Averages & Spread
SPEC S4 • STATISTICS
Mean, median, mode, range — and the estimated mean from grouped data. A high-priority topic on every paper.
Why We Need Averages
An "average" is a single number that represents a typical value in a dataset. There are three common averages and one measure of spread — each tells a different story.
The four measures
Mean — the arithmetic average (sum ÷ count).
Median — the middle value when data is in order.
Mode — the most frequent value.
Range — a measure of spread (not an average): largest − smallest.
MCE approach
Method — pick the right measure for the data and the question. Calculate — order the data, sum, divide; use midpoints for grouped data. Evaluate — check the answer is reasonable and state the units (and say "estimated" if grouped).
Four Measures on One Dataset
Take the data: 3, 5, 4, 7, 5, 4, 9, 11, 5, 7. Ordered: 3, 4, 4, 5, 5, 5, 7, 7, 9, 11.
MeasureValueHow
Mean660 ÷ 10 = 6
Median5Average of 5th and 6th = (5 + 5) ÷ 2
Mode5Appears three times — most frequent
Range811 − 3
The Mean — and Why Outliers Matter
Mean
mean = (sum of values) ÷ (number of values)
The mean uses every data value, which is usually a strength — but it is sensitive to outliers (values much larger or smaller than the rest). A single extreme value can drag the mean far from where the data actually sits.
Set A: 4, 5, 6, 7, 8
Mean 6, Median 6 — they agree.
Set B: 4, 5, 6, 7, 28
Mean 10, Median still 6 — the mean has been pulled up by the outlier.
Key takeaway
Replacing a single value (8 → 28) shifted the mean from 6 to 10 but left the median at 6. That's why the median is a "robust" measure when outliers are present.
Median & Mode
Median
The middle value when the data is listed in order.
If n is odd, the median is the single middle value — position (n + 1) ÷ 2.
If n is even, the median is the mean of the two middle values — positions n ÷ 2 and n ÷ 2 + 1.
Quick examples
{3, 5, 7, 8, 9} → odd, n = 5, middle position = 3, median = 7.
{3, 5, 7, 8} → even, n = 4, middle positions = 2nd and 3rd, median = (5 + 7) ÷ 2 = 6.
Mode
The most frequent value. A dataset can have:
One mode (e.g. {3, 5, 5, 7, 9} → mode = 5)
Multiple modes (e.g. {3, 4, 4, 7, 7} → modes are 4 and 7)
No mode if every value appears equally often (e.g. {2, 3, 4, 5})
Mode is the only average that works on categorical data
You cannot find the mean or median of {red, red, blue, green, blue, blue}, but you can say the mode is blue.
Worked Example — All Four From a List01
Ten students scored the following marks out of 12 on a quiz: 3, 5, 4, 7, 5, 4, 9, 11, 5, 7. Find the mean, median, mode and range. [4 marks]
MMethod. Order the data first. Mean = sum ÷ 10. Median = average of the 5th and 6th values (n = 10 is even). Mode = most frequent value. Range = largest − smallest.
Calculate.
Ordered: 3, 4, 4, 5, 5, 5, 7, 7, 9, 11.
•
Mean = (3 + 4 + 4 + 5 + 5 + 5 + 7 + 7 + 9 + 11) ÷ 10 = 60 ÷ 10 = 6 .
•
Median = (5th + 6th) ÷ 2 = (5 + 5) ÷ 2 = 5 .
•
Mode = 5 (appears three times).
•
Range = 11 − 3 = 8 .
EEvaluate. Median and mode coincide at 5; the mean is slightly higher (6) because of the larger values (9, 11) pulling it up — a good sense check.
Choosing the Right Average
Different averages suit different situations. Knowing which one to use (and why) is often a 1-mark comparison question.
UseWhenWhy
MeanData is roughly symmetric with no extreme outliersUses every value — most informative
MedianData has outliers or is skewed (e.g. incomes, house prices)Robust — unaffected by extreme values
ModeData is categorical (colours, brands), or you want the "most common"The only average that works for non-numerical data
"Advantage" and "disadvantage" questions
If a question asks "what is one advantage of using the median here?" the answer usually involves outliers: "The median is not affected by the outlier of 28, so it represents the typical value better than the mean."
Mean From an Ungrouped Frequency Table
When data appears in a frequency table (with specific numerical values, not class intervals), don't list out every value — multiply each value by its frequency.
Formula
mean = Σ(f × x) ÷ Σfwhere x is each value, f is its frequency, and Σ means "sum of".
Three-step routine
Add a new column for f × x (value times its frequency).
Total the f column (that's Σf — the number of data values).
Total the f × x column, then divide by Σf.
Don't divide by the number of rows
A classic error: dividing Σ(fx) by the number of rows in the table instead of by Σf. If the table has 5 rows but the total frequency is 25, you divide by 25, not 5.
02Worked Example — Ungrouped Frequency Table
The table shows the number of goals scored by a football team in 25 matches. Work out the mean number of goals per match. [3 marks]
Goals (x)Frequency (f)
04
16
28
35
42
MMethod. Add an f × x column; sum both the f column and the f × x column; divide.
Calculate.
Mean = 45 ÷ 25 = 1.8 goals per match.
EEvaluate. 1.8 is between 0 and 4, and closer to 2 (the modal value) — the answer is reasonable.
Grouped Data — Why "Estimated" Mean?
When data is presented in class intervals (like 0 ≤ t < 5 or 20 < h ≤ 30), the exact values inside each class are unknown. You can only approximate.
Midpoint rule
Use the midpoint of each class as the representative value. The midpoint of a ≤ x < b is (a + b) ÷ 2.
Examples of midpoints
0 ≤ t < 10 → midpoint = (0 + 10) ÷ 2 = 5
10 ≤ t < 20 → midpoint = 15
20 ≤ t < 30 → midpoint = 25
Why it's an estimate
A class like 0 ≤ t < 10 could contain any mix of values between 0 and 10. Using the midpoint (5) assumes they're spread evenly — which they may not be. That's why the answer is an estimate of the true mean, not the true mean itself.
The Estimated Mean Formula
Estimated mean
estimated mean = Σ(f × midpoint) ÷ Σf
Four-step routine
Work out the midpoint of each class.
Multiply each midpoint by its frequency (f × midpoint).
Total the f column (Σf) and the f × midpoint column.
Divide: Σ(f × midpoint) ÷ Σf.
Always write "estimated"
On grouped data, the exact mean is unknowable — so the word "estimated" must appear in your answer. Writing "the mean is 10.25 min" can lose a communication mark; "the estimated mean is 10.25 min" is what examiners want to see.
03Worked Example — Estimated Mean (Grouped)
The table shows the time taken by 40 students to solve a puzzle. Work out an estimate for the mean time. [4 marks]
Time t (min)Frequency
0 ≤ t < 56
5 ≤ t < 1014
10 ≤ t < 1512
15 ≤ t < 208
Total40
MMethod. Use midpoints 2.5, 7.5, 12.5, 17.5. Compute Σ(f × midpoint) ÷ Σf. Σf = 40.
Calculate.
Estimated mean = 410 ÷ 40 = 10.25 min.
EEvaluate. 10.25 min lies within the data range (0 to 20 min) and near the centre of the distribution. Note it sits just above the modal class 5 ≤ t < 10 — which makes sense because the longer-time classes (10–20 min) pull the mean slightly higher. Estimated mean = 10.25 minutes.
Modal Class & Median Class
For grouped data you cannot give a single mode or median value — but you can identify the class they live in.
Modal class
The class interval with the highest frequency.
For the WE3 table, the modal class is 5 ≤ t < 10 (frequency 14, highest).
Median class
The class containing the middle value (the (n ÷ 2)th value for grouped data).
For n = 40, the median lies at position 20. Running totals: 6, 20, 32, 40 — so the 20th value sits in 5 ≤ t < 10.
Class, not value
The modal class is the whole interval (e.g. "5 ≤ t < 10"), not the midpoint (7.5) or the frequency (14). Writing just "7.5" or "14" loses marks.
Outliers
Outlier
A data value that is much larger or much smaller than the rest of the data. Outliers can arise from measurement errors, unusual events, or genuinely extreme cases.
Effect on the four measures
Mean — heavily affected. One extreme value can pull it a long way.
Median — barely affected. Only the position of the middle matters, not the extremes.
Mode — not affected at all (unless the outlier is repeated).
Range — heavily affected. The outlier becomes the max or min, inflating the range.
What to write in a "comment" question
If asked to compare averages or comment on which is "most appropriate": "There is an outlier at 28, so the median (6) represents the typical value better than the mean (10)."
Comparing Two Distributions
When a question asks you to compare two datasets, you always need two statements: one about an average, and one about the spread. One of each is required for full marks — both statements must refer to the context.
The two-sentence template
Average: "On average, Class A scored higher than Class B because the [median / mean] for A was [x] but for B was [y]."
Spread: "Class A's scores were more consistent / spread out than Class B's because the range for A was [x] but for B was [y]."
At Higher tier, the interquartile range (IQR) is often used instead of the range — it's introduced in Lesson 6.03.
"More consistent" means smaller spread
A smaller range or IQR means the values cluster more tightly — which is more consistent. A common error: writing that a bigger range means more consistent. That's the opposite of what it means.
Context is non-negotiable
Saying "A has a bigger median" does not earn marks. You must say what the data is — "Class A scored higher on average" or "the heights of sunflowers in plot A were more varied". Numerical comparison alone is incomplete.
Examiner Alerts
Alert 1 — Class boundary instead of midpoint
Using 0, 10, 20, 30 (the class bounds) instead of midpoints 5, 15, 25, 35 is the single most common grouped-mean error. Always use the midpoint — half-way between the two bounds.
Alert 2 — Dividing by the number of classes
After calculating Σ(f × midpoint), students sometimes divide by the number of rows in the table (often 3, 4, or 5) instead of by the total frequency Σf (e.g. 30 or 40). Divide by the total frequency.
Alert 3 — Forgetting "estimated"
On grouped data, always state "estimated mean". Omitting the word "estimated" can lose a communication mark because the answer isn't exact — it relies on midpoint approximations.
Alert 4 — Modal/median class as a number
The modal class is the interval (e.g. "5 ≤ t < 10"). It is not the midpoint (7.5) or the frequency (14). Report the interval.
Practice Question
A farmer recorded the ages of 50 sheep. The grouped frequency table is shown below.
Age a (years)Frequency
0 ≤ a < 212
2 ≤ a < 418
4 ≤ a < 614
6 ≤ a < 86
Total50
(a) Write down the modal class. [1 mark]
(b) Work out an estimate for the mean age of the sheep. [3 marks]
(c) Explain why your answer to part (b) is an estimate rather than the exact mean. [1 mark]
MMethod. (a) Highest frequency = modal class. (b) midpoints × frequencies, sum, divide by 50. (c) grouped data hides the exact values.
Calculate.
(a) Highest frequency is 18, so modal class = 2 ≤ a < 4.
(b) Midpoints: 1, 3, 5, 7. Products: 12 × 1 = 12, 18 × 3 = 54, 14 × 5 = 70, 6 × 7 = 42. Σ = 12 + 54 + 70 + 42 = 178. Σf = 50.
Estimated mean = 178 ÷ 50 = 3.56 years.
(c) The exact age of each sheep is not known — only the class it falls in. Using midpoints assumes the ages are evenly distributed within each class, so the result is an estimate, not the true mean.
EEvaluate. 3.56 years sits inside the modal class 2 ≤ a < 4 — consistent. Answer has units (years) and is labelled "estimate".
Summary — MCE Checkpoint
Key formulas
mean = (sum of values) ÷ (number of values)
mean from frequency table = Σfx ÷ Σf
estimated mean (grouped) = Σ(f × midpoint) ÷ Σf
The four measures
Mean — sum ÷ count. Uses every value; sensitive to outliers.
Median — middle value when ordered. Robust to outliers.
Mode — most frequent value. Works on categorical data.
Range — largest − smallest. A measure of spread , not an average.
Examiner traps
Use midpoints , not class boundaries.
Divide by total frequency , not number of classes.
Always say "estimated" on grouped-data answers.
Modal class is the interval , not its midpoint or frequency.
"More consistent" = smaller spread (range, or IQR at Higher).
Comparisons need both an average and a spread, with context.
✓ You should now be able to
Find the mean, median, mode and range from a list of values.
Choose the most appropriate average for a given dataset.
Calculate the mean from an ungrouped frequency table using Σfx ÷ Σf.
Calculate the estimated mean from a grouped table using midpoints.
Identify the modal class and median class for grouped data.
Explain the effect of outliers on each measure.
Compare two distributions using an average and a measure of spread, with context.
Next: Why We Need Averages