658 0.9486 0.9261 0.8981 0.8645 14 0.9994 0.9986 0.9970 0.9943 0.9897 0.9827 0.9726 0.9585 0.9400 0.9165 15 0.9998 0.9995 0.9988 0.9976 0.9954 0.9918 0.9862 0.9780 0.9665 0.9513 16 0.9999 0.9998 0.9996 0.9990 0.9980 0.9963 0.9934 0.9889 0.9823 0.9730 17 1.0000 0.9999 0.9998 0.9996 0.9992 0.9984 0.9970 0.9947 0.9911 0.9857 18 1.0000 1.0000 0.9999 0.9999 0.9997 0.9993 0.9987 0.9976 0.9957 0.9928 19 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9995 0.9989 0.9980 0.9965 20 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9998 0.9996 0.9991 0.9984 21 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9998 0.9996 0.9993 22 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9999 0.9997

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Percentage Points of the 2 Distribution

The values in the table are those which a random variable with the 2 distribution on degrees of freedom exceeds with the probability shown.

0.995 0.990 0.975 0.950 0.900 0.100 0.050 0.025 0.010 0.0051 0.000 0.000 0.001 0.004 0.016 2.705 3.841 5.024 6.635 7.8792 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.5973 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.8384 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.8605 0.412 0.554 0.831 1.145 1.610 9.236 11.070 12.832 15.086 16.7506 0.676 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.812 18.5487 0.989 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.475 20.2788 1.344 1.646 2.180 2.733 3.490 13.362 15.507 17.535 20.090 21.9559 1.735 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666 23.589

10 2.156 2.558 3.247 3.940 4.865 15.987 18.307 20.483 23.209 25.18811 2.603 3.053 3.816 4.575 5.580 17.275 19.675 21.920 24.725 26.75712 3.074 3.571 4.404 5.226 6.304 18.549 21.026 23.337 26.217 28.30013 3.565 4.107 5.009 5.892 7.042 19.812 22.362 24.736 27.688 29.81914 4.075 4.660 5.629 6.571 7.790 21.064 23.685 26.119 29.141 31.31915 4.601 5.229 6.262 7.261 8.547 22.307 24.996 27.488 30.578 32.80116 5.142 5.812 6.908 7.962 9.312 23.542 26.296 28.845 32.000 34.26717 5.697 6.408 7.564 8.672 10.085 24.769 27.587 30.191 33.409 35.71818 6.265 7.015 8.231 9.390 10.865 25.989 28.869 31.526 34.805 37.15619 6.844 7.633 8.907 10.117 11.651 27.204 30.144 32.852 36.191 38.58220 7.434 8.260 9.591 10.851 12.443 28.412 31.410 34.170 37.566 39.99721 8.034 8.897 10.283 11.591 13.240 29.615 32.671 35.479 38.932 41.40122 8.643 9.542 10.982 12.338 14.042 30.813 33.924 36.781 40.289 42.79623 9.260 10.196 11.689 13.091 14.848 32.007 35.172 38.076 41.638 44.18124 9.886 10.856 12.401 13.848 15.659 33.196 36.415 39.364 42.980 45.55825 10.520 11.524 13.120 14.611 16.473 34.382 37.652 40.646 44.314 46.92826 11.160 12.198 13.844 15.379 17.292 35.563 38.885 41.923 45.642 48.29027 11.808 12.879 14.573 16.151 18.114 36.741 40.113 43.194 46.963 49.64528 12.461 13.565 15.308 16.928 18.939 37.916 41.337 44.461 48.278 50.99329 13.121 14.256 16.047 17.708 19.768 39.088 42.557 45.722 49.588 52.33630 13.787 14.953 16.791 18.493 20.599 40.256 43.773 46.979 50.892 53.672

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Critical Values for Correlation Coefficients

These tables concern tests of the hypothesis that a population correlation coefficient is 0. The values in the tables are the minimum values which need to be reached by a sample correlation coefficient in order to be significant at the level shown, on a one-tailed test.

Product Moment Coefficient Sample Level

Spearman’s Coefficient

0.10 0.05 Level 0.025

0.01 0.005 0.05 Level 0.025

0.01

0.8000 0.9000 0.9500 0.9800 0.9900 4 1.0000 - - 0.6870 0.8054 0.8783 0.9343 0.9587 5 0.9000 1.0000 1.0000 0.6084 0.7293 0.8114 0.8822 0.9172 6 0.8286 0.8857 0.9429 0.5509 0.6694 0.7545 0.8329 0.8745 7 0.7143 0.7857 0.8929 0.5067 0.6215 0.7067 0.7887 0.8343 8 0.6429 0.7381 0.8333 0.4716 0.5822 0.6664 0.7498 0.7977 9 0.6000 0.7000 0.7833 0.4428 0.5494 0.6319 0.7155 0.7646 10 0.5636 0.6485 0.7455 0.4187 0.5214 0.6021 0.6851 0.7348 11 0.5364 0.6182 0.7091 0.3981 0.4973 0.5760 0.6581 0.7079 12 0.5035 0.5874 0.6783 0.3802 0.4762 0.5529 0.6339 0.6835 13 0.4835 0.5604 0.6484 0.3646 0.4575 0.5324 0.6120 0.6614 14 0.4637 0.5385 0.6264 0.3507 0.4409 0.5140 0.5923 0.6411 15 0.4464 0.5214 0.6036 0.3383 0.4259 0.4973 0.5742 0.6226 16 0.4294 0.5029 0.5824 0.3271 0.4124 0.4821 0.5577 0.6055 17 0.4142 0.4877 0.5662 0.3170 0.4000 0.4683 0.5425 0.5897 18 0.4014 0.4716 0.5501 0.3077 0.3887 0.4555 0.5285 0.5751 19 0.3912 0.4596 0.5351 0.2992 0.3783 0.4438 0.5155 0.5614 20 0.3805 0.4466 0.5218 0.2914 0.3687 0.4329 0.5034 0.5487 21 0.3701 0.4364 0.5091 0.2841 0.3598 0.4227 0.4921 0.5368 22 0.3608 0.4252 0.4975 0.2774 0.3515 0.4133 0.4815 0.5256 23 0.3528 0.4160 0.4862 0.2711 0.3438 0.4044 0.4716 0.5151 24 0.3443 0.4070 0.4757 0.2653 0.3365 0.3961 0.4622 0.5052 25 0.3369 0.3977 0.4662 0.2598 0.3297 0.3882 0.4534 0.4958 26 0.3306 0.3901 0.4571 0.2546 0.3233 0.3809 0.4451 0.4869 27 0.3242 0.3828 0.4487 0.2497 0.3172 0.3739 0.4372 0.4785 28 0.3180 0.3755 0.4401 0.2451 0.3115 0.3673 0.4297 0.4705 29 0.3118 0.3685 0.4325 0.2407 0.3061 0.3610 0.4226 0.4629 30 0.3063 0.3624 0.4251 0.2070 0.2638 0.3120 0.3665 0.4026 40 0.2640 0.3128 0.3681 0.1843 0.2353 0.2787 0.3281 0.3610 50 0.2353 0.2791 0.3293 0.1678 0.2144 0.2542 0.2997 0.3301 60 0.2144 0.2545 0.3005 0.1550 0.1982 0.2352 0.2776 0.3060 70 0.1982 0.2354 0.2782 0.1448 0.1852 0.2199 0.2597 0.2864 80 0.1852 0.2201 0.2602 0.1364 0.1745 0.2072 0.2449 0.2702 90 0.1745 0.2074 0.2453 0.1292 0.1654 0.1966 0.2324 0.2565 100 0.1654 0.1967 0.2327

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Random Numbers

86 13 84 10 07 30 39 05 97 96 88 07 37 26 04 89 13 48 19 20

60 78 48 12 99 47 09 46 91 33 17 21 03 94 79 00 08 50 40 16

78 48 06 37 82 26 01 06 64 65 94 41 17 26 74 66 61 93 24 97

80 56 90 79 66 94 18 40 97 79 93 20 41 51 25 04 20 71 76 04

99 09 39 25 66 31 70 56 30 15 52 17 87 55 31 11 10 68 98 23

56 32 32 72 91 65 97 36 56 61 12 79 95 17 57 16 53 58 96 36

66 02 49 93 97 44 99 15 56 86 80 57 11 78 40 23 58 40 86 14

31 77 53 94 05 93 56 14 71 23 60 46 05 33 23 72 93 10 81 23

98 79 72 43 14 76 54 77 66 29 84 09 88 56 75 86 41 67 04 42

50 97 92 15 10 01 57 01 87 33 73 17 70 18 40 21 24 20 66 62

90 51 94 50 12 48 88 95 09 34 09 30 22 27 25 56 40 76 01 59

31 99 52 24 13 43 27 88 11 39 41 65 00 84 13 06 31 79 74 97

22 96 23 34 46 12 67 11 48 06 99 24 14 83 78 37 65 73 39 47

06 84 55 41 27 06 74 59 14 29 20 14 45 75 31 16 05 41 22 96

08 64 89 30 25 25 71 35 33 31 04 56 12 67 03 74 07 16 49 32

86 87 62 43 15 11 76 49 79 13 78 80 93 89 09 57 07 14 40 74

94 44 97 13 77 04 35 02 12 76 60 91 93 40 81 06 85 85 72 84

63 25 55 14 66 47 99 90 02 90 83 43 16 01 19 69 11 78 87 16

11 22 83 98 15 21 18 57 53 42 91 91 26 52 89 13 86 00 47 61

01 70 10 83 94 71 13 67 11 12 36 54 53 32 90 43 79 01 95 15

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Percentage Points of Student’s t Distribution

The values in the table are those which a random variable with Student’s t distribution on degrees of freedom exceeds with the probability shown.

0.10 0.05 0.025 0.01 0.005 1 3.078 6.314 12.706 31.821 63.657 2 1.886 2.920 4.303 6.965 9.925 3 1.638 2.353 3.182 4.541 5.841 4 1.533 2.132 2.776 3.747 4.604 5 1.476 2.015 2.571 3.365 4.032 6 1.440 1.943 2.447 3.143 3.707 7 1.415 1.895 2.365 2.998 3.499 8 1.397 1.860 2.306 2.896 3.355 9 1.383 1.833 2.262 2.821 3.250 10 1.372 1.812 2.228 2.764 3.169 11 1.363 1.796 2.201 2.718 3.106 12 1.356 1.782 2.179 2.681 3.055 13 1.350 1.771 2.160 2.650 3.012 14 1.345 1.761 2.145 2.624 2.977 15 1.341 1.753 2.131 2.602 2.947 16 1.337 1.746 2.120 2.583 2.921 17 1.333 1.740 2.110 2.567 2.898 18 1.330 1.734 2.101 2.552 2.878 19 1.328 1.729 2.093 2.539 2.861 20 1.325 1.725 2.086 2.528 2.845 21 1.323 1.721 2.080 2.518 2.831 22 1.321 1.717 2.074 2.508 2.819 23 1.319 1.714 2.069 2.500 2.807 24 1.318 1.711 2.064 2.492 2.797 25 1.316 1.708 2.060 2.485 2.787 26 1.315 1.706 2.056 2.479 2.779 27 1.314 1.703 2.052 2.473 2.771 28 1.313 1.701 2.048 2.467 2.763 29 1.311 1.699 2.045 2.462 2.756 30 1.310 1.697 2.042 2.457 2.750 32 1.309 1.694 2.037 2.449 2.738 34 1.307 1.691 2.032 2.441 2.728 36 1.306 1.688 2.028 2.435 2.719 38 1.304 1.686 2.024 2.429 2.712 40 1.303 1.684 2.021 2.423 2.704 45 1.301 1.679 2.014 2.412 2.690 50 1.299 1.676 2.009 2.403 2.678 55 1.297 1.673 2.004 2.396 2.668 60 1.296 1.671 2.000 2.390 2.660 70 1.294 1.667 1.994 2.381 2.648 80 1.292 1.664 1.990 2.374 2.639 90 1.291 1.662 1.987 2.369 2.632

100 1.290 1.660 1.984 2.364 2.626 110 1.289 1.659 1.982 2.361 2.621 120 1.289 1.658 1.980 2.358 2.617

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Percentage Points of the F Distribution

The values in the table are those which a random variable with the F distribution on 1 and 2 degrees of freedom exceeds with probability 0.05 or 0.01.

Probability 1

2

1 2 3 4 5 6 8 10 12 24

0.05

1 161.4 199.5 215.7 224.6 230.2 234.0 238.9 241.9 243.9 249.1 254.3 2 18.51 19.00 19.16 19.25 19.30 19.33 19.37 19.40 19.41 19.46 19.50 3 10.13 9.55 9.28 9.12 9.01 8.94 8.85 8.79 8.74 8.64 8.53 4 7.71 6.94 6.59 6.39 6.26 6.16 6.04 5.96 5.91 5.77 5.63 5 6.61 5.79 5.41 5.19 5.05 4.95 4.82 4.74 4.68 4.53 4.37 6 5.99 5.14 4.76 4.53 4.39 4.28 4.15 4.06 4.00 3.84 3.67 7 5.59 4.74 4.35 4.12 3.97 3.87 3.73 3.64 3.57 3.41 3.23 8 5.32 4.46 4.07 3.84 3.69 3.58 3.44 3.35 3.28 3.12 2.93 9 5.12 4.26 3.86 3.63 3.48 3.37 3.23 3.14 3.07 2.90 2.71

10 4.96 4.10 3.71 3.48 3.33 3.22 3.07 2.98 2.91 2.74 2.54 11 4.84 3.98 3.59 3.36 3.20 3.09 2.95 2.85 2.79 2.61 2.40 12 4.75 3.89 3.49 3.26 3.11 3.00 2.85 2.75 2.69 2.51 2.30 14 4.60 3.74 3.34 3.11 2.96 2.85 2.70 2.60 2.53 2.35 2.13 16 4.49 3.63 3.24 3.01 2.85 2.74 2.59 2.49 2.42 2.24 2.01 18 4.41 3.55 3.16 2.93 2.77 2.66 2.51 2.41 2.34 2.15 1.92 20 4.35 3.49 3.10 2.87 2.71 2.60 2.45 2.35 2.28 2.08 1.84 25 4.24 3.39 2.99 2.76 2.60 2.49 2.34 2.24 2.16 1.96 1.71 30 4.17 3.32 2.92 2.69 2.53 2.42 2.27 2.16 2.09 1.89 1.62 40 4.08 3.23 2.84 2.61 2.45 2.34 2.18 2.08 2.00 1.79 1.51 60 4.00 3.15 2.76 2.53 2.37 2.25 2.10 1.99 1.92 1.70 1.39

120 3.92 3.07 2.68 2.45 2.29 2.18 2.02 1.91 1.83 1.61 1.25

3.84 3.00 2.60 2.37 2.21 2.10 1.94 1.83 1.75 1.52 1.00

0.01

1 4052. 5000. 5403. 5625. 5764. 5859. 5982. 6056. 6106. 6235. 6366. 2 98.50 99.00 99.17 99.25 99.30 99.33 99.37 99.40 99.42 99.46 99.50 3 34.12 30.82 29.46 28.71 28.24 27.91 27.49 27.23 27.05 26.60 26.13 4 21.20 18.00 16.69 15.98 15.52 15.21 14.80 14.55 14.37 13.93 13.45 5 16.26 13.27 12.06 11.39 10.97 10.67 10.29 10.05 9.89 9.47 9.02 6 13.70 10.90 9.78 9.15 8.75 8.47 8.10 7.87 7.72 7.31 6.88 7 12.20 9.55 8.45 7.85 7.46 7.19 6.84 6.62 6.47 6.07 5.65 8 11.30 8.65 7.59 7.01 6.63 6.37 6.03 5.81 5.67 5.28 4.86 9 10.60 8.02 6.99 6.42 6.06 5.80 5.47 5.26 5.11 4.73 4.31

10 10.00 7.56 6.55 5.99 5.64 5.39 5.06 4.85 4.17 4.33 3.91 11 9.65 7.21 6.22 5.67 5.32 5.07 4.74 4.54 4.40 4.02 3.60 12 9.33 6.93 5.95 5.41 5.06 4.82 4.50 4.30 4.16 3.78 3.36 14 8.86 6.51 5.56 5.04 4.70 4.46 4.14 3.94 3.80 3.43 3.00 16 8.53 6.23 5.29 4.77 4.44 4.20 3.89 3.69 3.55 3.18 2.75 18 8.29 6.01 5.09 4.58 4.25 4.01 3.71 3.51 3.37 3.00 2.57 20 8.10 5.85 4.94 4.43 4.10 3.87 3.56 3.37 3.23 2.86 2.42 25 7.77 5.57 4.68 4.18 3.86 3.63 3.32 3.13 2.99 2.62 2.17 30 7.56 5.39 4.51 4.02 3.70 3.47 3.17 2.98 2.84 2.47 2.01 40 7.31 5.18 4.31 3.83 3.51 3.29 2.99 2.80 2.66 2.29 1.80 60 7.08 4.98 4.13 3.65 3.34 3.12 2.82 2.63 2.50 2.12 1.60

120 6.85 4.79 3.95 3.48 3.17 2.96 2.66 2.47 2.34 1.95 1.38

6.63 4.61 3.78 3.32 3.02 2.80 2.51 2.32 2.18 1.79 1.00

If an upper percentage point of the F distribution on 1 and 2 degrees of freedom is f , then the corresponding lower percentage point of the F distribution on 2 and 1 degrees of freedom is 1/ f.

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A Level Mathematics - Weebly · 2020-02-22 · A Level Mathematics Sample Assessment Materials DRAFT Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) First teaching from