What are the common methods of analysing data in Cambridge IGCSE Mathematics?

In Cambridge IGCSE Mathematics, common methods of analysing data include measures of central tendency (mean, median, mode), measures of dispersion (range, interquartile range, standard deviation), and graphical representations such as histograms, bar charts, and pie charts. These methods help in summarizing and interpreting data effectively.

How do you calculate the mean, median, and mode in a data set?

To calculate the mean, add all the numbers in the data set and divide by the number of values. The median is the middle value when the data is arranged in order, or the average of the two middle values if there is an even number of data points. The mode is the value that appears most frequently in the data set.

What is the importance of using graphical methods in data analysis?

Graphical methods, such as histograms, bar charts, and pie charts, are important in data analysis because they provide a visual representation of data, making it easier to identify patterns, trends, and outliers. These methods are particularly useful in Cambridge IGCSE Mathematics for interpreting and presenting data clearly and effectively.

How do you interpret the range and interquartile range in data analysis?

The range is the difference between the highest and lowest values in a data set, providing a basic measure of dispersion. The interquartile range (IQR) is the difference between the upper quartile (Q3) and the lower quartile (Q1), representing the spread of the middle 50% of the data. In Cambridge IGCSE Mathematics, these measures help in understanding the variability and consistency of data.

Why is standard deviation an important measure in data analysis?

Standard deviation is important because it quantifies the amount of variation or dispersion in a data set. A low standard deviation indicates that data points are close to the mean, while a high standard deviation suggests a wide spread around the mean. In Cambridge IGCSE Mathematics, understanding standard deviation helps in assessing the reliability and consistency of data.

Why is "Forgetting to sort before finding the median" a common mistake in Methods of Analysing Data?

Why it happens: Speed. How to avoid it: Always sort the list ascending before reading the middle.

Why is "Computing $\bar{x} = \sum fx / n$ instead of $\sum fx / \sum f$" a common mistake in Methods of Analysing Data?

Why it happens: Confusing n (total values) with the count of distinct values. How to avoid it: Denominator is f — total frequency — not the number of rows. Source: 0580/42 — recurring

Why is "Using class boundaries instead of midpoints in grouped data" a common mistake in Methods of Analysing Data?

Why it happens: Picking the lower bound as x. How to avoid it: Class [10, 20) has midpoint 10 + 20/2 = 15.

Why is "Calling the grouped-data mean 'exact'" a common mistake in Methods of Analysing Data?

Why it happens: Forgetting the original values are lost. How to avoid it: Always say ESTIMATED mean.

Cambridge IGCSE Mathematics (0580)

Methods of Analysing Data

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Short Notes - Methods of Analysing Data

Detailed Study Notes

Detailed notes on Statistics for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Mean, Median, Mode and Range — Cambridge IGCSE 0580 Maths Extended (2026)

The four statistics every Cambridge candidate must compute fluently — for raw lists, frequency tables, and grouped data. Cambridge wants the right method AND the right interpretation.

At a glance

Mean $= \dfrac{\text{sum}}{\text{count}}$ .
Median: middle value when sorted; average two middles for even count.
Mode: most common value (can be none, one, or several).
Range: max $-$ min.
Frequency table: $\bar{x} = \dfrac{\sum f x}{\sum f}$ .
Grouped data: use MIDPOINTS as $x$ values; the mean is an ESTIMATE.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E11.1 — Calculate the mean, median, mode and range for individual data.
E11.2 — Calculate an estimate of the mean for grouped data.
E11.3 — Identify the modal class from a grouped frequency distribution.

Mean, median, mode, range — raw data

Mean: sum/count. Median: middle when sorted. Mode: most common. Range: max - min.

Mean. Add up all the values, divide by how many there are. $\bar{x} = \dfrac{\sum x}{n}.$

Median. Sort the values from low to high. The median is the middle.

$n$ ODD: median is the $\dfrac{n+1}{2}$ -th value.
$n$ EVEN: median is the average of the $\dfrac{n}{2}$ -th and $(\dfrac{n}{2} + 1)$ -th values.

Mode. The value(s) that occur most often. A data set can be:

Unimodal: one mode.
Bimodal: two values tied for highest frequency.
No mode: every value occurs the same number of times.

Range. Maximum minus minimum.

Worked. Data: $4, 7, 2, 9, 4, 5, 8$ .

Sort: $2, 4, 4, 5, 7, 8, 9$ .
$n = 7$ , median = $\dfrac{8}{2} = 4$ th value = $5$ .
Mean: $\dfrac{2 + 4 + 4 + 5 + 7 + 8 + 9}{7} = \dfrac{39}{7} \approx 5.57$ .
Mode: $4$ (appears twice).
Range: $9 - 2 = 7$ .

Once the list is sorted, the median is the middle box, the mode is the most repeated value, and the range spans lowest to highest.

Mean = sum / count.
Median: sort, find the middle.
Mode: most common.
Range: max $-$ min.
$n$ even: median is the average of the two middles.

Mean from a frequency table

$\bar{x} = \dfrac{\sum f x}{\sum f}$ . Multiply each value by its frequency.

When data is summarised in a frequency table: $\bar{x} = \dfrac{\sum f x}{\sum f}.$

Worked. Number of pets per household:

Pets ( $x$ )	Frequency ( $f$ )
0	5
1	12
2	8
3	3
4	2

$\sum f = 5 + 12 + 8 + 3 + 2 = 30$ .
$\sum f x = (0)(5) + (1)(12) + (2)(8) + (3)(3) + (4)(2) = 0 + 12 + 16 + 9 + 8 = 45$ .
Mean: $\dfrac{45}{30} = 1.5$ pets per household.

Median from a frequency table. Find the cumulative frequency. Locate the value at position $\dfrac{n+1}{2}$ (or the average of $\dfrac{n}{2}$ and $\dfrac{n}{2}+1$ for $n$ even). Use the cumulative column to find which $x$ value sits there.

Mode from a frequency table. The value with the highest frequency.

Mean: $\dfrac{\sum f x}{\sum f}$ .
Use cumulative frequency to find the median.
Mode: highest-frequency value.
Set up an extra column $f x$ for clarity.

Estimating the mean for grouped data

Use the MIDPOINT of each class as $x$ . The result is an ESTIMATE — the original values are unknown.

When data is grouped into class intervals (e.g. $0 < t \le 10$ , $10 < t \le 20$ , …) we don't know the exact values within each class. We approximate by assuming each value sits at the midpoint.

Steps.

Find the midpoint of each class: $x = \dfrac{\text{lower} + \text{upper}}{2}$ .
Compute $f x$ for each row.
Estimate of mean: $\dfrac{\sum f x}{\sum f}$ .

Worked. Time spent on homework (minutes):

Class	Midpoint $x$	Frequency $f$	$f x$
$0 < t \le 30$	15	4	60
$30 < t \le 60$	45	12	540
$60 < t \le 90$	75	8	600
$90 < t \le 120$	105	6	630

$\sum f = 30$ , $\sum f x = 1830$ .
Estimate of mean: $\dfrac{1830}{30} = 61$ minutes.

Why "estimate"? We assumed every value in $0 < t \le 30$ was exactly $15$ — usually false. The result is an APPROXIMATION.

Modal class. The class with the HIGHEST frequency. Cambridge expects you to write the full class interval (e.g. " $30 < t \le 60$ "), not just the midpoint.

The modal class is the interval with the tallest bar — quote the full interval, not the midpoint.

Midpoint as the $x$ value.
Estimate of mean: $\dfrac{\sum f x}{\sum f}$ .
Modal class: highest-frequency interval.
Always write 'estimate' for grouped data.

Comparing data sets

Compare AVERAGES (mean / median) and SPREADS (range / IQR) — Cambridge wants both.

When asked to compare two data sets, Cambridge wants comments on TWO aspects:

1. Average / centre. Use mean or median.

Higher mean: data set is on average higher.
Outliers can pull the mean — median is more robust.

2. Spread / consistency. Use range or interquartile range (IQR — see Cumulative Frequency).

Smaller range: more consistent / less variable.
Smaller IQR: middle 50% is more tightly grouped.

Worked. Two classes' test scores:

Class A: mean $= 65$ , range $= 30$ .
Class B: mean $= 65$ , range $= 12$ .

Comment: "Both classes have the same mean ( $65$ ), so on average they performed equally. However, Class B has a much smaller range ( $12$ vs $30$ ), meaning Class B's scores are more consistent."

Same mean (65), but Class B's smaller range shows its scores are more consistent.

Cambridge mark scheme expectation. TWO comparison statements: one about average, one about spread. Each must reference the actual numerical statistics — vague statements like "Class B is more consistent" without citing the smaller range lose marks.

Compare TWO things: average AND spread.
Average: mean or median.
Spread: range or IQR.
Cite the actual statistics, not just descriptions.

Quick recap

Mean: $\dfrac{\sum x}{n}$ (or $\dfrac{\sum f x}{\sum f}$ for frequency tables).
Median: middle of sorted data.
Mode: most common value.
Range: max $-$ min.
Grouped data: use midpoints; mean is an ESTIMATE.
Comparison: comment on average AND spread.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Mean — sum divided by count: $\bar{x} = \dfrac{\sum x}{n}$ .
Median — middle value of sorted data.
Mode — most common value.
Range — maximum minus minimum.
Modal class — the class interval with the highest frequency in grouped data.
Estimate of the mean — for grouped data, computed using class midpoints.

How it’s examined

Statistics appears every Paper 2 (3-4 marks) and Paper 4 (5-7 marks). Cambridge typically combines a frequency-table mean with a comparison or interpretation question. Examiner reports flag two errors: (i) midpoints not used for grouped data, (ii) only one comparison statement (average without spread, or vice versa) when two are required.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E11.1-11.3); 0580/22 May/Jun 2024 — Q11 (mean from frequency table); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Methods of Analysing Data

Step-by-step solutions to past-paper-style questions on methods of analysing data, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Methods of Analysing Data

Almost every methods of analysing data exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the mean / median / mode / range — from a list, a frequency table, or grouped data, with one formula to apply.

How to approach it

Sort before reading a median; use $\bar{x} = \tfrac{\sum fx}{\sum f}$ for a frequency table, taking class midpoints as $x$ for grouped data; identify the modal class as the one with the highest frequency.

Common trap

Examiner reports flag dividing $\sum fx$ by the number of rows instead of $\sum f$ , using a class boundary instead of the midpoint, and forgetting to call a grouped-data mean 'estimated'.

Multi-step problem

Recognise it by

The mean is given and a missing value is wanted — the average formula must be rearranged and combined with the other data.

How to approach it

Multiply the mean by $n$ to recover the total $\sum x$ , subtract the sum of the known values, and the remainder is the unknown.

Common trap

Examiner reports flag dividing the recovered total by the wrong count — the unknown value still contributes to $\sum x$ , so use the full $n$ .

Word problem

Recognise it by

A real-world context gives summary statistics for two data sets and asks you to compare them — no calculation is set out for you.

How to approach it

Make one comparison of centre (mean) and one of spread (range), and phrase both explicitly in the context of the variable being measured.

Common trap

Examiner reports flag bare statements like 'B is higher' with no reference to the mean, the range, or the variable, and giving only one comparison when two are required.

1Mean, median, mode and range
CoreDirect calculation• averages
Question
Find the mean, median, mode and range of: $4, 6, 7, 7, 9, 12$ .
Step-by-step solution
1. Step 1
  Mean.
  $\bar{x} = \dfrac{4+6+7+7+9+12}{6} = \dfrac{45}{6} = 7.5$
2. Step 2
  Median (already sorted, $n = 6$ → average of middle two).
  $\text{Median} = \dfrac{7 + 7}{2} = 7$
3. Step 3
  Mode (most frequent).
  $\text{Mode} = 7$
4. Step 4
  Range.
  $\text{Range} = 12 - 4 = 8$
Answer
Mean $7.5$ , median $7$ , mode $7$ , range $8$ .
2Mean from a frequency table
ExtendedDirect calculation• frequency table
Question
Number of pets per household: $0$ (frequency $5$ ), $1$ (frequency $8$ ), $2$ (frequency $4$ ), $3$ (frequency $3$ ). Find the mean.
Step-by-step solution
1. Step 1
  $\bar{x} = \dfrac{\sum fx}{\sum f}$ .
  $\bar{x} = \dfrac{0(5) + 1(8) + 2(4) + 3(3)}{5+8+4+3} = \dfrac{0+8+8+9}{20} = \dfrac{25}{20}$
2. Step 2
  Compute.
  $= 1.25$
Answer
$1.25$ pets
3Estimated mean (grouped data)
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q10• grouped
Question
Times taken (minutes): $[10, 20)$ freq $4$ ; $[20, 30)$ freq $9$ ; $[30, 40)$ freq $7$ . Estimate the mean.
Step-by-step solution
1. Step 1
  Use class midpoints.
  $x = 15, 25, 35$
2. Step 2
  $\bar{x} = \dfrac{\sum fx}{\sum f}$ .
  $\bar{x} = \dfrac{4(15) + 9(25) + 7(35)}{20} = \dfrac{60 + 225 + 245}{20}$
3. Step 3
  Compute.
  $= \dfrac{530}{20} = 26.5$
Answer
Estimated mean $\approx 26.5$ minutes.
Examiner tip
Always say 'ESTIMATED' mean for grouped data — exact values aren't known.
4Modal class
ExtendedDirect calculation• modal class
Question
Using the same grouped data, identify the modal class.
Step-by-step solution
1. Step 1
  Highest frequency among the classes is $9$ .
  $\text{Modal class} = [20, 30)$
Answer
$[20, 30)$
5Mean of a simple list
CoreDirect calculation• mean, list
Question
Find the mean of $3,\ 8,\ 11,\ 6,\ 12,\ 5,\ 4$ .
Step-by-step solution
1. Step 1
  Add the values.
  $3 + 8 + 11 + 6 + 12 + 5 + 4 = 49$
2. Step 2
  Divide by count ( $n = 7$ ).
  $\bar{x} = \dfrac{49}{7} = 7$
Answer
$\bar{x} = 7$
6Median + mode + range from a list
CoreDirect calculation• median, mode, range
Question
Find the median, mode and range of $2,\ 5,\ 7,\ 5,\ 9,\ 2,\ 5,\ 8$ .
Step-by-step solution
1. Step 1
  Sort the data: $2, 2, 5, 5, 5, 7, 8, 9$ .
2. Step 2
  With $n = 8$ , median is the mean of the 4th and 5th values: $5$ and $5$ .
  $\text{Median} = \dfrac{5+5}{2} = 5$
3. Step 3
  Mode = most frequent value = $5$ (appears $3$ times).
4. Step 4
  Range $= 9 - 2 = 7$ .
Answer
Median $= 5$ , mode $= 5$ , range $= 7$ .
7Mean from an ungrouped frequency table
ExtendedDirect calculation• Adapted from 0580/42 Feb/Mar 2024 Q8• frequency table, mean
Question
Number of children per family: $0$ (freq $4$ ), $1$ (freq $9$ ), $2$ (freq $12$ ), $3$ (freq $5$ ), $4$ (freq $2$ ), $5$ (freq $1$ ). Find the mean number of children per family.
Step-by-step solution
1. Step 1
  Compute $\sum fx$ .
  $\sum fx = 0(4) + 1(9) + 2(12) + 3(5) + 4(2) + 5(1) = 0 + 9 + 24 + 15 + 8 + 5 = 61$
2. Step 2
  Compute $\sum f$ .
  $\sum f = 4 + 9 + 12 + 5 + 2 + 1 = 33$
3. Step 3
  Mean.
  $\bar{x} = \dfrac{61}{33} \approx 1.85$
Answer
$\bar{x} \approx 1.85$ children per family.
8Estimated mean from grouped data (extra context)
ExtendedDirect calculation• grouped, midpoint
Question
Heights (cm): $[140, 150) = 6$ ; $[150, 160) = 15$ ; $[160, 170) = 18$ ; $[170, 180) = 8$ ; $[180, 190) = 3$ . Estimate the mean height.
Step-by-step solution
1. Step 1
  Midpoints: $145, 155, 165, 175, 185$ .
2. Step 2
  Compute $\sum fx$ .
  $\sum fx = 6(145) + 15(155) + 18(165) + 8(175) + 3(185)$
3. Step 3
  Evaluate.
  $= 870 + 2325 + 2970 + 1400 + 555 = 8120$
4. Step 4
  Total $\sum f = 50$ .
  $\bar{x} \approx \dfrac{8120}{50} = 162.4$
Answer
Estimated mean $\approx 162.4\,\text{cm}$ .
9Find an unknown given the mean (Challenge)
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q9• unknown, challenge
Question
Six numbers have a mean of $12$ . Five of them are $8,\ 10,\ 14,\ 9,\ 17$ . Find the sixth number.
Step-by-step solution
1. Step 1
  Use $\bar{x} = \dfrac{\sum x}{n}$ . Rearrange.
  $\sum x = \bar{x} \times n = 12 \times 6 = 72$
2. Step 2
  Sum the five known values.
  $8 + 10 + 14 + 9 + 17 = 58$
3. Step 3
  The sixth value $= 72 - 58 = 14$ .
Answer
The sixth number is $14$ .
Examiner tip
The examiner report flags candidates who divide $72$ by $5$ — forgetting that the sixth value contributes to the total. Multiply mean by $n$ first.
10Compare two data sets using mean and spread (Challenge)
ChallengeWord problem• compare, challenge
Question
Class A's test scores have mean $58$ and range $24$ . Class B's scores have mean $62$ and range $42$ . Make TWO comparisons in context.
Step-by-step solution
1. Step 1
  Comparison 1 (centre): Class B has a higher mean, so on average Class B performed better.
2. Step 2
  Comparison 2 (spread): Class A has a smaller range, so Class A's scores are more consistent / less spread out.
Answer
Class B has a higher mean (higher average score), but Class A is more consistent (smaller range).
Examiner tip
Comparison questions need BOTH a centre comparison AND a spread comparison, ALWAYS in context. The examiner report consistently flags bare statements such as 'B is higher' with no reference to mean, range or the variable.

Key Formulae — Methods of Analysing Data

The formulae you need to memorise for methods of analysing data on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Mean of a list
$\bar{x} = \dfrac{\sum x}{n}$
$n$
number of values
When to use
Raw lists of data.
Mean from a frequency table
$\bar{x} = \dfrac{\sum fx}{\sum f}$
$f$
frequency of each value
$x$
value (or class midpoint)
When to use
Frequency tables and (using midpoints) grouped data.
Median position
$\text{Position} = \dfrac{n + 1}{2}$
When to use
Find which (sorted) value is the median. If position is non-integer, average the two surrounding values.

Key Definitions and Keywords — Methods of Analysing Data

Definitions to memorise and the exact keywords mark schemes credit for methods of analysing data answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Mean
Examiner keyword
Sum of values divided by the number of values; the arithmetic average.
Median
Examiner keyword
The middle value once data is sorted. Average the two middle values when $n$ is even.
Mode
Examiner keyword
The most frequent value. Multiple modes are possible; if all unique, no mode.
Range
Examiner keyword
Highest value minus lowest value.
Modal class
Examiner keyword
In grouped data, the class with the highest frequency.

Common Mistakes and Misconceptions — Methods of Analysing Data

The traps other students keep falling into on methods of analysing data questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to sort before finding the median
Why it happens
Speed.
How to avoid it
Always sort the list ascending before reading the middle.
✕Computing $\bar{x} = \sum fx / n$ instead of $\sum fx / \sum f$
0580/42 — recurring
Why it happens
Confusing $n$ (total values) with the count of distinct values.
How to avoid it
Denominator is $\sum f$ — total frequency — not the number of rows.
✕Using class boundaries instead of midpoints in grouped data
Why it happens
Picking the lower bound as $x$ .
How to avoid it
Class $[10, 20)$ has midpoint $\dfrac{10 + 20}{2} = 15$ .
✕Calling the grouped-data mean 'exact'
Why it happens
Forgetting the original values are lost.
How to avoid it
Always say ESTIMATED mean.