GCE AS Mathematics Summer 010 Mark Schemes Issued: October 010
NORTHERN IRELAND GENERAL CERTIFICATE OF SECONDARY EDUCATION (GCSE) AND NORTHERN IRELAND GENERAL CERTIFICATE OF EDUCATION (GCE) Introduction MARK SCHEMES (010) Foreword Mark Schemes are published to assist teachers and students in their preparation for examinations. Through the mark schemes teachers and students will be able to see what examiners are looking for in response to questions and exactly where the marks have been awarded. The publishing of the mark schemes may help to show that examiners are not concerned about finding out what a student does not know but rather with rewarding students for what they do know. The Purpose of Mark Schemes Examination papers are set and revised by teams of examiners and revisers appointed by the Council. The teams of examiners and revisers include experienced teachers who are familiar with the level and standards expected of 16 and 18-year-old students in schools and colleges. The job of the examiners is to set the questions and the mark schemes; and the job of the revisers is to review the questions and mark schemes commenting on a large range of issues about which they must be satisfied before the question papers and mark schemes are finalised. The questions and the mark schemes are developed in association with each other so that the issues of differentiation and positive achievement can be addressed right from the start. Mark schemes therefore are regarded as a part of an integral process which begins with the setting of questions and ends with the marking of the examination. The main purpose of the mark scheme is to provide a uniform basis for the marking process so that all the markers are following exactly the same instructions and making the same judgements in so far as this is possible. Before marking begins a standardising meeting is held where all the markers are briefed using the mark scheme and samples of the students work in the form of scripts. Consideration is also given at this stage to any comments on the operational papers received from teachers and their organisations. During this meeting, and up to and including the end of the marking, there is provision for amendments to be made to the mark scheme. What is published represents this final form of the mark scheme. It is important to recognise that in some cases there may well be other correct responses which are equally acceptable to those published: the mark scheme can only cover those responses which emerged in the examination. There may also be instances where certain judgements may have to be left to the experience of the examiner, for example, where there is no absolute correct response all teachers will be familiar with making such judgements. The Council hopes that the mark schemes will be viewed and used in a constructive way as a further support to the teaching and learning processes. iii
CONTENTS Page Unit C1 1 Unit C 7 Unit F1 15 Unit Unit S1 1 v
ADVANCED SUBSIDIARY (AS) General Certificate of Education 010 Mathematics Assessment Unit C1 assessing Module C1: Core Mathematics 1 [AMC11] WEDNESDAY 9 JUNE, AFTERNOON Standardising Meeting Version MARK SCHEME 1
GCE ADVANCED/ADVANCED SUBSIDIARY (AS) MATHEMATICS Introduction The mark scheme normally provides the most popular solution to each question. Other solutions given by candidates are evaluated and credit given as appropriate; these alternative methods are not usually illustrated in the published mark scheme. The marks awarded for each question are shown in the right-hand column and they are prefixed by the letters M, W and MW as appropriate. The key to the mark scheme is given below: M W indicates marks for correct method. indicates marks for working. MW indicates marks for combined method and working. The solution to a question gains marks for correct method and marks for an accurate working based on this method. Where the method is not correct no marks can be given. A later part of a question may require a candidate to use an answer obtained from an earlier part of the same question. A candidate who gets the wrong answer to the earlier part and goes on to the later part is naturally unaware that the wrong data is being used and is actually undertaking the solution of a parallel problem from the point at which the error occurred. If such a candidate continues to apply correct method, then the candidate s individual working must be followed through from the error. If no further errors are made, then the candidate is penalised only for the initial error. Solutions containing two or more working or transcription errors are treated in the same way. This process is usually referred to as follow-through marking and allows a candidate to gain credit for that part of a solution which follows a working or transcription error. Positive marking: It is our intention to reward candidates for any demonstration of relevant knowledge, skills or understanding. For this reason we adopt a policy of following through their answers, that is, having penalised a candidate for an error, we mark the succeeding parts of the question using the candidate s value or answers and award marks accordingly. Some common examples of this occur in the following cases: (a) a numerical error in one entry in a table of values might lead to several answers being incorrect, but these might not be essentially separate errors; (b) readings taken from candidates inaccurate graphs may not agree with the answers expected but might be consistent with the graphs drawn. When the candidate misreads a question in such a way as to make the question easier only a proportion of the marks will be available (based on the professional judgement of the examining team). [Turn over
1 (i) Midpoint P = (1, ) (ii) Grad m = 7 = 10 = 5 4 6 Perp. grad = 5 y + = (x 1) 5 x 5y 1 = 0 6 (i) f( 1) = a b + 6 = 1 M a + b = 9 (ii) f() = 7a 7 + b + 6 = 0 9a + b = 7 (iii) 9a + b = 7 a + b = 9 8a = 16 a = b = 11 7 (i) y > Α = (5, ) > x (ii) y > > x Α = (5, 4) (iii) y > 5, 4) > x 6
(x )(x + ) x 4 (a) x + 1 (x + 1) (x )(x + ) (x + 1) x + 1 x (x + ) (b) ( 7) ( 7 + ) ( 7 ) ( 7 + ) 7 + 6 7 7 7 4 = 7 1 MW (c) x x+ 1 = x+ 1 x = x + 1= x x = 14 dy 5 (i) = 4x 6x MW dx (ii) dy = 4x 6x = 0 dx x ( 4x 6) = 0 x = 0 or x = d y dx = 1x 1x d y x = = 7 18 ve min at x = dx dy x = 1 = 4 6 = 10 dx dy x = +1 = 4 6 = Point of inflection at x = 0 dx W 9 44 [Turn over
6 (i) y > y 10 = x > x (ii) y > y = x + 1 1 1 y 10 = x > x (iii) x + 1 = 10 x x + 1x 10 = 0 (x )(x + 5) = 0 x = or x = 5 W y = 15 or y = 9 5
7 (a) (i) x(x + 10) = 800 x + 0x 800 = 0 x + 10x 400 = 0 (b) 10 ± 100 + 1600 (ii) x = x = 10 ± 1700 Width = x = 10 + 10 17 Length = x + 10 = 5+ 5 17 b 4ac> 0 9 4p( 5 p) > 0 4p 0p+ 9> 0 ( p 1)( p 9) 1 9 p = or p = 1 p< or p> 4 1 M W 1 8 (i) P = l + r (ii) 400 = l + r l = 00 r (iii) A = lr A = (00 r)r A = 400r r da = 400 4 r dr 400 4 r = 0 r = 100 d A = 4 ve dr max l = 100 m 11 Total 75 66 [Turn over
ADVANCED General Certificate of Education 010 Mathematics Assessment Unit C assessing Module C: Core Mathematics [AMC1] THURSDAY 7 MAY, MORNING MARK SCHEME 77
GCE ADVANCED/ADVANCED SUBSIDIARY (AS) MATHEMATICS Introduction The mark scheme normally provides the most popular solution to each question. Other solutions given by candidates are evaluated and credit given as appropriate; these alternative methods are not usually illustrated in the published mark scheme. The marks awarded for each question are shown in the right-hand column and they are prefixed by the letters M, W and MW as appropriate. The key to the mark scheme is given below: M W indicates marks for correct method. indicates marks for working. MW indicates marks for combined method and working. The solution to a question gains marks for correct method and marks for an accurate working based on this method. Where the method is not correct no marks can be given. A later part of a question may require a candidate to use an answer obtained from an earlier part of the same question. A candidate who gets the wrong answer to the earlier part and goes on to the later part is naturally unaware that the wrong data is being used and is actually undertaking the solution of a parallel problem from the point at which the error occurred. If such a candidate continues to apply correct method, then the candidate s individual working must be followed through from the error. If no further errors are made, then the candidate is penalised only for the initial error. Solutions containing two or more working or transcription errors are treated in the same way. This process is usually referred to as follow-through marking and allows a candidate to gain credit for that part of a solution which follows a working or transcription error. Positive marking: It is our intention to reward candidates for any demonstration of relevant knowledge, skills or understanding. For this reason we adopt a policy of following through their answers, that is, having penalised a candidate for an error, we mark the succeeding parts of the question using the candidate s value or answers and award marks accordingly. Some common examples of this occur in the following cases: (a) a numerical error in one entry in a table of values might lead to several answers being incorrect, but these might not be essentially separate errors; (b) readings taken from candidates inaccurate graphs may not agree with the answers expected but might be consistent with the graphs drawn. When the candidate misreads a question in such a way as to make the question easier only a proportion of the marks will be available (based on the professional judgement of the examining team). 88 [Turn over
1 (i) u n + = 1 1+ u n u u u 1 = = = 1 4 u 4 = 6 7 u 5 = 14 1 MW (ii) converges and oscillates MW 5 (i) centre (0, ) MW radius = g + f c = 5 = 5 (ii) gradient of radius y x y x 1 1 = + = 0 4 gradient of tangent = 7 4 99
1 (i) Area of sector r 1 = 9 4 9 = cm 8 1 (ii) Area of triangle ab sin c 1 Area of triangle = 1 1 sin 60 = 0.4 cm 9π Volume of gold = 04. 01. 8 = 0. 10 cm 7 4 (i) a = 5 ar = r = 5 x = 5 9 x = = 5 5 9 x = 5 (ii) r < 1 (iii) S ` = a 1 r = 5 5 = 1 1 6 1010 [Turn over
5 (i) (1 + x) 4 4 4 = 1 + 4 x + (x) + (x) MW 1 1 = 1 + 1x + 54x + 108x (ii) 1 + 1x + 66x = 1 + 1x + 54x + 108x 0 = 108x 1x 0 = 1x (9x 1) 1 x 7 9 1111
6 (a) x dx x = x + + c MW (b) Area = = 0 a 4x x dx 4 4x x 4 4 = 4 a a 4 a 0 M MW if areas equal then a 4 a = 0 4 16 a Þ0 or a = 4a or Area = 4x x dx = 0 4 4 4x x 4 4 0 56 = 64 = 1 1 Area = x x dx a 4 4 a 4 4x x = 4 4 4 4 a 4 = 4 a a 1 1 4 = 0 M MW 4 if areas equal then 1 1 4 = a + a + 1 1 4 a = 16 11 11 [Turn over
7 (a) sin x + 8 cos x = 0 (1 cos x) + 8 cos x = 0 cos x + 8 cos x = 0 cos x 8 cos x = 0 ( cos x + 1)(cos x ) = 0 1 cos x = or cos x = MW x = ±1.91 c no sol n: MW (b) N N D θ 7 H 0 40 MW 45 C Distance d = 7 + 45 7 45 cos 110 d = 97.1 nm sin sin 110 Bearing = 45 97.1 = 5.8 bearing 180 + 0 5.8 = 184 17 11
8 (a) x = 7 log x = log 7 x log = log 7 log 7 x = log = 0.886 (b) log x + log x + log x = 1 log x + log x + 6 log x = 1 1 log x = 9 x = 1.9 or log x + log x + log x 6 = 1 log x 9 = 1 x 9 = 10 x = 1.9 (c) log x log y = 6 x log = 6 y x 6 = y x = y 1 x = z y y = xz 15 Total 75 1414 [Turn over
ADVANCED SUBSIDIARY (AS) General Certificate of Education 010 Mathematics Assessment Unit F1 assessing Module FP1: Further Pure Mathematics 1 [AMF11] THURSDAY 4 JUNE, MORNING Standardising Meeting Version MARK SCHEME 1515
GCE ADVANCED/ADVANCED SUBSIDIARY (AS) MATHEMATICS Introduction The mark scheme normally provides the most popular solution to each question. Other solutions given by candidates are evaluated and credit given as appropriate; these alternative methods are not usually illustrated in the published mark scheme. The marks awarded for each question are shown in the right-hand column and they are prefixed by the letters M, W and MW as appropriate. The key to the mark scheme is given below: M W indicates marks for correct method. indicates marks for working. MW indicates marks for combined method and working. The solution to a question gains marks for correct method and marks for an accurate working based on this method. Where the method is not correct no marks can be given. A later part of a question may require a candidate to use an answer obtained from an earlier part of the same question. A candidate who gets the wrong answer to the earlier part and goes on to the later part is naturally unaware that the wrong data is being used and is actually undertaking the solution of a parallel problem from the point at which the error occurred. If such a candidate continues to apply correct method, then the candidate s individual working must be followed through from the error. If no further errors are made, then the candidate is penalised only for the initial error. Solutions containing two or more working or transcription errors are treated in the same way. This process is usually referred to as follow-through marking and allows a candidate to gain credit for that part of a solution which follows a working or transcription error. Positive marking: It is our intention to reward candidates for any demonstration of relevant knowledge, skills or understanding. For this reason we adopt a policy of following through their answers, that is, having penalised a candidate for an error, we mark the succeeding parts of the question using the candidate s value or answers and award marks accordingly. Some common examples of this occur in the following cases: (a) a numerical error in one entry in a table of values might lead to several answers being incorrect, but these might not be essentially separate errors; (b) readings taken from candidates inaccurate graphs may not agree with the answers expected but might be consistent with the graphs drawn. When the candidate misreads a question in such a way as to make the question easier only a proportion of the marks will be available (based on the professional judgement of the examining team). 1616 [Turn over
1 (i) A I = 0 7 = 0 ( )( ) 1 = 0 4 + + 1 = 0 5 = 0 ( 5)( + 5) = 0 = ±5 (ii) 7 x y = 5 x y which gives x + 7y = 5x 7y = x x y = 5y x = 7y Hence an eigenvector is 7 Therefore a corresponding unit eigenvector is 1 58 7 9 (i) 9 a 1 1 = 0 a 1 1 9( + ) ( + a ) + a(1 + a 1) = 0 6a + a = 0 a( 6 + a) = 0 a = 0, 6 W (ii) 9x + y + 6z = 0 x y z = 0 5x + y + z = 0 + gives 6x + 0 + 0 = 0 Hence x = 0 Therefore using either or gives y = z Check equation which gives 0 6z + 6z = 0 The general solution is (0, t, t) 11 1717
(i) p q r s t u p p q r s t u q q p s r u t r r t p u q s s s u q t p r t t r u p s q u u s t q r p (ii) Identity is p s = p and hence s has period (iii) Inverse of r is r since r = p MW6 (iv) {p, s, t} 1 1818 [Turn over
4 (i) Centre is (, 5) Gradient of radius joining centre and the point (, 7) 7 5 1 = = + The tangent is perpendicular to the radius. Hence gradient of tangent is Equation of tangent is y 7 = (x ) which gives y = x +11 (ii) Substituting the point (, 5) into the equation of the tangent gives 5 = () + 11 5 = 6 + 11 Hence (, 5) lies on the tangent. (iii) y > (, 7) (, 5) (, 5) > x By symmetry the gradient of the other tangent is. Using the point (, 5) we obtain y 5 = (x ) y = x 6 + 5 y = x 1 11 1919
5 (a) (i) M = 1 0 0 1 MW (ii) Image = 1 0 0 1 = 7 7 Hence image is the point ( 7, ) (b) (i) Area of T = det N Area of S det N = 7 8 = 15 Hence area of T = 15 = 45 cm (ii) 1 x y = 4 7 mx my which gives x + mx = y 4x 7mx = my Divide to give 1 + m 1 = 4 7m m m m = 4 7m m 6m + 4 = 0 m m + = 0 Factorise to give (m )(m 1) = 0 Hence m =, 1 W 15 00 [Turn over
6 (i) + i = + Modulus = 1+ i = 1+ Modulus = 1 ( ) arg + i = tan π Hence argument = 4 1 ( ) arg 1 + i = tan π Hence argument = 1 (ii) Im > + Q(z 1 + z ) R(z ) P(z 1 ) 0 1 S 1+ > Re 11
(iii) Angle POR = π π 4 π = 1 π Angle QOS = + 1 π 4 1 = 7π 4 Q + O 1+ S Hence 7π tan = + 4 1+ 17 Total 75 [Turn over
ADVANCED SUBSIDIARY (AS) General Certificate of Education 010 Mathematics Assessment Unit assessing Module : Mechanics 1 [AM1] TUESDAY 18 MAY, MORNING MARK SCHEME
GCE ADVANCED/ADVANCED SUBSIDIARY (AS) MATHEMATICS Introduction The mark scheme normally provides the most popular solution to each question. Other solutions given by candidates are evaluated and credit given as appropriate; these alternative methods are not usually illustrated in the published mark scheme. The marks awarded for each question are shown in the right-hand column and they are prefixed by the letters M, W and MW as appropriate. The key to the mark scheme is given below: M W indicates marks for correct method. indicates marks for working. MW indicates marks for combined method and working. The solution to a question gains marks for correct method and marks for an accurate working based on this method. Where the method is not correct no marks can be given. A later part of a question may require a candidate to use an answer obtained from an earlier part of the same question. A candidate who gets the wrong answer to the earlier part and goes on to the later part is naturally unaware that the wrong data is being used and is actually undertaking the solution of a parallel problem from the point at which the error occurred. If such a candidate continues to apply correct method, then the candidate s individual working must be followed through from the error. If no further errors are made, then the candidate is penalised only for the initial error. Solutions containing two or more working or transcription errors are treated in the same way. This process is usually referred to as follow-through marking and allows a candidate to gain credit for that part of a solution which follows a working or transcription error. Positive marking: It is our intention to reward candidates for any demonstration of relevant knowledge, skills or understanding. For this reason we adopt a policy of following through their answers, that is, having penalised a candidate for an error, we mark the succeeding parts of the question using the candidate s value or answers and award marks accordingly. Some common examples of this occur in the following cases: (a) a numerical error in one entry in a table of values might lead to several answers being incorrect, but these might not be essentially separate errors; (b) readings taken from candidates inaccurate graphs may not agree with the answers expected but might be consistent with the graphs drawn. When the candidate misreads a question in such a way as to make the question easier only a proportion of the marks will be available (based on the professional judgement of the examining team). 44 [Turn over
1 6 sin 0 = P cos 10 P =.05 N Q = P sin 10 + 6 cos 0 Q = 5.7 N 7 (i) v(m s 1 ) 0 10 MW t 5 10 t(s) (ii) at = change in velocity 5 t = 0 10 t = 4 s (iii) Total distance = area under graph 10 + 0 5 + 15 4 + 0 MW = 80 m 8 55
(i) change in momentum = 0.05 (00 450) = 1.5 Ns (ii) I = F 0.00 = 1.5 F = 650 N 4 4 (i) R 10 10 6 R = 6 + 10 cos 10 R = 14 N or R 6 sin 60 10 + 6 cos 60 R = (10 + 6 cos 60 ) + (6 sin 60 ) R = 14 N (ii) F = ma 14 = 4 a a =.5 m s (iii) u = 0 a =.5 s = ut + 1 at t = s = 1.5 9 s =? = 15.8 m 7 66 [Turn over
5 (i) P Q P Q MW 90g 7g (ii) P + Q = 90g + 7g P + P = 16g P = 54g N (59) Q = 108g N (1058) (iii) Moments about P 7g 4 + 90g x = 108g 6 MW x = 4 m 11 6 (i) a = 4t 10t v = 4t 10t dt 5 4 = t t + c 5 5 at t = 0 v = c = 5 4 5 v = t t + (ii) 5 4 5 s = t t + dt 5 t t 5 = + t+ d at t = 0 s = 0 d = 0 5 t t 5 s = + t 77
(iii) when returns to start s = 0 5 t t 5t + = 0 5 4t t + 15t = 0 5 t 4t 15t = 0 4 t( t 4t 15) = 0 ( + )( ) = t t 5 t 0 t = 0 no sol n t = t = s returns to start after s 11 7 (i) + A B m m u 0 0 v momentum before = momentum after mu + 0= 0+ mv MW V = u 88 [Turn over
(ii) + B C m m u 0 v u momentum before = momentum after m u u + 0 = m V+ m u = V + u u V = B will collide with A again as B now moving towards A which is at rest. MW (iii) + A B m m u m V momentum before = momentum after m u + 0 = m V V = u u Speed in opposite direction to initial direction of A 11 99
8 (i) a T T µ R R 1 R a µ 1 R 1 g 5g 40 60 MW (ii) P F = ma T μ 1 R 1 g sin 40 = a W Q F = ma 5g sin 60 T μ R = 5a R 1 = g cos 40 R = 5g cos 60 T 0. g cos 40 g sin 40 = a 5g sin 60 T 0.15g cos 60 = 5a MW 5g sin 60 0.9g cos 40 g sin 40 0.5g cos 60 = 8a a = 1.79 m s W T 0.9g cos 40 g sin 40 = 1.79 T = 1.0 N 16 Total 75 00 [Turn over
ADVANCED SUBSIDIARY (AS) General Certificate of Education Summer 010 Mathematics Assessment Unit 1 assessing Module S1: Statistics 1 [AMS11] FRIDAY 4 JUNE, MORNING MARK SCHEME 1 [Turn over
GCE ADVANCED/ADVANCED SUBSIDIARY (AS) MATHEMATICS Introduction The mark scheme normally provides the most popular solution to each question. Other solutions given by candidates are evaluated and credit given as appropriate; these alternative methods are not usually illustrated in the published mark scheme. The marks awarded for each question are shown in the right hand column and they are prefixed by the letters M, W and MW as appropriate. The key to the mark scheme is given below: M W indicates marks for correct method. indicates marks for working. MW indicates marks for combined method and working. The solution to a question gains marks for correct method and marks for an accurate working based on this method. Where the method is not correct no marks can be given. A later part of a question may require a candidate to use an answer obtained from an earlier part of the same question. A candidate who gets the wrong answer to the earlier part and goes on to the later part is naturally unaware that the wrong data is being used and is actually undertaking the solution of a parallel problem from the point at which the error occurred. If such a candidate continues to apply correct method, then the candidate s individual working must be followed through from the error. If no further errors are made, then the candidate is penalised only for the initial error. Solutions containing two or more working or transcription errors are treated in the same way. This process is usually referred to as follow-through marking and allows a candidate to gain credit for that part of a solution which follows a working or transcription error. Positive marking: It is our intention to reward candidates for any demonstration of relevant knowledge, skills or understanding. For this reason we adopt a policy of following through their answers, that is, having penalised a candidate for an error, we mark the succeeding parts of the question using the candidate s value or answers and award marks accordingly. Some common examples of this occur in the following cases: (a) a numerical error in one entry in a table of values might lead to several answers being incorrect, but these might not be essentially separate errors; (b) readings taken from candidates inaccurate graphs may not agree with the answers expected but might be consistent with the graphs drawn. When the candidate misreads a question in such a way as to make the question easier only a proportion of the marks will be available (based on the professional judgement of the examining team). [Turn over
1 Midvalues: 4.5, 74.5, 14.5, 174.5, 4.5, (75) From calculator n = 00 fx = 700 fx = 441650 x = 16.5 σ n = 58.574 = 58.6 (sf) 5 (i) The outcome of one trial does not influence another (ii) possible outcomes probability of success is constant (iii) Let X be r.v. the number of people carrying a donor card X ~ B (6, 0.5) ( ) 6 P(X = ) = (0.5) (0.65) = 0.5 (sf) (iv) P(X ) = 1 P(X < ) = 1 P(X = 0, 1, ) ( ) ( ) 6 = 1 [ (0.5) 0 (0.65) 6 6 + (0.5) 1 (0.65) 0 1 5 6 + ( (0.5) (0.65) 4 ] MW ) = 1 [0.0754 + 0.466 + 0.800 ] = 0.59... = 0.5 (s.f.) 10 [Turn over
(i) Let X be r.v. No. of misprints per page X ~ Po (.6) e P(X = ) =.6! = 0.51 (s.f.) (ii) P(X ) = 1 P(X < ) = 1 P(X = 0, 1, ) [ ] = 1 e.6.6 0 +.6 1 +.6 MW 0! 1!! = 1 6.98e.6 = 0.48 (sf) (iii) P(X = 5 X ) = P(X = 5 X ) P(X ) P(X = 5) = P(X ) e P (X = 5) =.6.6 5 = 0.075 5! 0.075 P(X = 5 X ) = = 0.15 (s.f.) 11 0.48 4 (i) x 5 7 9 11 1 15 P(X = x) 1 16 16 16 4 16 16 16 1 16 MW4 (ii) Distribution is symmetrical. 9 is central value of X M (iii) E(X ) = 1 + 5 + 7 + 9 4 + 11 16 16 16 16 16 + 1 16 + 15 1 16 = 91 Var (X) = E (X ) [E(X)] = 91 9 = 10 10 4 [Turn over
5 (i) 1 0 k (x x ) dx = 1 M x 1 k x [ 4 ] 4 0 = 1 ( ) 1 1 k = 1 k = 4 4 1 1 (ii) P(0 X ) = 0 (4x 4x ) dx 1 = [x x 4 ] 0 1 1 7 = = 16 16 (iii) E(X ) = 1 0 x (4x 4x ) dx = 1 0 (4x 4x 5 ) dx [ ] ( ) = x 4 x 6 1 0 = 1 (0) 1 = Var (X) = E(X ) [E (X)] 1 8 = ( ) 11 = (0.048) 1 15 5 5 [Turn over
6 (i) X ~ N(4000, σ ) P(X > 405) = 0.40 P(X < 405) = 0.7580 405 4000 P(Z < ) = 0.7580 σ 5 σ 5 σ = Ф 1 (0.7580) = 0.7 σ = 50 4110 4000 (ii) P(X > 4110) = P(Z > ) = P(Z >.) 50 = 1 Ф (.) = 1 0.9861 = 0.019 888 4000 P(X < 888) = P(Z < ) = P(Z <.4) 50 = 1 Ф (.4) = 1 0.9875 = 0.015 P(X > 4110 or X < 888) = 0.019 + 0.015 = 0.064 =.64% 1 6 [Turn over
7 (a) (i) The events can t both happen P(A B) = 0 (ii) It is certain that at least one event occurs P(A B) = 1 (b) P(G) = 0.4 P(F) = 0.5 P(G F) = 0.61 P(F G) = P(G) + P(F) P(G F) = 0.4 + 0.5 0.61 = 0.14 P(G) P(F) = 0.4 0.5 = 0.14 = P(F G) Hence Yes the events are independent (c) (i) P(D) = x P(W D) = 0.75 P(W D) = 0.6 P(W D) = P(D) P(W D) = 0.75x P(W D) = P(W) + P(D) P(W D) 0.6 = P(W) + x 0.75x P(W) = 0.6 0.5x (ii) P(W) = 1.5x 1.5x = 0.6 0.5x 1.5x = 0.6 x = 0.4 14 Total 75 7 [Turn over