One average cannot describe every important feature of a data set. Mean, median and mode answer different questions about centre. Range and other measures of spread tell us how variable the values are. A useful comparison therefore needs both a typical value and an account of variation.
This thirty-first Secondary 4 Mathematics Learning Guide develops mean, median, mode, range and data comparison as one statistical reasoning system. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
The guide focuses on choosing appropriate summary measures, handling frequency data, estimating grouped means, recognising the effect of outliers and writing supported comparisons between data sets.
Mean uses every value
The arithmetic mean is:
mean = total of values / number of values.
Because every observation contributes, the mean responds to changes anywhere in the data set. This makes it informative, but also sensitive to unusually large or small values.
Worked Example 1 | Mean from raw data
Find the mean of 8, 11, 12, 14, 15.
Total=60 and number of values=5.
Mean=60/5=12.
Median depends on ordered position
The median is the middle value after the data are placed in order. If there is an even number of observations, the median is the mean of the two central values.
Unlike the mean, the median depends mainly on position rather than the exact magnitude of extreme values.
Worked Example 2 | Median with even sample size
Find the median of 3, 5, 7, 11, 13, 20.
The middle two values are 7 and 11.
Median=(7+11)/2=9.
Mode identifies the most frequent value
The mode is the value occurring most often. A data set may have one mode, more than one mode, or no mode if every value occurs with the same frequency.
Mode is especially useful when the most common category or value matters more than arithmetic balance.
Worked Example 3 | Multiple modes
For 2, 3, 3, 5, 7, 7, 9, the values 3 and 7 each appear twice and all others once.
The data set is bimodal: 3 and 7.
Range measures total spread from minimum to maximum
Range=max−min. It is simple and quick, but it uses only the two extreme values and ignores how the remaining data are distributed.
Worked Example 4 | Same mean, different range
Compare A={8,9,10,11,12} and B={2,6,10,14,18}.
Both means are 10.
Range A=12−8=4.
Range B=18−2=16.
A supported comparison is: both groups have the same mean, but B is much more spread out by range.
Outliers can pull the mean without moving the median much
Consider 10,11,12,13,14. Mean=12 and median=12. Replace 14 by 54. The new mean becomes 20, while the median remains 12.
This shows why the median is often more resistant to extreme observations.
Worked Example 5 | Choose a representative measure
Five monthly repair bills are $42, $45, $47, $49 and $310. Which measure better represents a typical month: mean or median?
Mean=(42+45+47+49+310)/5=493/5=$98.60.
Median=$47.
The $310 bill is unusual and pulls the mean upward. If the question is about a typical ordinary month, the median gives a more resistant centre. If total long-run financial cost matters, the mean may still be important.
Frequency tables compress repeated values
For discrete values with frequencies, the mean is found using Σfx/Σf, where x is a value and f its frequency.
Worked Example 6 | Mean from a frequency table
| x | f | fx |
|---|---|---|
| 1 | 2 | 2 |
| 2 | 3 | 6 |
| 3 | 4 | 12 |
| 4 | 1 | 4 |
Σf=10 and Σfx=24.
Mean=24/10=2.4.
The mode is 3 because it has the highest frequency, 4.
Weighted mean gives different observations different influence
A weighted mean is appropriate when components contribute unequally. Assessment components with different percentages are a common example.
Worked Example 7 | Weighted assessment score
A course grade uses 30% coursework and 70% examination. A student scores 80 for coursework and 68 for the examination.
Weighted score=0.30(80)+0.70(68)=24+47.6=71.6.
The ordinary mean of 80 and 68 would be 74, but it would ignore the different weights.
Grouped data usually produce an estimated mean
When exact values inside each class interval are unknown, use the class midpoint as a representative value. Because every observation in the class is temporarily replaced by the midpoint, the resulting mean is an estimate.
Worked Example 8 | Estimated grouped mean
| Interval | Frequency | Midpoint |
|---|---|---|
| 0–10 | 4 | 5 |
| 10–20 | 7 | 15 |
| 20–30 | 5 | 25 |
Estimated total=4(5)+7(15)+5(25)=20+105+125=250.
Total frequency=16.
Estimated mean=250/16=15.625.
The answer should be described as estimated because the raw values were grouped.
Comparing two data sets needs centre and spread
A strong comparison usually says something about where the data are centred and how variable they are.
For example: “Group A has a higher median than Group B, but Group A also has a larger range.” This distinguishes typical performance from consistency.
Worked Example 9 | Centre versus consistency
Class X has median 72 and range 18. Class Y has median 68 and range 10.
Class X has the higher median, while Class Y is less spread out by range.
It would be unsupported to say “Class X is better in every way” or “every student in X scored higher”. The summaries do not prove those claims.
Changing every value by a constant
If the same number k is added to every data value, the mean, median and mode each increase by k. The range stays unchanged because both maximum and minimum increase by the same amount.
Worked Example 10 | Translation of a data set
A data set has mean 24 and range 13. Every value is increased by 7. Find the new mean and range.
New mean=31; new range=13.
Multiplying every value changes both centre and spread
If every value is multiplied by positive factor k, the mean, median and range are all multiplied by k. The mode values are also scaled by k.
This is useful when converting units, such as metres to centimetres.
Worked Example 11 | Unit conversion and summaries
Lengths measured in metres have mean 1.8 m and range 0.6 m. Express these summaries in centimetres.
Mean=180 cm; range=60 cm.
Missing value from a known mean
If the mean and number of observations are known, the total is mean×number of observations. This can recover a missing value.
Worked Example 12 | Recover a missing observation
Five numbers have mean 18. Four are 12,15,19 and 24. Find the fifth.
Total required=5×18=90. Known total=70.
Missing value=90−70=20.
Combined means require combined totals
You cannot usually average two group means directly unless the groups have equal sizes. Recover each group total first.
Worked Example 13 | Combined mean
Group A has 20 students with mean 64. Group B has 30 students with mean 70. Find the combined mean.
Total A=20×64=1280.
Total B=30×70=2100.
Combined total=3380, combined size=50.
Combined mean=3380/50=67.6.
Mean and median can reveal skew but not prove the full shape
If the mean is much higher than the median, unusually large values may be pulling the mean upward. If the mean is much lower, unusually small values may be influential.
But summary statistics alone do not show the entire distribution. Use diagrams such as dot plots, histograms or box plots when shape matters.
Connect to Histograms, Statistical Diagrams and Misleading Data and Cumulative Frequency, Box Plots and Standard Deviation for deeper representation and spread work.
Common failure modes
| Error | Cause | Repair |
|---|---|---|
| Finds median before ordering data | Position concept ignored | Sort first |
| Calls highest value the mode | Magnitude confused with frequency | Mode means most frequent |
| Uses ordinary average for weighted components | Weights ignored | Multiply each component by its weight |
| Calls grouped mean exact | Midpoint approximation forgotten | Label it estimated |
| Compares medians only | Spread omitted | Add range, IQR or standard deviation when available |
| Averages group means directly | Different group sizes ignored | Recover totals first |
Independent practice
- Find mean, median and range of 4,7,9,10,15.
- Find the mode of 2,2,4,5,5,5,8.
- Six numbers have mean 13. Five are 8,10,12,15,17. Find the sixth.
- A test is 40% project and 60% exam. Scores are 75 and 82. Find the weighted score.
- Group A has 15 students with mean 60. Group B has 25 students with mean 72. Find combined mean.
- A data set has mean 12 and range 8. Every value is multiplied by 3. Find the new mean and range.
Explained answers
1. Mean=45/5=9; median=9; range=15−4=11.
2. 5.
3. Total required=78; known total=62; missing value=16.
4. 0.4(75)+0.6(82)=30+49.2=79.2.
5. Total=15(60)+25(72)=2700; size=40; combined mean=67.5.
6. New mean=36; new range=24.
Teaching sequence: centre, spread, evidence
Begin with raw lists so the distinction among mean, median, mode and range is visible. Then introduce outliers and ask which summary changes most.
Move next to frequency and grouped data. Finish with comparative statements that require one centre measure and one spread measure, with claims limited to what the statistics actually support.
Final thought
A statistical summary is useful only when it answers the question being asked. Mean balances all values. Median locates the centre position. Mode finds the most common value. Spread tells us whether that centre is tightly or loosely representative.
Compare where the data sit, then compare how widely they move around that centre.
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