S2 Data presentation and interpretation — revision question pack

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

Checked against Edexcel 9MA0 section S2. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

S2.1 · Interpret diagrams for single-variable data, including understanding that area in a histogram represents frequency; connect to probability distributions.

Explanation

In a histogram, frequency is proportional to bar area and frequency density is $\frac{\text{frequency}}{\text{class width}}$ ; unequal class widths make bar height alone misleading.
Frequency polygons show class patterns, box plots summarise centre and spread, and cumulative frequency diagrams support estimates of medians, quartiles and counts below a value.
For a continuous probability density histogram, total area is $1$ and the area above an interval is the probability of an observation in that interval.
Read class boundaries and axis scales before calculating.
A common error is to use frequency density as though it were frequency.

In a histogram, bar area represents frequency, so height is frequency density.

Worked example

A cumulative frequency diagram represents $80$ observations. The cumulative frequencies at $x=10$ and $x=20$ are $18$ and $54$ respectively. Estimate the number of observations satisfying $10<x\leq20$ .

1.The cumulative frequency at $20$ counts all $54$ observations up to that value, including the $18$ already counted up to $10$ .
2.Subtract to isolate the interval: $54-18=36$ .

Answer: $36$

Common mistakes

Don't use bar height rather than class width times frequency density to compare histogram frequencies.
Don't read cumulative frequencies directly as class frequencies instead of subtracting the boundary totals.

Exam tip

For cumulative-frequency intervals, subtract the cumulative totals at the two boundaries and respect endpoint inequalities.

Tier 1 · Easy

1.
A histogram class is $12\leq t<17$ and has frequency density $3.6$ . Find the frequency in this class.
(2)
(Total for Question 1 is 2 marks)
2.
Two bars in a histogram represent equal frequencies. The first class has width $6$ and frequency density $4$ . The second class has width $9$ . Find its frequency density.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
In a histogram, the class $10\leq x<15$ has frequency density $4$ , and the class $15\leq x<25$ has frequency density $2.5$ . Find the frequency in each class and the total frequency represented by these two bars.
(4)
(Total for Question 1 is 4 marks)
2.
On a cumulative frequency diagram for $80$ observations, the graph segment from $(30,22)$ to $(42,58)$ is a straight line. Estimate the median and the $65$ th percentile.
(4)
(Total for Question 2 is 4 marks)
3.
The vertical scale is missing from a histogram. The class $0\leq x<5$ has frequency $30$ and a bar height of $8$ grid units. The class $5\leq x<11$ has a bar height of $4$ grid units. Find the frequency density represented by one grid unit and the frequency in the second class.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
The probability density histogram for a continuous random variable $X$ has constant heights $0.12$ on $0\leq x<3$ , $0.08$ on $3\leq x<8$ , and $0.12$ on $8\leq x\leq10$ . Verify that it defines a probability distribution and find $P(1<X<6)$ .
(5)
(Total for Question 1 is 5 marks)
2.
A histogram of $60$ observations has frequency densities $3.2$ , $2.0$ and $k$ in the classes $0\leq x<5$ , $5\leq x<12$ and $12\leq x<20$ respectively. Find $k$ . Hence find the probability that a randomly selected observation from the data is at least $10$ .
(6)
(Total for Question 2 is 6 marks)
3.
A histogram has frequency densities $2$ , $5$ and $3$ in the classes $0\leq x<5$ , $5\leq x<9$ and $9\leq x<15$ respectively. Estimate the median and the interquartile range, assuming observations are distributed uniformly within each class.
(6)
(Total for Question 3 is 6 marks)
4.
Two histograms use the same class boundaries and both vertical axes show frequency density. In the class $10\leq x<15$ , sample A has $80$ observations in total and frequency density $4$ , while sample B has $120$ observations in total and frequency density $5$ . Calculate the frequency and the percentage of each sample in this class. Hence explain why comparing the two bar heights alone gives the wrong conclusion about which sample has the larger proportion in the class.
(5)
(Total for Question 4 is 5 marks)
5.
A histogram of $70$ observations has frequency densities $3$ , $5$ , $2$ and $4$ in the classes $0\leq x<4$ , $4\leq x<10$ , $10\leq x<a$ and $a\leq x<20$ , respectively, where $10<a<20$ . Find $a$ . Assuming observations are distributed uniformly within each class, estimate the median and find the probability that a randomly selected observation is at least $14$ .
(7)
(Total for Question 5 is 7 marks)

S2.2 · Interpret scatter diagrams and regression lines for bivariate data, including recognising distinct sections of the population (regression calculations excluded); interpret correlation informally; correlation does not imply causation.

Explanation

A scatter diagram shows paired bivariate data; describe the direction and strength of its association and note any outliers or clusters.
A regression line estimates the mean response for a given explanatory-variable value and is most defensible within the observed data range.
Distinct clusters may represent different sections of the population, so one overall correlation or regression line can hide different within-group patterns.
Correlation measures association, not causation; a lurking variable, reverse causation or coincidence may explain the observed relationship.
For $y=ax^n$ , plot $\log y$ against $\log x$ to get gradient $n$ and intercept $\log a$ ; for $y=kb^x$ , plot $\log y$ against $x$ to get gradient $\log b$ and intercept $\log k$ .

Worked example

A scatter diagram of journey distance against journey time contains one cluster for bicycles and a separate cluster for cars. Explain why fitting one regression line to all journeys may be misleading.

1.Identify vehicle type as a grouping variable.
2.Since the relationship between distance and time is likely to differ between bicycles and cars, pooling the groups can produce a line that fits neither group well.
3.Analyse the sections separately.

Answer: The two clusters represent distinct sections of the population with different travel speeds.; A single line may mainly reflect the gap between the clusters rather than either within-group relationship.; Separate regression lines or separate analyses would be more informative.

Common mistakes

Don't extrapolate a regression line far beyond the observed data range.
Don't interpret an overall regression line despite distinct clusters representing different subpopulations.

Exam tip

Before using a regression line, inspect clusters and explain whether one relationship is credible for the whole population.

Tier 1 · Easy

1.
A scatter diagram of daily ice-cream sales against temperature shows a strong positive correlation. Interpret this and explain why it does not prove that higher temperature causes every increase in sales.
(2)
(Total for Question 1 is 2 marks)
2.
A scatter diagram contains one compact cluster for electric bicycles and a separate compact cluster for conventional bicycles. State why one regression line for all the points may be misleading.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Across a year, ice-cream sales and the number of reported sunburn cases have strong positive correlation. Explain why this does not establish that buying ice cream causes sunburn.
(3)
(Total for Question 1 is 3 marks)
2.
For recordings made at wind speeds between $5$ and $25$ km h $^{-1}$ , the regression line of sound level $s$ dB on wind speed $w$ km h $^{-1}$ is $s=68-1.4w$ . Find the predicted sound level when $w=18$ and interpret the gradient in context.
(3)
(Total for Question 2 is 3 marks)
3.
A scatter diagram plots advertising cost in pounds against sales income in pounds, and its fitted line has gradient $0.40$ . The sales income is then converted from pounds to pence, without changing any observations. State the new gradient and describe the effect of this conversion on the direction and strength of the correlation.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
For trees aged between $3$ and $18$ years, a regression line of trunk diameter $d$ cm on age $a$ years is $d=1.8a+4.2$ . A $12$ -year-old tree has diameter $29$ cm. Interpret the difference between the observed and predicted values and critique using the line to predict the diameter of a $40$ -year-old tree.
(4)
(Total for Question 1 is 4 marks)
2.
A scatter diagram compares hours of optional practice with assessment score for two courses. Students on the advanced course form a cluster with both higher practice hours and higher scores than students on the introductory course. Within each course cluster, score tends to decrease slightly as practice hours increase, but the pooled data show positive correlation. Explain why the pooled correlation and a single regression line could be misleading, and why the diagram does not show that extra practice lowers scores.
(5)
(Total for Question 2 is 5 marks)
3.
For observations with $1\leq x\leq9$ , the fitted equation predicting $y$ from $x$ is $y=2x-1$ . After the observation $(9,25)$ is omitted, the equation becomes $y=1.2x+1.0$ . Calculate the residual of $(9,25)$ from the second fit and compare the two predicted values at $x=8$ . Interpret what these results show, and state why the observation should not be deleted solely on this evidence.
(5)
(Total for Question 3 is 5 marks)
4.
For observations with $2\leq x\leq20$ , a regression line of $\log_{10}y$ on $\log_{10}x$ is $\log_{10}y=\log_{10}2+1.5\log_{10}x$ . Obtain a model of the form $y=ax^n$ , and use it to estimate $y$ when $x=8$ . Comment on using the model when $x=50$ , and explain why a strong linear correlation on the transformed plot would not by itself establish a causal relationship.
(6)
(Total for Question 4 is 6 marks)
5.
For a bivariate data set with $\bar{x}=10$ and $\bar{y}=30$ , the fitted equation for predicting $y$ from $x$ is $y=2x+10$ . The fitted equation for predicting $x$ from $y$ is $x=0.4y-2$ . Estimate $y$ when $x=12$ , stating which equation you use. Show that both equations pass through $(\bar{x},\bar{y})$ . Explain why rearranging the second equation gives a different estimate and is not the appropriate procedure here.
(5)
(Total for Question 5 is 5 marks)

S2.3 · Interpret measures of central tendency and variation, extending to standard deviation; be able to calculate standard deviation, including from summary statistics.

Explanation

The mean uses every value, while the median is resistant to extremes; choose a measure that suits the distribution and context.
Range and interquartile range measure spread using endpoints or quartiles; standard deviation measures typical spread about the mean using all observations.
For $n$ data values, use $\sigma=\sqrt{\frac{\sum x^2}{n}-\left(\frac{\sum x}{n}\right)^2}$ unless a different convention is stated.
When groups are combined, add $n$ , $\sum x$ and $\sum x^2$ before recalculating; averaging separate standard deviations is not valid.
Coding data by $y=(x-a)/b$ simplifies calculation, after which transform the statistics back; grouped-data percentiles require linear interpolation within the relevant class.

Worked example

For $20$ observations, $\sum x=310$ and $\sum x^2=5020$ . Calculate the population standard deviation to $3$ significant figures.

1.The mean is $310/20=15.5$ .
2.Hence $\sigma=\sqrt{5020/20-15.5^2}=\sqrt{251-240.25}=\sqrt{10.75}=3.278\ldots$ , so the standard deviation is $3.28$ .

Answer: $3.28$

Common mistakes

Don't average two group means without weighting them by their different sample sizes.
Don't use $S_{xx}/n$ as the standard deviation without taking the square root.

Exam tip

Substitute the stated divisor into the variance formula, keep the square root until the end and check the result is non-negative.

Tier 1 · Easy

1.
Calculate the mean and population standard deviation of $4,7,7,8,9$ . Give the standard deviation to $3$ significant figures.
(3)
(Total for Question 1 is 3 marks)
2.
Eight readings have mean $14$ . A ninth reading, $16$ , is added. Find the mean of all nine readings, giving your answer to $3$ significant figures.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A data set $x$ has mean $12$ and standard deviation $3$ . A new variable is defined by $y=5+2x$ . Find the mean and standard deviation of $y$ .
(3)
(Total for Question 1 is 3 marks)
2.
Machine A fills $10$ containers with $\sum x=200$ and $\sum x^2=4040$ . Machine B fills $8$ containers with $\sum x=164$ and $\sum x^2=3400$ . Calculate the mean and population standard deviation for each machine, and compare their consistency. Give standard deviations to $3$ significant figures.
(5)
(Total for Question 2 is 5 marks)
3.
A set of $16$ observations has mean $7.5$ and population standard deviation $2.25$ . Find $\sum x$ and $\sum x^2$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Group A has $12$ values with mean $18$ and $\sum x^2=3996$ . Group B has $8$ values with mean $24$ and $\sum x^2=4736$ . Find the mean and population standard deviation of all $20$ values.
(5)
(Total for Question 1 is 5 marks)
2.
For $25$ observations of $x$ , the coded variable $y=(x-50)/4$ satisfies $\sum y=30$ and $\sum y^2=100$ . Find the mean and population standard deviation of $x$ . One further observation, $x=62$ , is then added. Find the new mean and population standard deviation. Give all non-integer answers to $3$ significant figures.
(7)
(Total for Question 2 is 7 marks)
3.
Eight observations have mean $10$ and population standard deviation $3$ . Six of the observations are $5,10,11,12,12,12$ ; the other two are $u$ and $v$ , where $u\leq v$ . Find $u$ and $v$ .
(6)
(Total for Question 3 is 6 marks)
4.
A grouped frequency table has classes $0\leq x<10$ , $10\leq x<20$ , $20\leq x<30$ and $30\leq x<50$ , with respective frequencies $3$ , $5$ , $8$ and $4$ . Using class midpoints, estimate the mean and population standard deviation. Give the standard deviation to $3$ significant figures and explain why both answers are estimates rather than exact statistics for the raw data.
(6)
(Total for Question 4 is 6 marks)
5.
Two groups together contain $10$ observations with mean $16.4$ and population variance $28.44$ . Group A contains $4$ observations with mean $11$ and population variance $5$ . Find the mean and population standard deviation of the $6$ observations in group B. Use unrounded values in your working and give the standard deviation to $3$ significant figures.
(6)
(Total for Question 5 is 6 marks)

S2.4 · Recognise and interpret possible outliers in data sets and statistical diagrams; select or critique data presentation techniques in context; clean data, including dealing with missing data, errors and outliers.

Explanation

A common outlier rule flags values below $Q_1-1.5\operatorname{IQR}$ or above $Q_3+1.5\operatorname{IQR}$ , but context should guide the final decision.
Investigate a suspicious value against the original record before correcting or removing it; an unusual valid observation is not automatically an error.
Handle missing data transparently: record how many values are missing, avoid inventing unsupported values and consider whether missingness could bias conclusions.
Choose displays to suit the data and purpose: histograms for continuous grouped data, box plots for comparing distributions, and scatter diagrams for paired variables.

Worked example

A table of package masses contains one blank entry and one value $482$ among values near $48.2$ grams. Describe a defensible way to clean these two entries before analysis.

1.First distinguish a data-entry error from a genuine extreme value by consulting the source.
2.Do not silently divide $482$ by $10$ .
3.The blank contains no observed value, so omit it from calculations unless a justified imputation rule has been chosen, and document the decision so its possible bias is visible.

Answer: Check the original measurement record for both entries.; Correct $482$ to $48.2$ only if the source confirms a decimal-point error; otherwise retain and flag it or exclude it with a stated reason.; Treat the blank as missing rather than replacing it without evidence, and report the reduced sample size or justified imputation method.

Common mistakes

Don't replace every missing value with zero, changing both the centre and spread of the data.
Don't delete an unusual value automatically instead of investigating whether it is an error or genuine observation.

Exam tip

Document separate rules for missing values, transcription errors and plausible outliers before recalculating summaries.

Tier 1 · Easy

1.
For a data set, $Q_1=14$ and $Q_3=22$ . Use the $1.5\operatorname{IQR}$ rule to determine whether the value $36$ is a possible outlier.
(2)
(Total for Question 1 is 2 marks)
2.
A researcher wants to compare the median and spread of journey times for two train operators. State a suitable diagram and give one reason for your choice.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
For a data set, the lower quartile is $18$ and the upper quartile is $30$ . Use the $1.5\times\operatorname{IQR}$ rule to decide whether a value of $50$ is an outlier, and state what should be done before removing it.
(4)
(Total for Question 1 is 4 marks)
2.
Two schools have different numbers of students. A report uses two pie charts to compare their students' continuous weekly screen times. Critique this presentation and suggest a more informative display for comparing the distributions.
(4)
(Total for Question 2 is 4 marks)
3.
The ordered data are $4,6,7,8,9,10,11,12,14,15,29$ . Use the Edexcel discrete convention: round a quartile position up when it is not a whole number; when it is a whole number, average the value in that position and the next value. Find $Q_1$ , $Q_3$ and the outlier fences, and identify any possible outlier.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A data set of $40$ readings was summarised as $\sum x=504$ and $\sum x^2=6856.9$ . One reading was entered as $31$ but the source record confirms it should be $13$ . Calculate the corrected mean and population standard deviation, and state the likely effect of the error on the original spread.
(5)
(Total for Question 1 is 5 marks)
2.
Ten response times, in seconds, are $8,9,9,10,10,10,11,11,12,40$ . For these data, $Q_1=9$ and $Q_3=11$ . Use the $1.5\operatorname{IQR}$ rule to identify any possible outlier. Calculate the mean and median, then recommend which measure better represents a typical response time. The source confirms that the value $40$ is a genuine response during a network outage; state whether it should automatically be deleted.
(5)
(Total for Question 2 is 5 marks)
3.
The ordered data are $3,4,5,6,7,8,9,10,x,14,15,25$ , where $x$ is an integer and $10\leq x\leq14$ . Use the Edexcel discrete convention: round a quartile position up when it is not a whole number; when it is a whole number, average the value in that position and the next value. Find all values of $x$ for which $25$ is a possible outlier under the $1.5\operatorname{IQR}$ rule. State how the value $25$ should be treated if its source record confirms that it is genuine.
(6)
(Total for Question 3 is 6 marks)
4.
Fifteen ordered readings are $2,4,5,7,8,9,10,11,12,13,14,15,16,18,30$ . A sixteenth reading is missing from the file, but its source record shows only that it is an integer from $10$ to $13$ inclusive. Do not replace it by an average. Using the Edexcel discrete convention, show that $Q_1$ , $Q_3$ and the upper outlier fence are the same for every possible value of the missing reading. Hence decide whether $30$ is always a possible outlier, and state what should be recorded before the data are used.
(6)
(Total for Question 4 is 6 marks)
5.
A data extract lists measurements $41,42,43,44,45,46,47,47,48,49,50,510$ and one blank entry. The two values $47$ have the same record identifier, the source confirms that $510$ is a decimal-point error for $51.0$ , and the blank cannot be recovered. State how each issue should be handled. For the cleaned data, find the sample size, median, quartiles and any possible outliers using the Edexcel discrete convention and the $1.5\operatorname{IQR}$ rule.
(7)
(Total for Question 5 is 7 marks)

Answer key

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

S2.1 · Interpret diagrams for single-variable data, including understanding that area in a histogram represents frequency; connect to probability distributions.

Tier 1 · Easy

Mark scheme for S2.1 Tier 1 · Easy
Question	Scheme	Marks
1	$18$	2
(2 marks)		2
Notes
The class width is $17-12=5$ . Hence frequency $=$ class width $\times$ frequency density $=5\times3.6=18$ .
2	$\dfrac{8}{3}$	2
(2 marks)		2
Notes
Equal frequencies give equal bar areas. The first bar has area $6(4)=24$ , so the second density is $24/9=8/3$ .

Tier 2 · Standard

Mark scheme for S2.1 Tier 2 · Standard
Question	Scheme	Marks
1	Frequencies $20$ and $25$ Total frequency $45$	4
(4 marks)		4
Notes
Histogram frequency equals bar area, so multiply frequency density by class width. The first class has width $5$ and frequency $4(5)=20$ . The second has width $10$ and frequency $2.5(10)=25$ . Their total is $20+25=45$ .
2	Median $=36$ $65$ th percentile $=40$	4
(4 marks)		4
Notes
The median has cumulative frequency $40$ . This is $18/36=1/2$ of the way from $22$ to $58$ , so its value is $30+(1/2)(42-30)=36$ . The $65$ th percentile has cumulative frequency $0.65(80)=52$ , which is $30/36=5/6$ of the way along the segment, giving $30+(5/6)(12)=40$ .
3	One grid unit represents frequency density $0.75$ . The frequency in $5\leq x<11$ is $18$ .	4
(4 marks)		4
Notes
The first class has width $5$ , so its frequency density is $30/5=6$ . A height of $8$ grid units therefore represents density $6$ , giving $6/8=0.75$ density units per grid unit. The second bar has density $4(0.75)=3$ and width $6$ , so its area and frequency are $6(3)=18$ .

Tier 3 · Hard

Mark scheme for S2.1 Tier 3 · Hard
Question	Scheme	Marks
1	Total area $=1$ . $P(1<X<6)=0.48$ .	5
(5 marks)		5
Notes
The total area is $3(0.12)+5(0.08)+2(0.12)=0.36+0.40+0.24=1$ , so the histogram defines a probability distribution. The area from $1$ to $3$ is $2(0.12)=0.24$ , and the area from $3$ to $6$ is $3(0.08)=0.24$ . Therefore $P(1<X<6)=0.24+0.24=0.48$ .
2	$k=3.75$ $P(X\geq10)=\dfrac{17}{30}$	6
(6 marks)		6
Notes
The first two frequencies are $5(3.2)=16$ and $7(2.0)=14$ . The last class therefore has frequency $60-16-14=30$ , so $k=30/8=3.75$ . From $10$ to $12$ , the histogram area gives frequency $2(2.0)=4$ ; all $30$ observations in the final class also qualify. Hence the probability is $(4+30)/60=17/30$ .
3	Estimated median $=7.8$ Estimated lower quartile $=5.4$ and upper quartile $=11$ Estimated interquartile range $=5.6$	6
(6 marks)		6
Notes
The class frequencies are $5(2)=10$ , $4(5)=20$ and $6(3)=18$ , giving total frequency $48$ . The median is the $24$ th value, so interpolation in the second class gives $5+\frac{24-10}{20}(4)=7.8$ . The lower quartile is the $12$ th value, giving $5+\frac{12-10}{20}(4)=5.4$ . The upper quartile is the $36$ th value, in the third class, giving $9+\frac{36-30}{18}(6)=11$ . Hence the estimated IQR is $11-5.4=5.6$ .
4	Sample A has frequency $5(4)=20$ , which is $20/80=25\%$ of sample A. Sample B has frequency $5(5)=25$ , which is $25/120=20.8\%$ of sample B to $3$ significant figures. The bar for B is taller and contains more observations, but A has the larger proportion in the class. Frequency-density heights from samples of different total sizes do not directly compare relative frequencies; percentage-frequency density or calculated proportions are needed.	5
(5 marks)		5
Notes
Both class widths are $5$ , so the bar areas give frequencies $20$ and $25$ . Divide each frequency by its own sample total: $20/80=0.25$ and $25/120=0.20833\ldots$ . The different denominators reverse the comparison made from raw bar heights.
5	$a=16$ Estimated median $=8.6$ $P(X\geq14)=\dfrac27$	7
(7 marks)		7
Notes
The total frequency is $4(3)+6(5)+(a-10)(2)+(20-a)(4)=102-2a$ . Equating this to $70$ gives $a=16$ . The first two class frequencies are $12$ and $30$ , so the $35$ th value lies in $4\leq x<10$ and is estimated by $4+\frac{35-12}{30}(6)=8.6$ . From $14$ to $16$ the estimated frequency is $2(2)=4$ , and from $16$ to $20$ it is $4(4)=16$ . Hence $P(X\geq14)=(4+16)/70=2/7$ .

S2.2 · Interpret scatter diagrams and regression lines for bivariate data, including recognising distinct sections of the population (regression calculations excluded); interpret correlation informally; correlation does not imply causation.

Tier 1 · Easy

Mark scheme for S2.2 Tier 1 · Easy
Question	Scheme	Marks
1	Higher temperatures tend to be associated with higher ice-cream sales. Correlation alone does not establish causation; other variables such as sunshine, holidays or visitor numbers may affect both quantities.	2
(2 marks)		2
Notes
Translate positive correlation as a tendency for larger values of one variable to occur with larger values of the other. Then distinguish association from a causal conclusion by identifying a plausible lurking variable.
2	The clusters represent distinct sections of the population. One line may describe mainly the separation between bicycle types rather than the relationship within either type, so the groups should be considered separately.	2
(2 marks)		2
Notes
The bicycle type is a grouping variable. Pooling two separated groups can create an overall pattern that does not represent the within-group relationship, so separate analyses are more informative.

Tier 2 · Standard

Mark scheme for S2.2 Tier 2 · Standard
Question	Scheme	Marks
1	Correlation does not imply causation. A valid lurking variable is hot or sunny weather, season, holiday periods or visitor numbers, any of which could increase both ice-cream sales and sunburn cases. Evidence controlling for plausible lurking variables, or an appropriately designed study, would be needed to support a causal claim.	3
(3 marks)		3
Notes
The scatter pattern records association only. Temperature or sunshine can affect both variables: hot days encourage ice-cream purchases and increase exposure to ultraviolet radiation. This common cause can generate the correlation even if ice cream has no effect on sunburn.
2	Predicted sound level $=42.8$ dB. Within the observed range, the model predicts a decrease of $1.4$ dB in mean sound level for each increase of $1$ km h $^{-1}$ in wind speed.	3
(3 marks)		3
Notes
Substitute $w=18$ : $s=68-1.4(18)=42.8$ . The gradient $-1.4$ is the fitted change in the response variable for a one-unit increase in the explanatory variable; it describes association within the data range.
3	New gradient $=100(0.40)=40$ pence of sales income per pound of advertising cost. The direction remains positive. The strength of the correlation is unchanged because multiplying every response value by the same positive constant only changes its scale, not the scatter pattern.	3
(3 marks)		3
Notes
Converting pounds to pence multiplies every response value, and hence the fitted change in response, by $100$ . The gradient becomes $40$ . This positive rescaling does not alter which points lie relatively above or below one another or how tightly they follow a linear pattern, so it changes neither the positive direction nor the strength of the correlation.

Tier 3 · Hard

Mark scheme for S2.2 Tier 3 · Hard
Question	Scheme	Marks
1	The predicted diameter at $a=12$ is $25.8$ cm, so the difference between the observed and predicted values is $29-25.8=3.2$ cm. The tree is $3.2$ cm thicker than the value predicted by the line. A prediction at $40$ years is extrapolation far outside the observed age range and the linear relationship may not continue.	4
(4 marks)		4
Notes
Substitute $a=12$ to get the fitted value $d=1.8(12)+4.2=25.8$ . Residual $=$ observed $-$ fitted $=29-25.8=3.2$ cm. Since $40$ is well beyond the data range $3$ to $18$ , using the line there is unsupported extrapolation.
2	Course level creates two distinct sections of the population. The pooled positive correlation may be caused mainly by the separation of the two clusters and need not describe either course. A single regression line may fit neither within-course relationship well, so the courses should be analysed separately. The slight negative within-course association does not prove causation; prior attainment, difficulty or students seeking help could affect both practice time and score.	5
(5 marks)		5
Notes
The overall association mixes a between-course difference with two different within-course patterns. Course level is therefore a lurking grouping variable, and one pooled line conceals the structure. Even after separating the courses, an observational association cannot establish that practice causes lower scores because other variables can influence both quantities.
3	The second line predicts $11.8$ at $x=9$ , so the residual of $(9,25)$ is $25-11.8=13.2$ . At $x=8$ , the full-data prediction is $15$ and the prediction after omission is $10.6$ , a difference of $4.4$ . The observation has a large residual and substantially influences the fitted line. Influence does not prove a recording error; the source and context should be checked before retaining, correcting or excluding it.	5
(5 marks)		5
Notes
Using the line fitted without the observation isolates how unusual it is relative to the remaining pattern: $25-[1.2(9)+1.0]=13.2$ . The two fitted values at $x=8$ are $2(8)-1=15$ and $1.2(8)+1.0=10.6$ . Their difference shows that the point changes the fitted relationship materially. Statistical influence is a reason to investigate, not automatic evidence that the value is invalid.
4	$y=2x^{1.5}$ , so $a=2$ and $n=1.5$ . When $x=8$ , $y\approx2(8)^{1.5}=45.3$ to $3$ significant figures. $x=50$ is outside the observed range, so this would be unsupported extrapolation and the power relationship may not continue. Correlation on the transformed plot is association only; lurking variables, reverse causation or the data-generating process must be considered before making a causal claim.	6
(6 marks)		6
Notes
Taking powers of $10$ gives $y=2x^{1.5}$ . Substitution gives $2\sqrt{8^3}=45.254\ldots$ . The fitted relationship was supported only for $2\leq x\leq20$ , and a regression pattern does not identify a causal mechanism.
5	Use the regression line of $y$ on $x$ : the estimate is $y=2(12)+10=34$ . At $x=10$ , the first line gives $y=30$ ; at $y=30$ , the second gives $x=10$ , so both pass through $(10,30)$ . Rearranging $x=0.4y-2$ would give $y=35$ when $x=12$ . The two regression lines minimise deviations in different variables. To predict $y$ from a given $x$ , use the regression of $y$ on $x$ , not the algebraic inverse of the regression of $x$ on $y$ .	5
(5 marks)		5
Notes
The response being estimated is $y$ , so substitute $x=12$ into $y=2x+10$ . The mean point satisfies both supplied equations. Rearrangement of the other line gives $(12+2)/0.4=35$ , but that line was fitted to predict $x$ from $y$ and generally is not the inverse of the $y$ -on- $x$ line unless correlation is perfect.

S2.3 · Interpret measures of central tendency and variation, extending to standard deviation; be able to calculate standard deviation, including from summary statistics.

Tier 1 · Easy

Mark scheme for S2.3 Tier 1 · Easy
Question	Scheme	Marks
1	Mean $=7$ . Standard deviation $=1.67$ to $3$ significant figures.	3
(3 marks)		3
Notes
$\sum x=35$ and $\sum x^2=259$ , so $\bar{x}=35/5=7$ . Then $\sigma=\sqrt{259/5-7^2}=\sqrt{2.8}=1.673\ldots$ , giving $1.67$ .
2	$14.2$ to $3$ significant figures	2
(2 marks)		2
Notes
The original total is $8(14)=112$ . After adding $16$ , the total is $128$ , so the new mean is $128/9=14.222\ldots=14.2$ to $3$ significant figures.

Tier 2 · Standard

Mark scheme for S2.3 Tier 2 · Standard
Question	Scheme	Marks
1	Mean $29$ Standard deviation $6$	3
(3 marks)		3
Notes
Adding $5$ shifts every value and hence adds $5$ to the mean but does not change the spread. Multiplying by $2$ doubles both the mean contribution and the standard deviation. Thus the new mean is $5+2(12)=29$ and the new standard deviation is $\|2\|(3)=6$ .
2	Machine A: mean $20$ , standard deviation $2.00$ . Machine B: mean $20.5$ , standard deviation $2.18$ . Machine A is slightly more consistent because it has the smaller standard deviation.	5
(5 marks)		5
Notes
For A, $\bar{x}=200/10=20$ and $\sigma_A=\sqrt{4040/10-20^2}=2$ . For B, $\bar{x}=164/8=20.5$ and $\sigma_B=\sqrt{3400/8-20.5^2}=\sqrt{4.75}=2.179\ldots$ . Smaller standard deviation means less spread, so A is more consistent.
3	$\sum x=120$ $\sum x^2=981$	4
(4 marks)		4
Notes
The sum is $n\bar{x}=16(7.5)=120$ . Rearranging $\sigma^2=\sum x^2/n-\bar{x}^2$ gives $\sum x^2=n(\sigma^2+\bar{x}^2)=16(2.25^2+7.5^2)=16(61.3125)=981$ .

Tier 3 · Hard

Mark scheme for S2.3 Tier 3 · Hard
Question	Scheme	Marks
1	Combined mean $=20.4$ . Combined standard deviation $=4.52$ to $3$ significant figures.	5
(5 marks)		5
Notes
The combined sum is $12(18)+8(24)=408$ , so the combined mean is $408/20=20.4$ . The combined sum of squares is $3996+4736=8732$ . Therefore $\sigma=\sqrt{8732/20-20.4^2}=\sqrt{436.6-416.16}=\sqrt{20.44}=4.521\ldots$ , giving $4.52$ .
2	Original mean $=54.8$ and original standard deviation $=6.40$ . New mean $=55.1$ and new standard deviation $=6.43$ .	7
(7 marks)		7
Notes
For $y$ , the mean is $30/25=1.2$ and the variance is $100/25-1.2^2=2.56$ , so $\sigma_y=1.6$ . Since $x=4y+50$ , $\bar{x}=4(1.2)+50=54.8$ and $\sigma_x=4(1.6)=6.4$ . Also $\sum x=1370$ and $\sum x^2=76100$ . After adding $62$ , these become $1432$ and $79944$ for $26$ values. Thus the new mean is $1432/26=55.0769\ldots$ and the new standard deviation is $\sqrt{79944/26-(1432/26)^2}=6.4266\ldots$ , giving $55.1$ and $6.43$ .
3	$u=5$ and $v=13$	6
(6 marks)		6
Notes
The total sum is $8(10)=80$ , while the six known values sum to $62$ , so $u+v=18$ . The total sum of squares is $8(3^2+10^2)=872$ . The known squares sum to $678$ , so $u^2+v^2=194$ . Hence $2uv=(u+v)^2-(u^2+v^2)=324-194=130$ , giving $uv=65$ . Thus $u$ and $v$ are the roots of $t^2-18t+65=0=(t-5)(t-13)$ ; the stated order gives $u=5$ and $v=13$ .
4	Estimated mean $=22.5$ Estimated population standard deviation $=11.1$ to $3$ significant figures The calculations replace every observation in a class by its midpoint, so the within-class positions of the raw observations are unknown.	6
(6 marks)		6
Notes
Use midpoints $5,15,25,40$ . Then $\sum fx=3(5)+5(15)+8(25)+4(40)=450$ , so $\bar{x}=450/20=22.5$ . Also $\sum fx^2=3(25)+5(225)+8(625)+4(1600)=12600$ . Hence $\sigma=\sqrt{12600/20-22.5^2}=\sqrt{123.75}=11.124\ldots$ , giving $11.1$ . Midpoint substitution loses the unknown variation within each class.
5	Mean of group B $=20$ Population standard deviation of group B $=3.42$ to $3$ significant figures	6
(6 marks)		6
Notes
The combined sum is $10(16.4)=164$ and group A contributes $4(11)=44$ , so group B has sum $120$ and mean $20$ . From $\sigma^2=\sum x^2/n-\bar{x}^2$ , the combined sum of squares is $10(28.44+16.4^2)=2974$ , while group A contributes $4(5+11^2)=504$ . Thus group B has sum of squares $2470$ , variance $2470/6-20^2=35/3$ , and standard deviation $\sqrt{35/3}=3.41565\ldots=3.42$ .

S2.4 · Recognise and interpret possible outliers in data sets and statistical diagrams; select or critique data presentation techniques in context; clean data, including dealing with missing data, errors and outliers.

Tier 1 · Easy

Mark scheme for S2.4 Tier 1 · Easy
Question	Scheme	Marks
1	The upper boundary is $34$ , so $36$ is a possible outlier.	2
(2 marks)		2
Notes
The interquartile range is $22-14=8$ . The upper outlier boundary is $22+1.5(8)=34$ . Since $36>34$ , it is flagged as a possible outlier.
2	Use box plots drawn on a common scale. They display the median and interquartile range, so the centres and spreads of the two distributions can be compared directly.	2
(2 marks)		2
Notes
Box plots summarise each distribution using its median, quartiles and extremes. Putting them on the same scale makes the requested comparison direct.

Tier 2 · Standard

Mark scheme for S2.4 Tier 2 · Standard
Question	Scheme	Marks
1	$\operatorname{IQR}=12$ and the upper outlier boundary is $48$ . $50$ is an outlier. Check the original record and context: correct it only if an error is confirmed, retain it if genuine, or exclude it only with a justified and documented reason.	4
(4 marks)		4
Notes
The interquartile range is $30-18=12$ . The upper boundary is $Q_3+1.5\operatorname{IQR}=30+18=48$ , so $50>48$ is flagged as an outlier. Being an outlier does not prove the value is wrong: check the source and measurement conditions, then document any correction or exclusion.
2	Pie charts require screen time to be grouped into categories and do not show the detailed shape or spread of the continuous data. Unless the sample sizes are shown, equal-looking sectors can hide very different frequencies between the schools. Use comparative histograms with common class boundaries and frequency density, or box plots on a common scale. Histograms compare distribution shape; box plots compare median and spread.	4
(4 marks)		4
Notes
The variable is continuous, while pie charts reduce it to proportions in chosen categories and can conceal both sample size and distributional detail. Common-scale histograms preserve shape; common-scale box plots give a concise comparison of centre and spread.
3	$Q_1=7$ and $Q_3=14$ $\operatorname{IQR}=7$ ; the fences are $-3.5$ and $24.5$ . $29$ is the only possible outlier.	4
(4 marks)		4
Notes
Here $n=11$ . The lower-quartile position is $11/4=2.75$ , rounded up to the third value, so $Q_1=7$ . The upper-quartile position is $33/4=8.25$ , rounded up to the ninth value, so $Q_3=14$ . Thus the IQR is $7$ and the fences are $7-1.5(7)=-3.5$ and $14+1.5(7)=24.5$ . Only $29$ lies outside them.

Tier 3 · Hard

Mark scheme for S2.4 Tier 3 · Hard
Question	Scheme	Marks
1	Corrected mean $=12.15$ . Corrected standard deviation $=2.00$ . The error inflated the original spread.	5
(5 marks)		5
Notes
Replace the contribution of $31$ by $13$ : the corrected sum is $504-31+13=486$ , and the corrected sum of squares is $6856.9-31^2+13^2=6064.9$ . Thus $\bar{x}=486/40=12.15$ and $\sigma=\sqrt{6064.9/40-12.15^2}=\sqrt{4}=2.00$ . The erroneous value lay much farther from the centre, so it made the original spread larger.
2	$\operatorname{IQR}=2$ , with lower boundary $6$ and upper boundary $14$ , so $40$ is the only possible outlier. Mean $=13$ seconds and median $=10$ seconds. The median better represents a typical response time because it is resistant to the extreme value. $40$ should not automatically be deleted; it is genuine and should be retained or analysed separately according to whether outage performance is part of the population of interest.	5
(5 marks)		5
Notes
The IQR is $11-9=2$ . The lower fence is $9-1.5(2)=6$ and the upper fence is $11+1.5(2)=14$ , so only $40$ is flagged. The total is $130$ , giving mean $13$ , and the middle two values are both $10$ , giving median $10$ . The mean is pulled upward by the genuine outage value. An outlier rule flags a value for investigation; it does not by itself justify deletion.
3	$Q_1=(5+6)/2=5.5$ , $Q_3=(x+14)/2$ and $\operatorname{IQR}=(x+3)/2$ . The upper fence is $\dfrac{x+14}{2}+1.5\left(\dfrac{x+3}{2}\right)=\dfrac{5x+37}{4}$ . $25$ is a possible outlier when $25>(5x+37)/4$ , so $x<63/5=12.6$ . The integer values are $x\in\{10,11,12\}$ ; at $x=13$ , the upper fence is $25.5$ , so $25$ is not outside it. A genuine value should not be deleted automatically; retain it or analyse it separately according to the population and purpose, documenting the decision.	6
(6 marks)		6
Notes
With $n=12$ , $n/4=3$ and $3n/4=9$ are whole numbers. The quartiles are therefore the midpoints of the $3$ rd and $4$ th values, and of the $9$ th and $10$ th values: $Q_1=5.5$ and $Q_3=(x+14)/2$ . Hence $\operatorname{IQR}=(x+3)/2$ and the upper fence is $(5x+37)/4$ . The strict inequality $25>(5x+37)/4$ gives $x<12.6$ , so the stated integer range gives $10,11,12$ . A verified observation is not an error merely because it is unusual.
4	For $n=16$ , $Q_1$ is the mean of the $4$ th and $5$ th values, so $Q_1=(7+8)/2=7.5$ . For every allowed missing value, the $12$ th and $13$ th values are $14$ and $15$ , so $Q_3=14.5$ . The IQR is $7$ and the upper fence is $14.5+1.5(7)=25$ . $30>25$ , so $30$ is a possible outlier for every allowed value of the missing reading. Record the value as unresolved within the verified range, document that no unsupported imputation was made, and check the original source before any final correction or exclusion.	6
(6 marks)		6
Notes
Since $16/4=4$ and $3(16)/4=12$ are whole positions, average positions $4$ and $5$ , then positions $12$ and $13$ . Inserting any integer from $10$ to $13$ changes only the middle ordering: the relevant pairs remain $7,8$ and $14,15$ . The quartiles and fence are therefore invariant, so the outlier decision for $30$ is robust even though the missing reading itself remains unresolved.
5	Remove one of the duplicate $47$ records, correct $510$ to $51.0$ from the confirmed source, and treat the unrecoverable blank as missing rather than as zero. The cleaned ordered data are $41,42,43,44,45,46,47,48,49,50,51$ , so $n=11$ and the median is $46$ . $Q_1=43$ and $Q_3=49$ . The IQR is $6$ , giving fences $34$ and $58$ ; there are no possible outliers. The cleaning decisions and reduced sample size should be documented.	7
(7 marks)		7
Notes
A repeated identifier shows duplication, while the source record supplies evidence for the decimal correction. The blank supplies no observed value. After cleaning, the sixth value is the median. Since $n/4=2.75$ and $3n/4=8.25$ are not whole numbers, round up to positions $3$ and $9$ , giving quartiles $43$ and $49$ . The resulting fences contain all $11$ values.