Unit II · Lectures 9–10 of 40 · B.Sc. Data Science, 1st Year, Semester II
A measure of central tendency is a single value that attempts to describe a dataset by identifying the central or typical value within that set. It represents the centre point of the distribution — the value around which the data clusters. Also called a measure of location or an average.
Imagine placing all the data values as weights on a number line. The central tendency is the point where the number line balances perfectly — like the fulcrum of a seesaw. It is the one value that "stands for" the entire dataset.
Condenses thousands of observations into one representative value. E.g., "The average marks of the class is 72" is simpler than listing all 60 marks.
Enables comparison between groups. E.g., "Class A's average = 75, Class B's average = 68" — instantly tells which class performed better.
Used in business, policy, and research. E.g., "Average monthly sales = ₹5L" helps set targets and allocate resources.
Central tendency is the foundation for dispersion (standard deviation), skewness, correlation, and all advanced statistical measures.
We use averages daily without realising: average speed, average rainfall, average income, average temperature, batting average.
Government reports, census data, economic surveys — all use averages to summarise national-level data for citizens and policymakers.
Measures of central tendency (averages) are classified into two broad categories:
Based on all observations and derived through mathematical operations (addition, division). They consider the magnitude of every value.
Members:
• Arithmetic Mean (most common)
• Geometric Mean (for rates/ratios)
• Harmonic Mean (for speed/rates)
Based on the position of values when arranged in order. They do not depend on the actual magnitude of every value — only on where the central value falls.
Members:
• Median (middle value)
• Mode (most frequent value)
Click each average to explore its definition, formula, and a real-life example:
The arithmetic mean (or simply mean) is the sum of all observations divided by the total number of observations. It is the most commonly used measure of central tendency and is denoted by x̄ (x-bar) for a sample and μ (mu) for a population.
Imagine 5 friends have ₹10, ₹20, ₹30, ₹40, ₹50. They pool all the money and share it equally. Each person gets: (10+20+30+40+50)/5 = ₹150/5 = ₹30. The mean is the equal share each person would get.
The marks obtained by 8 students in a Statistics test are:
45, 52, 68, 71, 58, 63, 75, 49
Find the arithmetic mean.
Step 1: List all observations.
x₁=45, x₂=52, x₃=68, x₄=71, x₅=58, x₆=63, x₇=75, x₈=49
Step 2: Count the observations. n = 8.
Step 3: Sum all observations.
Σxᵢ = 45 + 52 + 68 + 71 + 58 + 63 + 75 + 49 = 481
Step 4: Divide by n.
x̄ = Σxᵢ / n = 481 / 8 = 60.125
| xᵢ | xᵢ − x̄ |
|---|---|
| 45 | 45 − 60.125 = −15.125 |
| 52 | 52 − 60.125 = −8.125 |
| 68 | 68 − 60.125 = +7.875 |
| 71 | 71 − 60.125 = +10.875 |
| 58 | 58 − 60.125 = −2.125 |
| 63 | 63 − 60.125 = +2.875 |
| 75 | 75 − 60.125 = +14.875 |
| 49 | 49 − 60.125 = −11.125 |
| Sum | Σ(xᵢ − x̄) = 0.000 ✓ |
The mean is where a number line would balance if data values were placed as weights. Click "Balance" to see it:
When data is grouped into class intervals, we do not know the exact values of individual observations. We assume that all observations in a class are concentrated at the class mark (midpoint) of that class. The mean is then calculated using the formula:
If 5 students scored in the range 40–50, we assume all 5 scored exactly 45 (the midpoint). Then the total marks of these 5 students = 5 × 45 = 225. Summing across all classes gives Σ(fᵢ · xᵢ). Dividing by N gives the mean.
Find the mean marks from the following frequency distribution:
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| f | 5 | 10 | 15 | 12 | 8 |
| Class | fᵢ | xᵢ (midpoint) | fᵢ · xᵢ |
|---|---|---|---|
| 0–10 | 5 | 5 | 25 |
| 10–20 | 10 | 15 | 150 |
| 20–30 | 15 | 25 | 375 |
| 30–40 | 12 | 35 | 420 |
| 40–50 | 8 | 45 | 360 |
| Total | Σfᵢ = 50 | Σfᵢxᵢ = 1330 |
The assumed mean method simplifies calculations when class marks have large values. Instead of multiplying large numbers, we choose a convenient assumed mean (A) — usually the class mark of the middle class — and calculate deviations dᵢ = xᵢ − A.
Choose A = 25 (midpoint of the middle class 20–30).
| Class | fᵢ | xᵢ | dᵢ = xᵢ − 25 | fᵢ · dᵢ |
|---|---|---|---|---|
| 0–10 | 5 | 5 | −20 | −100 |
| 10–20 | 10 | 15 | −10 | −100 |
| 20–30 | 15 | 25 | 0 | 0 |
| 30–40 | 12 | 35 | +10 | +120 |
| 40–50 | 8 | 45 | +20 | +160 |
| Total | 50 | Σfᵢdᵢ = 80 |
The step-deviation method goes one step further: it divides the deviations by the class width (h) to make the numbers even smaller. This is the most efficient method when class widths are equal.
A = 25, h = 10.
| Class | fᵢ | xᵢ | dᵢ = xᵢ − 25 | uᵢ = dᵢ / 10 | fᵢ · uᵢ |
|---|---|---|---|---|---|
| 0–10 | 5 | 5 | −20 | −2 | −10 |
| 10–20 | 10 | 15 | −10 | −1 | −10 |
| 20–30 | 15 | 25 | 0 | 0 | 0 |
| 30–40 | 12 | 35 | +10 | +1 | +12 |
| 40–50 | 8 | 45 | +20 | +2 | +16 |
| Total | 50 | Σfᵢuᵢ = 8 |
Click "Calculate" to watch the step-deviation method unfold row by row:
| Feature | Direct Method | Assumed Mean | Step-Deviation |
|---|---|---|---|
| Formula | x̄ = Σfx / Σf | x̄ = A + Σfd / Σf | x̄ = A + h × Σfu / Σf |
| Computation | Multiply f × x (large numbers) | Multiply f × d (smaller numbers) | Multiply f × u (smallest numbers) |
| Best for | Small values, few classes | Large class marks | Equal class widths, many classes |
| Complexity | Simple but tedious | Moderate | Simplest arithmetic |
| Result | Exact mean | Exact mean | Exact mean |
— Lecture 10 —
The weighted arithmetic mean is the mean calculated when different observations have different degrees of importance (called weights). Instead of treating all values equally, each value is multiplied by its weight before averaging. It is denoted by x̄w.
When computing your CGPA, a 4-credit subject has more weight than a 1-credit subject. A grade of A in a 4-credit course affects your CGPA four times more than an A in a 1-credit course. This is a weighted mean — the credits are the weights.
A student's marks and the credit weightage of each subject are given below. Find the weighted average marks.
| Subject | Marks (xᵢ) | Credits (wᵢ) |
|---|---|---|
| Statistics | 80 | 4 |
| Mathematics | 70 | 3 |
| Python | 90 | 3 |
| English | 60 | 2 |
| Economics | 75 | 2 |
Step 1: Multiply each mark by its weight:
80×4 = 320
70×3 = 210
90×3 = 270
60×2 = 120
75×2 = 150
Step 2: Sum of weighted marks:
Σ(wᵢ·xᵢ) = 320+210+270+120+150 = 1070
Step 3: Sum of weights:
Σwᵢ = 4+3+3+2+2 = 14
Step 4: Divide:
x̄w = 1070 / 14 = 76.43
The weighted average = 76.43.
The simple average (ignoring credits) would be:
(80+70+90+60+75) / 5 = 375/5 = 75.
The weighted mean is higher (76.43 vs 75) because the student scored higher in heavily weighted subjects (Statistics=80, Python=90 have 4 and 3 credits respectively).
Click "Compare" to see how weights shift the mean towards heavily-weighted subjects:
The combined mean (or pooled mean) is the overall mean of two or more groups taken together. If we know the mean and size of each group, we can compute the combined mean without needing the individual observations.
Batch A: 30 laddoos with average weight 50g. Batch B: 20 laddoos with average weight 60g. The combined average weight of all 50 laddoos = (30×50 + 20×60) / (30+20) = 2700/50 = 54g. The combined mean is a weighted mean where the group sizes are the weights.
Two sections of B.Sc. Data Science took the same Statistics test:
Section A: 35 students, average marks = 68
Section B: 25 students, average marks = 74
Find the combined average of both sections together.
Given:
n₁ = 35, x̄₁ = 68
n₂ = 25, x̄₂ = 74
Step 1: Calculate total marks for each section:
Section A: n₁x̄₁ = 35 × 68 = 2,380
Section B: n₂x̄₂ = 25 × 74 = 1,850
Step 2: Combined total marks:
2,380 + 1,850 = 4,230
Step 3: Total students:
35 + 25 = 60
Step 4: Combined mean:
x̄c = 4,230 / 60 = 70.5 marks
The combined mean (70.5) is not the simple average of 68 and 74 (which would be 71). It is closer to 68 because Section A has more students (35 > 25) and therefore pulls the combined mean more towards its own average.
This is exactly the weighted mean principle — larger groups get more influence.
Three factories produce the same item. Their daily production and defect rates are:
| Factory | Items Produced (nᵢ) | Avg Defects/item (x̄ᵢ) |
|---|---|---|
| P | 200 | 3.5 |
| Q | 300 | 2.8 |
| R | 500 | 4.2 |
Find the combined average defect rate.
Σ(nᵢ · x̄ᵢ) = (200 × 3.5) + (300 × 2.8) + (500 × 4.2)
= 700 + 840 + 2100 = 3,640
Σnᵢ = 200 + 300 + 500 = 1,000
x̄c = 3,640 / 1,000 = 3.64 defects per item
1. Easy to understand and calculate — everyone knows "average."
2. Uses all observations — considers every data point.
3. Unique — only one possible mean for a dataset.
4. Algebraically tractable — can be used in further calculations (variance, regression, etc.).
5. Best for symmetrical data — accurately represents the centre of a bell-shaped distribution.
6. Sampling stability — the mean varies least across different samples compared to median and mode.
7. Basis for inferential statistics — most advanced tests (t-test, ANOVA, regression) use the mean.
1. Affected by extreme values (outliers): One very high or low value can distort the mean. E.g., incomes of {2, 3, 4, 5, 100} → mean = 22.8, but most values are 2–5.
2. Cannot be used for open-end classes: Without exact class marks, the mean cannot be calculated.
3. Not suitable for qualitative data: Cannot compute the mean of {Red, Blue, Green}.
4. Can be misleading for skewed data: In a right-skewed distribution, the mean is pulled towards the tail and may not represent the "typical" value.
5. May not be an actual observation: The mean of {2, 3, 4} = 3 (which is a data point), but the mean of {2, 3} = 2.5 (not an actual data point).
6. Cannot be determined graphically: Unlike the median, the mean cannot be read directly from a graph.
Q1: The arithmetic mean of 10, 20, 30, 40, 50 is:
Q2: Σ(xᵢ − x̄) for any dataset always equals:
Q3: In the step-deviation method, uᵢ equals:
Q4: Group A has 40 students with mean 60. Group B has 60 students with mean 80. The combined mean is:
Q5: Which of the following is a limitation of the arithmetic mean?
Q6: The weighted mean is used when:
Q7: If every observation is multiplied by 5, the new mean is:
Q8: The assumed mean method uses the formula:
1. Find the arithmetic mean of: 12, 18, 25, 31, 44, 50, 37, 29. Verify that Σ(xᵢ − x̄) = 0.
2. Calculate the mean using ALL THREE methods (direct, assumed mean, step-deviation) for the following data:
| Class | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 |
|---|---|---|---|---|---|---|
| f | 8 | 12 | 20 | 15 | 10 | 5 |
Questions? Let's discuss!
Lecture 11–12: Median, Mode & Empirical Relationship
Lecture 13–14: Partition Values — Quartiles, Deciles, Percentiles
Lecture 15: Unit II Review & Problem-Solving
Unit II · Lectures 9–10 · BDS113 Introduction to Statistics · B.Sc. Data Science, Semester II