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Averages Mark Scheme

Q1.

(a) (b)(i) (b)(ii)

Qi. (a) MI 8+8+9+9+6+10 Averages Mark Scheme +10+4+10+10 10 Al 84 M1 for appropriate addition and division by 10 (condone one error) Al cao Q) (b) Mi 10-4 Al 6 Al cao Q) Q2. Question number Answer Additional guidance @ B1 for Method 1 is not appropriate as it does not take the different number of students into account. B1 for Method 2 is appropriate as there are different numbers of students in each class B1 for a complete assessment of the the mean of the three means. explanation B1 for a complete assessment of the appropriateness of finding the weighted mean. Condone; Yes. with a correct explanation. inappropriateness of finding Condone; No with a correct 2) ®) M1 for 28+ +32n Al 24 cao 28 63x + x +32 72 S5n = 64.1(= =n 24.395...) Make sure you mark part D as part a is also seen. The method must be correct. 2304 = 4068] T&I methods; award M1A1 for an answer of 24 only. [28 x 63 + 32 x 72 = 1764+ Q Q3. Question number Answer Additional guidance Mark @@ M1 for 150+ ... (23-13) Ml for.....+ = 17 Al =179 awrt x50 M1 for identifying 4th class and attempting to find median value within the class. Do not award for just writing 150. There must be some attempt to find the median within the class however erroneous. Don’t be too concemed with the mechanics of their attempt. M1 for use of correct fraction and class width (condone use of 23.5 for 23) Al for value rounding to 179 (or 181 if 23.5 used) NB The estimated mean is 180.43... We ‘MUST see a correct method here, or else sight of awrt 179.4 as evidence of the correct method. Note 180.43.. scores MOMOAO G) (aay Blft (46) teenagers spend more time (on average) than the average person on social media B1ft for correct comparison using their estimated median Comparison with the mean is BO. The comparison must be with values not the median class. @) Q6. Bl 7.2, 7.4, 7.6, 7.8, 8.0, 8.2 B1 for midpoints correct. (3) (a) Condone one error. (Can be Mi for Df =7.2 x 1+7.4x2+7.6x5+... implied by 218.8 seen) (218.8 + 28) M1 for sensible attempt at Lf (x must be a consistent value Al 78 within each class) Al for awrt 7.8 not from wrong method Q7. Answer Additional guidance Mark M1 56/54 x 100 M1 for correct calculation of chain base (6) index number Al 103.7 Al for awrt 103.7 M1 for ¥138.9 = 108 x 103.7" M1 ft for correct calculation of the Al ft 115.9 B1 ft for (average) rate of ‘increase’ per two years B1 ft... is “15.9°% geometric mean of 3 chain base index numbers Alft for awrt 115.9 B1 ft for correct contextual interpretation as tate of increasing per two years B1 ft for correct contextual interpretation of the value for their geometric Qs. (a) | Bl 13 B1 for correct evaluation of | (1) remaining counties (b) | M1 200+... M1 for identifying 2nd (3) class and attempting to find (28-15) median value within the Bch Gas 200; class. Al for use of correct Al = 386 awrt fraction and class width (condone use of 28.5 for 28) Al] for value rounding to 386 (or awrt 393 if 28.5 used) (c)@) | B1 mean greater (than median) B1 for equivalent statement (3) recognising mean will be greater than median (c)(@ii) | B1 median more appropriate with attempt to give a reason B1 for correct choice AND attempt to justify B1 ...due to (positive) skew B1 for stating data is skewed. This may be seen in either (® or (ii) Qs. Question | Answer Additional guidance Mark number (a) B2 for weighted mean (or Paul) AND | B2 for a complete assessment of the (2) reference to there being different appropriate choice with reason proportions / percentages of people earning each weekly amount OR OR B1 for an incomplete assessment of the B1 for weighted mean (or Paul) with appropriate choice attempt at reason (b) M1 for 0.2x260 + 0.35370 + M1 for a correct method to find the (2) 0.45x510 Al for 411 weighted mean Al for the correct answer Q12. Question number Answer Additional guidance Mark M1 2x72+3x84+5x88 (= 836) M1 = 90x(2+3+5+8) © 1620) M1 eee Al Yes, it is possible (if he achieves 98 (or greater)). M1 for using weighting with first 3 assignments M1 for attempt at total score required M1 for calculating score needed on final exam. Al for correct conclusion from correct supporting working Altemative: M1 2x72+3x84+5x88 (= 836) M1 °836° + 1008 (= 1636) 836+100x8 Ga34545) 8) Altemative: Ml 72x=+84x>+88 x =(= 46.44) M1 xx =90-'46.44" M1 for correct method to solve for x x= = x (90 — '46.44') 12) Q13. Question number Answer Additional _guidance Mark (a) M1M1 25X7+7.5X10+12.5X9+17.5X6+22.5x3 M1 for consistent use of Y fx with x within interval @) 7+10+9+6+3 (= 10.78...) Al for 10.8 (including ends) M1 for correct use of Y fx with x the mid-interval value (can be implied by 377.5) Al for awrt 10.8 (b) B1B1 for each of two limitations of conclusion eg. e = The data is just for Ben’s office e Estimated mean in part (a) as we don’t know the actual distances travelled/ have used midpoints ¢ Don’t know how the data for the newspaper article was collected/secondary data ¢@ Small sample size e Newspaper refers to ‘average’ — we don’t know that this is the mean ¢ We only know the average for 2001 and the year that Ben carried out his research. BIBI for each of two limitations of conclusion Q) Do not accept ‘Newspaper is out of date Source is unreliable Q14. Bl ft ...(at a rate) of *7.5°% per year follow through their answer to c(i) B1ft for complete correct contextual interpretation of their value for geometric mean. Question Answer Additional guidance Mark number (a) M1 for either ==" x 100 (= 85.75...) M1 for correct calculation of | (2) an chain-based index number. OR == x 100 (= 98.2...) May be implied by one Al for awrt 85.8 and awrt 98.2 correct answer. Al both correct (b)(i) MI for 93.8 X 92.8 X '85.8' x "98.2" M1 for correct calculation of | (2) the geometric mean of the four chain-based index Alft awrt 92.5 numbers Alft correct answer ft their (a) (by (i) Blft for (average) “decrease” .... BI ft for correct contextual (2) interpretation which must