Mark Scheme (Results) Summer 2007

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Dale Randall
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1 Mark (Results) Summer 007 GCE GCE Mathematics Statistics S (6684) Edexcel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WC1V 7BH

2 June Statistics S Mark 1(a) Continuous uniform distribution or rectangular distribution. f(x) 1 5 1/5,(0),5 0 by start at y axis 0 5 x (b) E(X) =.5 ft from their a and b, must be a number ft Var(X) = 1 (5 0) or attempt to use 5 1 f( ) d µ x x x - 0 use their f(x) = 5 or.08 o.e 1 awrt.08 (c) (d) P(X > 3) = 5 = 0.4 P(X = 3) = 0 times their 1/5 from diagram ft (1) (1) (Total 8)

3 One tail test Method 1 H o : λ = 5 (λ =.5) µ H 1 : λ > 5 (λ >.5) X ~ Po (.5) may use λ or P(X 7) = 1 P(X 6) [ P(X 5) = = ] att P(X 7) P(X 6) = P(X 6) = = = CR X 6 awrt < or 7 is in critical region or 7 is significant (Reject H 0.) There is significant evidence at the 5% significance level that the factory is polluting the river with bacteria. or The scientists claim is justified Total 7 Method H o : λ = 5 (λ =.5) may use λ or µ H 1 : λ > 5 (λ >.5) X ~ Po (.5) P(X < 7) [P(X < 5) = 0.891] att P(X < 7) P(X < 6) P(X < 6) = = CR X 6 wrt > or 7 is in critical region or 7 is significant (Reject H 0.) There is significant evidence at the 5% significance level that the factory is polluting the river with bacteria. or The scientists claim is justified

4 Two tail test Method 1 H o : λ = 5 (λ =.5) may use λ or µ H 1 : λ 5 (λ.5) X ~ Po (.5) P(X 7) = 1 P(X 6) [P(X 6) = = 0.040] att P(X 7) P(X 7) = P(X 7) = = = CR X 7 awrt < or 7 is in critical region or 7 is significant (Reject H 0.) There is significant evidence at the 5% significance level that the factory is polluting the river with bacteria. or The scientists claim is justified B0 Method H o : λ = 5 (λ =.5) may use λ or µ H 1 : λ 5 (λ.5) X ~ Po (.5) P(X < 7) [P(X < 6) = ] att P(X < 7) P(X < 7) P(X < 7) = = CR X 7 awrt > or 7 is in critical region or 7 is significant (Reject H 0.) There is significant evidence at the 5% significance level that the factory is polluting the river with bacteria. or The scientists claim is justified B0

5 3(a) X ~ Po (1.5) need Po and 1.5 (1) (b) Faulty components occur at a constant rate. any two of the 3 Faulty components occur independently or randomly. only need faulty Faulty components occur singly. once () (c) P(X = ) = P(X ) P(X 1) or e (1.5) 1.5 = = 0.51 awrt 0.51 () (d) X~ Po(4.5) 4.5 P(X 1) = 1 P(X = 0) = 1 e 4.5 = = awrt Total 8

6 4 Attempt to write down combinations at least one seen (5,5,5), (5,5,10) any order (10,10,5) any order, (10,10,10) (5,10,5), (10,5,5), (10,5,10), (5,10,10), all 8 cases considered. May be implied by 3 * ( 10,5,10) and 3 * (5,5,10) median 5 and Median = 5 P(M = m) = = = add at least two prob using ¼ and ¾. identified by having same median of 5 or 10 Allow no 3 for M Median = 10 P(M = m) = = = Total 7

7 5(a) If X ~ B(n,p) and n is large, n > 50 p is small, p < 0. then X can be approximated by Po(np) () (b) P( consecutive calls) = 0.01 = () (c) X ~B(5, 0.01) P(X >1) = 1 P(X = 1) P(X = 0) = 1 5(0.01)(0.99) 4 ( 0.99) 5 = = awrt (d) X ~B(1000, 0.01) Mean = np = 10 Variance = np(1 p) = 9.9 by correct mean and variance (e) X ~ Po(10) P(X > 6) = 1 P (X 6) = = awrt () Total 1

8 6 One tail test Method 1 H o : p = 0. H 1 : p > 0. X ~ B(5, 0.) P(X 3) = 1 P(X ) [P(X 3) = = ] att P(X 3) P(X 4) = P(X 4) = = = CR X 4 awrt > or 3 is not in critical region or 3 is not significant (Do not reject H 0.) There is insufficient evidence at the 5% significance level that there is an increase in the number of times the taxi/driver is late. Or Linda s claim is not justified Method H o : p = 0. H 1 : p > 0. Total 7 X ~ B(5, 0.) P(X < 3) = [P(X < 3) = 0.941] att P(X < 3) P(X < 4) P(X < 4) = CR X 4 awrt < or 3 is not in critical region or 3 is not significant (Do not reject H 0.) There is insufficient evidence at the 5% significance level that there is an increase in the number of times the taxi/driver is late. Or Linda s claim is not justified

9 Two tail test Method 1 H o : p = 0. H 1 : p 0. X ~ X ~ B(5, 0.) P(X 3) = 1 P(X ) [P(X 3) = = ] att P(X 3) P(X 4) = P(X 4) = = = CR X 4 awrt > or 3 is not in critical region or 3 is not significant (Do not reject H 0.) There is insufficient evidence at the 5% significance level that there is an increase in the number of times the taxi/driver is late. Or Linda s claim is not justified Method H o : p = 0. H 1 : p 0. B0 B0 X ~ X ~ B(5, 0.) P(X < 3) = [P(X < 3) = 0.941] att P(X < 3) P(X < 4) P(X < 4) = CR X 4 awrt < or 3 is not in critical region or 3 is not significant Do not reject H 0. There is insufficient evidence at the 5% significance level that there is an increase in the number of times the taxi/driver is late. Or Linda s claim is not justified Special Case If they use a probability of 1 7 throughout the question they may gain M0 A0. NB they must attempt to work out the probabilities using 1 7

10 7(a) i ii If X ~ B(n,p) and n is large or n > 10 or np > 5 or nq >5 p is close to 0.5 or nq >5 and np >5 then X can be approximated by N(np,np(1- p)) mean = np () variance = np(1 p) must be in terms of p () (b) X ~ N (60, 58.) or X ~ N (60, 7.63 ) 60, 58. P(X > 40) = P(X > 39.5) using 39.5 or = 1 P z <± 58. = 1 P( z < ) standardising 39.5 or 40 or 40.5 and their µ and σ = allow answers in range , dep on both M (5) (c) E(X) = 60 or ft from part (b) ft Expected profit = (000 60) = Total 1

11 8(a) f(x) (0), 4, 0.5 axis by start at y 0 4 x both patio must be straight (b) Mode is x = 3 (1) (c) x 1 F(x) = t dt (for 0 x 3 ) ignore limits for M = 1 x F(x) = 1 3 t dt; t dt (for 3 < x 4) need limit of 3 and variable upper 0 6 x must use limit of 0 limit; need limit 0 and 3 ; = x 1 4 x 3 (d) F( x) 0 x < 0 1 x 1 0 x x > 4 x - x - 3 <x middle pair ends ft F(m) = 0.5 either eq 1 x =0.5 1 eq for their 0<x<3 x = 6 =.45 6 or awrt.45 ft Total 14

Mark Scheme (Results) June 2008

Mark Scheme (Results) June 2008 Mark (Results) June 8 GCE GCE Mathematics (6684/) Edexcel Limited. Registered in England and Wales No. 44967 June 8 6684 Statistics S Mark Question (a) (b) E(X) = Var(X) = ( ) x or attempt to use dx µ