What is S2.4: Outliers and cleaning data?

A common outlier rule flags values below Q_1-1.5IQR or above Q_3+1.5IQR, but context should guide the final decision.

How many marks are S2.4 questions worth?

S2.4 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has S2.4 appeared in?

We have not yet indexed a verified past-paper appearance for S2.4 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

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.

Draft — not yet indexed

Outliers and cleaning data

Worked answers and methods for S2.4 on Edexcel A-level Maths 9MA0.

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.

Worked practice

Tier 1 · Easy

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)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	The upper boundary is $34$ , so $36$ is a possible outlier.	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 marks)

Tier 2 · Standard

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 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$\operatorname{IQR}=12$ 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
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.

(4 marks)

Tier 3 · Hard

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 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Corrected mean $=12.15$ . Corrected standard deviation $=2.00$ . The error inflated the original spread.	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.

(5 marks)

Tier 1 · Easy

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 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	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
Notes
Box plots summarise each distribution using its median, quartiles and extremes. Putting them on the same scale makes the requested comparison direct.

(2 marks)

Tier 2 · Standard

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 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	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
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.

(4 marks)

Tier 3 · Hard

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 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$\operatorname{IQR}=2$ , 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
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.

(5 marks)

Tier 2 · Standard

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 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$Q_1=7$ and $Q_3=14$ $\operatorname{IQR}=7$ ; the fences are $-3.5$ and $24.5$ . $29$ is the only possible outlier.	4
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.

(4 marks)

Tier 3 · Hard

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 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$Q_1=(5+6)/2=5.5$ , $Q_3=(x+14)/2$ 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
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.

(6 marks)

Tier 3 · Hard

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

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 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	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
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.

(6 marks)

Q10

Tier 3 · Hard

10.

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 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	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
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.

(7 marks)

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