2023 CCEA GCSE Statistics 2260 模擬試題連答案詳解

(a) The class with the highest frequency (14) is 20 <= t < 30, so this is the modal class.
(b) The midpoint of the class 30 <= t < 40 is \( \dfrac{30+40}{2} = 35 \). The frequency for this class is 8, so the point plotted is (35, 8).
(c) Time is continuous data, so there are no natural gaps between classes; a frequency polygon (formed by joining midpoints with straight lines) reflects this continuity better than separate bars, clearly shows the overall shape and trend of the distribution (e.g. where it peaks), and, unlike a bar chart, allows a second frequency polygon to be plotted on the same axes for direct visual comparison between two distributions.

(a) B1 for 20<=t<30.
(b) M1 for correct midpoint 35; A1 for coordinate (35,8).
(c) A1 for reference to data being continuous; A1 for reference to showing the shape/trend of the distribution; A1 for reference to ease of comparison with another distribution.

(a) This is opportunity (convenience) sampling, since the researcher simply selects whoever is conveniently available at that time and place. A limitation is that the sample is unlikely to be representative of the whole town, since it only includes people who happen to shop at that particular supermarket on a Tuesday morning, excluding people who work at that time, shop elsewhere, or shop on different days.
(b) This is cluster sampling, since the population is divided into geographically-based groups (clusters), a random selection of whole clusters is chosen, and everyone within the selected clusters is surveyed.
(c) This is quota sampling, since the researcher fills fixed quotas (20 males, 20 females) that match known characteristics of the population. It is not a form of random sampling because, within each quota, the researcher chooses which individuals to approach based on her own judgement, so not every member of the population has a known, non-zero, or equal chance of being selected, unlike in true random sampling.

(a) B1 for 'opportunity sampling' (accept 'convenience sampling'); B2 for a valid, developed limitation ([1] for a basic/vague limitation).
(b) B1 for 'cluster sampling'.
(c) B1 for 'quota sampling'; B3 for a full, correct explanation of why it is not random ([1]-[2] for a partial explanation, e.g. only stating researcher chooses who to ask without explaining the consequence for selection probability).

(a) Expected number \( = 0.08 \times 250 = 20 \) adults.
(b) Risk \( = \dfrac{48}{400} \times 100 = 12\% \).
(c) Town A has a known risk of 8%, while the sample from Town B suggests a risk of 12%, which is higher. However, since Town B's figure of 12% is estimated from just one sample of 400 adults rather than the whole population, this difference could simply be due to natural sampling variation rather than a genuine difference in the underlying risk between the two towns; a larger sample, or repeated samples, or a formal statistical test, would be needed before concluding that the allergy is truly more common in Town B.

(a) M1 for 0.08x250; A1 for 20.
(b) M1 for 48/400; A1 for 12%.
(c) A1 for correctly comparing 12% (Town B) with 8% (Town A); A1 for stating this alone does not prove a genuine difference; A1 for a valid reason (e.g. sampling variation, single sample, need for larger sample/test).

(a) The 20 ordered values are: 15,16,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,35,38. The median is the mean of the 10th and 11th values: \( \dfrac{25+26}{2} = 25.5 \) minutes.
(b) Q1 is the median of the lower 10 values (15 to 25): \( \dfrac{20+21}{2} = 20.5 \). Q3 is the median of the upper 10 values (26 to 38): \( \dfrac{30+31}{2} = 30.5 \). IQR \( = 30.5 - 20.5 = 10 \) minutes.
(c) Group A's median (25.5 minutes) is higher than Group B's median (22 minutes), so runners in Group A typically took longer to complete the run than runners in Group B. Group A's IQR (10 minutes) is larger than Group B's IQR (24-18=6 minutes), so the middle 50% of times in Group A are more spread out/variable than in Group B, whose times were more consistent.
(d) (i) Removing 15 and inserting 51 into the sorted data gives: 16,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,35,38,51. The new median is the mean of the 10th and 11th values: \( \dfrac{26+27}{2}=26.5 \) minutes, an increase from 25.5. (ii) The new range is \( 51-16 = 35 \) minutes, compared with the original range of \( 38-15=23 \) minutes, a large increase, since range is very sensitive to extreme values, whereas the median only changed slightly because it depends on the middle values of the ordered data.
(e) A genuine outlier could occur, for example, if one runner was an exceptionally fast, elite-level athlete, or if a runner stopped during the race to help an injured competitor, genuinely increasing their time.

(a) M1 for identifying 10th/11th values (25,26); A1 for median 25.5.
(b) M1 for correct method for Q1 (median of lower half); A1 for Q1=20.5; M1 for correct method for Q3 (median of upper half); A1 for Q3=30.5; A1 for IQR=10.
(c) A2 for correct median comparison in context (Group A higher/took longer) ([1] for stating values without context); A2 for correct IQR comparison in context (Group A more variable) ([1] for stating values without context); A1 for correct calculation of Group B's IQR (6).
(d) A1 for new median 26.5 with correct reasoning; A1 for new range 35; A1 for explaining why range changes much more than median (sensitivity to extreme values).
(e) B1 for a valid, plausible reason for a genuine outlier.

(a) Warning limits \( = 500 \pm 2(4) = 500 \pm 8 \), i.e. 492 ml to 508 ml. Action limits \( = 500 \pm 3(4) = 500 \pm 12 \), i.e. 488 ml to 512 ml.
(b) Checking each value against the action limits [488, 512]: 501, 503, 495, 489, 507 and 499 are all within the action limits; 510 is within the action limits but beyond the upper warning limit (508); 513 ml is beyond the upper action limit (512 ml). Since a sample mean has fallen beyond the action limit, this signals the process is out of control; the factory should stop the production line immediately and investigate/adjust the machine before continuing production.
(c) One assumption is that the process mean (500 ml) and standard deviation (4 ml) used to set the limits are accurate values obtained from a period when the process was known to be running correctly (in control).
(d) If an out-of-control signal is not identified and acted on promptly, the factory may continue producing bottles that are significantly overfilled or underfilled for an extended period, wasting raw materials (if overfilled) or risking under-delivery to customers and regulatory/legal issues (if underfilled), as well as damaging customer trust once the problem is discovered.

(a) M1 for correct method for warning limits; A1 for 492 and 508; M1 for correct method for action limits; A1 for 488 and 512.
(b) A1 for correctly identifying 513 ml as beyond the action limit; A1 for correctly stating no other value breaches the action limits; A1 for stating the process is out of control; A1 for a valid recommended action (stop and investigate the process).
(c) B1 for a valid assumption (e.g. limits based on a genuinely in-control historical process, or that data would be normally distributed when in control).
(d) B2 for a valid, developed consequence ([1] for a basic/vague consequence).

(a) \( 0.3(68) + 0.7(x) = 75 \Rightarrow 20.4 + 0.7x = 75 \Rightarrow 0.7x = 54.6 \Rightarrow x = 78 \).
(b) Since the mean of 5 numbers is 12, their sum is \( 5 \times 12 = 60 \). The sum of the four known numbers is \( 8+10+14+15=47 \). The fifth number is \( 60-47=13 \).
(c) The mean uses every value in the data set in its calculation, making it a more representative measure of the whole data set (when there are no extreme outliers) than the median, which only depends on the middle value(s); the mean is also more useful for further statistical calculations, such as finding the standard deviation.

(a) M1 for correct weighted mean equation set up; M1 for correctly simplifying (20.4+0.7x=75); M1 for correctly isolating 0.7x=54.6; A1 for correct rearrangement; A1 for x=78.
(b) M1 for total = 5x12=60; M1 for sum of four known values = 47; A1 for fifth number = 13.
(c) B2 for a valid, developed advantage ([1] for a basic/vague advantage).

(a) \( \sum f = 4+8+10+6+2 = 30 \). \( \sum fx = 5(4)+10(8)+15(10)+20(6)+25(2) = 20+80+150+120+50 = 420 \). Mean \( = \dfrac{420}{30} = 14 \).
(b) \( \sum fx^2 = 25(4)+100(8)+225(10)+400(6)+625(2) = 100+800+2250+2400+1250 = 6800 \). Standard deviation \( = \sqrt{\dfrac{6800}{30} - 14^2} = \sqrt{226.67 - 196} = \sqrt{30.67} = 5.54 \) (3 s.f.).
(c) The students' mean (14 texts) is higher than the adults' mean (9 texts), so students send more texts per day on average than adults. The students' standard deviation (5.54) is higher than the adults' standard deviation (3.2), so the number of texts sent by students is more variable (spread out) than the number sent by adults, whose texting habits are more consistent.

(a) M1 for Σf=30; M1 for Σfx=420; A1 for mean=14.
(b) M1 for Σfx²=6800; M1 for correct substitution into formula; M1 for 6800/30-196 = 30.67 (or equivalent unsimplified expression); A1 for correct square root method; A1 for SD=5.54 (3 s.f., own-figure rule applies).
(c) A2 for correct mean comparison in context ([1] for stating values without context); A2 for correct SD comparison in context ([1] for stating values without context).

(a) Using \( N = \dfrac{n_1 \times n_2}{m} \), where \( n_1=60 \) (first catch, tagged), \( n_2=90 \) (second catch) and \( m=18 \) (tagged fish recaptured): \( N = \dfrac{60 \times 90}{18} = \dfrac{5400}{18} = 300 \) fish.
(b) Two assumptions needed are: the population is closed, meaning there are no births, deaths, or migration into or out of the lake between the two catches; and the tagged fish mix back in randomly and evenly with the rest of the population, so that the second catch is representative of the whole lake (all fish equally likely to be caught).
(c) If some tagged fish died before the second catch, fewer tagged fish would be available to be recaptured, so the number of tagged fish found in the second sample (m) would tend to be smaller than it would be if all 60 tagged fish had survived. Since a smaller value of m makes the calculated estimate \( N = n_1n_2/m \) larger, this would cause the method to overestimate the true population of fish in the lake.

(a) M1 for correct formula N=n1n2/m; M1 for correct substitution; A1 for N=300.
(b) B1 for each of two valid, distinct assumptions (e.g. closed population; random mixing; tagging doesn't affect survival/catchability; equal catchability), up to [2].
(c) A1 for correctly identifying that m would be smaller than expected; A1 for correctly concluding the estimate would be too high (an overestimate), with valid reasoning linking to the formula.

(a) The differences in rank, d, are: A: -1, B: 1, C: -1, D: 1, E: -1, F: 1, so \( d^2 = 1 \) for each student, giving \( \sum d^2 = 6 \). With \( n=6 \): \( r_s = 1 - \dfrac{6(6)}{6(6^2-1)} = 1 - \dfrac{36}{6(35)} = 1 - \dfrac{36}{210} = 1 - 0.1714 = 0.829 \) (3 s.f.).
(b) A value of rs = 0.829 indicates a strong positive correlation between students' rank in Maths and their rank in Science: students who rank highly (i.e. do well) in Maths tend to also rank highly in Science, and vice versa.
(c) A strong correlation between two variables does not prove that one causes the other; there could be a third, underlying factor (such as general academic ability, effort, or study habits) that independently causes a student to perform well in both Maths and Science, without either subject directly causing success in the other.

(a) M1 for correct differences d; M1 for Σd²=6; M1 for correct substitution into formula; A1 for unsimplified value (1-36/210); A1 for rs=0.829 (3 s.f.).
(b) A2 for a correct, contextualised interpretation referencing strong positive correlation ([1] for a partial/vague interpretation).
(c) A1 for stating correlation does not imply causation; A1 for a valid, specific alternative explanation (e.g. a named plausible third factor).

(a) 159 cm and 171 cm are exactly 1 standard deviation below and above the mean (165 +/- 6). By the empirical rule, approximately 68% of values in a normal distribution lie within 1 standard deviation of the mean.
(b) 153 cm and 177 cm are exactly 2 standard deviations below and above the mean (165 +/- 12). By the empirical rule, approximately 95% of values lie within 2 standard deviations of the mean.
(c) \( z = \dfrac{174-165}{6} = \dfrac{9}{6} = 1.5 \).
(d) \( z = \dfrac{190-178}{7} = \dfrac{12}{7} = 1.71 \) (3 s.f.).
(e) The man's z-score (1.71) is greater than the woman's z-score (1.5). Since a z-score measures how many standard deviations a value is above (or below) its own population's mean, the man's height is relatively further above his own population's average height than the woman's height is above hers, so the man is relatively taller compared to his own population, even though his actual height in cm and his population's standard deviation both differ from the woman's.

(a) B2 for 68% (accept 68.2%).
(b) B2 for 95% (accept 95.4%).
(c) M1 for correct method (174-165)/6; A1 for correct unsimplified value; A1 for z=1.5.
(d) M1 for correct method (190-178)/7; A1 for correct unsimplified value; A1 for z=1.71 (3 s.f.).
(e) A1 for correctly identifying the man as relatively taller; A1 for correctly comparing the two z-scores (1.71 > 1.5); A2 for a correct, fully explained reason referencing what a z-score represents (ft from (c) and (d)).

(a) Percentage increase \( = \dfrac{48000-42000}{42000} \times 100 = \dfrac{6000}{42000}\times100 = 14.3\% \) (3 s.f.).
(b) By starting the vertical axis at £40,000 instead of £0, the visible portion of each bar only represents sales above £40,000 (£2,000 for Year 1 and £8,000 for Year 2), so the second bar appears four times taller than the first, even though the underlying increase in total sales is a much more modest 14.3%. This truncated axis exaggerates the visual impression of growth, making a moderate increase look dramatic.

(a) M1 for correct method (6000/42000); A1 for 14.3%.
(b) A1 for identifying that only the portion above £40,000 is shown/compared; A1 for correctly explaining this exaggerates the relative visual difference between the bars; A1 for linking this back to the true, much smaller, percentage increase found in (a).

(a) A suitable null hypothesis is: 'There is no difference in the average amount of sleep obtained by students who walk to school and students who are driven to school' (i.e. mode of travel to school and amount of sleep are independent/unrelated).
(b) An appropriate sampling frame would be the school's full register (list) of all currently enrolled students, from which a sample could then be randomly selected.
(c) Two problems with the question are: it is a leading question ('Don't you agree...'), which pressures respondents towards answering 'yes'; and it does not actually collect the factual data needed to test the hypothesis (travel mode and amount of sleep), instead asking for an opinion on health that is not directly relevant. A more suitable pair of questions would be, for example: 'How do you usually travel to school? (tick one): Walk / Driven / Other' and 'On an average school night, how many hours of sleep do you get? _____ hours', which are neutral, closed, and collect exactly the data needed.

(a) B2 for a correctly worded null hypothesis referencing no difference/independence between travel mode and sleep ([1] for a partial/vague statement).
(b) B2 for a valid sampling frame (e.g. school register/list of all students) ([1] for a vague/partial answer).
(c) B1 for identifying the question is leading/biased; B1 for identifying the question does not collect the relevant factual data; B3 for a genuinely improved, closed, unbiased rewritten question (or pair of questions) that would allow the hypothesis to be tested ([1]-[2] for a partial improvement).

(a) Warning limits \( = 1000 \pm 2(3) = 1000 \pm 6 \), i.e. 994 g to 1006 g. Action limits \( = 1000 \pm 3(3) = 1000 \pm 9 \), i.e. 991 g to 1009 g.
(b) Checking each value against the action limits [991,1009]: 999, 1004, 1007 and 996 are within the action limits (1007 is beyond the warning limit of 1006 but not the action limit); 988 g is below the lower action limit (991); 1010 g is above the upper action limit (1009). Both 988 g and 1010 g fall beyond the action limits, signalling the process is out of control; the operator should stop the production line and investigate/adjust the machine.
(c) A run of six sample means clustered this tightly around 1001-1003 g, with far less variation than the historical standard deviation of 3 g would suggest, could indicate a genuine improvement in the consistency of the filling process. However, it could equally suggest a problem with the monitoring equipment itself, such as a stuck or faulty weighing gauge that is repeatedly giving very similar readings rather than accurately measuring each sample, so this pattern should prompt a check of the measuring equipment as well as the filling process.
(d) One cost of stopping the line unnecessarily (a 'false alarm', where the process was actually still in control) is the lost production output and staff time while the line is halted and checked, reducing the factory's overall productivity and potentially causing delays in meeting customer orders.

(a) M1 for correct method for warning limits; A1 for 994 and 1006; M1 for correct method for action limits; A1 for 991 and 1009.
(b) A1 for correctly identifying 988g beyond the action limit; A1 for correctly identifying 1010g beyond the action limit; A1 for correctly stating no other values breach the action limits; A1 for stating the process is out of control; A1 for a valid recommended action.
(c) A1 for suggesting genuine improved consistency as one possibility; A1 for suggesting a fault with the measuring equipment as a second possibility; A1 for a valid recommended follow-up action (checking the equipment).
(d) B2 for a valid, developed cost of an unnecessary stoppage ([1] for a basic/vague cost).

(a) \( r = \dfrac{n\sum xy - \sum x \sum y}{\sqrt{\left(n\sum x^2-(\sum x)^2\right)\left(n\sum y^2-(\sum y)^2\right)}} = \dfrac{5(1242)-(30)(180)}{\sqrt{\left(5(220)-30^2\right)\left(5(7138)-180^2\right)}} \)
\( = \dfrac{6210-5400}{\sqrt{(1100-900)(35690-32400)}} = \dfrac{810}{\sqrt{200 \times 3290}} = \dfrac{810}{\sqrt{658000}} = \dfrac{810}{811.17} = 0.999 \) (3 s.f.).
(b) A PMCC of 0.999 indicates a very strong positive (linear) correlation between advertising spend and sales for these shops: as advertising spend increases, sales increase in an almost perfectly linear way.
(c) Overall, the time series shows an upward trend in sales over the 5 years, rising from £20,000 to £42,000. However, this was not a steady rise throughout: sales rose steadily for the first 3 years, then experienced a temporary fall in Year 4 (from £35,000 to £30,000), before recovering sharply and rising to a new high of £42,000 by Year 5.

(a) M1 for correct numerator (6210-5400=810); M1 for correct first bracket (1100-900=200); M1 for correct second bracket (35690-32400=3290); A1 for correct unsimplified expression; A1 for r=0.999 (3 s.f.).
(b) A2 for a correct, contextualised interpretation referencing very strong positive correlation ([1] for a partial/vague interpretation).
(c) A1 for correctly identifying the overall upward trend; A1 for correctly identifying the Year 4 dip; A1 for correctly identifying the sharp Year 5 recovery.

(a) Since 90 study French and 30 of these also study Spanish, 60 study French but not Spanish. Since 70 study Spanish and 30 of these also study French, 40 study Spanish but not French. The remaining students study neither: \( 200-(30+60+40)=70 \).
Two-way table:
| Studies Spanish | Doesn't study Spanish | Total
Studies French | 30 | 60 | 90
Doesn't study French | 40 | 70 | 110
Total | 70 | 130 | 200
(b) \( P(\text{French not Spanish}) = \dfrac{60}{200} = 0.3 \).
(c) \( P(\text{French}|\text{Spanish}) = \dfrac{30}{70} = \dfrac{3}{7} \approx 0.429 \).
(d) The frequency tree starts with 200 students, splitting into 90 (French) and 110 (Not French). The French branch (90) splits into 30 (Spanish) and 60 (Not Spanish). The Not French branch (110) splits into 40 (Spanish) and 70 (Not Spanish). The number of students studying neither language is the 'Not French, Not Spanish' branch: 70 students.
(e) \( P(\text{French}) = \dfrac{90}{200} = 0.45 \). \( P(\text{French}|\text{Spanish}) = \dfrac{30}{70} \approx 0.429 \). Since \( P(\text{French}|\text{Spanish}) \neq P(\text{French}) \) (0.429 is not equal to 0.45), studying French and studying Spanish are not independent events.

(a) B1 for each correct missing cell value (30 given/not required to derive; 60, 40, 70), up to [4] (accept correctly completed table in any equivalent layout).
(b) M1 for 60/200; A1 for 0.3.
(c) M1 for 30/70; A1 for 3/7 or 0.429 (accept equivalent).
(d) B1 for correct first-level split (90/110); B1 for correct second-level splits (30/60 and 40/70); B1 for correctly identifying neither=70.
(e) M1 for correctly calculating both P(French) and P(French|Spanish); A1 for correct conclusion (not independent) with valid comparison of the two values.

(a) The intersection (both sports) region contains 12 students. The 'football only' region contains \( 35-12=23 \) students. The 'basketball only' region contains \( 28-12=16 \) students. The number of students represented within the two circles is \( 23+16+12=51 \), so the region outside both circles (neither sport) contains \( 60-51=9 \) students.
(b) \( P(F \cup B) = \dfrac{51}{60} = 0.85 \) (or equivalently \( \dfrac{17}{20} \)).
(c) \( P(\text{football only}) = \dfrac{23}{60} \approx 0.383 \) (3 s.f.).

(a) B1 for both=12; B1 for football only=23; B1 for basketball only=16; B1 for neither=9.
(b) M1 for 51/60 (ft from (a)); A1 for 0.85 (or 17/20).
(c) M1 for 23/60 (ft from (a)); A1 for 0.383 (accept equivalent fraction/decimal).

(a) Cumulative frequencies: 12, 32, 56, 72, 80.
(b) The median is the \( \tfrac{80}{2}=40\text{th} \) value, which falls in the class 4<=d<6 (cumulative frequency rises from 32 to 56 in this class). Using linear interpolation: \( \text{median} = 4 + \dfrac{40-32}{56-32}\times 2 = 4 + \dfrac{8}{24}\times2 = 4.67 \) km (3 s.f.).
(c) Q1 is the \( \tfrac{80}{4}=20\text{th} \) value, which falls in the class 2<=d<4: \( Q_1 = 2 + \dfrac{20-12}{32-12}\times2 = 2+\dfrac{8}{20}\times2 = 2.8 \) km. Q3 is the \( \tfrac{3(80)}{4}=60\text{th} \) value, which falls in the class 6<=d<8: \( Q_3 = 6+\dfrac{60-56}{72-56}\times2 = 6+\dfrac{4}{16}\times2=6.5 \) km. IQR \( = 6.5-2.8=3.7 \) km.
(d) The rural school's median (8 km) is higher than the urban school's median (4.67 km), so rural students typically travel further to school than urban students. The rural school's IQR (12-5=7 km) is also higher than the urban school's IQR (3.7 km), so the distances travelled by rural students are more variable/spread out than those travelled by urban students, whose journeys are more consistent.
(e) Rural areas typically have a lower population density than urban areas, meaning homes and the nearest school are, on average, further apart, and there may be fewer schools to choose from locally.

(a) B2 for all correct cumulative frequencies (12,32,56,72,80) ([1] for at least 3 correct).
(b) M1 for correctly identifying the 40th value falls in class 4<=d<6; M1 for correct interpolation method; A1 for median=4.67 km (accept 4.66-4.67).
(c) M1 for correctly identifying the 20th value's class; A1 for Q1=2.8 km; M1 for correctly identifying the 60th value's class; A1 for Q3=6.5 km; A1 for IQR=3.7 km.
(d) A2 for correct median comparison in context (rural higher/travel further) ([1] for stating values without context); A2 for correct IQR comparison in context (rural more variable) ([1] for stating values without context, and for correctly calculating rural IQR=7).
(e) B1 for a valid, developed reason (e.g. lower population density, schools further apart, fewer local schools).

(a) Two conditions needed are: there is a fixed number of trials (here, 5 questions); each trial is independent of the others; each trial has only two possible outcomes (correct or incorrect); and the probability of success (correct guess, 0.25) is constant across all trials.
(b) \( P(X=2) = \binom{5}{2}(0.25)^2(0.75)^3 = 10 \times 0.0625 \times 0.421875 = 0.264 \) (3 s.f.).
(c) \( P(X \geq 1) = 1-P(X=0) = 1-(0.75)^5 = 1-0.237 = 0.763 \) (3 s.f.).

(a) B1 for each of two valid, distinct conditions, up to [2] (e.g. fixed number of trials; independence; only two outcomes; constant probability).
(b) M1 for correct binomial coefficient C(5,2)=10; M1 for correct powers 0.25² and 0.75³; A1 for correct unsimplified expression; A1 for P(X=2)=0.264 (3 s.f.).
(c) M1 for correct method 1-(0.75)^5; A1 for P(X>=1)=0.763 (3 s.f.).

(a) A chain base index of 105 for 2021 means the price rose by 5% between 2020 and 2021.
(b) A chain base index of 98 for 2022 means the price fell by 2% between 2021 and 2022.
(c) Price in 2021 \( = £40 \times 1.05 = £42 \).
(d) Continuing the chain: price in 2022 \( = £42 \times 0.98 = £41.16 \); price in 2023 \( = £41.16 \times 1.10 = £45.28 \) (2 d.p.). Overall percentage change from 2020 to 2023: \( \dfrac{45.28-40}{40}\times100 = 13.2\% \) (3 s.f.) increase.

(a) A2 for correct interpretation (price rose by 5%, 2020 to 2021) ([1] for a partial/vague interpretation).
(b) A2 for correct interpretation (price fell by 2%, 2021 to 2022) ([1] for a partial/vague interpretation).
(c) M1 for 40x1.05; A1 for £42 (ft from correct method even if (a) misread).
(d) M1 for correctly chaining through 2022 (£41.16) and 2023 (£45.28); A1 for overall percentage change 13.2% (own-figure rule applies throughout).

(a) \( \sum x = 12+15+18+20+14+16+19+22 = 136 \). Mean \( = \dfrac{136}{8} = 17 \).
(b) \( \sum x^2 = 144+225+324+400+196+256+361+484 = 2390 \). Standard deviation \( = \sqrt{\dfrac{2390}{8} - 17^2} = \sqrt{298.75-289} = \sqrt{9.75} = 3.12 \) (3 s.f.).
(c) For the first test: \( z = \dfrac{21-17}{3.12} = 1.28 \) (3 s.f.). For the second test: \( z = \dfrac{78-72}{5} = 1.2 \).
(d) The student's z-score for the first test (1.28) is higher than for the second test (1.2), meaning the score of 21 was relatively further above its class mean (in standard deviation terms) than the score of 78 was above its own class mean. The student therefore performed relatively better, compared with their respective class, in the first test.

(a) M1 for Σx=136; A1 for mean=17.
(b) M1 for Σx²=2390; M1 for correct substitution into formula; A1 for 9.75 (unsimplified value under root); A1 for SD=3.12 (3 s.f., own-figure rule applies).
(c) M1 for correct method for both z-scores; A1 for z(first test)=1.28 (ft from (a),(b)); A1 for z(second test)=1.2.
(d) A1 for correctly identifying the first test as relatively better, with valid reasoning comparing the two z-scores (ft from (c)).