Mark (Results) June 0 GCE Statistics S (6684) Paper
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EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edexcel Mathematics 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) should not be subdivided.. 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 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 oe or 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
June 0 6684 Statistics S Mark. The list of ID numbers (b) F ~ B(50,0.0) B B Notes: B for idea of list/register/database and identity numbers NB B0 if referring to the sample or 50 or only part of the population. (b) These must be in part (b) to gain the marks st B for Binomial distribution nd B for n = 50 and p = 0.0 or (50,0.0) NB (0.0, 50) is B0 Po() alone is B0B0 For a probability table st B Use of B(50,0.0) NB P(X = 0) = 0.64 nd B Table must have all 50 values and their probabilities. B () () GCE Statistics S (6684) June 0
. Poisson B () (b) H 0 : µ = 9 (or λ = 6) H : µ > 9 (or λ>6) B B X ~Po(9) and P(X > ) = - P(X < ) or P(X 4) = 0.9585 P(X 5) = 0.045 = -0.800 = 0.97 CR X 5 A (0.97 > 0.05) so not significant/ accept H 0 / Not in CR d he does not have evidence to switch on the speed restrictions (o.e) Aft (6) (c) Let Y = the number of vehicles in 0 s then Y ~Po(6) B Tables: P(Y < 0) = 0.9574 so P(Y > ) = 0.046 so needs vehicles A () 0 Notes: B for Poisson or Po. Ignore their value for the mean. (b) st B for H 0 : µ / λ = 9 or µ / λ = 6 nd B for H : µ / λ > 9 or µ / λ > 6 One tail st for writing or using - P(X < ) or writing P(X 4) = 0.9585 or P(X 5) = 0.045. May be implied by correct CR.or probability = 0.97 A for 0.97 or a correct CR. Allow X >4. NB P(X < ) = 0.800 on its own scores A nd dependent on the st being awarded. For a correct statement based on the table below. Do not allow non-contextual conflicting statements eg significant and accept H 0. Ignore comparisons. nd A for a correct contextualised statement. NB A correct contextual statement on its own scores A. 0.05 < p < 0.95 p < 0.05 or p > 0.95 nd not significant/ accept H 0 / Not in CR significant/ reject H 0 / In CR nd A Insufficient evidence to switch on the speed restrictions Sufficient evidence to switch on the speed restrictions Two tail st for writing or using - P(X < ) or writing P(X 5) = 0.9780 or P(X 6) = 0.0. May be implied by correct CR. or probability = 0.97 A for 0.97 or CR X 6. Allow X >5. NB P(X < ) = 0.800 on its own scores A nd dependent on the st being awarded. For a correct statement based on the table below. Do not allow non-contextual conflicting statements eg significant and accept H 0. Ignore GCE Statistics S (6684) June 0
comparisons. nd A for a correct contextualised statement. NB A correct contextual statement on its own scores A. 0.05 < p < 0.975 p < 0.05 or p > 0.975 nd not significant/ accept H 0 / Not in CR significant/ reject H 0 / In CR nd A Insufficient evidence to switch on the speed restrictions Sufficient evidence to switch on the speed restrictions (c) B for identifying Po(6) - may be implied by use of correct tables any one of the probs 0.9574 or 0.046 or 0.9799 or 0.00 may be implied by correct answer of A cao do not accept X > NB answer of with no working gains all three marks.. Mode = from graph B () GCE Statistics S (6684) June 0 (b) kx kx dx = 0.5 = 0.5 0 0 A So 7 0 0.5 k = 8 (using median = ) d A (4) (c) Height of triangle = = 8 Bft Area of triangle = ( a ) = so a = 5 cao A () (d) From graph distribution is negative skew (left tail is longer) B µ < median for negative skew so E(X) < Bd () [ N.B. E(X) = ] 4 0 Notes: (b) st for attempt to integrate f(x) (need x ). Integration must be in part (b) st A for correct integration. Ignore limits for these two marks. nd Dependent on the previous M mark being awarded. For use of correct limits and set equal to 0.5 - leading to a linear equation for k. No need to see 0 substituted. nd A for k = or exact equivalent 8 NB k = with no working gains M0A0M0A0 8 k = = without sight of integration is M0A0M0A0 9 8 B for correct height of triangle using their k. ie 9k. May be seen in working for area of triangle. (c) 9k Or correct gradient of line ie o.e. ( a)
(d) 4 for a correct linear equation for a, in the form ± ( a ) 9k = (Must see the halves) NB if they have stated their height and then used their height rather than 9k allow A cao NB stating a = 5 and then verifying area of the triangle = 0.5 is acceptable. NB a = 5 on its own is B0M0A0 SC Integration of both parts = or Integration of line = 0.5 leading to a 8a + 5 = 0 gets B and if they identify a = 5 A st B for identifying negative skew nd B dependent on previous B mark being awarded. For correct deduction E(X) < 9.5 7 0 7 5 = 6 awrt 0.8 A 5 (b) P(Longest > 9.5) = - P(all < 9.5) = 6 9 = or 0.4 A 6 (c) P(a stick < 7.6) = 0.6 0. = B Let Y = number of sticks (out of 6) <7.6 then Y~B(6, 0.) P(Y > 4) = - P(Y < 4) = - 0.9984 A = 0.006 or 65 Notes: (b) (c) for an expression for the probability e.g. dx 7 for ( a) or ( a) + ( a) a + ( a) a A awrt 0.4 B 0. may be implied by at least one correct probability st for writing or using B(6, p) may be implied by np x ( p) 6-x using their p and n > nd for writing or using P(Y < 4) or np 5 ( p) + p 6 (n is an integer > ) A cao 9.5 () () (4) 8 5. NB 0.006 with no working gets B0M0M0A0 X ~Po(5); P(X < ) = 0.650 A () GCE Statistics S (6684) June 0 4
(b) Let Y = the no.of planks with at most defects, Y ~Binomial Y~B(6, 0.65) Aft P(Y < ) = P(Y < ) 6 5 = 0.75 + 6 0.65 0.75 A = 0.4987. awrt 0.499 or 0.498 A (c) Let T = total number of defects on 6 planks, T ~Po(0) so T S ~Normal S~N(0, 0) A P(T < 8) = P(S < 7.5) 7.5 0 = P z < 0 (5) = P( Z < -.8 ) A = 0.0... awrt 0.0 or 0.0 A Notes: for identifying Po(5) - it should be clearly seen somewhere or implied A for correct probability. Allow 0.65 (b) st for writing or using the binomial - may be implied by use of nq x (- q) 6-x with n > st Aft for n = 6 and p = their may be implied by 6p(- p) 5 or (- p) 6 NB if they write B(6,) they get A nd for writing P(Y <) or P(Y = 0) + P(Y = ) or (- q) 6 + nq(- q) 5 with n > nd A (- p) 6 + 6p(- p) 5 where p = their rd A for awrt 0.499 SC use of a probability in the tables lose last two marks could get A M0 A0 (c) st for a normal approx st A for correct mean and sd nd for use of continuity correction, either 7.5 or 8.5 or 4.5 or 4.5 seen rd Standardising with their mean and their sd and 7.5 or 8 or 8.5 or 4.5 or 4 or 4.5 NB if they have not written down a mean and sd then they need to be correct in the standardisation to gain this mark. nd A for z = +.8 or better. May be awarded for.9] rd A for awrt 0.0 or 0.0 [NB no approximation gives 0.0077 ] SC using P(X<8.5) P(X<7.5) can get A M0A0A0 (6) 7.5 0 ± [NB no continuity correction z = 0 GCE Statistics S (6684) June 0 5
6. 0 H : p= 0.5 H : p 0.5 B B X ~ B(0, 0.5) P(X < ) = 0.0480 or CR: X = 0 (0.0480 > 0.05) not a significant result or do not reject H0 or not in CR there is no evidence of a change in the proportion of customers buying an item from the display. (b) H 0 : p= 0. H : p> 0. B Let S = the number who buy sandwiches, S ~B(0, 0.), ( ) S W ~ N 4, 9. A P(S > ) = P(W > 0.5) 0.5 4 x 0.5 4 = P Z > or =.86 9. 9. [= P(Z >.48..) ] = - 0.906 = 0.0694 x = 0. A < 0.0 so a significant result, there is evidence that more customers are purchasing Bft sandwiches or the shopkeepers claim is correct. (8) Notes: 4 st B for H 0 must use p nd B for H must use p st for writing or using B(0,0.5) may be implied by correct CR st A 0.0480 or X = 0. Allow X < 0. Ignore upper CR. NB Allow CR X < if using one tail test. nd A correct statement (see table below) Do not allow non-contextual conflicting statements eg significant and accept H 0. Ignore comparisons nd A for a correct statement in context. For context we need idea of change/decrease in number of customers buying from display may use different words. NB A correct contextual statement on its own scores A Two tail 0.05 < p < 0.975 or One tail 0.05 < p < 0.95 Two tail p < 0.05 or p > 0.975 or One tail p < 0.05 or p > 0.95 nd not significant/ accept H 0 / Not in CR or significant/ reject H 0 / In CR or contextual nd A contextual There is no evidence of a change/decrease in the proportion of customers buying an item from the display A Aft There is evidence of a change/decrease in the proportion of customers buying an item from the display. (b) st B both hypotheses correct must use p. st for a normal approx st A for correct mean and sd nd for use of continuity correction, either 0.5 or.5 or (x ± 0.5) seen rd standardising with their mean and their sd and 0.5, or.5 or x or (x ± 0.5)) 4 th for - tables value or.86 nd A for awrt 0.069 or x = 0. nd Bft For a correct conclusion in context using their probability and 0. For context we need idea of more customers buying sandwiches may use different words (6) GCE Statistics S (6684) June 0 6
One tail 0. < p < 0.9 or Two tail 0.05 < p < 0.95 One tail p < 0. or p > 0.9 or Two tail p < 0.05 or p > 0.95 nd not significant/ accept H 0 / Not in CR or significant/ reject H 0 / In CR or contextual contextual nd A There is no evidence of an increase in the proportion of customers buying sandwiches There is evidence of a change/increase in the proportion of customers buying sandwiches. SC using P(X<.5) P(X<0.5) can get B A M0A0B0 7 shape which does not go below the x-axis [condone missing patios] B Graph must end at the points (,0) and (5,0) and the points labelled at and 5 B (b) E(X) = (by symmetry) B (c) [ E( X ] = x f ( x) dx 4 ( 6x x 5x ) = dx 5 4 5 6 5 = x x x 4 5 = 6 65 65 6 5 65 4 4 5 = 9.8 (*) A () () A cso (4) (d) s.d. = 9.8 E( X ), = 0.8944 awrt 0.894 A () (e) F() = 0 ( a 5 + 9 ) = 0, leading to a = 7 A () (f) F(.9) = 0.449, F(.) = 0.55 A Since F( q ) = 0.5 and these values are either side of 0.5 then.9< q <. A () (g) Since the distribution is symmetric q =5-. =.7 cao B () (h) We know P( q =. < X <.7= q ) = 0.5 so kσ = 0.7 0.7 so k = = 0.786.. = awrt 0.78 0.894... A () GCE Statistics S (6684) June 0 7 7
Notes: (c) This part is a show that therefore we need to see all the steps in the working st for showing intention of doing x f( x) and attempt to multiply out bracket st A for correct integration, cao, ignore limits for this mark. nd for use of correct limits. Need to see evidence of subst both 5 and. nd A for cso leading to 9.8. Do not ignore subsequent working for this final A mark. (d) for a correct expression for standard deviation, must include... (e) A allow awrt 0.894, 0.8, 5 oe 5 for a correct method to find a. e.g F(5) = or f (x) = (f) for an attempt at F(.9) or F(.) or put F(x) = 0.5 (ft their value of a) st A for both values seen. awrt 0.45 and 0.5 find solutions awrt 6.76/6.75,.05, -0.064 nd A for comparison with 0.5 and stating Q state only.0 in range and stating Q lies between.9 and. lies between.9 and. (h) For k σ = awrt 0.7 A Allow awrt 0.78 NB a correct awrt 0.78 gains A 5 GCE Statistics S (6684) June 0 8
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