What is S6: Scatter graphs?

A scatter graph displays paired values.

How many marks are S6 questions worth?

In verified past-paper appearances, S6 questions on Edexcel GCSE Maths 1MA1 have been worth 2–4 marks.

Which past exam papers has S6 appeared in?

S6 has appeared in 9 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

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

Scatter graphs

Worked answers, methods and verified real exam appearances for S6 on Edexcel GCSE Maths 1MA1.

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.

Worked practice

Tier 1 · Easy

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)

Mark scheme

Mark scheme for question 1
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.

Tier 2 · Standard

For data with observed

x

-values from

5

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 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$y=31.2$ 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.

Tier 3 · Hard

A scatter graph compares a puppy's age

a

months with mass

m

kg for ages from

2

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 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	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.

Tier 1 · Easy

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 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	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

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 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	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.

Tier 3 · Hard

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 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	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.

Tier 2 · Standard

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 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	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

A scatter graph compares time since charging,

x

hours, with battery level,

y\%

, for times from

4

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 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	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.

Tier 3 · Hard

For data with observed

x

-values from

2

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 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	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.

Q10

Tier 3 · Hard

Two scatter graphs use the same observed

x

-range, from

3

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 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	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.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2022-11	2H	Q1	4	Allowed	Higher	QP MS
2021-11	3H	Q1	4	Allowed	Higher	QP MS
2019-11	3F	Q24	2	Allowed	Foundation	QP MS
2019-11	3H	Q3	2	Allowed	Higher	QP MS
2023-06	1F	Q25	3	Non-calculator	Foundation	QP MS
2022-11	2F	Q18	4	Allowed	Foundation	QP MS
2021-11	3F	Q21	4	Allowed	Foundation	QP MS
2024-11	2F	Q21	3	Allowed	Foundation	QP MS
2024-11	2H	Q2	3	Allowed	Higher	QP MS

Other points in S Statistics

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