Mark Scheme (Results) January 0 GCE Statistics S (84) Paper
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General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded f what they have shown they can do rather than penalised f omissions. Examiners should mark accding to the mark scheme not accding to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. 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 wthy of credit accding to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out wk should be marked UNLESS the candidate has replaced it with an alternative response.
EDEXCEL GCE MATHEMATICS General Instructions f Marking. The total number of marks f the paper is 75.. The Edexcel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded f 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. 3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used f crect ft cao crect answer only cso - crect solution only. There must be no errs in this part of the question to obtain this mark isw igne subsequent wking awrt answers which round to SC: special case oe equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are crect answer only (cao.), unless shown, f example, as A ft to indicate that previous wrong wking is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks.
General Principals f Ce Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark f solving 3 term quadratic:. Factisation ( x + bx + c) = ( x + p)( x + q), where pq = c, leading to x =... ( ax + bx + c) = ( mx + p)( nx + q), where pq = c and mn = a, leading to x =. Fmula Attempt to use crect fmula (with values f a, b and c), leading to x = 3. Completing the square Solving + bx + c = 0 x b : ( ) x ± ± q ± c, q 0, leading to x = Method marks f differentiation and integration:. Differentiation Power of at least one term decreased by. ( n. Integration Power of at least one term increased by. ( n x x ) + x n x n ) Use of a fmula Where a method involves using a fmula that has been learnt, the advice given in recent examiners repts is that the fmula should be quoted first. Nmal marking procedure is as follows: Method mark f quoting a crect fmula and attempting to use it, even if there are mistakes in the substitution of values. Where the fmula is not quoted, the method mark can be gained by implication from crect wking with values, but may be lost if there is any mistake in the wking.
January 0 84 Statistics S Mark Scheme Question Number Scheme Marks (a) 9 + 3 E( X ) = = (b) (9 3) Var( X ) = = 3 A (c) P( X > 7) = (9 7) = 3 (d) P(4< X <) P( X < X > 4) = P( X >4) = = 5 5 Notes A A A () () () (3) 8 (b) (9 3) (9 + 3) (c) ( 9 7) (7 3) 9 dx 7 7 dx 3 A Also acceptable 0. 3 &, 0.33 & and awrt 0.333 (d) A P(4 < X < ) P( X > 4) P(4 < X < ) P( X > 4) P( X P( X 5 < > ) 4) 5 P( X P( X 3 5 > > ) 4) P( X P( X 4 9 4 > > ) 4) 4 9 4 5 3 An answer of 5 gains all 3 marks. NB and are accepted in the above fmulae
Question Number Scheme Marks H 0 : p = 0.5 H : p > 0.5 X~B(30,0.5) Using crect Bin P(X ) = - P(X 0) P(X 9) = 0.950 P(X 0) = 0.0494 = - 0.978 = 0.04 CR X 0 A so significant/reject H 0 /in Critical region Evidence to suggest David s claim is increct The weather fecast produced by the local radio is better than those achieved by tossing/flipping a coin Notes dep st f H 0 : p = 0.5 nd f H : p > 0.5 SC If both hypotheses are crect but a different letter to p is used they get B0. If no letter is used they get B0 B0. A (7) 7 st writing using B(30,0.5) One tail nd f writing using - P(X 0) writing P(X 9) = 0.950 P(X 0) = 0.0494. May be implied by crect CR. probability = 0.04 A f 0.04 CR X 0/ X >9. NB P(X 0) = 0.978 on its own sces A 3 rd dependent on the nd being awarded. F a crect statement based on the table below. Do not allow non-contextual conflicting statements eg significant and accept H 0. Igne comparisons. nd A f a crect contextualised statement. NB A crect contextual statement on its own sces A. 0.05 < p < 0.95 p < 0.05 p > 0.95 3 rd not significant/ accept H 0 / Not in CR significant/ reject H 0 / In CR nd A David's claim is crect weather fecast produced by the local radio is no better than those achieved by tossing/flipping a coin David's claim increct weather fecast produced by the local radio is better than those achieved by tossing/flipping a coin Two tail st f writing using - P(X < 0) writing P(X 0)= 0.978 P(X ) = 0.04. May be implied by crect CR. probability = 0.97 A f 0.04 CR X / X >0. NB P(X < 0) = 0.978 on its own sces A 3 rd dependent on the nd being awarded. F a crect statement based on the table below. Do not allow non-contextual conflicting statements eg significant and accept H 0. Igne comparisons. nd A f a crect contextualised statement. NB A crect contextual statement on its own sces A. 0.05 < p < 0.975 p < 0.05 p > 0.975 3 rd not significant/ accept H 0 / Not in CR significant/ reject H 0 / In CR nd A David's.claim is crect weather fecast produced by the local radio is no better than those achieved by tossing/flipping a coin David's claim increct weather fecast produced by the local radio is better than those achieved by tossing/flipping a coin Question Scheme Marks
Number 3 (a) 0 P( X = 0) = 0.85 from tables = 0.99 awrt 0.97 A (b) P ( X > 3) = P ( X 3) = 0.477 = 0.353 awrt 0.35 A (c) n 0.5 = 5 n = 33 34 A (d) - P(X = 0) > 0.95 ( 0.85) n > 0.95. A 0.85 n < 0.05 n >8.4 n = 9 A () () () (3) 9 Notes (a) (p) 0 with 0 < p < (b) writing using - P( X 3) (c) np = 5 0 < p < (d) writing using - P(X = 0) > 0.95 P(X = 0) < 0.05 (also accepted are = > instead of > and = instead of <) P(X 0) is equivalent to P(X =0) A writing using - (0.85) n > 0.95 (0.85) n < 0.05 (also accepted are > instead of > and instead of <). Any value of n may be used A cao NB an answer of 8.4 gets A A0 An answer of 9 gets A A unless it follows from clearly increct wking.
Question Number Scheme Marks 4 (a) Poisson () (b) Hits occur singly in time Hits are independent Hits occur randomly Hits occur at a constant rate () (c) X~ Po(5) P(X = 0) = P(X 0) - P(X 9) e 5 5 0 0! = 0.983-0.98 = 0.08 awrt 0.08 (d) X~ Po(0) (e) P( X 5) = P( X 4) = 0.95 = 0.0835 awrt 0.0835 A X~ Po(50) Approximated by N(50,50) 70.5 50 P( X > 70) = P Z > 50 A (3) (3) (b) (c) (d) (e) = P( Z >.899...) A = 0.998 = 0.009 awrt 0.009 A Notes st Any one of the 3 statements - no context required. NB It must be a constant (mean) rate and not a constant probability a constant mean. nd A different statement with context of hits. NB random and independent are the same statement. If only one mark awarded give the st. Never award B0 writing using Po(5) writing using P(X 0) - P(X 9) writing using Po(0) writing using - P(X 4) e 5 5 0 0! st f a nmal approximation nd f crect mean and sd (may be seen in standardisation fmula st f attempting a continuity crection (7 ± 0.5) nd Standardising using their mean and their sd and using [9.5, 70, 70.5, 7 7.5] allow ± z NB if they have not written down a mean and sd then they need to be crect in the standardisation to gain this mark. 70.5 50 st A f z = + awrt.9 better. May be awarded f ± 50 3rd f - tables value SC using P(X< 70.5/7.5) P(X<9.5/70.5) can get M0A0 M0A0 (7)
Question Number Scheme Marks (a) 5 (a) X ~ B(0,0.075) (b) Approximated by Po(9) P( X > 3) = P( X 3) = 0.0 = 0.9788 awrt 0.979 A P(At least 4 defective components in each box) =P( X >3) P( X >3) = 0.9788 = 0.95804944 awrt 0.958 A Notes Writing use of B(0,0.075) may be implied by using Po(9) N(9,8.35) st writing use of Poisson st A writing use of Po(9) nd f writing using - P( X 3) this may be implied by an awrt 0.97 using nmal approximation. A (5) () 7 (b) ((their (a)) 0.979 0.9788 0.98 Question Number Scheme Marks
(a) f(x) k-0.5 0.5 shape labels 0 k x () (b) k x dx = (c) k x x = k k = 0 o.e. A A cso k = ( + 5 ) (4) 0, x < 0 x, 0 x < F( x) = x x +, x k, x > k Note: Wking f the AA k x dx + C = x x ; + AA st and last () (A;A) (d) P(0.5 < X <.5) = F(.5) F(0.5) = 0.875-0.5 = 0.5 (e) Median is x = A () (f) Mode is x = k ( + 5) awrt. Negative skew Median<mode from graph me values are to the right. d () () 8 (a) Notes st Crect shape with straight lines. Must all be above the x-axis nd A fully crect graph with the labels, k, 0.5, k - 0.5 seen in the crect places. Allow the use of ( + 5) /awrt. instead of k.
(b) (c) k st x dx = 0.5 k x dx + 0.5 = igne limits k x dx + k dx = ( k 0.5 + 0.5)( k ) = 0. 5 any crect method of finding the area st A f a quadratic equation in the fm a(k k -) = 0 ak ak = a. where a is a constant. nd crect method f solving a quadratic of the fm ak - bk + c = 0 where a,b,c 0. There must be at least one crect step befe the final answer. Allow substituting in k into a quadratic of the fm ak - bk + c = 0. nd A cso f k = ( + 5 ) st f second line. Do not penalise the use of < instead of and vice versa k f use of x dx + C igne limits. F use they must have x x st A crect integration x x nd A C = (d) (e) (f) NB AA may be implied by crect 3rd line in F(x) nd f 3rd line. Statement of the fm x x ± C. Do not penalise the use of < instead of and vice versa. Allow k value of k. C may equal 0. 3rd f first and last line. Do not penalise the use of instead of < and instead of >. Allow k value of k Using F(.5) - F(0.5)..5 must be put into the third line of the c.d.f. and 0.5 must be put into the second line of the c.d.f.. +. 5 x dx x dx need to attempt integration, at least one x n x n+ 0.5 seeing 0.5 + 0.375 any crect method of finding the area.. (NB if they have not used + C C = 0 they will get 0.5. This will get A0). An answer of 0.5 from an increct method gains M0 A0. If it is not clear which one is the mode and which one is the median assume the median is the first answer and mode the second. negative/negative skew(ness). Do not allow negative crelation. dependent on previous B mark being awarded. Reason must follow from their values diagram. Question Scheme Number 7 (a) (i) The range of values/region/area/set of values of the test statistic that would lead you to reject H 0 (a) (ii) The probability of increctly rejecting H 0 Probability of rejecting H 0 when H 0 is true Marks ()
(b) (i) X ~Po(8) P( X 4) = 0.099 P( X 3) = 0.044 Critical region [0,3] (b) (ii) awrt 0.044 (3) (c) H 0 : λ = 8 ( µ =8) H : λ > 8 ( µ >8) P(X 3) = - P(X ) P(X 3) = 0.958 P(X 4) = 0.034 = - 0.93 = 0.038 CR X 4 A so insufficient evidence to reject H 0 /not significant/ not in critical region dep There in insufficient evidence of an increase/change in the rate/number of sales per A month the estate agents claim is increct (5) Notes 0 (a)(i) (ii) (b) Allow accept H instead of reject H 0. It must be clear which hypothesis gets rejected/accepted. Allow equivalent wding. Writing using Po(8). May be implied by crect critical region. A allow 0 X 3 CR 3 X 3. Any letter may be used but not P(X 3 ). This must be on its own. (c) both hypotheses crect. Must use λ µ. One tail st f writing using - P(X < ) writing P(X 3) = 0.958 P(X 4) = 0.034. May be implied by crect CR. probability = 0.038 A f 0.038 X 4. Allow X >3. NB P(X < ) = 0.93 on its own sces A nd dependent on the st being awarded. F a crect statement based on the table below. Do not allow noncontextual conflicting statements eg not significant and reject H 0. Igne comparisons. nd A f a crect contextualised statement. NB A crect contextual statement on its own sces A. 0.05 < p < 0.95 p < 0.05 p > 0.95 nd not significant/ accept H 0 / Not in CR significant/ reject H 0 / In CR nd A Insufficient evidence of an increase/change in the rate/number of sales per month Sufficient evidence of an increase/change in the rate/number of sales per month Two tail st f writing using - P(X < ) writing P(X 4) = 0.987 P(X 5) = 0.073. May be implied by crect CR. probability = 0.038 A f 0.038 X 5. Allow X >4. NB P(X < ) = 0.93 on its own sces A nd dependent on the st being awarded. F a crect statement based on the table below. Do not allow noncontextual conflicting statements eg not significant and reject H 0.Igne comparisons. nd A f a crect contextualised statement. NB A crect contextual statement on its own sces A. 0.05 < p < 0.975 p < 0.05 p > 0.975 nd not significant/ accept H 0 / Not in CR significant/ reject H 0 / In CR nd A Insufficient evidence of an increase/change in the rate/number of sales per month A Sufficient evidence of an increase/change in the rate/number of sales per month
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