Unit III · Lecture 17 of 40 · B.Sc. Data Science, 1st Year, Semester II
A measure of central tendency (mean, median, mode) tells us the centre of the data — but it tells us nothing about how the data is spread out around that centre. Two datasets can have the same mean but be fundamentally different in their variability.
Consider two classes that both scored an average of 60 marks in an exam:
Class A: 58, 59, 60, 61, 62 → Mean = 60 → Scores are tightly clustered around the mean.
Class B: 20, 40, 60, 80, 100 → Mean = 60 → Scores are widely spread across the range.
Both classes have the same average, but their performance patterns are completely different. Class A is consistent; Class B has extreme variation. The mean alone cannot capture this difference. We need a measure of dispersion.
Dispersion (also called scatter, spread, or variability) refers to the extent to which individual data values deviate from a central value (usually the mean or median). A measure of dispersion quantifies how much the data varies.
Imagine two archers shooting at a target. Both hit the bullseye on average (same centre). But Archer A's arrows are tightly grouped near the centre (low dispersion), while Archer B's arrows are scattered all over the target (high dispersion). Measures of dispersion tell us how "tight" or "scattered" the data is around the centre.
Common Measures of Dispersion:
The Mean Absolute Deviation (MAD) is the arithmetic mean of the absolute deviations of each observation from a central value (usually the mean or median). It tells us, on average, how far each data point is from the centre — regardless of direction.
"Deviation" = the difference between a data value and the central value (xᵢ − x̄).
"Absolute" = we take the absolute value |xᵢ − x̄| so that negative deviations don't cancel out positive ones.
"Mean" = we average all these absolute deviations by dividing by the number of observations.
Suppose your office is 10 km from home. Some days you travel 8 km (shortcut), other days 12 km (detour), and sometimes exactly 10 km. The MAD of your daily distances from 10 km tells you, on average, how much your commute deviates from the expected 10 km — regardless of whether you went shorter or longer.
MAD = (1/n) × Σ |xᵢ − x̄|
Where:
• xᵢ = each individual observation (i = 1, 2, 3, …, n)
• x̄ = arithmetic mean of the data
• |xᵢ − x̄| = absolute deviation of each observation from the mean
• n = total number of observations
• Σ = summation over all observations
MAD = (1/N) × Σ fᵢ |xᵢ − x̄|
Where:
• fᵢ = frequency of the i-th class / value
• xᵢ = mid-point (or value) of the i-th class
• x̄ = weighted mean = Σfᵢxᵢ / Σfᵢ
• N = Σfᵢ = total frequency
Find the Mean Absolute Deviation of the following marks obtained by 5 students: 10, 15, 20, 25, 30
Step 1: Find the Mean
x̄ = (10 + 15 + 20 + 25 + 30) / 5 = 100 / 5 = 20
Step 2: Calculate Deviations and Absolute Deviations
| xᵢ (Marks) | Deviation (xᵢ − x̄) | |xᵢ − x̄| |
|---|---|---|
| 10 | 10 − 20 = −10 | 10 |
| 15 | 15 − 20 = −5 | 5 |
| 20 | 20 − 20 = 0 | 0 |
| 25 | 25 − 20 = +5 | 5 |
| 30 | 30 − 20 = +10 | 10 |
| Σ|xᵢ − x̄| = | 30 | |
On average, each student's marks deviate by 6 marks from the mean of 20.
The daily maximum temperatures (°C) recorded in a city for a week were: 32, 35, 28, 30, 34, 36, 29. Find the MAD about the mean.
Step 1: Find the Mean
x̄ = (32 + 35 + 28 + 30 + 34 + 36 + 29) / 7 = 224 / 7 = 32°C
Step 2: Calculate Deviations and Absolute Deviations
| xᵢ (°C) | xᵢ − x̄ | |xᵢ − x̄| |
|---|---|---|
| 32 | 0 | 0 |
| 35 | +3 | 3 |
| 28 | −4 | 4 |
| 30 | −2 | 2 |
| 34 | +2 | 2 |
| 36 | +4 | 4 |
| 29 | −3 | 3 |
| Σ|xᵢ − x̄| = | 18 | |
On average, the daily temperature deviates by approximately 2.57°C from the weekly mean of 32°C.
The number of goals scored by a football team in 20 matches is given below. Find the MAD.
| Goals (xᵢ) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Frequency (fᵢ) | 3 | 5 | 7 | 3 | 2 |
Step 1: Find the Weighted Mean
N = Σfᵢ = 3 + 5 + 7 + 3 + 2 = 20
Σfᵢxᵢ = (3×0) + (5×1) + (7×2) + (3×3) + (2×4) = 0 + 5 + 14 + 9 + 8 = 36
x̄ = 36 / 20 = 1.8
Step 2: Calculate fᵢ × |xᵢ − x̄| for each class
| xᵢ | fᵢ | xᵢ − x̄ | |xᵢ − x̄| | fᵢ × |xᵢ − x̄| |
|---|---|---|---|---|
| 0 | 3 | −1.8 | 1.8 | 5.4 |
| 1 | 5 | −0.8 | 0.8 | 4.0 |
| 2 | 7 | +0.2 | 0.2 | 1.4 |
| 3 | 3 | +1.2 | 1.2 | 3.6 |
| 4 | 2 | +2.2 | 2.2 | 4.4 |
| Σ fᵢ|xᵢ − x̄| = | 18.8 | |||
On average, the number of goals per match deviates by 0.94 goals from the mean of 1.8 goals.
The marks obtained by 40 students are grouped as follows. Find the MAD about the mean.
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| No. of Students (fᵢ) | 4 | 8 | 14 | 9 | 5 |
Step 1: Find Mid-points (xᵢ) and the Mean
Mid-points: 5, 15, 25, 35, 45
N = 4 + 8 + 14 + 9 + 5 = 40
Σfᵢxᵢ = (4×5) + (8×15) + (14×25) + (9×35) + (5×45) = 20 + 120 + 350 + 315 + 225 = 1030
x̄ = 1030 / 40 = 25.75
| Marks | xᵢ | fᵢ | fᵢxᵢ | |xᵢ − x̄| | fᵢ|xᵢ − x̄| |
|---|---|---|---|---|---|
| 0–10 | 5 | 4 | 20 | 20.75 | 83.00 |
| 10–20 | 15 | 8 | 120 | 10.75 | 86.00 |
| 20–30 | 25 | 14 | 350 | 0.75 | 10.50 |
| 30–40 | 35 | 9 | 315 | 9.25 | 83.25 |
| 40–50 | 45 | 5 | 225 | 19.25 | 96.25 |
| Totals: | 1030 | 359.00 | |||
On average, each student's marks deviate by approximately 8.98 marks from the mean of 25.75.
While MAD is most commonly calculated about the mean, it can also be calculated about the median. In fact, the median minimises the sum of absolute deviations, making MAD about the median a theoretically important measure.
MAD (about median) = (1/n) × Σ |xᵢ − Mₑ|
Where Mₑ is the median of the data.
Using the same data: 10, 15, 20, 25, 30
Median (Mₑ) = 20 (the middle value)
|10−20| + |15−20| + |20−20| + |25−20| + |30−20| = 10 + 5 + 0 + 5 + 10 = 30
MAD (about median) = 30 / 5 = 6
In this case, the mean and median are both 20, so MAD about mean = MAD about median. This will not always be the case.
• When the data is symmetric (mean ≈ median)
• When you want a measure tied to the arithmetic mean
• Most common in introductory statistics
• When the data is skewed or has outliers
• The median is more robust to extreme values
• MAD about median is always ≤ MAD about mean
Enter your own data (comma-separated numbers) and watch the MAD being calculated step by step:
Compare two datasets with the same mean but different MAD values. Click each dataset to visualise its spread:
48, 49, 50, 51, 52
30, 40, 50, 60, 70
1. Simple to understand: The concept of "average distance from the mean" is intuitive and easy to explain to non-statisticians.
2. Easy to compute: Requires only basic arithmetic — addition, subtraction, and division. No squaring or square roots needed.
3. Uses all observations: Unlike the range, MAD considers every data point, providing a more complete picture of variability.
4. Less affected by outliers: Extreme values have a smaller effect on MAD than on variance or standard deviation (because deviations are not squared).
5. Rigidly defined: MAD has a unique value for any given dataset — it is not dependent on subjective choices.
1. Not algebraically tractable: MAD cannot be easily manipulated mathematically. It does not decompose neatly into within-group and between-group components.
2. Not suitable for further statistical analysis: Most advanced statistical techniques (hypothesis testing, regression, ANOVA) use variance and standard deviation, not MAD.
3. Ignores direction of deviation: By taking absolute values, MAD loses information about whether values tend to be above or below the mean.
4. Not widely used in practice: Standard deviation and variance are the dominant measures of dispersion in most fields. MAD is primarily used as a teaching tool and in specific applications (e.g., forecasting error measurement).
5. Open-end classes are problematic: For grouped data with open-end intervals (e.g., "50 and above"), MAD cannot be computed without assumptions.
| Feature | Range | MAD | Standard Deviation (σ) |
|---|---|---|---|
| Definition | Maximum − Minimum | Average of |xᵢ − x̄| | √(Average of (xᵢ − x̄)²) |
| Uses all data? | No (only 2 values) | Yes | Yes |
| Outlier sensitivity | Very high | Moderate | High (due to squaring) |
| Ease of calculation | Easiest | Easy | Moderate |
| Algebraic tractability | Poor | Poor | Excellent |
| Used in further analysis? | Rarely | Rarely | Extensively (hypothesis testing, regression, etc.) |
| Units | Same as data | Same as data | Same as data |
| Best for | Quick, rough comparison | Intuitive understanding of spread | Rigorous statistical analysis |
The Coefficient of MAD is a relative measure of dispersion that allows comparison between datasets with different units or different means. It is a dimensionless ratio.
Coefficient of MAD = MAD / x̄
It is expressed as a ratio (not a percentage). A smaller coefficient indicates more consistency; a larger coefficient indicates more variability.
Factory A: Mean daily output = 200 units, MAD = 10 units
Coefficient of MAD = 10 / 200 = 0.05
Factory B: Mean daily output = 500 units, MAD = 30 units
Coefficient of MAD = 30 / 500 = 0.06
Even though Factory B has a larger absolute MAD (30 vs 10), Factory A has a slightly smaller coefficient (0.05 vs 0.06), meaning Factory A is relatively more consistent in its output.
Enter two datasets and compare their means, MADs, and coefficients of MAD side by side:
Q1: The Mean Absolute Deviation is always:
Q2: If all observations in a dataset are identical, the MAD is:
Q3: Why do we take the absolute value of deviations when computing MAD?
Q4: The data is: 5, 10, 15. The mean is 10. What is the MAD?
Q5: Which of the following is a limitation of MAD?
Q6: MAD about the median is always:
Q7: If y = 3x + 5, and MAD(x) = 4, then MAD(y) is:
Q8: The Coefficient of MAD is useful because:
1. Calculate the Mean Absolute Deviation of the following dataset: 4, 8, 6, 10, 12
2. The weekly pocket expenses (in ₹) of 6 students are: 150, 200, 180, 220, 170, 250. Find:
(a) The mean expense
(b) The MAD about the mean
(c) Interpret the result in one sentence
3. Find the MAD for the following frequency distribution:
Value: 2, 4, 6, 8, 10
Frequency: 3, 5, 8, 4, 2
4. Find the MAD for the following grouped data:
Class: 0–10, 10–20, 20–30, 30–40
Frequency: 5, 10, 15, 10
5. Two cricketers scored the following runs in 5 innings:
Cricketer A: 30, 50, 40, 60, 20
Cricketer B: 35, 45, 40, 55, 25
(a) Who has the higher mean?
(b) Who has the lower MAD?
(c) Who is more consistent? Explain.
6. Explain why the standard deviation is generally preferred over MAD in advanced statistical analysis, despite MAD being simpler to compute.
7. BONUS: Prove that Σ(xᵢ − x̄) = 0 for any dataset. Then explain why this property makes the absolute value necessary in the MAD formula.
Questions? Let's discuss!
Lecture 18: Variance and Standard Deviation
Lecture 19: Standard Deviation — Properties & Shortcut Methods
Lecture 20: Coefficient of Variation & Comparison of Distributions
Unit III · Lecture 17 · BDS113 Introduction to Statistics · B.Sc. Data Science, Semester II