S Statistics — revision question pack

6 specification points · notes, questions, answers and worked methods

Checked against Edexcel 1MA1 section S. Review basis: the qualification registry sourced from the Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics (1MA1) specification; registry verification recorded 9 July 2026.

How this checking works

Loading your tier…

Answer ALL questions.

Write your answers in the spaces provided.

You must write down all the stages in your working.

S1 · Infer properties of populations or distributions from a sample, while knowing the limitations of sampling

Explanation

A population is the whole group being studied; a sample is the smaller group from which data are collected.
If the sample is representative, its proportion or mean can be used to estimate a population value.
Scale a sample proportion by the population size when estimating a count.
The result is an estimate, not a certainty.
Small samples, convenience sampling, under-coverage and non-response can make the sample unrepresentative and the inference unreliable.

Worked example

In a random sample of $75$ residents, $27$ support a proposal. Estimate how many of the town's $1250$ residents support it and state the assumption needed.

1.Sample proportion $=\dfrac{27}{75}=0.36$ .
2.Estimated population count $=0.36\times1250=450$ .
3.The estimate assumes that the sample is representative of the town's residents.

Answer: About $450$ residents, provided the sample is representative.

Common mistakes

Don't fall into the trap of using the sample frequency as the population estimate without scaling.
Don't fall into the trap of assuming a large but biased sample must be representative.

Exam tip

State a limitation in context, explaining how it could make the sample unrepresentative.

Tier 1 · Easy

1
In a random sample of $50$ library users, $30$ prefer later opening. Estimate how many of the library's $1000$ users prefer later opening.
(2)
(Total for Question 1 is 2 marks)
2
A council wants the views of all $6000$ residents about a cycle lane and surveys $80$ visitors to a sports centre. Write down the population and give one reason why the sample may be unrepresentative.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
A random sample of $80$ items from a production run of $800$ contains $18$ items with a surface mark. Estimate the number in the whole run with a surface mark and state one limitation of the estimate.
(3)
(Total for Question 1 is 3 marks)
2
Two random samples estimate support for a new crossing. In a sample of $25$ , $14$ people support it. In a sample of $200$ , $104$ people support it. Use the more reliable sample to estimate how many of $3000$ residents support the crossing. Give a reason for your choice of sample.
(3)
(Total for Question 2 is 3 marks)
3
A recycling team uses a sample of $160$ households to estimate that $1050$ of the district's $3000$ households use a food-waste collection. Work out how many households in the sample used the collection. Give one reason why the estimate of $1050$ may not equal the true district total.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A service invites a random sample of $450$ customers to answer a survey. Only $270$ reply, and $189$ of the replies support a change. Use the replies to estimate the number of supporters among all $12000$ customers, then explain a serious limitation.
(4)
(Total for Question 1 is 4 marks)
2
In a representative sample of $120$ households, $78$ own at least one pet. In a separate representative sample of $90$ pet-owning households, $54$ own a dog. Use both samples to estimate the percentage of all households that own a dog. Explain why using $54/90$ as the estimate would be wrong.
(4)
(Total for Question 2 is 4 marks)
3
A club emails $320$ randomly selected members. Of the $224$ who reply, $98$ intend to renew their membership. Without making an assumption about the members who did not reply, work out the least and greatest possible percentages of the $320$ sampled members who intend to renew. Explain why using $98/224$ to describe all club members may be unreliable.
(4)
(Total for Question 3 is 4 marks)
4
A coach operator surveys $150$ passengers who booked their journeys using its app. Of these passengers, $96$ say the service was satisfactory. Use the sample to estimate how many of the operator's $3200$ weekly passengers are satisfied. Explain why this sampling method could make the estimate unreliable.
(4)
(Total for Question 4 is 4 marks)
5
A leisure centre wants the views of all its adult members. Method A uses a random-number generator to select $120$ members from the complete membership list and arranges a time for each selected member to answer. Method B telephones members between $10$ am and $2$ pm on a weekday until $120$ members have answered. State which method is more likely to give a representative sample. Explain why, referring to a group Method B is likely to exclude.
(3)
(Total for Question 5 is 3 marks)

S2 · Interpret and construct frequency tables, bar charts, pie charts and pictograms (categorical data), vertical line charts (ungrouped discrete numerical data), tables and line graphs (time series)

Explanation

Match the display to the data. Use separated bars for categories, a vertical line chart for ungrouped discrete numbers, and joined points in time order for a time series.
A pictogram needs a clear key.
For a pie chart, $\text{sector angle}=\dfrac{\text{frequency}}{\text{total}}\times360^\circ$ .
Any graph needs clear labels, an even scale and accurate plotting.
Bar widths and gaps should be consistent, and unrelated categories should not be joined by lines.

Worked example

A $126^\circ$ sector of a pie chart represents $42$ students. Find the total number of students.

1.The sector represents the fraction $\dfrac{126}{360}$ of the total.
2.Total $=42\div\dfrac{126}{360}=42\times\dfrac{360}{126}=120$ .

Answer: $120$ students.

Common mistakes

Don't fall into the trap of using frequency directly as an angle instead of scaling to $360^\circ$ .
Don't fall into the trap of joining categorical bars or plotting a time series out of order.

Exam tip

After calculating pie-chart sectors, check that all sector angles add to $360^\circ$ .

Tier 1 · Easy

1
The data are $2,3,2,5,4,3,2,4,5,2$ . Construct a frequency table for the values $2$ , $3$ , $4$ and $5$ .
(2)
(Total for Question 1 is 2 marks)
2
A vertical line chart shows the number of goals scored by a team in each match. The lines at $0$ , $1$ , $2$ and $3$ goals have frequencies $3$ , $8$ , $6$ and $4$ respectively. Write down the frequency for $2$ goals and the modal number of goals.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
In a survey of $240$ journeys, $54$ are made by bicycle. Work out the angle of the bicycle sector in a pie chart.
(2)
(Total for Question 1 is 2 marks)
2
A café records the number of hot drinks sold on four consecutive Mondays as $42$ , $46$ , $44$ and $51$ . Write down the four points for a time-series graph, using Monday number on the horizontal axis, and describe the overall trend.
(2)
(Total for Question 2 is 2 marks)
3
A survey of $48$ pupils records their favourite after-school activity. Sport has frequency $14$ , music has frequency $10$ and drama has frequency $8$ . The remaining pupils choose art. A bar chart uses a scale of $1\,\text{cm}$ for $2$ pupils. Work out the frequency for art, write down the height of each of the four bars, and state whether the bars should touch.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1
Quarterly sales are Q₁: $320$ , Q₂: $410$ , Q₃: $390$ , Q₄: $520$ , and the following Q₁: $350$ . State the five points to plot on a time-series line graph. Work out the percentage change from the first Q₁ to the following Q₁ and explain why comparing Q₄ directly with the following Q₁ could mislead.
(4)
(Total for Question 1 is 4 marks)
2
A pie chart represents $300$ votes. Three sectors have angles $72^\circ$ , $108^\circ$ and $126^\circ$ . Work out the angle of the fourth sector and the frequency represented by each of the four sectors.
(4)
(Total for Question 2 is 4 marks)
3
A bar chart compares $64$ bookings on Monday with $76$ bookings on Tuesday. The vertical axis begins at $60$ , so the visible parts of the bars have heights of $4$ units and $16$ units. Sam says, ‘Tuesday had four times as many bookings as Monday.’ Is Sam correct? Work out the percentage increase from Monday to Tuesday and explain how the axis makes the chart misleading.
(4)
(Total for Question 3 is 4 marks)
4
A time-series graph uses the number of months after January 2024 on the horizontal axis and the meter reading on the vertical axis. The readings are $120$ in January 2024, $150$ in April 2024, $132$ in October 2024 and $168$ in July 2025. Write down the four coordinates that should be plotted, in date order.
(4)
(Total for Question 4 is 4 marks)
5
A pie chart represents the after-school clubs chosen by $240$ pupils. The school has exactly these four clubs: coding, drama, art and music. The coding sector has angle $72^\circ$ and represents $48$ pupils. There are $32$ pupils who choose drama. The numbers who choose art and music are in the ratio $5 : 3$ . Work out the angles of the art, music and drama sectors.
(4)
(Total for Question 5 is 4 marks)

S3 · Construct and interpret diagrams for grouped discrete and continuous data, i.e. histograms with equal and unequal class intervals and cumulative frequency graphs [Higher only]

Explanation

Higher tier only. In a histogram, bar area represents frequency and bar height is frequency density: $\text{frequency density}=\dfrac{\text{frequency}}{\text{class width}}$ .
Therefore $\text{frequency}=\text{frequency density}\times\text{class width}$.
Use the class boundaries for each width, draw touching bars, and label both axes.
For a cumulative frequency graph, plot the lower boundary with cumulative frequency $0$ , then each upper class boundary against its running total.
Estimate the median at $n/2$ , the lower quartile at $n/4$ and the upper quartile at $3n/4$ .

Worked example

The classes $10<x\leq25$ and $25<x\leq35$ are adjacent. The first has frequency $30$ ; the second has histogram height $1.4$ . Find the first bar height, the second frequency and the cumulative-frequency points.

1.First width $=15$ , so its height is $30\div15=2$ .
2.Second width $=10$ , so its frequency is $1.4\times10=14$ .
3.The cumulative-frequency points are $(10,0)$ , $(25,30)$ and $(35,44)$ .

Answer: First height $2$ ; second frequency $14$ ; cumulative points $(10,0)$ , $(25,30)$ and $(35,44)$ .

Common mistakes

Don't fall into the trap of using frequency as the bar height instead of dividing by class width.
Don't fall into the trap of plotting cumulative frequencies at class midpoints instead of upper boundaries.

Exam tip

Write the frequency-density formula beside the histogram before calculating any height or area.

Tier 1 · Easy

1
A histogram class is $15<x\leq20$ and has frequency $20$ . Work out the frequency density.
(2)
(Total for Question 1 is 2 marks)
2
A histogram class has width $8$ and frequency density $2.5$ . Work out the frequency in the class.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Grouped data have classes $0<x\leq10$ , $10<x\leq25$ and $25<x\leq40$ with frequencies $12$ , $30$ and $18$ . Calculate the three frequency densities and identify the tallest histogram bar.
(3)
(Total for Question 1 is 3 marks)
2
In a histogram, the class $5<x\leq15$ has frequency $24$ and its bar is $6\,\text{cm}$ high. The bar for $15<x\leq21$ is $7.5\,\text{cm}$ high. Work out the frequency in the second class.
(3)
(Total for Question 2 is 3 marks)
3
Three adjacent classes in a histogram have class widths $4$ , $6$ and $10$ . Their bars measure $6\,\text{cm}$ , $5\,\text{cm}$ and $3\,\text{cm}$ high respectively. There are $84$ values altogether and there are no other classes. Work out the frequency in each class.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1
A cumulative frequency graph for $80$ values passes through $(10,8)$ , $(20,26)$ , $(30,50)$ , $(40,70)$ and $(50,80)$ . Use linear interpolation between the given points to estimate the median and the interquartile range.
(5)
(Total for Question 1 is 5 marks)
2
The class $12<x\leq b$ has frequency $32$ and frequency density $2$ . The next class is $b<x\leq36$ and has frequency $18$ . The cumulative frequency at $x=12$ is $15$ . Work out $b$ , the cumulative frequencies at $x=b$ and $x=36$ , and the frequency density of the second class.
(5)
(Total for Question 2 is 5 marks)
3
For $50$ values, a student plots the cumulative-frequency points $(0,0)$ , $(10,8)$ , $(20,23)$ , $(30,19)$ , $(40,42)$ and $(50,50)$ . Explain why the point $(30,19)$ cannot be correct. The frequency in the class $20<x\leq30$ is $12$ . Work out the correct point at $x=30$ , the frequency in the class $30<x\leq40$ , and the class containing the median.
(4)
(Total for Question 3 is 4 marks)
4
The histogram class $4<x\leq16$ has frequency $36$ . The adjacent class $16<x\leq b$ has the same bar area and a bar twice as tall. The class $b<x\leq34$ has a bar one third as tall as the second bar. The bars use the same vertical scale. Work out $b$ and the total frequency in the three classes.
(5)
(Total for Question 4 is 5 marks)
5
Higher only: A histogram has classes $0<x\leq8$ , $8<x\leq20$ , $20<x\leq32$ and $32<x\leq42$ . Their frequency densities are $1.5$ , $2.5$ , $3$ and $1.2$ respectively. Assuming values are evenly spread within each class, estimate how many values are greater than $26$ . Also estimate the value below which $70\%$ of the data lie.
(5)
(Total for Question 5 is 5 marks)

S4 · Interpret, analyse and compare data-set distributions via graphs (incl. box plots), central tendency (median, mean, mode, modal class) and spread (range, outliers, quartiles, inter-quartile range)

Explanation

A measure of central tendency describes a typical value: the mean uses every value, the median is the ordered middle, and the mode or modal class is most frequent.
Spread describes variation: range is maximum minus minimum.
At Higher tier, box plots also show quartiles and the interquartile range $Q_3-Q_1$ .
To compare distributions, make one contextual statement about centre and one about spread.
Outliers can strongly affect the mean and range, so the median and interquartile range may be more representative.

Worked example

Delivery service A has median time $38$ minutes and range $22$ minutes. Service B has median time $43$ minutes and range $12$ minutes. Compare the distributions.

1.A has the lower median, so its deliveries are typically quicker.
2.B has the smaller range, so its delivery times are more consistent.

Answer: Service A is typically quicker, but service B has more consistent delivery times.

Common mistakes

Don't fall into the trap of giving comparison figures without stating what they mean in context.
Don't fall into the trap of finding the median before putting raw data in order.

Exam tip

A full comparison usually needs both a typical-value statement and a spread statement.

Tier 1 · Easy

1
For the data $4,6,6,9,10$ , work out the mean and the range.
(2)
(Total for Question 1 is 2 marks)
2
For the data $2,4,4,6,9,10,12$ , write down the mode and the median.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
Data set A has median $24$ and range $33$ . Data set B has median $25$ and range $29$ . Compare the two distributions.
(2)
(Total for Question 1 is 2 marks)
2
Five readings have mean $18$ . A sixth reading of $24$ is added. Work out the new mean and explain why it is greater than $18$ .
(3)
(Total for Question 2 is 3 marks)
3
A data set has mean $12$ , median $11$ and range $9$ . Every value in the data set is doubled and then $3$ is added. Work out the mean, median and range of the new data set.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
The data are $12,13,13,14,15,15,16,52$ . Work out the mean, median and range. Decide which of the mean or median better describes a typical value for these data. Give a reason.
(4)
(Total for Question 1 is 4 marks)
2
The ordered data are $6,9,11,x,18,20$ . Their mean is $13$ . Work out $x$ , the median and the range.
(4)
(Total for Question 2 is 4 marks)
3
Seven integer values are written in order of size: $3,7,a,11,b,16,20$ . Their mean is $11$ and their only mode is $7$ . Work out $a$ , $b$ and the range.
(4)
(Total for Question 3 is 4 marks)
4
The values $4$ , $7$ , $10$ and $13$ have frequencies $2$ , $5$ , $k$ and $1$ respectively. The mean is $8$ . Work out $k$ , the median, the mode and the range.
(4)
(Total for Question 4 is 4 marks)
5
Higher only: Nine values are written in order: $4,7,9,12,15,18,x,26,31$ . The upper quartile is $24$ . The median is excluded when the lower and upper halves are formed. Work out $x$ , the lower quartile and the interquartile range.
(4)
(Total for Question 5 is 4 marks)

S5 · Apply statistics to describe a population

Explanation

Statistics can summarise a population or estimate its features from representative samples.
When groups have different sizes, combine their means using a weighted mean: add each group size multiplied by its mean, then divide by the total population.
When different subgroups have different sample proportions, estimate each subgroup separately before adding.
A simple average of group means is only valid when the groups are the same size.

Worked example

A town has $900$ northern and $600$ southern residents. In representative samples, $12$ of $40$ northern residents and $20$ of $50$ southern residents cycle to work. Estimate the town total and percentage.

1.North estimate $=900\times\dfrac{12}{40}=270$ .
2.South estimate $=600\times\dfrac{20}{50}=240$ .
3.Total $=270+240=510$ ; percentage $=\dfrac{510}{1500}\times100=34\%$ .

Answer: About $510$ residents, or $34\%$ of the town.

Common mistakes

Don't fall into the trap of averaging two group means without using the group sizes.
Don't fall into the trap of applying one subgroup's sample proportion to the whole population.

Exam tip

Turn each mean back into a total first; combine totals, then divide once.

Tier 1 · Easy

1
A representative sample of parcels has mean mass $2.4\,\text{kg}$ . Estimate the total mass of $250$ parcels in the population.
(2)
(Total for Question 1 is 2 marks)
2
The mean weekly electricity use for a representative sample of $64$ households is $118\,\text{kWh}$ . Estimate the mean weekly electricity use for all households in the village.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
A population contains $120$ junior members with mean attendance $6.5$ sessions and $80$ senior members with mean attendance $8$ sessions. Work out the mean attendance for the whole population.
(3)
(Total for Question 1 is 3 marks)
2
A company has $16$ part-time employees whose mean working time is $15$ hours per week. The full-time employees have a mean working time of $39$ hours per week. The mean working time for all the employees is $23$ hours per week. Work out the number of full-time employees.
(3)
(Total for Question 2 is 3 marks)
3
A school has $2400$ pupils. The mean number of days absent last term was $1.5$ per pupil. Of the pupils, $25\%$ had no absence. Work out the mean number of days absent for the pupils who had at least one day absent.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A town has $1800$ residents in the north and $1200$ in the south. In representative samples, $15$ of $60$ northern residents and $20$ of $50$ southern residents cycle to work. Estimate the total number and percentage of the town's residents who cycle to work.
(5)
(Total for Question 1 is 5 marks)
2
A company has $280$ office employees and $420$ warehouse employees. Overall, $32\%$ of the employees work part-time. Of the office employees, $20\%$ work part-time. Work out the percentage of the warehouse employees who work part-time.
(4)
(Total for Question 2 is 4 marks)
3
A biologist records the masses of fish from a lake in two random samples. Sample A has $40$ fish with a mean mass of $620$ g. Sample B has $60$ fish with a mean mass of $570$ g. Work out the mean mass of all $100$ sampled fish. Give one reason why this combined mean is a better estimate of the mean mass of fish in the lake than the mean of sample A alone. You must show all your working.
(4)
(Total for Question 3 is 4 marks)
4
A survey estimates the number of oak trees affected by a disease in three woods. Wood A has $900$ oak trees; $21$ of a random sample of $30$ are affected. Wood B has $600$ oak trees; $16$ of a random sample of $25$ are affected. Wood C has $500$ oak trees; $14$ of a random sample of $n$ are affected. Using these samples gives an estimated total of $1264$ affected oak trees. Work out $n$ . State which wood has the lowest estimated proportion of affected oak trees.
(4)
(Total for Question 4 is 4 marks)
5
A representative random sample of $80$ one-square-metre plots has a mean of $6.4$ seedlings per plot. Use the sample to estimate the number of seedlings in a habitat of area $1250\,\text{m}^2$ . A later census finds $7680$ seedlings. Work out the percentage error in the estimate as a percentage of the census figure, and state whether the estimate is too high or too low. Write the percentage to two decimal places.
(4)
(Total for Question 5 is 4 marks)

S6 · Use and interpret scatter graphs of bivariate data; recognise correlation, know it does not indicate causation; draw estimated lines of best fit; make predictions; interpolate/extrapolate with caution

Explanation

A scatter graph displays paired values.
An upward pattern shows positive correlation, a downward pattern negative correlation, and no clear pattern no correlation.
Draw a line of best fit through the centre of the points with a roughly balanced spread above and below.
Use it for estimates: interpolation stays within the observed data range, while extrapolation goes beyond it and is less reliable.
Correlation does not prove causation because another variable, reverse causation or coincidence may explain the relationship.

An illustrative scatter graph with an upward trend and a straight line of best fit passing through the centre of the points.

Worked example

A scatter graph compares weekly revision time with test score for revision times from $1$ to $8$ hours. A sensible line of best fit gives a score of about $72$ at $6$ hours and about $84$ when extended to $10$ hours. Interpret both estimates.

1. $6$ hours is inside the observed range, so $72$ is an interpolation and is reasonably reliable.
2. $10$ hours is outside the observed range, so $84$ is an extrapolation and is less reliable because the trend may not continue.
3.The positive correlation does not prove that extra revision alone caused the higher scores; prior attainment could affect both variables.

Answer: $72$ is the more reliable interpolation; $84$ is a less reliable extrapolation, and the graph does not establish causation.

Common mistakes

Don't fall into the trap of saying that correlation proves one variable causes the other.
Don't fall into the trap of using a line of best fit far outside the observed range without warning.

Exam tip

Answer the command precisely: for 'describe the relationship', write 'as $x$ increases, $y$ tends to increase/decrease', not just 'positive/negative'. For a prediction, identify interpolation or extrapolation and comment on reliability.

Tier 1 · Easy

1
A scatter graph shows that as daily temperature increases, ice-cream sales usually increase. State the type of correlation and explain why the graph alone does not prove that temperature is the only cause of higher sales.
(2)
(Total for Question 1 is 2 marks)
2
A scatter graph compares altitude with air temperature and shows a downward pattern. Write down the type of correlation. Give one reason why using the graph to estimate the air temperature at an altitude beyond the plotted points would be unreliable.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
For data with observed $x$ -values from $5$ to $20$ , a line of best fit is $y=1.8x+6$ . Estimate $y$ when $x=14$ . Explain why using the line at $x=35$ is less reliable.
(3)
(Total for Question 1 is 3 marks)
2
A scatter graph contains $12$ points. A student's line of best fit runs across the full plotted range, but it is made from two straight sections joined at a corner and $10$ of the $12$ points lie above it. Write down two errors in the student's line of best fit.
(2)
(Total for Question 2 is 2 marks)
3
A scatter graph uses wing length in centimetres on the horizontal axis and mass in grams on the vertical axis. One bird has wing length $180\,\text{mm}$ and mass $75\,\text{g}$ . The line of best fit is $y=3x+20$ . Write down the coordinates of this bird on the graph. Work out the value predicted by the line and how far the recorded mass is above or below that value.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A scatter graph compares a puppy's age $a$ months with mass $m$ kg for ages from $2$ to $12$ months. Its estimated line of best fit is $m=0.42a+1.8$ . Estimate the mass at $9$ months and at $16$ months. Comment on the reliability of both estimates and on whether the graph proves that age alone causes the change in mass.
(5)
(Total for Question 1 is 5 marks)
2
For primary-school pupils, a scatter graph of shoe length against reading score shows strong positive correlation. Explain why this does not show that longer shoes cause better reading, suggest a variable that could affect both measurements, and explain why a prediction for an adult would be unreliable.
(3)
(Total for Question 2 is 3 marks)
3
A scatter graph compares time since charging, $x$ hours, with battery level, $y\%$ , for times from $4$ to $18$ hours. An estimated line of best fit has gradient $-3$ and passes through $(10,52)$ . Work out the equation of the line in the form $y=mx+c$ . Use the line to estimate the time when the battery level is $40\%$ . State whether this estimate is interpolation or extrapolation.
(4)
(Total for Question 3 is 4 marks)
4
For data with observed $x$ -values from $2$ to $14$ , a line of best fit is $y=4.5x+12$ . A point has coordinates $(8,54)$ . Work out how many units above or below the line's prediction this point lies. Another point has $x=13$ and lies $18$ units below the line's prediction. Work out its observed $y$ -value. State which point lies farther from the line.
(4)
(Total for Question 4 is 4 marks)
5
Two scatter graphs use the same observed $x$ -range, from $3$ to $12$ . Their estimated lines of best fit are $y=5x+19$ for group A and $y=3x+35$ for group B. Work out the value of $x$ at which the lines predict the same value of $y$ , and find this value of $y$ . At $x=10$ , state which group has the greater predicted value and by how much. State whether the prediction at $x=10$ is interpolation or extrapolation.
(4)
(Total for Question 5 is 4 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

S1 · Infer properties of populations or distributions from a sample, while knowing the limitations of sampling

Tier 1 · Easy

Mark scheme for S1 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$600$ users	2	The sample proportion is $30/50=0.6$ . Apply this to the population: $0.6\times1000=600$ .
2	The population is all $6000$ residents. Visitors to the sports centre may have different views from residents who do not use it (accept any valid reason that the sports-centre sample may be biased).	2	The group whose views the council wants is all $6000$ residents, so that is the population. The sample comes from only one location and may over-represent people who use the sports centre, so it may not represent all residents.

Tier 2 · Standard

Mark scheme for S1 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$180$ items. The sample may differ from the population by chance, so the estimate need not equal the true number.	3	The sample is one tenth of the production run, so scale $18$ by $10$ to get $180$ . Because only a sample was inspected, sampling variation remains even if the selection was random.
2	$1560$ residents. The sample of $200$ is more reliable because a larger random sample is likely to have less sampling variation.	3	Use the larger random sample. Its support proportion is $\dfrac{104}{200}=0.52$ , so the estimate is $0.52\times3000=1560$ . A larger random sample is generally less affected by chance variation than a sample of $25$ .
3	$56$ households. Households that use the food-waste collection may be more or less likely to appear in the sample — for example, uptake may vary between streets or seasons — so the sample proportion may differ from the district's (accept any valid sampling limitation).	3	The estimated population proportion is $1050/3000=0.35$ . Apply this proportion to the sample: $0.35\times160=56$ . Even a representative sample gives an estimate, because another sample could contain a different proportion.

Tier 3 · Hard

Mark scheme for S1 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Estimated supporters $=8400$ . Non-response bias may make the replies unrepresentative because the $180$ non-responders may have different views.	4	Among replies, the support proportion is $189/270=0.7$ , giving $0.7\times12000=8400$ . However, the estimate assumes responders and non-responders have similar opinions; the low response rate may break that assumption.
2	Estimated percentage $=39\%$ . $54/90$ estimates the proportion of pet-owning households that own a dog, not the proportion of all households that own a dog.	4	Estimate the proportion of households that own a pet as $78/120=0.65$ . Estimate the proportion of pet-owning households that own a dog as $54/90=0.6$ . Therefore the estimated proportion of all households that own a dog is $0.65\times0.6=0.39$ , or $39\%$ . The second sample alone has pet-owning households as its population, so its $60\%$ cannot be applied to all households.
3	Least possible percentage $=30.625\%$ (or $30.6\%$ to 1 d.p.); greatest possible percentage $=60.625\%$ (or $60.6\%$ to 1 d.p.). The members who replied may have different intentions from those who did not reply, so the replies may be unrepresentative of all club members.	4	There are $320-224=96$ non-responders. The least possible number intending to renew is $98$ , giving $98/320\times100=30.625\%$ . The greatest is $98+96=194$ , giving $194/320\times100=60.625\%$ . This wide range shows the possible effect of non-response bias.
4	Estimated number satisfied $=2048$ . Passengers who book at a ticket counter or by telephone are excluded, and their experience or opinion of the service may differ from that of app users.	4	The sample proportion satisfied is $96/150=0.64$ , so the estimate is $0.64\times3200=2048$ . The sampling frame contains only app users and excludes passengers using other booking methods, so it may not represent all weekly passengers.
5	Method A is more likely to give a representative sample. Method B is likely to exclude members who are at work, at college or otherwise unavailable during weekday daytime; their views may differ from those of members who are available then.	3	Choose Method A first. Selecting at random from the complete membership list gives every adult member a chance of selection. Method B favours members who are available during weekday daytime and can omit working members, students and others who are away then, so it is more likely to be biased.

S2 · Interpret and construct frequency tables, bar charts, pie charts and pictograms (categorical data), vertical line charts (ungrouped discrete numerical data), tables and line graphs (time series)

Tier 1 · Easy

Mark scheme for S2 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$2:4$ , $3:2$ , $4:2$ , $5:2$ .	2	Tally each value once. The frequencies $4+2+2+2=10$ match the number of data values.
2	The frequency for $2$ goals is $6$ . The modal number of goals is $1$ .	2	Read the height of the line at $2$ goals to get frequency $6$ . The greatest frequency is $8$ , at $1$ goal, so the mode is $1$ .

Tier 2 · Standard

Mark scheme for S2 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$81^\circ$	2	The bicycle fraction is $54/240$ . Multiply by a full turn: $\dfrac{54}{240}\times360^\circ=81^\circ$ .
2	$(1,42)$ , $(2,46)$ , $(3,44)$ and $(4,51)$ . The number of hot drinks sold shows an overall increase (also accept noting the fall from $46$ to $44$ between Mondays $2$ and $3$ ).	2	Pair each Monday number with its sales value and plot the points in time order. Although sales fall from $46$ to $44$ between Mondays $2$ and $3$ , they rise from $42$ to $51$ overall.
3	Art frequency $=16$ . Bar heights: sport $7\,\text{cm}$ , music $5\,\text{cm}$ , drama $4\,\text{cm}$ and art $8\,\text{cm}$ . The bars should not touch because the activities are categories.	4	The art frequency is $48-(14+10+8)=16$ . Divide each frequency by $2$ to use the stated scale, giving heights $7$ , $5$ , $4$ and $8\,\text{cm}$ . A categorical bar chart has gaps between its bars.

Tier 3 · Hard

Mark scheme for S2 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$(\text{Q1},320),(\text{Q2},410),(\text{Q3},390),(\text{Q4},520),(\text{next Q1},350)$ . Percentage increase $=9.375\%$ . Q₄ and Q₁ may have different seasonal patterns, so that comparison may not represent an underlying fall.	4	Plot the values in time order and join consecutive points. Like-for-like Q₁ change is $350-320=30$ , so the percentage change is $30/320\times100=9.375\%$ . Comparing different quarters confounds the change with possible seasonality.
2	Fourth angle $=54^\circ$ . The frequencies are $60$ , $90$ , $105$ and $45$ , corresponding respectively to $72^\circ$ , $108^\circ$ , $126^\circ$ and $54^\circ$ .	4	The missing angle is $360^\circ-(72^\circ+108^\circ+126^\circ)=54^\circ$ . Since $300/360=\dfrac{5}{6}$ , multiply each angle by $\dfrac{5}{6}$ : the frequencies are $60$ , $90$ , $105$ and $45$ . They add to $300$ .
3	No, Tuesday did not have four times as many bookings as Monday. The percentage increase is $18.75\%$ . Starting the axis at $60$ makes the visible heights $4$ and $16$ , exaggerating the difference; the actual values are $64$ and $76$ .	4	The increase is $76-64=12$ , so the percentage increase is $12/64\times100=18.75\%$ . The ratio of the actual values is $76/64=1.1875$ , not $4$ . The fourfold visual ratio comes only from subtracting the truncated-axis baseline of $60$ .
4	The coordinates are $(0,120)$ , $(3,150)$ , $(9,132)$ and $(18,168)$ , in date order.	4	The dates are $0$ , $3$ , $9$ and $18$ months after January 2024. Pair each of these data values with its meter reading to give $(0,120)$ , $(3,150)$ , $(9,132)$ and $(18,168)$ .
5	Art sector $=150^\circ$ , music sector $=90^\circ$ and drama sector $=48^\circ$ .	4	After the $48$ coding pupils and $32$ drama pupils, $240-48-32=160$ pupils remain. Split $160$ in the ratio $5 : 3$ to get $100$ art pupils and $60$ music pupils. Each pupil represents $360^\circ/240=1.5^\circ$ , so the three required angles are $150^\circ$ , $90^\circ$ and $48^\circ$ .

S3 · Construct and interpret diagrams for grouped discrete and continuous data, i.e. histograms with equal and unequal class intervals and cumulative frequency graphs [Higher only]

Tier 1 · Easy

Mark scheme for S3 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$4$	2	The class width is $20-15=5$ . Frequency density is $20/5=4$ .
2	$20$	1	Frequency is the area of the bar: $\text{frequency}=\text{class width}\times\text{frequency density}=8\times2.5=20$ .

Tier 2 · Standard

Mark scheme for S3 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	Frequency densities: $1.2$ , $2$ , $1.2$ . The class $10<x\leq25$ has the tallest bar.	3	Divide each frequency by its class width: $12/10=1.2$ , $30/15=2$ and $18/15=1.2$ . The greatest density, $2$ , gives the tallest bar.
2	$18$	3	For $5<x\leq15$ , the class width is $10$ , so the frequency density is $24/10=2.4$ . A height of $6\,\text{cm}$ represents density $2.4$ , so $1\,\text{cm}$ represents $0.4$ . The second density is $7.5\times0.4=3$ , and its width is $6$ , giving frequency $3\times6=18$ .
3	The frequencies are $24$ , $30$ and $30$ , respectively.	4	Histogram frequency is proportional to bar area. The relative areas are $4\times6=24$ , $6\times5=30$ and $10\times3=30$ , in the ratio $4 : 5 : 5$ . The $14$ ratio parts represent $84$ values, so each part represents $6$ . The frequencies are therefore $24$ , $30$ and $30$ .

Tier 3 · Hard

Mark scheme for S3 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Median $\approx25.8$ . Lower quartile $\approx16.7$ and upper quartile $=35$ . Interquartile range $\approx18.3$ .	5	The median is the $40$ th value, between cumulative frequencies $26$ and $50$ : $20+\dfrac{40-26}{50-26}\times10\approx25.8$ . The lower quartile is the $20$ th value: $10+\dfrac{20-8}{26-8}\times10\approx16.7$ . The upper quartile is the $60$ th value: $30+\dfrac{60-50}{70-50}\times10=35$ . Hence $IQR\approx35-16.7=18.3$ .
2	$b=28$ . Cumulative frequency $=47$ at $x=28$ and $65$ at $x=36$ . The second frequency density is $2.25$ .	5	The first class width is $32/2=16$ , so $b=12+16=28$ . Add the first frequency to get cumulative frequency $15+32=47$ at $x=28$ , then add $18$ to get $65$ at $x=36$ . The second width is $36-28=8$ , so its frequency density is $18/8=2.25$ .
3	A cumulative frequency cannot decrease as $x$ increases, but the plotted value falls from $23$ to $19$ . The correct point is $(30,35)$ . The frequency in $30<x\leq40$ is $7$ . The median is in the class $20<x\leq30$ .	4	The cumulative frequency is already $23$ at $x=20$ , so it cannot be $19$ at the larger boundary $x=30$ . Add the class frequency $12$ to get $23+12=35$ , giving $(30,35)$ . The next class frequency is $42-35=7$ . The $25$ th and $26$ th values lie after cumulative frequency $23$ and no later than cumulative frequency $35$ , so the median lies in $20<x\leq30$ .
4	$b=22$ and the total frequency is $96$ .	5	The first class has width $12$ , so its frequency density is $36/12=3$ . The second bar has density $6$ and, because it has the same area, frequency $36$ . Its width is therefore $36/6=6$ , giving $b=16+6=22$ . The third density is $6/3=2$ and its width is $34-22=12$ , so its frequency is $24$ . The total is $36+36+24=96$ .
5	Estimated number of values greater than $26$ is $30$ . $70\%$ of the data lie below an estimated value of $27$ .	5	The class frequencies are $8(1.5)=12$ , $12(2.5)=30$ , $12(3)=36$ and $10(1.2)=12$ , so there are $90$ values. Above $26$ , the part of the third class has estimated frequency $(32-26)\times3=18$ , and the final class contributes $12$ , giving $30$ . The $70\%$ point is the $0.70\times90=63$ rd value. The cumulative frequency is $42$ at $x=20$ , so interpolate $21$ values into the third class: $20+\dfrac{63-42}{36}\times12=27$ .

S4 · Interpret, analyse and compare data-set distributions via graphs (incl. box plots), central tendency (median, mean, mode, modal class) and spread (range, outliers, quartiles, inter-quartile range)

Tier 1 · Easy

Mark scheme for S4 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	Mean $=7$ ; range $=6$ .	2	The total is $4+6+6+9+10=35$ , so the mean is $35/5=7$ . The range is $10-4=6$ .
2	Mode $=4$ ; median $=6$ .	2	$4$ occurs most often, so it is the mode. The data are ordered and the fourth of the seven values is $6$ , so the median is $6$ .

Tier 2 · Standard

Mark scheme for S4 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	B has the slightly higher median: $25$ compared with $24$ . B has the smaller range: $29$ compared with $33$ , so its values are less spread out overall.	2	Compare the typical values using the medians, then compare spread using the ranges. B has both the larger centre and the smaller spread.
2	New mean $=19$ . The added reading is greater than the original mean, so it raises the mean.	3	The original total is $5\times18=90$ . The new total is $90+24=114$ , so the new mean is $114/6=19$ . Since $24$ is above the original mean, the mean increases.
3	Mean $=27$ ; median $=25$ ; range $=18$ .	3	The mean and median both undergo the same transformation as every value: $2\times12+3=27$ and $2\times11+3=25$ . Doubling every value doubles the range, while adding the same amount to every value does not change the range, so the new range is $2\times9=18$ .

Tier 3 · Hard

Mark scheme for S4 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Mean $=18.75$ , median $=14.5$ , range $=40$ . The median is more representative because $52$ is an outlier that pulls up the mean.	4	The total is $150$ , giving mean $150/8=18.75$ . The median is $(14+15)/2=14.5$ and the range is $52-12=40$ . The isolated value $52$ has a strong effect on the mean but not on the median, so the median better represents the main cluster.
2	$x=14$ ; median $=12.5$ ; range $=14$ .	4	The total of all six values is $6\times13=78$ . The known values total $6+9+11+18+20=64$ , so $x=78-64=14$ . The median is the mean of the third and fourth values: $(11+14)/2=12.5$ . The range is $20-6=14$ .
3	$a=7$ , $b=13$ and the range is $17$ .	4	For $7$ to be the only mode, it must occur again, so $a=7$ . A mean of $11$ for seven values gives a total of $77$ . The known values, including $a=7$ , total $64$ , so $b=77-64=13$ . This is consistent with the stated order. The range is $20-3=17$ .
4	$k=4$ , median $=7$ , mode $=7$ and range $=9$ .	4	The total frequency is $8+k$ and the total of the values is $2(4)+5(7)+10k+13=56+10k$ . Therefore $56+10k=8(8+k)$ , so $2k=8$ and $k=4$ . There are $12$ values; the sixth and seventh are both $7$ , so the median is $7$ . The greatest frequency is $5$ , so the mode is $7$ , and the range is $13-4=9$ .
5	$x=22$ , lower quartile $=8$ and interquartile range $=16$ .	4	After excluding the median $15$ , the upper half is $18,x,26,31$ . Its median is $(x+26)/2=24$ , so $x=22$ , which is consistent with the stated order. The lower half is $4,7,9,12$ , so the lower quartile is $(7+9)/2=8$ . The interquartile range is $24-8=16$ .

S5 · Apply statistics to describe a population

Tier 1 · Easy

Mark scheme for S5 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$600\,\text{kg}$	2	Apply the sample mean to all $250$ parcels: $2.4\times250=600\,\text{kg}$ .
2	$118\,\text{kWh}$	1	Use the sample mean as the estimate of the population mean. Because the sample is representative, the estimated mean for all households is $118\,\text{kWh}$ .

Tier 2 · Standard

Mark scheme for S5 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$7.1$ sessions	3	The total attendances represented are $120\times6.5=780$ and $80\times8=640$ . Divide their sum by all $200$ members: $(780+640)/200=7.1$ .
2	$8$ full-time employees	3	Let $n$ be the number of full-time employees. The total weekly hours give $16\times15+39n=23(16+n)$ . Therefore $240+39n=368+23n$ , so $16n=128$ and $n=8$ .
3	$2$ days	3	The total number of absence days is $2400\times1.5=3600$ . The number of pupils with at least one day absent is $75\%$ of $2400$ , which is $1800$ . Their mean is therefore $3600/1800=2$ days.

Tier 3 · Hard

Mark scheme for S5 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Estimated total $=930$ residents. Estimated percentage $=31\%$ .	5	For the north, estimate $1800\times15/60=450$ . For the south, estimate $1200\times20/50=480$ . The total is $450+480=930$ from a population of $3000$ , so the percentage is $930/3000\times100=31\%$ . Calculating by subgroup correctly accounts for their different sizes.
2	$40\%$	4	There are $280+420=700$ employees, so $0.32\times700=224$ work part-time. Of the office employees, $0.20\times280=56$ work part-time. Therefore $224-56=168$ warehouse employees work part-time, giving $168/420\times100=40\%$ .
3	$590$ g The combined sample is larger, so its mean is less affected by sampling variation and is likely to be closer to the population mean.	4	The total mass in sample A is $40\times620=24{,}800$ g and in sample B is $60\times570=34{,}200$ g. The combined total is $59{,}000$ g across $100$ fish, so the combined mean is $59{,}000\div100=590$ g. A larger sample gives a more reliable estimate of the population mean.
4	$n=28$ . Wood C has the lowest estimated proportion of affected oak trees.	4	The estimates for woods A and B are $900\times21/30=630$ and $600\times16/25=384$ . Wood C therefore contributes $1264-630-384=250$ to the estimated total. Its estimated affected proportion is $250/500=0.5$ , so $14/n=0.5$ and $n=28$ . The estimated proportions for A, B and C are $0.7$ , $0.64$ and $0.5$ , respectively, so Wood C has the lowest proportion.
5	Estimated number $=8000$ seedlings. The percentage error is $4.17\%$ (or $4.2\%$ to 1 d.p.) and the estimate is too high.	4	The estimated density is $6.4$ seedlings per square metre, giving $6.4\times1250=8000$ seedlings. The estimate exceeds the census by $8000-7680=320$ . Relative to the census, the percentage error is $320/7680\times100=4.1666\ldots\%\approx4.17\%$ , so the estimate is too high.

S6 · Use and interpret scatter graphs of bivariate data; recognise correlation, know it does not indicate causation; draw estimated lines of best fit; make predictions; interpolate/extrapolate with caution

Tier 1 · Easy

Mark scheme for S6 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	Positive correlation. Correlation does not establish causation; another factor such as weekends or visitor numbers could affect sales.	2	An upward association is positive correlation. Then identify a plausible confounding variable to show why the paired data alone cannot isolate a causal effect.
2	Negative correlation. Beyond the plotted points the pattern may not continue, so the estimate is extrapolation and unreliable.	2	A downward pattern from left to right is negative correlation. The graph only gives evidence for altitudes within the plotted range; assuming the same pattern beyond it is extrapolation, which may not hold.

Tier 2 · Standard

Mark scheme for S6 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$y=31.2$ when $x=14$ . $x=35$ is outside the observed range, so this would be extrapolation and the trend may not continue.	3	Substitute $x=14$ : $y=1.8(14)+6=25.2+6=31.2$ . Since $14$ lies inside the data range this is interpolation, whereas $35$ lies outside it.
2	It should be one straight line, not two sections joined at a corner. It does not pass through the centre of the points, with a roughly equal number of points on each side.	2	A line of best fit should be a single straight line through the centre of the scatter. The corner makes the student's line unsuitable, and having $10$ of $12$ points above it shows that the points are not balanced around the line.
3	The coordinates are $(18,75)$ . The predicted mass is $74\,\text{g}$ , so the recorded mass is $1\,\text{g}$ above the line.	3	Convert $180\,\text{mm}$ to $18\,\text{cm}$ , so the point is $(18,75)$ . Substituting $x=18$ gives $y=3(18)+20=74$ . Since $75-74=1$ , the recorded mass is $1\,\text{g}$ above the line.

Tier 3 · Hard

Mark scheme for S6 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	At $9$ months, estimated mass $=5.58\,\text{kg}$ . At $16$ months, estimated mass $=8.52\,\text{kg}$ . The $9$ -month estimate is interpolation and is more reliable; the $16$ -month estimate is extrapolation. The association does not prove age alone causes the change because factors such as breed or diet may also affect mass.	5	Substitute into the line: for $a=9$ , $m=0.42(9)+1.8=5.58$ ; for $a=16$ , $m=0.42(16)+1.8=8.52$ . The first age lies within $2$ to $12$ , while the second lies outside. The graph shows association only and does not control other variables.
2	Correlation does not prove that longer shoes cause better reading. Age could affect both shoe length and reading score (accept another valid variable affecting both). An adult is outside the range of the pupil data, so the prediction would be extrapolation and the trend may not continue.	3	The graph shows association only. Older pupils tend to have both longer feet and more developed reading skills, so age is a confounding variable. An adult is outside the group and data range studied, making any estimate an unreliable extrapolation.
3	The line is $y=-3x+82$ . Estimated time $=14$ hours. Interpolation, because $14$ hours is within the observed range from $4$ to $18$ hours.	4	Write $y=-3x+c$ . Using $(10,52)$ gives $52=-30+c$ , so $c=82$ and the line is $y=-3x+82$ . Set $y=40$ : $40=-3x+82$ , so $3x=42$ and $x=14$ . This lies inside the observed range, so it is interpolation.
4	The point $(8,54)$ lies $6$ units above the line's prediction. The observed value at $x=13$ is $52.5$ . The point with $x=13$ lies farther from the line.	4	At $x=8$ , the line predicts $4.5(8)+12=48$ . Since $54-48=6$ , the point is $6$ units above the prediction. At $x=13$ , the line predicts $4.5(13)+12=70.5$ , so a point $18$ units below it has observed value $70.5-18=52.5$ . Since $18>6$ , the second point is farther from the line.
5	The lines give the same prediction at $x=8$ , where $y=59$ . At $x=10$ , group A has the greater predicted value by $4$ . The predictions at $x=10$ are interpolations.	4	Set the predictions equal: $5x+19=3x+35$ , so $2x=16$ and $x=8$ . Substitution gives $y=59$ . At $x=10$ , group A predicts $69$ and group B predicts $65$ , so A is greater by $4$ . Since $10$ lies within the observed range $3$ to $12$ , both predictions are interpolations.