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
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January 010 668 Statistics S Mark Q1 (a) X ~ B(0,0.05) (b) P( X = 0) = 0.95 0 = 0.58859 or 0.585 using tables. (c) P ( X > ) =1 - P ( X ) = 1 0.997 = 0.006 () () () (d) Mean = 0 0.05 = 1 Variance = 0 0.05 0.95 = 0.95 () Total [8] Q1 (a) 1 st for binomial nd for 0 and 0.05 o.e These must be in part (a) (b) for finding (p) 0 0< p <1 this working needs to be seen if answer incorrect to gain the awrt 0.58 or 0.59. (c) for writing 1 - P ( X ) or 1 [ P(X = 0) + P(X = 1) + P(X = ) + P(X = ) + P(X = )] or 1 0.997 or 1 0.9568 awrt 0.006 or.6, do not accept a fraction e.g. 6/10000 10 (d) 1 st for 1 nd for 0.95 NB In parts b, c and d correct answers with no working gain full marks GCE Mathematics Statistics S (668) January 010
Q (a) P(X < 0) = F(0) 1 = = 6 (b) df( x) f( x) = dx 1 x f ( x ) = 6 0 otherwise (c) Continuous Uniform (Rectangular) distribution (d) Mean = 1 () () (1) Q ( ) Variance is = () 1 (e) P( X = 1) = 0 (1) Total [10] (a) for attempting to find F(0) by a correct method eg subst 0 into F(x) or 0 6 1 dx 0 x + 1 1 Do NOT award for dx or both of which give the correct 6 answer by using F(x) as the pdf 1/ o.e or awrt 0. Correct answer only with no incorrect working gets (b) for attempting to differentiate F(x). (for attempt it must have no xs in) for the first line. Condone < signs for the second line. They must have 0 x < and x > only. (c) must have continuous and uniform or Rectangular (d) for mean = 1 [ ± ( b a)] for attempt to use, they must subst in values and not just quote the 1 formula, or using x ( their f ( x)) ( their mean), including limits. Must get x when they integrate. cao. (e) cao GCE Mathematics Statistics S (668) January 010
Q (a) Y ~ Po(0.5) P(Y=0) = e -0.5 = 0.7788 (b) X ~ Po(0.) P(Robot will break down) = 1 P( X = 0) = 1 e 0. = 1 0.0670 = 0.97 (c) 0. e (0.) PX P(X ( = ) ) = = 0.056 () () () (d) 0.97 or answer to part (b) as Poisson events are independent ft dep () Total [10] Q (a) for seeing or using Po(0.5) -a for finding P(Y=0) either by e, where a is positive (a needn t equal their λ ) or using tables if their value of λ is in them Beware common Binomial error using, p = 0. 05 gives 0.778 but scores B0 M0 A0 awrt 0.779 (b) for stating or a clear use of Po(0.) in part (b) or (c) for writing or finding 1 P(X =0) awrt 0. (c) -λ e λ for finding P(X =) e.g with their value of λ in! or if their λ is in the table for writing P(X < ) - P(X < 1) awrt 0.056 (d) 1 st their answer to part(b) correct to sf or awrt 0. nd need the word independent. This is dependent on them gaining the first SC Use of Binomial. Mark parts a and b as scheme. They could get (a) B0,M0,A0 (b) B0 A0 n In part c allow for C (p) (1-p) n- with their n and their p. They could get (c),a0 DO NOT GIVE for p ( x ) p( x 1) In (d) they can get the first only. They could get (d) B0 GCE Mathematics Statistics S (668) January 010
Q (a) kx ( x + )d x + kx d = 0 1 1 k x x + x + 0 [ kx] 1 (=1) or k x x + x + k (=1) 0 9k = 1 1 k = **given** cso 9 (b) x 1 For 0 < x, F( x) = ( t t+ )dt 0 9 1 1 9 x x x = + For < x, F( x) x = kdt+ x 1 = 0 x 0 dep () 1 ( 6 ) 0 x x + x < x 7 F(x) = x 1 F( x) = < x 1 x > ft (6) (c) x x E( X ) = ( x x+ )dt+ d 0 9 x 1 1 1 = x x x x 9 + + 6 0 (d) F(m) = 0.5 9 = or.16 or awrt. 1 1 F(.6) = (.6.6 + 6.6) = awrt 0.8 7 F(.7) = 1 (.7.7 + 6.7) = awrt 0.5 7 Hence median lies between.6 and.7 da () Total [17] () GCE Mathematics Statistics S (668) January 010
Q (a) 1 st attempting to integrate at least one part (at least one x n x ) (ignore limits) 1 st Correct integration. Limits not needed. nd dependent on the previous M being awarded. Adding the two answers together, putting equal to 1 and have the correct limits. nd cso (b) 1 st Att to integrate 1 ( t t + ) (at least one x n x n+1 ). Ignore 9 limits for method mark 1 st 1 x x + x allow use of t. Must have used/implied use of limit of 0. 9 This must be on its own without anything else added x nd attempting to find k + (must get kt or kx ) 1 and they must use the correct limits and add ( t ) or use + C and use F() = 1 nd x 1 must be correct t + or 0 9 1 st middle pair followed through from their answers. condone them using < or < incorrectly they do not need to match up nd 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 0<x is correct they can get otherwise M0 A0. If <x is correct they can get otherwise M0 A0. you cannot award ft if they show no working unless the middle parts are correct. (c) (d) 1 st attempting to use integral of x f(x) on one part 1 st Correct Integration for both parts added together. Ignore limits. nd cao or awrt. 1 st for using F(X) = 0.5. This may be implied by subst into F(X) and comparing answers with 0.5. nd for substituting both.6 and.7 into their F(X) 0.5 or their F(X) 1 st awrt 0.8 and 0.5 if using their F(X). and awrt 0.0 and 0.0 or if using their F(X) 0.5 Other values possible. You may need to check their values for their correct equation NB these last two marks are on epen but mark as nd for conclusion but only award if it follows from their numbers. Dependent on previous A mark being awarded SC using calculators for sign of a suitable equation for awrt.66 provided equation is correct correct comment GCE Mathematics Statistics S (668) January 010
Q5 (a) X ~ Po(10) P( X < 9) = P( X 8) = 0.8 (b) Y ~ Po(0) Y is approximately N(0,0) P( Y > 50) = 1 P( Y 50) 50.5 0 = 1 P Z < 0 = 1 P Z < ( 1.660.. ) = 1 0.9515 = 0.085 N.B. Calculator gives 0.087. Poisson gives 0.056 (but scores nothing) Q5 (a) for using Po(10) for attempting to find P( X 8) : useful values P( X 9) is 0.579(M0), using Po(6) gives 0.87, (). 1 awrt 0. but do not accept () (6) Total [9] (b) 1 st for identifying the normal approximation 1 st for [mean = 0] and [sd = 0 or var = 0] NB These two marks are on epen These first two marks may be given if the following are seen in the standardisation formula : 0 and 0 or awrt 6. nd for attempting a continuity correction (50 or 0 + 0.5 is acceptable) rd for standardising using their mean and their standard deviation and using either 9.5, 50 or 50.5. (9.5, 0, 0.5) accept + nd 50.5 0 correct z value awrt +1.66 or this may be awarded if see + or 0 9.5 0 + 0 rd awrt sig fig in range 0.08 0.085 GCE Mathematics Statistics S (668) January 010
Q6 (a) The set of values of the test statistic for which the null hypothesis is rejected in a hypothesis test. () (b) X~B(0,0.) P( X ) = 0.009 P( X ) = 0.001 P( X 16) = 1 0.996 = 0.006 P( X 17) = 1 0.9979 = 0.001 Critical region is (0 ) x or 16 x( 0) (5) (c) Actual significance level 0.001+0.006=0.0085 or 0.85% (1) (d) 15 (it) is not in the critical region Bft, 1, 0 not significant No significant evidence of a change in p = 0. accept H 0, (reject H 1 ) P ( x 15) = 0. 0169 () Total [10] Q6 (a) 1 st for values/ numbers nd for reject the null hypothesis o.e or the test is significant (b) for using B(0,0.) P x = 0. nd 0.006 1 st ( ) 001 rd for (X) < or (X) < They get A0 if they write P(X < / X < ) th (X) > 16 or (X) > 15 They get A0 if they write P(X > 16 X > 15 NB these are but mark as 16< X < etc is accepted To describe the critical regions they can use any letter or no letter at all. It does not have to be X. (c) correct answer only (d) Follow through 15 and their critical region for any one of the 5 correct statements up to a maximum of B for any incorrect statements GCE Mathematics Statistics S (668) January 010
Q7 (a) x 1p p P( X = x) 1 1 µ = 1 + = 7 or 1 or 1.75 1 7 σ = 1 + = or 0.1875 16 (b) (1,1,1), (1,1,) any order, (1,,) any order, (,,) () (c) (1,,1) (,1,1) (,1,) (,,1) all 8 cases considered. May be implied by * (1,1,) and *(1,,) x 1 5 P( X = x) 1 1 1 = 1 1 = 1 = 1 6 9 6 7 6 = 7 6 () (6) Total [11] Q7 (a) 1.75 oe for using ( x p) µ 0.1875 oe (b) ignore repeats (c) 1 st correct means (allow repeats) 1 st for p for either of the ends 1st for 1/6or awrt 0.016 and 7/6 or awrt 0. nd p (1 p) for either of the middle two 0 < p < 1 May be awarded for finding the probability of the samples with mean of either / or 5/. nd for 9/6 (or /6 three times) and 7/6 (or 9/6 three times) accept awrt dp. rd fully correct table, accept awrt dp. GCE Mathematics Statistics S (668) January 010
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