Mark (Results) June 008 GCE Mark (Final) GCE Mathematics (6683/01) Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH
June 008 6683 Statistics S1 Mark Q1 0.95 Positive Test 0.0 Disease (0.05) Negative Test 0.03 Positive Test (0.98) No Disease (0.97) Negative Test Tree without probabilities or labels 0.0(Disease), 0.95(Positive) on correct branches 0.03(Positive) on correct branch. (b) P(Positive Test) =0.0 0.95 + 0.98 0.03 ft = 0.0484 (c) 0.98 0.03 P( Do not have disease Postive test) = 0.0484 = 0.607438.. awrt 0.607 (d) Test not very useful OR High probability of not having the disease for a person with a positive test :All 6 branches. Bracketed probabilities not required. (b) for sum of two products, at least one correct from their diagram ft follows from the probabilities on their tree 11 for correct answer only or 500 (c) for conditional probability with numerator following from their tree and denominator their answer to part (b). 147 also for. 4 Total 9 [] [1]
Q 50 [1] (b) Q 1 = 45 Q = 50.5 ONLY Q 3 = 63 (c) Mean = 1469 5.46486.. 8 = awrt 5.5 8113 1469 Sd = 8 8 =1.164. or 1.38716 for divisor n-1 awrt 1. or 1.4 (d) 5.46.. 50 = awrt 0.0 or 0.1 sd (e) 1. mode/median/mean Balmoral>mode/median/mean Abbey. Balmoral sd < Abbey sd or similar sd or correct comment from their values, Balmoral range<abbey range, Balmoral IQR>Abbey IQR or similar IQR 3. Balmoral positive skew or almost symmetrical AND Abbey negative skew, Balmoral is less skew than Abbey or correct comment from their value in (d) 4. Balmoral residents generally older than Abbey residents or equivalent. Only one comment of each type max 3 marks (c) for their 1469 between 1300 and 1600, divided by 8, for awrt 5.5.. Please note this is on Epen use of correct formula including sq root awrt 1. or 1.4 Correct answers with no working award full marks. (d) for their values correctly substituted Accept 0. as a special case of awrt 0.0 with 0 missing (e) Technical terms required in correct context in lines 1 to 3 e.g. average and spread B0 1 correct comment B0B0 correct comments B0 3 correct comments [] Total 13
Q3 1 p + 1 0. + 0.15 + 3 0.15 = 0.55 d p = 0.4 p+ q+ 0. + 0.15 + 0.15 = 1 q = 0.1 [5] (b) Var( X) = ( 1) p+ 1 0. + 0.15 + 3 0.15, 0.55, =.55 0.305 =.475 awrt.5 (c) E(X-4) =E(X)-4 =.9 [] Total 11 for at least correct terms on LHS Division by constant e.g. 5 then M0 d dependent on first for equate to 0.55 and attempt to solve. Award for p=0.4 with no working for adding probabilities and equating to 1. All terms or equivalent required e.g. p+q=0.5 Award for q=0.1 with no working (b) attempting E(X ) with at least correct terms for fully correct expression or.55 Division by constant at any point e.g. 5 then M0 for subtracting their mean squared for awrt.5 Award awrt.5 only with no working then 4 marks (c) for x(their mean) -4 Award marks for -.9 with no working
Q4 401.3 S tt = 109.81 = 186.6973 15 awrt 187 5.08 S vv = 4.3356 = 0.40184 15 awrt 0.40 401.3 5.08 S tv = 677.971 = 6.9974 awrt 7.00 15 (b) 6.9974 r = 186.6973 0.40184 ft = 0.807869 awrt 0.808 (c) t is the explanatory variable as we can control temperature but not frequency of noise or equivalent comment [] (d) High value of r or r close to 1 or Strong correlation [1] (e) 6.9974 b = = 0.03748 awrt 0.0375 186.6973 5.08 401.3 a = b 0.669874 15 15 = awrt 0.669 (f) t= 19, v=0.669874+0.03748x19=1.381406 awrt 1.4 [1] Total 15 any one attempt at a correct use of a formula. Award full marks for correct answers with no working. Epen order of awarding marks as above. (b) for correct formula and attempt to use ft for their values from part 677.971 NB Special Case for A0 109.81 4.3356 awrt 0.808 Award 3 marks for awrt 0.808 with no working (c) are independent. Second mark requires some interpretation in context and can be statements such as temperature effects / influences pitch or noise temperature is being changed BUT B0 for temperature is changing (e) their values the right way up for awrt 0.0375 attempt to use correct formula with their value of b awrt 0.669 (f) awrt 1.4
Q5 A B 30 3 1 10 100 5 100 0 3 closed intersecting curves with labels 100 100,30 1,10,3,5 Box C (b) (c) (d) P(Substance C) 100 + 100 + 10 + 5 35 47 = = = or exact equivalent 300 300 60 10 10 P( All 3 A) = = or exact equivalent 30 + 3+ 10 + 100 143 0 1 P(Universal donor) = = or exact equivalent 300 15 0 not required. Fractions and exact equivalent decimals or percentages. (b) For adding their positive values in C and finding a probability ft for correct answer or answer from their working (c) their 10 divided by their sum of values in A ft for correct answer or answer from their working (d) for their 0 divided by 300 correct answer only ft [] ft [] cao [] Total 10
Q6 F(4)=1 (4 + k) = 5 k = 1 as k > 0 (b) x 3 4 P(X=x) 9 7 9 5 5 5 for use of F(4) = 1 only If F()=1 and / or F(3)=1 seen then M0. F()+F(3)+F(4)=1 M0 for k=1 and ignore k= -9 (b) ft follow through their k for P(X=) either exact or 3sf between 0 and 1 inclusive. correct answer only or exact equivalent correct answer only or exact equivalent [] ft Total 5
Q7 (b) (c) 53 50 z = Attempt to standardise P(X>53)=1-P(Z<1.5) 1-probability required can be implied =1-0.933 =0.0668 P( X x ) = 0.01 0 x0 50 =.363 x 0 = 45.3474 awrt 45.3 or 45.4 [5] P( weigh more than 53kg and 1 less) = 3 0.0668 (1 0.0668) ft = 0.0149487.. awrt 0.01 Total 1 for using 53,50 and, either way around on numerator 1- any probability for mark 0.0668 cao (b) can be implied or seen in a diagram or equivalent with correct use of 0.01 or 0.99 for attempt to standardise with 50 and numerator either way around for ±. 363 Equate expression with 50 and to a z value to form an equation with consistent signs and attempt to solve awrt 45.3 or 45.4 (c) for 3, p (1 p) for any value of p ft for p is their answer to part without 3 awrt 0.01 or 0.015