Central Tendency — Mean, Median, and Mode

🎯 Learning Objectives

• Compute the mean, median, and mode for raw data and frequency distributions
• Explain the properties and limitations of each measure
• Choose the appropriate measure of central tendency based on data type and distribution shape
• Identify skewness and understand how it affects the mean vs. median
• Recognize outliers and their disproportionate effect on the mean

📋 Prerequisites

• Data Types & Scales (01-data-types-scales) — knowing which measures are valid for which scales
• Categorical Data (02-categorical-frequency) — familiarity with frequency tables
• Basic arithmetic: addition, division, ordering

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📖 Core Content

4.1 Intuition: Finding the "Center" of Data

1. The average (mean): Add up all salaries and divide by 100
2. The middle (median): Sort all salaries and pick the one in the middle
3. The most common (mode): Find the salary that appears most frequently Each gives you a different kind of "typical" value. The interesting part is that they can tell very different stories — especially when the data is skewed (not symmetric).

Everyday analogy: At a party, people guess the "average" age. If there are 10 children (age 10), 10 adults (age 35), and 2 grandparents (age 70), the average is about 33, the median is between 10 and 35 (about 22), and the mode is 10 and 35 (two modes). Three different "typical" values, three different stories! 🔑 Key Insight: There's no single "correct" measure of central tendency. The right choice depends on your data and what you want to communicate.

4.2 The Mean (Arithmetic Average)

4.2.1 Intuition

4.2.2 Formal Definition

$$\mu = \frac{\sum_{i=1}^{N} x_i}{N}$$ $$\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$$

• $x_i$ = each individual value
• $N$ = population size
• $n$ = sample size
• $\sum$ = "sum of"

Notation alert: μ (mu) is a parameter (population). x̄ (x-bar) is a statistic (sample). The formula is the same — only the letter changes.

4.2.3 Worked Example

$$\bar{x} = \frac{70 + 85 + 90 + 65 + 80}{5} = \frac{390}{5} = 78$$

4.2.4 Properties of the Mean

4.2.5 The Weighted Mean

$$\bar{x}_w = \frac{\sum_{i=1}^{n} w_i x_i}{\sum_{i=1}^{n} w_i}$$ $$\bar{x}_w = \frac{0.40 \times 75 + 0.60 \times 85}{0.40 + 0.60} = \frac{30 + 51}{1} = 81$$

4.3 The Median

4.3.1 Intuition

4.3.2 Formal Definition

• Odd n: The median is the value at position $(n+1)/2$
• Even n: The median is the average of the two middle values (positions $n/2$ and $n/2 + 1$)

4.3.3 Worked Examples

• Sort: ✓ already sorted
• n = 5 (odd)
• Position = $(5+1)/2 = 3$
• 3rd value = 80
• Median = 80 Example B (even count): Scores: 65, 70, 80, 85, 90, 95
• Sort: ✓ already sorted
• n = 6 (even)
• Positions: $6/2 = 3$ and $6/2 + 1 = 4$
• 3rd value = 80, 4th value = 85
• Median = $(80 + 85)/2 = \textbf{82.5}$

4.3.4 Properties of the Median

4.4 The Mode

4.4.1 Intuition

4.4.2 Formal Definition

• No mode: All values appear once
• One mode (unimodal): One value appears most frequently
• Two modes (bimodal): Two values tie for highest frequency
• Multiple modes (multimodal): More than two values tie

4.4.3 Worked Example

4.4.4 Properties of the Mode

4.5 Comparing Mean, Median, and Mode

Relationship in Different Distribution Shapes

• The mean is pulled toward the long tail (toward extreme values)
• The median stays at the middle position
• The mode stays at the peak (most common value)

4.6 Which Measure to Use?

4.7 The Effect of Outliers

• Mean: (5 + 6 + 5.5 + 6.5 + 50) / 5 = 73 / 5 = 14.6
• Median: Sorted: 5, 5.5, 6, 6.5, 50 → 6
• Mode: 5 (appears once... actually all values are unique except) → No clear mode Interpretation:
• The mean says the "average" salary is ₹14.6 lakhs — but no one except the founder earns close to that!
• The median says the middle salary is ₹6 lakhs — which reflects what most employees earn
• The mode has no clear value If we remove the outlier (50):
• Mean becomes (5 + 6 + 5.5 + 6.5) / 4 = 5.75
• Median becomes (5.5 + 6) / 2 = 5.75 Without the outlier, mean and median are almost identical.

Moral: When outliers are present, the median is usually a better measure of "typical" than the mean.

4.8 Mean from a Frequency Table

$$\bar{x} = \frac{\sum_{i=1}^{k} f_i \times m_i}{n}$$

• $f_i$ = frequency of class i
• $m_i$ = midpoint of class i
• $n$ = total frequency
• $k$ = number of classes Example: Ages of participants at an event

$$\bar{x} \approx \frac{642.5}{25} = 25.7 \text{ years}$$

Note: This is an estimate because we've lost the exact values. The true mean of the raw data may differ slightly.

4.9 Worked Examples

Example 1: All Three Measures (Easy)

$$\bar{x} = \frac{55 + 60 + 65 + 70 + 75 + 80 + 85 + 90 + 95}{9} = \frac{675}{9} = 75$$

• Sorted (already): 55, 60, 65, 70, 75, 80, 85, 90, 95
• n = 9 (odd), position = (9+1)/2 = 5
• 5th value = 75 Mode: All values appear exactly once → No mode Result: Mean = Median = 75. This is symmetric data.

Example 2: Outlier Effect (Medium)

$$\bar{x} = \frac{0 + 55 + 60 + 65 + 70 + 75 + 80 + 85 + 90 + 95}{10} = \frac{675}{10} = 67.5$$

• Sorted: 0, 55, 60, 65, 70, 75, 80, 85, 90, 95
• n = 10 (even), positions = 5 and 6
• 5th = 70, 6th = 75
• Median = (70 + 75)/2 = 72.5 Comparison:

Example 3: Skewed Income Data (Harder)

20, 22, 25, 25, 28, 30, 32, 35, 38, 200 (CEO) a) Compute mean and median. b) Which better represents "typical" employee salary? c) How would you describe the skew? Solution: a) Mean:

$$\bar{x} = \frac{20 + 22 + 25 + 25 + 28 + 30 + 32 + 35 + 38 + 200}{10} = \frac{455}{10} = 45.5$$

• Sorted: 20, 22, 25, 25, 28, 30, 32, 35, 38, 200
• n = 10 (even), positions = 5 and 6
• 5th = 28, 6th = 30
• Median = (28 + 30)/2 = 29 So the median is ₹29 thousand. b) Which is more representative? The median (₹29K) is much more representative. Only the CEO (₹200K) earns more than ₹38K. The mean (₹45.5K) is inflated by the CEO's extreme salary and doesn't reflect what most employees earn. c) Skew: This is right-skewed (positive skew). Most values cluster at the low end (₹20K-₹38K), with one extreme value pulling a long tail to the right. We confirm: Mode (<25) < Median (29) < Mean (45.5).

4.10 Edge Cases & Gotchas

All Values Are the Same

Symmetric Bimodal Distribution

• Mean = (1 + 1 + 2 + 3 + 4 + 4) / 6 = 15/6 = 2.5
• Median: sorted 1, 1, 2, 3, 4, 4 → (2+3)/2 = 2.5
• Modes: 1 and 4 (bimodal) The mean and median don't reveal the bimodality — they suggest a normal distribution when the data actually has two clusters. Always visualize!

The Median for Ordinal Data

• Codes: 2, 5, 3, 4, 1, 4, 4, 3, 2, 5
• Sorted: 1, 2, 2, 3, 3, 4, 4, 4, 5, 5
• Median: positions 5,6 → (3+4)/2 = 3.5 → between Neutral and Agree The median response is between Neutral and Agree — a valid conclusion. The mean (3.3) would be invalid because we're assuming equal intervals.

4.11 Why This Matters

• Business: "What's the average revenue per customer?"
• Education: "What's the average test score?"
• Healthcare: "What's the average recovery time?"
• Policy: "What's the median income of a region?" They also form the foundation for almost every other statistical concept:
• Variance (next topic) uses the mean
• Correlation (Week 4) uses the mean
• Hypothesis testing (Stats 2) uses the mean
• Regression uses the mean Understanding which measure to use and why is the difference between good analysis and misleading analysis.

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📐 Key Formulas / Concepts

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⚠️ Common Pitfalls

Pitfall 1: Using the Mean for Skewed Data

Pitfall 2: Forgetting the Mode Is the Only Measure for Nominal Data

Pitfall 3: Confusing Mean and Median

Pitfall 4: Thinking the Mean Is Always a Value in the Dataset

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📝 Practice Questions

Q1: Basic Mean Calculation

Find the mean of: 12, 15, 20, 22, 25, 28, 30

<details> <strong>Solution</strong>

Step 1: Sum = 12 + 15 + 20 + 22 + 25 + 28 + 30 = 152 Step 2: n = 7 Step 3: Mean = 152 / 7 ≈ 21.71

$\boxed{\bar{x} \approx 21.71}$

</details> > **Q2: Median with Even Count** > > Find the median of: 3, 7, 8, 12, 15, 18, 20, 22 > > <details> <strong>Solution</strong> > > **Step 1:** Sort (already sorted). **Step 2:** n = 8 (even). **Step 3:** Position 1 = 8/2 = 4, Position 2 = 8/2 + 1 = 5. **Step 4:** 4th value = 12, 5th value = 15. **Step 5:** Median = (12 + 15) / 2 = 13.5 > > $\boxed{\text{Median} = 13.5}$ </details> > **Q3: Identifying Modes** > > Find the mode(s) for each dataset: > > a) 2, 3, 5, 3, 4, 3, 6, 3 b) 1, 2, 2, 3, 4, 4, 5, 6 c) 10, 20, 30, 40, 50 > > <details> <strong>Solution</strong> > > **a)** Value 3 appears 4 times, others appear once or twice. **Mode = 3** (unimodal) > > **b)** Values 2 and 4 each appear twice. **Modes = 2 and 4** (bimodal) > > **c)** All values appear exactly once. **No mode**. </details> > **Q4: Weighted Mean** > > A student's grades: Homework = 85 (weight 10%), Quizzes = 78 (weight 20%), Midterm = 92 (weight 30%), Final = 88 (weight 40%). Find the weighted mean. > > <details> <strong>Solution</strong> > > **Step 1:** Convert percentages to decimals. > > **Step 2:** $\bar{x}_w = \frac{0.10(85) + 0.20(78) + 0.30(92) + 0.40(88)}{0.10 + 0.20 + 0.30 + 0.40}$ > > **Step 3:** Numerator = 8.5 + 15.6 + 27.6 + 35.2 = 86.9 > > **Step 4:** Denominator = 1.0 > > **Step 5:** Weighted mean = 86.9 / 1.0 = 86.9 > > $\boxed{\bar{x}_w = 86.9}$ </details> > **Q5: Mean from Frequency Table** > >

Find the estimated mean.

<details> <strong>Solution</strong>

$\bar{x} \approx \frac{360}{20} = 18$

$\boxed{\bar{x} \approx 18}$

</details> > **Q6: Effect of Adding a Constant** > > If you add 5 to every value in a dataset, what happens to the mean, median, and mode? > > <details> <strong>Solution</strong> > > **All three increase by 5.** > > - **Mean:** If $\bar{x} = \frac{\sum x}{n}$, then new mean $= \frac{\sum (x + 5)}{n} = \frac{\sum x + 5n}{n} = \bar{x} + 5$ > - **Median:** Every value shifts right by 5, so the middle value also shifts by 5 > - **Mode:** The most frequent value each increases by 5 > > **General rule:** Adding a constant to all data shifts all measures of central tendency by that constant. </details> > **Q7: Multiplying by a Constant** > > If you multiply every value in a dataset by 3, what happens to the mean, median, and mode? > > <details> <strong>Solution</strong> > > **All three are multiplied by 3.** > > - **Mean:** $\frac{\sum 3x}{n} = 3 \times \frac{\sum x}{n} = 3\bar{x}$ > - **Median:** Every value triples, so the middle value triples > - **Mode:** The most frequent value triples > > **General rule:** Multiplying all data by a constant multiplies all measures of central tendency by that constant. </details> > **Q8: Choosing the Right Measure** > > For each scenario, recommend a measure of central tendency and explain why: > > a) Housing prices in a city (most homes ₹30-50 lakhs, some mansions ₹10+ crores) b) Favorite color of 500 respondents c) Daily website visitors over a month d) Cancer survival times (most patients die within 2 years, some live 20+ years) > > <details> <strong>Solution</strong> > > **a) Housing prices — Median** > > - Distribution is heavily right-skewed (a few mansions pull the mean high) > - Median better represents the "typical" home price > > **b) Favorite color — Mode** > > - Nominal data — mean and median are invalid > - Mode tells us the most popular color > > **c) Daily visitors — Mean** > > - Website traffic is typically roughly symmetric day-to-day > - Mean gives the most precise estimate using all data > - (Unless there are extreme spikes — then consider median) > > **d) Cancer survival — Median** > > - Heavily right-skewed (most patients die relatively quickly, a few live much longer) > - Mean would be misleadingly high > - Median better represents "typical" survival time </details> > **Q9: Outlier Impact** > > Dataset: 10, 12, 13, 15, 16, 18, 20, 22, 25, 100 > > a) Compute the mean and median. b) Identify the outlier. c) Remove the outlier and recompute. How much did each change? > > <details> <strong>Solution</strong> > > **a) With outlier:** > > - Sum = 10 + 12 + 13 + 15 + 16 + 18 + 20 + 22 + 25 + 100 = 251 > - Mean = 251/10 = 25.1 > - Sorted: 10, 12, 13, 15, 16, 18, 20, 22, 25, 100 > - Median = (16 + 18)/2 = 17 > > **b) Outlier:** 100 (far above the rest) > > **c) Without outlier (100 removed):** > > - Sum = 151 > - n = 9 > - Mean = 151/9 ≈ 16.78 > - Sorted: 10, 12, 13, 15, 16, 18, 20, 22, 25 > - Median = 16 > > **Changes:** > >

The mean was far more affected by the single outlier than the median.

</details> > **Q10: Symmetric Distribution** > > A symmetric dataset has mean = 50 and median = 50. If the largest value increases by 20, what happens to the mean and median? > > <details> <strong>Solution</strong> > > **Mean:** Increases (by 20/n). Adding 20 to one value increases the sum by 20, so the mean increases by 20/n. > > **Median:** Stays the same (assuming the increased value was already above the median and the change doesn't affect the middle position). > > **Example with n = 5:** > > Original: 40, 45, 50, 55, 60 > > - Mean = 250/5 = 50 > - Median = 50 > > New (largest increases by 20): 40, 45, 50, 55, 80 > > - Mean = 270/5 = 54 (↑ 4) > - Median = 50 (unchanged) > > $\boxed{\text{Mean increases; Median unchanged}}$ </details> * * * ## 🔗 Cross-References - **Next topic:** [Dispersion and Percentiles](/notes/01-foundation-bsma1002-stats-1-week03-04-dispersion-percentiles) — central tendency alone isn't enough; we also need to measure spread - **Previous:** [Data Types & Scales](/notes/01-foundation-bsma1002-stats-1-week01-01-data-types-scales) — understanding which measures are valid - **Week 4 (Correlation):** The mean is used in calculating Pearson's r - **Week 9-10 (Random Variables):** Expectation (E\[X\]) is the mean of a distribution - **BSMA1004 (Stats 2):** Hypothesis testing centers on comparing means [Join Discord](https://discord.gg/gE2m4Qrdqv) [Previous**Categorical Data**](/notes/01-foundation-bsma1002-stats-1-week02-02-categorical-frequency)[Next**Dispersion & Percentiles**](/notes/01-foundation-bsma1002-stats-1-week03-04-dispersion-percentiles)