S Statistics — notes on one page

Every Edexcel GCSE Maths 1MA1 specification point in S, read straight through: 6 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

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

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

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.

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)

Definition. Specification S2: interpret and construct tables, charts and diagrams, including frequency tables, bar charts, pie charts and pictograms for categorical data, vertical line charts for ungrouped discrete numerical data, tables and line graphs for time series data, and know their appropriate use.

Choose the display that matches the data: separated bars or a pictogram with a key for categories, a vertical line chart for ungrouped discrete values, and points joined in time order for a time series.
In a pie chart each sector angle is $\dfrac{\text{frequency}}{\text{total}}\times360^\circ$ ; reversing the formula recovers a frequency or the total from an angle.
Complete a two-way table from the totals: fill the cells you can, then subtract.
Reading a chart for a percentage means frequency divided by total, times $100$ .
Every chart needs labelled axes, an even scale, a key where needed and accurate plotting; a criticism question expects specific faults such as missing labels or sector angles that do not match the data.

Worked example

A pie chart shows how

120

students travel to school. The sector for walking has an angle of

126^\circ

. Work out how many of the students walk to school.

(2)

(Total for Question 1 is 2 marks)

Working

1.The sector is $\dfrac{126}{360}$ of the whole pie chart.
2.Apply that fraction to the total: $\dfrac{126}{360}\times120$ .
3.Simplify: $\dfrac{126}{360}=\dfrac{7}{20}$ , and $\dfrac{7}{20}\times120=42$ .

Answer: $42$ students

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for $\dfrac{126}{360}\times120$ or $120\div360\times126$ or $126\div3$	M1	oe
	for $42$	A1	cao

Mark-scheme wording to use

“makes some comment about the labels … comments about the inaccuracy of the angles in the pie chart” — Pearson 1MA1, November 2022 Paper 3F, Q16
“for process to find total German eg 200 – 104 – 70 (= 26) … for complete process to find boys choosing Spanish eg 90 – (60 + ("26" – 18))” — Pearson 1MA1, November 2022 Paper 3H, Q3
““2” ÷ 40 × 100 … cao” — Pearson 1MA1, November 2018 Paper 3H, Q4(a)

What the examiner wants. Precise criticism and accurate plotting. The November 2022 Paper 3F scheme rejects vague comments such as “the pie chart is wrong” or “the angles do not add up to $360$ ”, and the November 2024 Paper 3H report says points joined with a curve, or the first and last points joined directly, lost the diagram mark.

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 $360^\circ$ .

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]

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.

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)

Definition. Specification S4: interpret, analyse and compare the distributions of data sets from univariate empirical distributions through appropriate graphical representation involving discrete, continuous and grouped data, including box plots, and through appropriate measures of central tendency (median, mean, mode and modal class) and spread (range, including consideration of outliers, quartiles and inter-quartile range).

A comparison scores when it names a measure of centre and a measure of spread, quotes the figures, and says what they mean in context. From a stem-and-leaf diagram, count to the middle value for the median and subtract the smallest from the largest for the range.
Higher tier: on a cumulative frequency graph read the median at half the total and the quartiles at a quarter and three quarters; the interquartile range $Q_3-Q_1$ measures the middle half and ignores outliers.
A box plot shows both directly.
Estimate a mean from a grouped table with midpoints, $\sum fx\div\sum f$ ; with two known group means, multiply each by its size to recover totals.
Compare median with median and range with range.

Worked example

For a group of Year 9 girls, the least height is

152

cm, the median height is

166

cm and the greatest height is

176

cm. The heights, in cm, of

15

Year 9 boys are

156,\ 158,\ 161,\ 163,\ 165,\ 166,\ 169,\ 171,\ 172,\ 174,\ 175,\ 178,\ 179,\ 181,\ 183

. Compare the distribution of the girls' heights with the distribution of the boys' heights.

(3)

(Total for Question 1 is 3 marks)

Working

1.The boys' median is the $8$ th of the $15$ ordered values: $171$ cm.
2.The boys' range is $183-156=27$ cm; the girls' range is $176-152=24$ cm.
3.Compare centres: the boys' median height of $171$ cm is greater than the girls' $166$ cm, so the boys are typically taller.
4.Compare spreads: the boys' range of $27$ cm is greater than the girls' $24$ cm, so the boys' heights are more spread out.

Answer: Boys' median $171$ cm and range $27$ cm; the boys are typically taller than the girls and their heights are more varied.

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for identification of the median (171) or the range (27) of the boys, or the range (24) of the girls	B1
	for a correct comparison of medians and a correct comparison of ranges supported by correct figures, eg the median height for boys (171) is greater than the median for girls (166) and the range for boys (27) is greater than the range for girls (24)	C2	At least one comparison must be in context. (C1 for a correct comparison of medians or of ranges that could ft their figures)

Mark-scheme wording to use

“At least one comparison must be in context referring to height or quoting cm. Simply quoting values for median, range is insufficient; they must be compared” — Pearson 1MA1, June 2020 Paper 1H, Q7
“B1 for correctly identifying the median of (School) A as 57 C1 (ft, dep on value for the median being stated) for making a correct comparison of median eg ‘the median of (School) A is higher than the median of (School) B’” — Pearson 1MA1, November 2024 Paper 3H, Q6
“for a first step in the process … for a complete process eg (50 × 167.6 - 20 × 182) ÷ 30” — Pearson 1MA1, November 2022 Paper 3H, Q8

What the examiner wants. Stated values and a direction. The November 2024 Paper 3H report says the median of $57$ had to be clearly stated, not just circled on the stem-and-leaf diagram, and that a comparison must say which group is greater; simply giving the difference between the values was not enough.

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.

S5 · Apply statistics to describe a population

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.

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

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.