2025 CCEA GCSE Statistics 2260 Practice Paper with Answers

(a) Ordered data: 23, 25, 28, 31, 32, 34, 34, 35, 37, 38, 41, 43, 45, 47, 62.
Stem and leaf diagram (key: 2 | 3 means 23 minutes):
2 | 3 5 8
3 | 1 2 4 4 5 7 8
4 | 1 3 5 7
5 |
6 | 2

(b) With n=15 values ordered, the median is the 8th value: 35 minutes. Range \( = 62-23 = 39 \) minutes.

(c) Lower half (first 7 values): 23,25,28,31,32,34,34 — lower quartile LQ = 4th value = 31. Upper half (last 7 values): 37,38,41,43,45,47,62 — upper quartile UQ = 4th value = 43. \( IQR = UQ-LQ = 43-31 = 12 \) minutes.

(d) Upper outlier bound \( = UQ+1.5\times IQR = 43+1.5(12) = 43+18 = 61 \). Since 62 > 61, the value 62 IS classified as an outlier.

Final answer: median = 35, range = 39, IQR = 12, and 62 is an outlier.

(a) M1 for stems 2-6 and leaves correctly sorted in ascending order within each row; A1 for all 15 leaves correctly placed; A1 for a correct key. (b) A1 median = 35; A1 range = 39. (c) M1 correct identification of LQ (=31) and UQ (=43) by splitting the ordered data into halves; A1 IQR = 12. (d) MA1 correct calculation of the upper bound (61) AND correct conclusion that 62 is an outlier (both required for the mark).

(a) The question is leading because its phrasing ('Don't you agree that...') suggests the 'correct'/expected answer is 'yes', pressuring respondents towards agreeing rather than giving their genuine, independent opinion; this would bias the results towards showing artificially strong support for eating five portions a day.

(b) For example: 'How many portions of fruit and vegetables do you think an adult should eat per day? (0 / 1-2 / 3-4 / 5 or more)' — a neutrally-worded closed question offering fixed response options without suggesting a particular answer.

(c) \( IQR = UQ-LQ = 6-3 = 3 \) portions. \( \text{range} = \text{max}-\text{min} = 9-1 = 8 \) portions.

(d) 20%.

(e) Percentage point change \( = 40\%-30\% = 10 \) percentage points (an increase).

Final answer: (c) IQR=3, range=8; (d) 20%; (e) +10 percentage points.

(a) A1 correctly identifies the question suggests/pressures towards a particular ('yes') answer; A1 correctly links this to biased/unrepresentative results. (b) A1 valid closed question with neutral wording and fixed response categories. (c) A1 IQR = 3; A1 range = 8. (d) A1 20%. (e) M1 correct method (2024 value minus 2023 value); A1 correct value 10; A1 correctly states this as an increase/'percentage point' change (not a % change).

(a) First toss: P(H) = 0.6, P(T) = 0.4. From each of these, second toss: P(H) = 0.6, P(T) = 0.4 again (the coin has no memory), giving four branch pairs: H then H (0.6, 0.6); H then T (0.6, 0.4); T then H (0.4, 0.6); T then T (0.4, 0.4).

(b) HH: \( 0.6\times0.6=0.36 \); HT: \( 0.6\times0.4=0.24 \); TH: \( 0.4\times0.6=0.24 \); TT: \( 0.4\times0.4=0.16 \). (Check: \( 0.36+0.24+0.24+0.16=1.00 \), correct.)

(c) Exactly one head occurs via outcomes HT or TH: \( P(\text{exactly one head}) = 0.24+0.24 = 0.48 \)

(d) \( P(\text{at least one head}) = 1-P(TT) = 1-0.16 = 0.84 \)

(e) Expected number of repetitions \( = P(\text{exactly one head})\times200 = 0.48\times200 = 96 \)

Final answer: (c) 0.48; (d) 0.84; (e) 96.

(a) A1 correct probabilities (0.6, 0.4) on first-toss branches; A1 correct probabilities (0.6, 0.4) repeated on both pairs of second-toss branches; A1 all four branch pairs clearly labelled H/T. (b) A1 all four outcomes (HH, HT, TH, TT) correctly listed; A1 all four probabilities correctly calculated and summing to 1. (c) M1 correctly identifies HT and TH as the relevant outcomes; M1 correct addition method; A1 \( P=0.48 \). (d) MA1 correct method (\( 1-P(TT) \)) and correct value 0.84. (e) M1 correct method (probability × 200); A1 answer = 96.

(a) The price rose every year over the five-year period, from £40 in 2019 to £54 in 2023 — the price shows a steady, continuous upward (increasing) trend.

(b) Index \( = \dfrac{\text{price in year}}{\text{price in base year}}\times100 \). 2020: \( \dfrac{42}{40}\times100=105.00 \). 2021: \( \dfrac{45}{40}\times100=112.50 \). 2022: \( \dfrac{50}{40}\times100=125.00 \). 2023: \( \dfrac{54}{40}\times100=135.00 \).

(c) An index of 125.00 for 2022 means that the price of the item in 2022 was 25% higher than its price in the base year, 2019.

(d) Index numbers allow price changes to be compared easily as percentages relative to a fixed base year, making it simple to see and compare the SIZE of relative changes over time (or between different items with very different original prices) without needing to work with the original, differently-scaled values each time.

Final answer: (b) 2020=105.00, 2021=112.50, 2022=125.00, 2023=135.00.

(a) A1 correctly describes a continuous/steady increase; A1 correctly quotes the overall change from £40 to £54 (or equivalent supporting detail). (b) M1 correct formula/method shown; A1 2020 and 2021 correct; A1 2022 correct; A1 2023 correct (all four to 2 d.p.). (c) MA1 correct interpretation: price 25% higher than in 2019 (both the percentage AND the correct reference to the base year required). (d) A1 basic advantage stated; A1 fully-developed explanation (e.g. easy comparison of relative change/percentage terms, or comparability across different items/scales).

(a) The number of pets owned is discrete data, since it can only take whole-number (integer) values (a household cannot own, for example, 2.5 pets). A stepped cumulative frequency diagram is appropriate because the cumulative frequency only changes (jumps) at each specific discrete value, remaining constant in between, rather than increasing smoothly and continuously as it would for grouped continuous data.

(b) Cumulative frequency table:
Pets: 0, 1, 2, 3, 4, 5
Cumulative frequency: 8, 20, 30, 36, 39, 40

(c) With n=40, the median is the average of the 20th and 21st values. From the cumulative frequency table, the 20th value is the last value in the 'up to 1 pet' group (cumulative frequency reaches exactly 20 at 1 pet), so the 20th value = 1; the 21st value falls in the next group (2 pets), so the 21st value = 2. Median \( = \dfrac{1+2}{2} = 1.5 \) pets.

(d) 3rd decile position \( = 0.3\times40 = 12 \)th value; this falls within the '1 pet' group (cumulative frequency reaches 20 at 1 pet, and the 12th value lies between the 9th and 20th values), so the 3rd decile = 1. 90th percentile position \( = 0.9\times40 = 36 \)th value; this falls exactly at the boundary of the '3 pets' group (cumulative frequency reaches 36 at 3 pets), so the 90th percentile = 3.

(e) A 90th percentile of 3 means that 90% of the households in the sample own 3 pets or fewer (and correspondingly, only 10% of households own more than 3 pets).

Final answer: (c) median = 1.5; (d) 3rd decile = 1, 90th percentile = 3.

(a) A1 correctly identifies discrete; A1 correct reasoning linking discreteness to the stepped (rather than smooth curve) shape. (b) A1 correct cumulative frequencies for pets 0-2 (8, 20, 30); A1 correct cumulative frequencies for pets 3-5 (36, 39, 40). (c) M1 correct identification of the 20th and 21st values using the table; A1 median = 1.5. (d) M1 correct position calculation for the 3rd decile (12th value) and correct value = 1; A1 correct position calculation for the 90th percentile (36th value) and correct value = 3 (M1 A1, or MA1 for each if combined). (e) MA1 correct interpretation referencing 90% of households owning 3 or fewer pets.

(a) Points plotted accurately, with hours studied on the x-axis and exam score on the y-axis; as x increases, y also clearly tends to increase, showing strong positive correlation.

(b) \( r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{255.5}{\sqrt{42.0\times1565.9}}=\dfrac{255.5}{\sqrt{65\,767.8}}=\dfrac{255.5}{256.45}=0.996 \) (3 d.p.). Since r is very close to +1, this indicates a very strong positive linear correlation between hours studied and exam score.

(c) \( b=\dfrac{S_{xy}}{S_{xx}}=\dfrac{255.5}{42.0}=6.083 \) (3 d.p.). \( \bar{x}=\dfrac{2+3+4+5+6+7+8+9}{8}=\dfrac{44}{8}=5.5 \), \( \bar{y}=\dfrac{40+45+55+58+65+70+78+82}{8}=\dfrac{493}{8}=61.625 \). \( a=\bar{y}-b\bar{x}=61.625-6.083(5.5)=61.625-33.46=28.17 \) (2 d.p.). So \( y=28.17+6.08x \).

(d) The gradient, b ≈ 6.08, means that, on average, each additional hour of studying is associated with an increase of approximately 6.08 marks in the examination score.

Final answer: \( r=0.996 \); \( y=28.17+6.08x \).

(a) A1 correctly labelled axes and scale; A1 all 8 points plotted accurately; A1 correct description (strong positive correlation). (b) M1 correct substitution into the PMCC formula; A1 \( r=0.996 \); A1 correct interpretation (very strong positive linear correlation). (c) M1 correct method for b (\( S_{xy}/S_{xx} \)); A1 correct \( \bar{x} \), \( \bar{y} \) and correctly-derived value of a; A1 fully correct equation \( y=28.17+6.08x \) (accept values rounded to 2-3 d.p.). (d) A1 correct contextual interpretation of the gradient (marks gained per extra hour studied).

(a) A control chart with weight (g) on the vertical axis and sample number on the horizontal axis, with horizontal lines drawn at: target = 50 g (centre line), warning lines at 45 g and 55 g, and action lines at 40 g and 60 g; all 10 sample means plotted as points at their correct sample number and value, usually joined by straight line segments.

(b) Sample 7 (58 g) lies between the upper warning line (55 g) and the upper action line (60 g) — a warning breach. Sample 9 (62 g) lies beyond the upper action line (60 g) — an action breach. All other samples (1-6, 8, 10) lie between the two warning lines and so are within normal, expected variation.

(c) Following sample 7 (a single warning breach), the process should continue to be monitored closely, but no immediate corrective action is required from a single warning-line breach alone (a warning signals the process may be starting to drift and should be watched, but is not yet conclusive evidence of a problem). Following sample 9 (an action breach, 62 g, beyond the action line), the factory should stop the production process and investigate immediately, since this indicates the process mean has very likely shifted out of control and boxes may not be meeting their target weight specification.

(d) Warning and action lines allow production staff to distinguish normal, expected random variation in sample weights (falling within the warning lines) from signs that the process may be drifting out of control (warning breach) or has definitely gone out of control (action breach), enabling problems to be identified and corrected promptly — helping to maintain consistent product quality (e.g. ensuring customers reliably receive the advertised weight of product) while avoiding unnecessary process stoppages for normal, minor fluctuations.

Final answer: Sample 7 triggers a warning; Sample 9 triggers an action (process should be stopped and investigated).

(a) A1 correctly labelled axes; A1 all five reference lines (target, 2 warning, 2 action) correctly drawn and labelled; A1 all 10 points plotted accurately. (b) A1 correctly identifies sample 7 as a warning breach; A1 correctly identifies sample 9 as an action breach; A1 correctly confirms all other samples are within the warning lines. (c) A1 correct action following sample 7 (continue/monitor, no immediate stoppage) with valid reasoning; A1 correct action following sample 9 (stop and investigate the process); A1 valid, clearly-expressed reasoning for the sample 9 decision. (d) A1 basic explanation of the purpose of warning/action lines; A1 fully-developed explanation referencing distinguishing normal variation from a genuine process problem and maintaining product quality/consistency.

(a) \( \text{mean} = \dfrac{\Sigma fx}{\Sigma f} = \dfrac{4(5)+9(15)+12(25)+7(35)+3(45)}{35} = \dfrac{20+135+300+245+135}{35} = \dfrac{835}{35} = 23.86 \) years (2 d.p.).

(b) This is only an estimate because the exact individual ages within each class interval are not known; the calculation assumes every person in a class interval has an age exactly equal to that interval's midpoint, which is unlikely to be true for every individual (their actual ages are simply spread somewhere within each interval).

(c) Adding a constant (3 years) to every value increases the mean by exactly that same constant (new mean = old mean + 3), because the whole distribution shifts up by 3 years; however, the standard deviation is UNCHANGED, because adding a constant to every value does not change how spread out the values are relative to one another (the shape and spread of the distribution stays the same, it is simply shifted).

(d) Multiplying every value by a constant (12, to convert years to months) multiplies the mean by that same constant (new mean = old mean × 12); the standard deviation is also multiplied by that same constant (new SD = old SD × 12), because scaling every value also scales the spread of the data by the same factor.

Final answer: (a) mean estimate = 23.86 years.

(a) M1 correct method (Σfx/Σf) with correct Σfx = 835 shown; A1 mean = 23.86 (or 23 6/7) years. (b) A1 correctly identifies individual ages within a class are unknown; A1 correctly explains the midpoint assumption underlying the estimate. (c) A1 correctly states mean increases by 3 (to new mean); A1 correctly states SD is unchanged, with valid reasoning. (d) A1 correctly states mean is multiplied by 12; A1 correctly states SD is also multiplied by 12, with valid reasoning.

(a) A 4-point moving average is appropriate because there are 4 quarters in each year, so averaging every 4 consecutive values smooths out the seasonal variation between quarters, revealing the underlying trend.

(b) First 4-point MA (Q1-22 to Q4-22): \( \dfrac{40+55+70+50}{4}=53.75 \). Second 4-point MA (Q2-22 to Q1-23): \( \dfrac{55+70+50+45}{4}=55.00 \).

(c) Continuing, the uncentred 4-point MAs are: 53.75, 55.00, 56.25, 57.50, 58.75. Since a 4-point moving average falls between two quarters (an even number of points), pairs of consecutive uncentred MAs are averaged ('centring') to align each value with an actual quarter: \( \frac{53.75+55.00}{2}=54.375 \) (Q3-22); \( \frac{55.00+56.25}{2}=55.625 \) (Q4-22); \( \frac{56.25+57.50}{2}=56.875 \) (Q1-23); \( \frac{57.50+58.75}{2}=58.125 \) (Q2-23).

(d) The trend line values (54.375, 55.625, 56.875, 58.125) increase steadily from Q3-22 to Q2-23, indicating that, once seasonal quarterly fluctuations are smoothed out, underlying sales are showing a steady upward (increasing) trend over this period.

(e) Forecasting beyond the available data (extrapolation) assumes that the existing trend and seasonal pattern will continue unchanged into the future, which may not be true — unforeseen factors (e.g. a new competitor, economic downturn, change in consumer tastes) could cause actual future sales to deviate significantly from a simple extrapolation of the trend line.

(a) A1 correct explanation referencing 4 quarters per year/smoothing seasonal variation. (b) A1 first MA = 53.75; A1 second MA = 55.00. (c) M1 correct method (averaging successive uncentred MA pairs); A1 Q3-22 = 54.375; A1 Q4-22 = 55.625; A1 Q1-23 = 56.875 and Q2-23 = 58.125 (both required for this final mark). (d) A1 correctly plots/describes both series; A1 correct comment identifying a steady increasing underlying trend. (e) A1 valid limitation of extrapolation (assumes trend/pattern continues; unforeseen factors could change the actual outcome).

(a) A symmetric, bell-shaped curve, centred on and symmetric about the mean, 100 mm, with the curve tailing off (approaching but never quite reaching the horizontal axis) on either side.

(b) \( z = \dfrac{x-\mu}{\sigma} = \dfrac{130-100}{15} = \dfrac{30}{15} = 2.0 \)

(c) Since \( |z|=2.0 \), which is less than 3, a bolt of length 130 mm would NOT be classed as an anomaly (its length, while somewhat unusual, is not extreme enough to be flagged, since values more than 3 standard deviations from the mean are considered very unusual).

(d) Approximately 95% of values lie within two standard deviations of the mean, i.e. between \( 100-2(15)=70 \) mm and \( 100+2(15)=130 \) mm. This leaves approximately \( 100\%-95\%=5\% \) of values outside this range, split (by symmetry) equally between the two tails: \( 5\%\div2=2.5\% \) of bolts have length greater than 130 mm.

(e) 85 mm and 115 mm are exactly one standard deviation below and above the mean respectively (\( 100-15=85 \), \( 100+15=115 \)); since 68% of values lie within one standard deviation of the mean, approximately 68% of bolts have lengths between 85 mm and 115 mm.

Final answer: (b) z=2.0; (c) not an anomaly; (d) 2.5%; (e) 68%.

(a) A1 correct symmetric bell shape; A1 mean correctly marked at the centre/axis of symmetry. (b) M1 correct substitution into \( z=(x-\mu)/\sigma \); A1 z = 2.0. (c) A1 correctly states not an anomaly (ECF from (b)); A1 valid reasoning (|z|<3). (d) M1 correctly identifies 130 mm as 2 SD above the mean; M1 correct method for the tail probability (100-95, halved); A1 2.5%. (e) A1 correctly identifies 85 and 115 as one SD below/above the mean; A1 correct value, 68%.

(a) Both male and female average screen time increased steadily every year from 2020 to 2023 (male: 18 to 24 hours; female: 16 to 25 hours).

(b) In 2020, 2021 and 2022, male teenagers had higher average screen time than female teenagers (e.g. 18 vs 16 hours in 2020); however, female screen time increased at a faster rate each year, and by 2023 female screen time (25 hours) had overtaken male screen time (24 hours) — this is the point at which the relationship between the two groups changes.

(a) A1 correctly describes both groups increasing steadily; A1 correct supporting start/end values for at least one group. (b) A1 correctly identifies male initially higher than female; A1 correctly identifies female overtakes male by 2023.

1. The vertical axis is truncated (does not start at zero), which visually exaggerates the size of the differences between the three years' profits, making relatively small actual changes look much larger than they really are. 2. The starting value of the truncated axis is not clearly labelled, so a reader glancing at the chart may not even realise the axis has been truncated, increasing the risk of being misled. 3. The bars are drawn with different widths, which is inappropriate for a bar chart (widths should be equal) and could visually suggest a difference in importance or magnitude between the years beyond what the height (value) alone represents.

A1 each for any three distinct, correctly identified problems (truncated/non-zero axis; axis start value not clearly labelled; unequal bar widths); max [3]. All other valid problems (e.g. no scale markings at all, misleading 3D effect if described) credited.

(a) A suitable hypothesis: 'Students who cycle to school have better cardiovascular fitness (e.g. a lower resting heart rate, or a better result in a standard fitness test) than students who are driven to school.'

(b) This would be primary data, since the student would need to collect new, original data specifically for this investigation (it does not already exist in a usable, pre-collected form). An appropriate method would be to carry out a standard fitness test (e.g. a timed run, or measuring resting heart rate/recovery rate after exercise) on a sample of students, alongside recording their usual mode of travel to school.

(c) A stratified sample, selecting proportionally from each year group (or from 'cyclists' and 'driven' groups identified in advance), would be appropriate, because it would ensure the sample fairly represents students across different year groups/ages, reducing the risk that age-related fitness differences bias the comparison between the two travel-mode groups.

(d) A relevant confounding variable to control or record would be the amount of other physical exercise/sport each student does outside of travelling to school (e.g. participation in sports clubs), since a student's overall fitness level could be affected by this rather than (or as well as) their mode of travel to school, potentially confounding the results if not accounted for. (Other valid answers: age, gender, distance travelled, general health conditions.)

(a) A1 valid hypothesis relating mode of travel to fitness; A1 hypothesis is specific/measurable (references a particular fitness measure or clear comparison). (b) A1 correctly identifies primary data; A1 valid, specific data collection method described. (c) A1 valid sampling method named (e.g. stratified, systematic, random); A1 correct reasoning for why this method is appropriate in this context. (d) A1 valid confounding variable identified; A1 correct reasoning for why it could confound the results if not controlled/recorded.

(a) Cumulative frequency table (using upper class boundaries):
Upper boundary: 10, 20, 30, 40, 50, 60
Cumulative frequency: 3, 11, 26, 46, 56, 60

(b) A smooth curve (or series of straight-line segments) plotted through the points (10,3), (20,11), (30,26), (40,46), (50,56), (60,60), with height on the horizontal axis and cumulative frequency on the vertical axis.

(c) With n=60: median position = 30th value. This falls in the 30-40 class (cumulative frequency reaches 26 at height 30, and 46 at height 40), so by linear interpolation: \( \text{median} = 30+\dfrac{30-26}{20}\times10 = 30+2.0 = 32.0\ \text{cm} \). LQ position = 15th value, in the 20-30 class (cumulative frequency 11 to 26): \( LQ = 20+\dfrac{15-11}{15}\times10 = 20+2.67 = 22.67\ \text{cm} \). UQ position = 45th value, in the 30-40 class (cumulative frequency 26 to 46): \( UQ = 30+\dfrac{45-26}{20}\times10 = 30+9.5 = 39.5\ \text{cm} \).

(d) \( IQR = UQ-LQ = 39.5-22.67 = 16.83\ \text{cm} \). Box plot: minimum = 0, LQ = 22.67, median = 32.0, UQ = 39.5, maximum = 60, drawn as a box from LQ to UQ (with a line at the median) and whiskers extending to the minimum and maximum values.

(e) The distance from the median to the lower quartile (\( 32.0-22.67=9.33\ \text{cm} \)) is greater than the distance from the median to the upper quartile (\( 39.5-32.0=7.5\ \text{cm} \)); since the lower half of the middle 50% of the data is more spread out than the upper half, the distribution shows a slight negative skew (a longer 'tail' towards the lower height values).

Final answer: median = 32.0 cm, LQ = 22.67 cm, UQ = 39.5 cm, IQR = 16.83 cm; slightly negatively skewed.

(a) A1 correct upper class boundaries; A1 correct cumulative frequencies (3, 11, 26, 46, 56, 60). (b) A1 correctly labelled axes; A1 all six points plotted accurately; A1 smooth curve or line segments drawn through the points. (c) M1 correct interpolation method shown for at least one quartile; A1 median = 32.0 (accept 31-33); A1 LQ = 22.67 (accept 22-23.5); A1 UQ = 39.5 (accept 39-40). (d) A1 IQR = 16.83 (ECF); A1 box correctly drawn from LQ to UQ with median line; A1 whiskers correctly drawn to given minimum (0) and maximum (60). (e) M1 correct comparison of (median-LQ) with (UQ-median); A1 correct values compared (9.33 vs 7.5); A1 correct conclusion: slight negative skew, with valid reasoning.

(a) \( \text{mean} = \dfrac{\Sigma fx}{\Sigma f} = \dfrac{6(2.5)+14(7.5)+18(12.5)+9(17.5)+3(22.5)}{50} = \dfrac{15+105+225+157.5+67.5}{50} = \dfrac{570}{50} = 11.40\ \text{kg}\)

(b) Error \( = 11.40-11.62 = -0.22\ \text{kg} \) (the grouped-data estimate is 0.22 kg lower than the true mean). This error arises because the calculation assumes every parcel within a class interval weighs exactly the midpoint value, when in reality the individual parcel weights are spread throughout each interval (not concentrated exactly at the midpoint), so some information about the exact distribution of weights within each class is lost when the data is grouped.

Final answer: (a) 11.40 kg; (b) error = -0.22 kg.

(a) M1 correct method (Σfx/Σf) with correct Σfx = 570 shown; A1 correct mean = 11.40 kg. (b) A1 correct error calculation, -0.22 kg (ECF from (a)); A1 valid reason (midpoint assumption/loss of information about spread within classes) correctly explained.

(a) The area of a circle is proportional to the SQUARE of its radius (\( A=\pi r^2 \)); since we want the AREA (not the radius itself) to be proportional to the number of respondents, the radius must be proportional to the SQUARE ROOT of the number of respondents, so that when it is squared (to give the area), the area comes out correctly proportional to the sample size.

(b) Since \( r\propto\sqrt{n} \): \( r_B = r_A\times\sqrt{\dfrac{n_B}{n_A}} = 4\times\sqrt{\dfrac{450}{200}} = 4\times\sqrt{2.25} = 4\times1.5 = 6.0\ \text{cm} \)

(c) \( \text{angle} = \dfrac{90}{200}\times360^\circ = 0.45\times360^\circ = 162^\circ \)

Final answer: (b) 6.0 cm; (c) 162°.

(a) A1 correctly identifies area proportional to radius squared (\( A=\pi r^2 \)); A1 correctly explains radius must be proportional to the square root of n for area to be correctly proportional to n. (b) M1 correct method, \( r_B=r_A\sqrt{n_B/n_A} \); A1 correct ratio \( \sqrt{450/200}=1.5 \); A1 \( r_B=6.0\ \text{cm} \). (c) M1 correct method (fraction × 360°); A1 \( 162^\circ \).

(a) Uncentred 4-point MAs: \( \frac{20+45+60+25}{4}=37.5 \); \( \frac{45+60+25+24}{4}=38.5 \); \( \frac{60+25+24+50}{4}=39.75 \); \( \frac{25+24+50+68}{4}=41.75 \); \( \frac{24+50+68+28}{4}=42.5 \).

(b) Centred (averaging successive pairs): \( \frac{37.5+38.5}{2}=38.0 \) (Q3-22); \( \frac{38.5+39.75}{2}=39.125 \) (Q4-22); \( \frac{39.75+41.75}{2}=40.75 \) (Q1-23); \( \frac{41.75+42.5}{2}=42.125 \) (Q2-23).

(c) The centred moving average trend values (38.0, 39.125, 40.75, 42.125) increase steadily over the four quarters shown, indicating a steady underlying upward trend in sales, once seasonal quarterly variation has been smoothed out.

(d) Two assumptions required: (i) that the underlying (smoothed) trend seen in the data will continue in the same direction and at broadly the same rate into the future; and (ii) that the seasonal pattern of variation between quarters (e.g. higher sales in summer quarters) will remain broadly the same in future years as it was in the data collected.

(a) M1 correct method shown; A1 first three MAs correct (37.5, 38.5, 39.75); A1 remaining two MAs correct (41.75, 42.5). (b) M1 correct centring method; A1 first two centred values correct (38.0, 39.125); A1 remaining two correct (40.75, 42.125). (c) A1 correctly identifies a steady increasing trend. (d) A1 each for two valid, distinct assumptions (trend continues; seasonal pattern remains the same); max [2].

(a) Venn diagram regions: football only = 10, basketball only = 8, swimming only = 12, F∩B only = 5, F∩S only = 4, B∩S only = 3, F∩B∩S = 2, none = 6. Sum: \( 10+8+12+5+4+3+2+6=50 \), which matches the total of 50 students surveyed, confirming the diagram is consistent.

(b) \( |F| = 10+5+4+2 = 21 \) students play football. \( |S| = 12+4+3+2 = 21 \) students swim.

(c) \( P(F\cap B\cap S) = \dfrac{2}{50} = 0.04 \)

(d) \( P(S\mid F) = \dfrac{P(S\cap F)}{P(F)} = \dfrac{(4+2)/50}{21/50} = \dfrac{6}{21} = \dfrac{2}{7} \approx 0.286 \) (3 s.f.)

(e) For independence, we require \( P(S\cap F) = P(S)\times P(F) \). \( P(S)\times P(F) = \dfrac{21}{50}\times\dfrac{21}{50} = \dfrac{441}{2500} = 0.1764 \). \( P(S\cap F) = \dfrac{6}{50} = 0.12 \). Since \( 0.12 \neq 0.1764 \), the events 'plays football' and 'swims' are NOT statistically independent.

Final answer: (b) |F|=21, |S|=21; (c) 0.04; (d) 2/7 ≈ 0.286; (e) not independent.

(a) A1 all eight regions correctly placed; A1 correct sum shown (50); A1 diagram correctly structured with three overlapping circles/universal set boundary described. (b) A1 |F|=21; A1 |S|=21. (c) A1 \( P(F\cap B\cap S)=2/50=0.04 \). (d) M1 correct identification of \( P(S\cap F)=6/50 \); M1 correct conditional probability method; A1 \( P(S|F)=2/7\approx0.286 \). (e) M1 correct method, comparing \( P(S\cap F) \) with \( P(S)\times P(F) \); A1 correct value of \( P(S)\times P(F)=0.1764 \); A1 correct value of \( P(S\cap F)=0.12 \); A1 correct conclusion, not independent, since the two values differ.

(a) A chain-base index compares each period's value with the value in the IMMEDIATELY PRECEDING period (the base changes/updates every year), showing the year-on-year percentage change; a fixed-base index, by contrast, compares every period's value with the value in a single, fixed base year that does not change.

(b) Chain-base index for 2021 = 110 means the 2021 price is 110% of the 2020 price: \( \text{price}_{2021} = 20.00\times\dfrac{110}{100} = £22.00 \)

(c) Chain-base index for 2022 = 105 means the 2022 price is 105% of the 2021 price: \( \text{price}_{2022} = 22.00\times\dfrac{105}{100} = £23.10 \)

(d) Chain-base index for 2023 = 108 means the 2023 price is 108% of the 2022 price: \( \text{price}_{2023} = 23.10\times\dfrac{108}{100} = £24.95 \) (2 d.p.)

(e) Geometric mean \( = \sqrt[3]{1.10\times1.05\times1.08} = \sqrt[3]{1.2474} = 1.0765 \) (4 d.p.). This corresponds to an average annual growth rate of approximately \( (1.0765-1)\times100 = 7.65\% \) — i.e. if the price had grown at this same, constant rate every year instead of the three different actual rates, it would have reached the same final 2023 price from the 2020 starting price.

Final answer: (b) £22.00; (c) £23.10; (d) £24.95; (e) geometric mean growth rate ≈ 7.65% per year.

(a) A1 correct description of chain-base (compares with previous period); A1 correct description of fixed-base (compares with one fixed base period), clearly distinguishing the two. (b) M1 correct method (price × index/100); A1 £22.00. (c) M1 correct method applied to the 2021 price (ECF); A1 £23.10. (d) M1 correct method applied to the 2022 price (ECF); A1 £24.95. (e) M1 correct method, cube root of the product of the three growth factors; A1 correct geometric mean growth rate ≈7.65% (accept 7.6-7.7%), with correct interpretation as an average annual rate.

(a) First draw: P(Red) = 5/8, P(Blue) = 3/8. If the first marble is red (7 marbles remain: 4 red, 3 blue): second draw P(Red|Red) = 4/7, P(Blue|Red) = 3/7. If the first marble is blue (7 marbles remain: 5 red, 2 blue): second draw P(Red|Blue) = 5/7, P(Blue|Blue) = 2/7.

(b) \( P(RR) = \dfrac{5}{8}\times\dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14} \)

(c) \( P(BB) = \dfrac{3}{8}\times\dfrac{2}{7} = \dfrac{6}{56} = \dfrac{3}{28} \). \( P(\text{at least one red}) = 1-P(BB) = 1-\dfrac{3}{28} = \dfrac{25}{28} \)

(d) Given the first marble was blue, 7 marbles remain in the bag: 5 red and 2 blue (one blue marble has been removed and not replaced). So \( P(\text{2nd is red}\mid\text{1st is blue}) = \dfrac{5}{7} \). This differs from the probability of drawing red on the first draw (5/8) because, since the marbles are drawn WITHOUT replacement, removing the first (blue) marble changes the total number of marbles remaining in the bag (from 8 to 7) and the composition of the bag, so the probabilities for the second draw are conditional on (dependent on) the outcome of the first draw.

Final answer: (b) 5/14; (c) 25/28; (d) 5/7.

(a) A1 correct first-draw probabilities (5/8, 3/8); A1 correct second-draw probabilities following red (4/7, 3/7); A1 correct second-draw probabilities following blue (5/7, 2/7); A1 diagram structure correctly shows four branch pairs with denominators changing to 7. (b) M1 correct method; A1 \( P(RR)=5/14 \). (c) M1 correctly identifies \( 1-P(BB) \) as the method; M1 correct calculation of P(BB)=3/28; A1 \( P(\text{at least one red})=25/28 \). (d) A1 correct value, 5/7; A1 correct explanation referencing the changed total/composition of the bag due to sampling without replacement.

(a) Risk (smoker) \( = \dfrac{45}{200}\times100 = 22.5\% \). Risk (non-smoker) \( = \dfrac{20}{400}\times100 = 5.0\% \)

(b) Relative risk \( = \dfrac{\text{risk (smoker)}}{\text{risk (non-smoker)}} = \dfrac{22.5}{5.0} = 4.5 \). This means a smoker is 4.5 times as likely to develop the condition as a non-smoker.

(c) Absolute risk difference \( = 22.5\%-5.0\% = 17.5 \) percentage points. This means that, for every 100 smokers (compared with 100 non-smokers), approximately 17.5 more people would be expected to develop the condition as a direct/additional result of (association with) smoking.

(d) \( X\sim B(10,\,0.225) \), where X is the number of the 10 sampled smokers who develop the condition, n=10 is the sample size, and p=0.225 is the probability of the condition for each (independent) smoker.

(e) \( P(X=2) = \binom{10}{2}(0.225)^2(0.775)^8 = 45\times0.050625\times0.13031 = 0.296 \) (3 s.f.)

(f) The binomial model assumes each smoker's chance of developing the condition is independent of every other sampled smoker and identical (constant at 22.5%) for each; in reality, this may not be entirely realistic, since individual risk could vary between smokers (e.g. depending on how much/how long they have smoked, or other individual health/genetic factors), meaning the true probability is not identical for every smoker as the model assumes.

Final answer: (a) 22.5%, 5.0%; (b) RR=4.5; (c) 17.5 percentage points; (e) P(X=2)≈0.296.

(a) A1 risk (smoker) = 22.5%; A1 risk (non-smoker) = 5.0%. (b) M1 correct method (ratio of the two risks); A1 RR = 4.5; A1 correct contextual interpretation (smokers 4.5 times as likely). (c) A1 correct value, 17.5 percentage points; A1 correct contextual interpretation. (d) A1 correctly states \( B(10,0.225) \); A1 correctly identifies/defines both parameters n and p in context. (e) M1 correct binomial coefficient \( \binom{10}{2}=45 \); M1 correct substitution into the binomial formula; A1 \( P(X=2)=0.296 \) (accept 0.294–0.298). (f) A1 valid assumption identified (constant/identical probability, or independence between smokers) with correct reasoning as to why it may not hold exactly in this context.