Mark Scheme (Results) June 2008

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1 Mark (Results) June 8 GCE GCE Mathematics (6684/) Edexcel Limited. Registered in England and Wales No

2 June Statistics S Mark Question (a) (b) E(X) = Var(X) = ( ) x or attempt to use dx µ = = = 8 = 8.3 awrt P(X ) = ( ) = or or. B (3) () (c) =.3 or or o.e. 3 () (d) P(X 8) or P(X > 8) P(X 8) P(X 8 X ) = P(X ) = = (3) alternative remaining time ~ U[,] or U[,] P(X 3 or 8 ) = (Total ) (a) B cao (a b) using the correct formula and subst in or or for an attempt at the integration they must increase the power of x by and subtract their E(X) squared. cao (b) for P(X ) or P(X < ) cao (c) (their b). If the answer is incorrect we must see this. No need to check with your calculator cao (d) writing P(X 8) (may use > sign). If they do not write P(X 8) then it must be clear from their working that they are finding it.. on its own with no working gets M For attempting to use a correct conditional probability.

3 / Full marks for / on its own with no incorrect working Alternative for P(X 3 ) or P(X 8 ) may use > sign using either U[,] or U[,] /

4 Question X ~B(,.8) Y ~ N (8, 4.36) B B B [P(X > ) = P(X > )]. 8 = P z ± 4.36 = P( z > -. ) using. or. or 49. or 48. standardising.,,.,48.,49, 49. and their µ and σ for =.937 alternative X ~B(,.4) Y ~ N (4, 4.36) (7) B B B [P(X < ) = P(X < 49)] using. or. or 49. or = P z ± standardising.,,.,48.,49, 49. and their µ and σ for 4.36 = P( z <. ) =.937 (Total 7) The first 3 marks may be given if the following figures are seen in the standardisation formula :- 8 or 4, 4.36 or 4.36 or 4.4 or awrt Otherwise B normal B 8 or 4 B 4.36 using. or. or 49. or 48.. ignore the direction of the inequality. standardising.,,.,48., 49, 49. and their µ and σ. They may use 4 or 4.36 or 4.4 or awrt 4.94 for σ or the of their variance.. 8 ±.. may be awarded for ± 4.36 or ± o.e awrt.936

5 Question 3(a) X ~ Po (9) may be implied by calculations in part a or b P( X 3) =. P(X 6) =. CR X 3; X 6 ; (3) (b) P(rejecting Ho) =. +. =.43 or.433 cao () Total (a) for using Po (9) other values you might see which imply Po (9) are.,.4,.978,.98,.9889,.,.6 or may be assumed by at least one correct region. for X 3 or X < 4 condone c or CR instead of X for X 6 or X > They must identify the critical regions at the end and not just have them as part of their working. Do not accept P(X< 3) etc gets A (b) if they use. and. they can gain these marks regardless of the critical regions in part a. If they have not got the correct numbers they must be adding the values for their critical regions.(both smaller than.) You may need to look these up. The most common table values for lambda = 9 are in this table x awrt.43 or.433 Special case If you see.43 /.433 and then they go and do something else with it eg.43 award A

6 Question 4(a) X ~ B(,.) () (b) E(X) =. =. B Var (X) =. (.) =.497 B () (c) X ~ Po (.) P( X ) =.884 d (4) Total 8 (a) for Binomial, fully correct These cannot be awarded unless seen in part a (b)b cao B also allow.,.497,.4973, do not allow. (c) for Poisson for using Po (.) this is dependent on the previous M mark. It is for attempting to find P( X ) awrt.884 Special case If they use normal approximation they could get M A A if they use. in their standardisation. NB exact binomial is.883

7 Question (a) (b) (c) X ~ B(,.) P ( X = 8 ) = P (X 8) P(X 7) or =.6964.! ( p ) 8!7! 8 ( p) 7 =.964 awrt.96 P (X 4) = P(X 3) =.76 B B () () =.984 () (d) H o : p =. H : p >. X ~ B(,.) P(X 3) = P(X ) [P(X ) =.984 =.76] att P(X 3) =.9963 P(X 3) =.9963 =.37 =.37 CR X 3 awrt.37/ CR X 3.37 <. 3 > 3 Reject H or it is significant or a correct statement in context from their values B B There is sufficient evidence at the % significance level that the coin is biased in favour of heads Or There is evidence that Sues belief is correct (6) (a) B for Binomial B for and. must be in part a This need not be in the form written (b) attempt to find P ( X = 8) any method. Any value of p awrt.96 Answer only full marks (c) for - P (X < 3). awrt.98

8 (d) B for correct H. must use p or π B for correct H must be one tail must use p or π attempt to find P(X 3) correctly. E.g. P(X<) correct probability or CR To get the next marks the null hypothesis must state or imply that (p) =. for correct statement based on their probability or critical region or a correct contextualised statement that implies that. not just 3 is in the critical region. This depends on their being awarded for rejecting H. Conclusion in context. Must use the words biased in favour of heads or biased against tails or sues belief is correct. NB this is a B mark on EPEN. They may also attempt to find P(X < 3) =.9963 and compare with.99

9 Question 6(a) Calls occur singly any two of the 3 Calls occur at a constant rate only need calls Calls occur independently or randomly. once B B () (b) (i) X~ Po(4.) used or seen in (i) or (ii) P ( X = ) = P (X ) P(X 4) =.79.3 =.78 (3) (ii) (c) P ( X > 8 ) = P (X 8) =.997 =.43 H o : λ = 9 (λ = 8) may use λ or µ H : λ > 9 (λ > 8) B () X ~ Po (9) may be implied B P(X 4) = P(X 3) [P(X 4) =.96 =.739] att P(X 4) P(X ) =.96 P(X ) =.98 =.4 =.739 CR X awrt >. 4 Accept H. or it is not significant or a correct statement in context from their values There is insufficient evidence to say that the number of calls per hour handled by the agent has increased. (6) (a) B B They must use calls at least once. Independently and randomly are the same reason. Award the first B if they only gain mark. Special case if they don t put in the word calls but write two correct statements award BB (b) correct answers only score full marks (i) Po (4.) may be implied by them using it in their calculations in (i) or (ii) e λ λ for P(X < ) P( X < 4) or! only awrt.7

10 (ii) for P (X 8) only awrt.43 (c) B both. Must be one tail test. They may use λ or µ and either 9 or 8 and match H and H Po (9) may be implied by them using it in their calculations. attempt to find P(X 4) eg P(X < 3) or P(X < 4) correct probability or CR To get the next marks the null hypothesis must state or imply that (λ) = 9 or 8 for a correct statement based on their probability or critical region or a correct contextualised statement that implies that.. This depends on their being awarded for accepting H. Conclusion in context. Must have calls per hour has not increased. Or the rate of calls has not increased. Any statement that has the word calls in and implies the rate not increasing e.g. no evidence that the rate of calls handled has increased Saying the number of calls has not increased gains A as it does not imply rate NB this is an A mark on EPEN They may also attempt to find P(X < 4) =.96 and compare with.9

11 Question 7(a) d x x = x 4 = oe attempt to integrate both parts kx dx kx 4 = 4k k oe both answer correct 4 + 4k k = adding two answers and putting = 4 4 k 3 = 4 4 k = * ddep on previous M (4) (b) x dx = = 4 d x x 3 6 x = 6 x = 3 = 3 or.4 attempt to integrate xf(x) for one part /6 E(X) 3 = + 6 (c) 6 = = =.46 (4) x F(x) = t dt (for x ) ignore limits for M = x 4 must use limit of x 3 F(x) = dt; dt + t (for < x ) need limit of and variable upper limit; need limit and ; = 4 x +

12 F( x) x < x 4 x 4 x+ <x x > middle pair ends B ft B (7) (d) F(m) =. either eq 4 m + =. eq for their <x< m = 4 6 or.7 or awrt.7 ft (3) (e) negative skew B This depends on the previous B being awarded. One of the following statements which must be compatible with negative skew and their figures. If they use mode then they must have found a value for it Mean < Median Mean < mode Mean < median (< mode) Median < mode Sketch of the pdf. db () (a) attempting to integrate both parts both answers correct dependent on the previous M being awarded.. adding the two answers together cso (b) attempting to use integral of x f(x) on one part /6 3/ awrt.4 (c) Att to integrate t (they need to increase the power by ). Ignore limits for method mark x allow use of t. must have used/implied use of limit of. This must be 4 on its own without anything else added x 3 att to integrate t dt and correct limits.

13 dt t + Att to integrate using limits and. no need to see them put in. x 3 they must add this to their t dt. may be given if they add /4 Alternative method for these last two M marks 3 for att to dt and putting + C t use of F() = to find C 4 x + must be correct B middle pair followed through from their answers. condone them using < or < incorrectly they do not need to match up B end pairs. condone them using < or <. They do not need to match up NB if they show no working and just write down the distribution. If it is correct they get full marks. If it is incorrect then they cannot get marks for any incorrect part. So if <x< is correct they can get otherwise M A. if 3<x<4 is correct they can get otherwise M AA. you cannot award Bft if they show no working unless the middle parts are correct. (d) either of their x or 4 x + =. 4 for their F(X) <x< =. cao If they add both their parts together and put =. they get M I they work out both parts separately and do not make the answer clear they can get A (e) B negative skew only B Dependent on getting the previous B. their reason must follow through from their figures.

Mark Scheme (Results) January 2010

Mark Scheme (Results) January 2010 Mark (Results) January 010 GCE Statistics S (668) Edexcel Limited. Registered in England and Wales No. 96750 Registered Office: One90 High Holborn, London WC1V 7BH Edexcel is one of the leading examining