Mark (Results) January 00 GCE Statistics S (668) Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WCV 7BH
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January 00 668 Statistics S Mark Q (a) Red 4 Red Blue Blue Green Red Green M A A () 4 Green Red Blue Q (a) P(Blue bead and a green bead) equivalent) = + = 4 4 6 (or any exact M A Total [5] M for shape and labels: branches followed by,, with some R, B and G seen Allow branches followed by,, if 0 probabilities are seen implying that,, intended Allow blank branches if the other probabilities imply probability on blanks is zero Ignore further sets of branches st A for correct probabilities and correct labels on st set of branches. nd A for correct probabilities and correct labels on nd set of branches. (accept 0., 0.67 etc or better here) M for identifying the cases BG and GB and adding products of probabilities. These cases may be identified by their probabilities e.g. + 4 4 NB 6 (or exact equivalent) with no working scores / Special Case With Replacement (This oversimplifies so do not apply Mis-Read: max mark /5) (a) B for branches followed by,, with correct labels and probabilities of, 4, 4 on each. M for identifying, possibly correct cases and adding products of probabilities but A0 for wrong answer + 4 4 4 4 will be sufficient for MA0 here but + would score M0 4 GCE Mathematics Statistics S (668) January 00
Q (a) Median is B () Q = 4, Q = 40, IQR = 6 B B Bft () Q IQR=4 6 = 8 M So 7 is only outlier Aft Box Outlier Whisker Bft B Bft Q (accept either whisker) () Total [9] st B for Q = 4 and nd B for Q = 40 rd Bft for their IQR based on their lower and upper quartile. Calculation of range (40 7 = ) is B0B0B0 Answer only of IQR = 6 scores /. For any other answer we must see working in or on stem and leaf diagram M for evidence that Q -IQR has been attempted, their 8 (>7) seen or clearly attempted is sufficient A ft must have seen their 8 and a suitable comment that only one person scored below this. st Bft for a clear box shape and ft their Q, Q and Q readable off the scale. Allow this mark for a box shape even if Q = 40, Q = 7 and Q = are used nd B for only one outlier appropriately marked at 7 rd Bft for either lower whisker. If they choose the whisker to their lower limit for outliers then follow through their 8. ( There should be no upper whisker unless their Q < 40, in which case there should be a whisker to 40) A typical error in is to draw the lower whisker to 7, this can only score BB0B0 GCE Mathematics Statistics S (668) January 00
Q (a).75 or 4, 5.5 or 5.50 or 5 B B Mean birth weight = 484.7 500 = & awrt. M A 5889.5 484 Standard deviation = = 0.409... or s = 0.47... M Aft A 500 500 () 40 Q =.00 + 0.5 =.457... (allow 40.5...5) M A 80 (e) Mean(.)<Median(.5) (or very close) Bft Negative Skew (or symmetrical) dbft Total [] Q M for a correct expression for mean. Answer only scores both. (e) M for a correct expression (ft their mean) for sd or variance. Condone mis-labelling eg sd= with no square root or no labelling st Aft for a correct expression (ft their mean) including square root and no mis-labelling Allow st A for = 0.77... = 0.4... nd A for awrt 0.4. Answer only scores / M for a correct expression (allow 40.5 i.e. use of n + ) but must have.00, 80 and 0.5 A for awrt.5 provided M is scored. NB.5 with no working scores 0/ as some candidates think mode is.5. st Bft for a comparison of their mean and median (may be in a formula but if +(mean - median) is calculated that s OK. We are not checking the value but the sign must be consistent.) Also allow for use of quartiles provided correct values seen: Q =.0, Q =.47 [They should get ( 0. = ) Q Q < Q Q( = 0.) and say (slight) negative skew or symmetric] nd dbft for a compatible comment based on their comparison. Dependent upon a suitable, correct comparison. Mention of correlation rather than skewness loses this mark. GCE Mathematics Statistics S (668) January 00
4 (a) S D closed curves and 4 in centre Evidence of subtraction M M 4 4 6 4,6,4 4,7, Labels on loops, 6 and box A A B 7 6 4 (a) N P(None of the options)= 6 = 4 Bft 80 45 P(Networking only)= 7 80 P(All options/technician)= 4 = M A 40 0 Total [9] nd M There may be evidence of subtraction in outer portions, so with 4 in the centre then 5, 40 8 (instead of,6,4) along with, 9, can score this mark but A0A0 N.B. This is a common error and their 6 becomes 8 but still scores B0 in part (a) 6 Bft for 80 or any exact equivalent. Can ft their 6 from their box. If there is no value for their 6 in the box only allow this mark if they have shown some working. Bft ft their 7. Accept any exact equivalent M If a probability greater than is found in part score M0A0 P( S D N) for clear sight of and an attempt at one of the probabilities, ft their values. P( S N) Allow P(all S N ) = 4 or to score M A0. 6 9 Allow a correct ft from their diagram to score MA0 e.g. in,,9 case in (a): 4 44 or is MA0 A ratio of probabilities with a product of probabilities on top is M0, even with a correct formula. A for 4 or 40 0 or an exact equivalent Allow 4 or to score both marks if this follows from their diagram, otherwise some 40 0 explanation (method) is required. Bft (5) () () GCE Mathematics Statistics S (668) January 00
Q5 (a) k + 4 k + 9k = M 4k = k = 4 **given** cso A P( X ) = P( X = ) or P(X = ) + P(X = ) M = k = or 0.9857... awrt 0.99 A 4 E(X) = k + k 4 + k 9 or 6k M 6 8 4 = = or (or exact equivalent) A 4 7 7 8 Var( X) = k+ 4 k 4 + 9 k 9, M M 7 Var( X ) = Var( X ) M Q5 (a) 9 = or 0.87755... awrt 0.88 A 49 (4) Total [0] M for clear attempt to use p( x ) =, full expression needed and the must be clearly seen. This may be seen in a table. Acso for no incorrect working seen. The sum and = must be explicitly seen somewhere. A verification approach to (a) must show addition for M and have a suitable comment e.g. therefore k = 4 for A cso M for - P(X < ) or P(X = ) + P(X = ) A for awrt 0.99. Answer only scores / M for a full expression for E(X) with at least two terms correct. NB If there is evidence of division (usually by ) then score M0 A for any exact equivalent - answer only scores / st M for clear attempt at E( X ), need at least terms correct in k+ 4 4k+ 9 9k or E( X ) =7 nd M for their E( X ) ( their µ ) rd M for clearly stating that Var( - X) = Var(X), wherever seen A accept awrt 0.88. All M marks are required. Allow 4/4 for correct answer only but must be for Var( X). GCE Mathematics Statistics S (668) January 00
Q6 Q6 (a) 8 S pp = 0697 = 770 7 M A 4 8 4 0986 S tp = 4948 = 69, S tt = 88 = 569.4857... or 7 7 7 A A 69 r = 770 569.4857.. M Aft = 0.7075 awrt (0.70) A + (f) (Pmcc shows positive correlation.) Older patients have higher blood pressure Points plotted correctly on graph: - each error or omission (within one square of correct position) * see diagram below for correct points (f) Line drawn with correct intercept, and gradient Bft B (+) (e) 69 b = =.509466... 569.4857.. M A 8 4 a= b = 45.4674... 7 7 M p = 45.5 +.5t A (4) (g) t = 40, p= 05.84... from equation or graph. awrt 06 M A Total [8] (a) M for at least one correct expression st A for S pp = 770, nd A for S tp = 69 or 70, rd A for S tt = awrt 570 M for attempt at correct formula and at least one correct value (or correct ft) M0 for 4948 0697 88 Aft All values correct or correct ft. Allow for an answer of 0.7 or 0.70 Answer only: awrt 0.70 is /, answer of 0.7 or 0.70 is / B B (4) () () B for comment in context that interprets the fact that correlation is positive, as in scheme. Must mention age and blood pressure in words, not just t and p. Record point incorrect as BB0 on epen. [NB overlay for (60, 5) is slightly wrong] (e) (f) (g) st M for use of the correct formula for b, ft their values from (a) st A allow.5 or better nd M for use of y bx with their values nd A for full equation with a = awrt 45.5 and b = awrt.5. Must be p in terms of t, not x and y. st Bft ft their intercept (within one square). You may have to extend their line. nd B for correct gradient i.e. parallel to given line (Allow square out when t = 80) M for clear use of their equation with t = 40 or correct value from their graph. A for awrt 06. Correct answer only (/) otherwise look for evidence on graph to award M GCE Mathematics Statistics S (668) January 00
Q6 + (f) Diagram for Q6 + (f) GCE Mathematics Statistics S (668) January 00
Q7 (a) (a) 5% P( X < 54) = 0.05 54 µ µ 54 M =.6449 or =.6449 B µ = 54 +.6449 **given** A cso () 7 µ = 0. 544 or 7 µ = 0.544 (allow z = 0.5 or better here but must be in an equation) B Solving gives = 8.976075 (awrt 8.0) and µ = 67.6487 (awrt 68) M A A (4) 60 µ P(Taller than 60cm) = P Z > M = P( Z < 0.97994) B = 0.8 awrt 0.8 A () Total [] nd B for 54 and 7 marked but 54 must be < µ and 7 > µ. But µ need not be marked. 54 µ 7 µ Allow for and marked on appropriate sides of the peak. rd B the 5% and 0% should be clearly indicated in the correct regions i.e. LH tail and RH tail. M ( 54 µ ) for ± =z value (z must be recognizable e.g..64,.65,.96 but NOT 0.599 etc) B for +.6449 seen in a line before the final answer. Acso for no incorrect statements (in µ, ) equating a z value and a probability or incorrect signs e.g. 54 µ 54 µ = 0.05 or =. 6449 or P( Z µ 54 ) =. 6449 54 µ 7 B for a correct nd equation (NB 7 µ = 0.55 is B0, since z is incorrect) M for solving their two linear equations leading to µ =... or =... st A for = awrt 8.0, nd A for µ = awrt 68 [NB the 68 can come from false working. These A marks require use of correct equation from, and a z value for 0.544 in ] NB use of z = 0.5 will typically get =8. and µ = 67.67 and score BMA0A No working and both correct scores 4/4, only one correct scores 0/4 Provided the M is scored the As can be scored even with B0 (e.g. for z =0.55) M for attempt to standardise with 60, their µ and their (> 0). Even allow with symbolsµ and. B for z = awrt + 0.9 No working and a correct answer can score / provided and µ are correct to sf. 0% < bell shaped, must have inflexions 54,7 on axis 5% and 0% B B B () GCE Mathematics Statistics S (668) January 00
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