Measures of Central Tendency
Five people in a room have incomes of $40,000; $45,000; $50,000; $55,000; and $1,000,000. The mean is $238,000 — a figure that suggests everyone is wealthy, when only one person is. The median of $50,000 tells a very different and more accurate story. Knowing when to use mean, median or mode is one of the most important skills in statistical literacy.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
House prices in a suburb: $600K, $620K, $650K, $680K, $700K, $750K, $2,500K.
Before reading on — estimate the mean and median. Which better represents what a typical home costs? Write your gut feeling.
Three measures summarise where the centre of a data set lies. Each tells a different story.
Mean $\bar{x} = \dfrac{\sum x}{n}$ — the arithmetic average. Affected by outliers.
Median — the middle value when data is ordered. Not affected by outliers. Better for skewed data.
Mode — the most frequently occurring value. The only measure that works for categorical data.
Mean from frequency table: $\bar{x} = \dfrac{\sum f \times x}{\sum f}$
Key facts
- Definitions of mean, median, mode
- Frequency table calculations
- Effect of outliers on each measure
Concepts
- Why different measures suit different data
- How outliers distort the mean
- When each measure is most appropriate
Skills
- Calculate all three measures
- Find mean from frequency tables
- Choose the best measure for a context
The mean is the sum of all values divided by the number of values:
$$\bar{x} = \frac{\sum x}{n}$$Example: Test scores: 72, 85, 90, 65, 88
$$\bar{x} = \frac{72 + 85 + 90 + 65 + 88}{5} = \frac{400}{5} = 80$$
From a frequency table use $\bar{x} = \dfrac{\sum f \times x}{\sum f}$:
| Score ($x$) | Frequency ($f$) | $f \times x$ |
|---|---|---|
| 70 | 3 | 210 |
| 80 | 5 | 400 |
| 90 | 2 | 180 |
| Total | 10 | 790 |
$$\bar{x} = \frac{790}{10} = 79$$
What to write in your book
- Mean = $\sum x \div n$ for raw data.
- Mean from frequency table = $\sum fx \div \sum f$.
- Outliers pull the mean toward them — use median when outliers are present.
Quick check: Scores are 4, 5, 5, 6, 6, 6, 7. What is the mean?
Median — the middle value when data is ordered from smallest to largest.
Odd number of values: the median is the exact middle value.
Example: 12, 15, 18, 22, 25 → Median = 18
Even number of values: the median is the average of the two middle values.
Example: 12, 15, 18, 22 → Median = $\dfrac{15 + 18}{2} = 16.5$
Mode — the value that appears most frequently.
Example: 3, 5, 5, 7, 8, 5, 9 → Mode = 5
A data set can be:
- Unimodal: one mode (most common).
- Bimodal: two modes.
- Multimodal: more than two modes.
- No mode: all values appear equally often.
The mode is the only measure of centre that applies to categorical data.
What to write in your book
- Median: order the data, find the middle. For even $n$: average the two middle values.
- Mode: most frequent value. Only measure that works for categorical data.
- Median is resistant to outliers; mean is not.
True or false: The median of the data set 10, 12, 14, 16 is 13.
Worked examples · reveal each step
Daily sales ($): 450, 520, 480, 600, 500, 2000, 550, 510, 490, 530. Find the mean, median and mode. Which best represents typical daily sales?
Test marks: 60 scored by 4 students, 70 by 8, 75 by 10, 80 by 6, 90 by 2. Calculate the mean mark.
An outlier is a value far from the rest of the data. Outliers pull the mean toward them but do not affect the median.
Example — Salaries: $40K, $45K, $50K, $55K, $200K
Mean = $\dfrac{40+45+50+55+200}{5} = \$78K$ (misleading). Median = $50K (representative).
Skewed distributions:
- Right-skewed (positive skew): Long tail to the right. Mean > Median. Example: income distribution, house prices.
- Left-skewed (negative skew): Long tail to the left. Mean < Median.
- Symmetric: Mean ≈ Median ≈ Mode. Bell-shaped curves.
What to write in your book
- Right-skewed: mean > median (tail goes right).
- Left-skewed: mean < median (tail goes left).
- Symmetric: mean ≈ median ≈ mode.
- Use median for skewed data or when outliers are present.
Fill the gap: In a right-skewed distribution, the mean is the median.
Common errors · traps that cost marks
What to write in your book
- Always order data before finding the median.
- Even $n$: median = average of the two middle values.
- Outlier present? Use median, not mean.
Match each situation to the best measure of centre:
Quick-fire practice · 2 activities
Find mean, median and mode for: 5, 7, 8, 8, 10, 12, 15. Also find mean from the frequency table: score 4 (freq 2), score 5 (freq 5), score 6 (freq 3).
A company reports a mean salary of $120K. (a) Why might the median be more informative for job seekers? (b) Give an example where mode is the most useful measure. (c) Explain why "average house price" in the news usually means median, not mean.
Top 3 list: Give THREE real-world contexts where choosing the wrong measure of centre (mean vs median vs mode) would lead to misleading conclusions. For each, state which measure was misused and why.
Mean $= (600+620+650+680+700+750+2500) \div 7 = 6500 \div 7 \approx \$928.6K$. Median $= \$680K$ (the 4th of 7 values). The mean is heavily inflated by the $2.5M outlier and suggests the suburb is far more expensive than it really is for most buyers. The median of $680K is the better measure — half the houses are cheaper and half are more expensive, giving a fair picture of what a typical house costs.
What changed in your understanding? What did you estimate correctly? What surprised you?
Pick your answer, then rate your confidence — that tells the system what to drill next.
Q1. The data set is: 14, 18, 22, 22, 25, 30, 35, 80. What is the median?
Q2. A frequency table shows: score 4 (freq 2), score 5 (freq 5), score 6 (freq 3). What is the mean score?
Q3. A company CEO earns $2,000,000 and 99 employees earn $50,000 each. The mean salary is approximately:
Q4. In a right-skewed distribution, which statement is correct?
Q5. Which measure of centre is most appropriate when reporting the most popular dress size stocked by a retailer?
SA 1. (a) Calculate mean, median and mode for: 14, 18, 22, 22, 25, 30, 35, 80. (b) Which measure best represents this data? (c) Calculate the mean after removing the outlier. How much does it change? (2 marks)
SA 2. A frequency table shows test marks: 60 (4 students), 70 (8), 75 (10), 80 (6), 90 (2). (a) Calculate the mean mark. (b) Find the median mark. (c) Why might the median be a fairer measure to report to parents? (2 marks)
SA 3. A company's CEO earns $2M, while 99 employees earn $50K each. (a) Calculate the mean and median salary. (b) Which would the CEO quote in a press release? Which would the union quote? Explain. (c) A politician claims "average wages have risen 5%" using the mean, but the median rose only 1%. Explain how this discrepancy can occur and what it reveals about income distribution. (3 marks)
Comprehensive answers (click to reveal)
MC 1 — C: Ordered data: 14, 18, 22, 22, 25, 30, 35, 80. $n = 8$ (even): median $= (22+25) \div 2 = 23.5$.
MC 2 — B: $\sum fx = 4(2)+5(5)+6(3) = 8+25+18 = 51$. $\sum f = 10$. Mean $= 51 \div 10 = 5.1$.
MC 3 — A: Mean $= (2\,000\,000 + 99 \times 50\,000) \div 100 = (2\,000\,000 + 4\,950\,000) \div 100 = \$69\,500$.
MC 4 — D: In a right-skewed distribution the long tail pulls the mean to the right of the median: mean > median.
MC 5 — C: Dress size is categorical. Mode (most frequent value) is the appropriate measure.
SA 1 (2 marks): (a) Mean $= 266 \div 8 = 33.25$; Median $= 23.5$; Mode $= 22$ [0.5]. (b) Median (23.5) or mode (22); mean is inflated by outlier 80 [0.5]. (c) Without 80: mean $= 186 \div 7 \approx 26.57$; change $= 33.25 - 26.57 = 6.68$ [1].
SA 2 (2 marks): (a) $\sum fx = 240+560+750+480+180 = 2210$; $\bar{x} = 2210 \div 30 \approx 73.67$ [0.5]. (b) Median position $= 15.5$th value; cumulative frequencies: 4, 12, 22 — 15.5th falls in 75 group; Median $= 75$ [0.5]. (c) Mean is pulled down by the 4 students scoring 60; median shows that half the class scored 75 or above [1].
SA 3 (3 marks): (a) Mean $= \$69\,500$; Median $= \$50\,000$ [1]. (b) CEO quotes mean ($69.5K) to appear generous; union quotes median ($50K) to show most workers earn much less [1]. (c) High earners received large raises, pulling mean up, while typical workers saw minimal change; reveals right-skewed income distribution where gains concentrate at the top [1].
Drill 1: Mean $= 65 \div 7 \approx 9.29$; Median $= 8$; Mode $= 8$. Frequency table mean $= 51 \div 10 = 5.1$.
Drill 2: (a) CEO's high salary inflates mean; median better represents what most employees earn. (b) Most popular clothing/shoe size. (c) House prices are right-skewed by mansions; median is not distorted by these outliers.
Five timed questions on mean, median, mode, outliers and skewed distributions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering questions on central tendency. Pool: lesson 2.
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