Big picture
Two classes can have the same average score but very different distributions of scores. Measures of dispersion describe how widely observations spread around a central value. They complement averages and help distinguish consistent performance from a mixture of very high and very low values.
This chapter develops range, mean deviation, variance and standard deviation for individual and frequency data. Every calculation should identify the data model and its units. Grouped data use representative class marks, so their numerical measures are generally approximations to those from the ungrouped observations.
Learning goals
Calculate and interpret range and mean deviation.
Determine variance and standard deviation from individual data.
Use frequency and grouped-data methods accurately.
Apply shortcut and coding formulas without changing their meaning.
Compare variability and explain transformations of data.
Core ideas
1. Central values and dispersion
The arithmetic mean is x̄ = Σx/n for n observations. With frequencies fᵢ, x̄ = Σfᵢxᵢ/N, where N = Σfᵢ. The median divides ordered data into two equal-count parts. These locate the centre but do not measure spread.
Consider datasets {4, 4, 4} and {1, 4, 7}. Both have mean 4, but the first has no dispersion. The second has observations three units below and above its mean. The signed deviations from the mean sum to zero, so simply averaging x − x̄ cannot measure spread. Mean deviation uses absolute values; variance uses squares. Each avoids cancellation in a different way.
2. Range
The range is maximum observation minus minimum observation. For {12, 15, 17, 20, 24}, range = 24 − 12 = 12 units. It is simple and uses the original units, but depends only on the extremes and can be strongly affected by one outlier.
If the same dataset gains an observation 60, the range becomes 60 − 12 = 48, even though the original central observations are unchanged. A range does not show how observations are distributed between endpoints. Two datasets can have the same range while one is clustered centrally and the other concentrated near the extremes. Use it as a rough description, not a complete summary.
3. Mean deviation for individual data
Mean deviation about a value a is MD(a) = Σ|xᵢ − a|/n. Common choices are mean and median. For data {2, 4, 7, 7}, mean = 20/4 = 5. Absolute deviations from 5 are 3, 1, 2, 2, totalling 8. Therefore MD(mean) = 8/4 = 2.
The ordered-data median is (4 + 7)/2 = 5.5. Absolute deviations from 5.5 are 3.5, 1.5, 1.5, 1.5, again totalling 8, so MD(median) = 2. Equality here does not make mean and median interchangeable in every dataset. Mean absolute deviation is minimised about a median; in even-sized datasets there can be an interval of minimising centres. Always state the centre used.
4. Mean deviation for frequency data
With frequencies, MD(a) = Σfᵢ|xᵢ − a|/N. The frequency multiplies each deviation because the corresponding value occurs repeatedly. For values 1, 3, 5 with frequencies 2, 3, 1, N = 6 and Σfx = 2 + 9 + 5 = 16. The mean is 8/3.
Absolute deviations are 5/3, 1/3 and 7/3. Weighted total = 2 × 5/3 + 3 × 1/3 + 1 × 7/3 = 20/3. Thus MD(mean) = (20/3)/6 = 10/9 ≈ 1.111. The median is 3, since both middle observations are 3. Weighted absolute deviations about 3 total 2 × 2 + 3 × 0 + 1 × 2 = 6, giving MD(median) = 1. Frequencies must not be ignored when finding either the centre or deviations.
5. Variance and standard deviation
For the complete dataset considered as a population, variance σ² = Σ(xᵢ − x̄)²/n. Standard deviation σ is the nonnegative square root of variance. Variance has squared units, while standard deviation has the original units. This chapter uses denominator n or N; an n − 1 sample estimator belongs to a different inferential context.
For {2, 4, 7, 7}, mean = 5. Squared deviations are 9, 1, 4, 4, totalling 18. Thus variance = 18/4 = 4.5 and standard deviation = √4.5 ≈ 2.121. A variance cannot be negative, and it is zero exactly when all observations are equal. Because squaring emphasises large deviations, variance and standard deviation are sensitive to extreme values.
6. Shortcut variance formula
Expanding the squares gives σ² = Σx²/n − x̄². With frequencies, σ² = Σfx²/N − (Σfx/N)². This is algebraically the same measure as the direct formula, not a new type of variance.
For values 1, 3, 5 with frequencies 2, 3, 1, Σfx² = 2 × 1 + 3 × 9 + 1 × 25 = 54. Hence σ² = 54/6 − (8/3)² = 9 − 64/9 = 17/9 ≈ 1.889, and σ = √17/3 ≈ 1.374. Keep exact fractions or adequate precision until the final answer. Rounding the mean too early can distort the subtraction and, with badly rounded large numbers, even produce a false negative result.
7. Grouped continuous data
For a class interval with lower and upper limits l and u, the class mark is x = (l + u)/2. Treat observations in that class as represented by x and apply frequency formulas. For classes 0–10, 10–20 and 20–30 with frequencies 2, 4 and 2, class marks are 5, 15 and 25.
N = 8, Σfx = 2 × 5 + 4 × 15 + 2 × 25 = 120, so mean = 15. Squared deviations from 15 are 100, 0 and 100. Weighted sum is 2 × 100 + 4 × 0 + 2 × 100 = 400. Therefore grouped variance = 400/8 = 50 and standard deviation = √50 ≈ 7.071. This approximates the original data's dispersion because exact within-class positions are unknown.
8. Assumed mean and step deviation
Choose an assumed mean, A, and write dᵢ = xᵢ − A. Then x̄ = A + Σfd/N and σ² = Σfd²/N − (Σfd/N)². Subtracting a constant does not change spread, which explains this simplification.
For equally spaced class marks, use uᵢ = (xᵢ − A)/h, where h is the common width. Then x̄ = A + hΣfu/N and σ² = h²[Σfu²/N − (Σfu/N)²]. In the preceding grouped example, choose A = 15 and h = 10, so u values are −1, 0, 1. Σfu = 0 and Σfu² = 4; hence variance = 10² × 4/8 = 50, matching the direct result. The standard deviation is |h| times the coded standard deviation, not h² times it.
9. Transformations and relative comparison
Adding a constant c to every observation changes the mean by c but leaves range, variance and standard deviation unchanged. Multiplying all observations by b multiplies the mean by b, variance by b² and standard deviation by |b|. If Y = 3X + 10 and X has mean 5 and variance 4, Y has mean 25, variance 36 and standard deviation 6.
The coefficient of variation, CV = 100σ/x̄, compares relative dispersion for positive meaningful means. A class with mean 60 and standard deviation 6 has CV = 10%; another with mean 80 and standard deviation 12 has CV = 15%. The first is relatively more consistent by this measure. CV is an extension comparison tool and is unsuitable when the mean is zero or lacks a meaningful ratio-scale interpretation, such as arbitrary Celsius temperature baselines.
10. Combined data and interpretation
For two groups, the combined mean is (N₁x̄₁ + N₂x̄₂)/(N₁ + N₂). Combined variance must account for within-group spread and differences between group means; merely averaging the two standard deviations is incorrect.
Suppose two observations in each group have means 10 and 14, and both groups have variance 1. Combined mean is (2 × 10 + 2 × 14)/4 = 12. Each group mean is 2 from the combined mean, adding 4 to its within-group variance. Combined variance = [2(1 + 4) + 2(1 + 4)]/4 = 5 and combined standard deviation √5 ≈ 2.236. These extension calculations demonstrate why merging groups can increase dispersion even if their individual variability is small.
A reliable approach to questions
Identify individual, discrete-frequency or grouped data.
Calculate N and the appropriate central value before deviations.
Choose absolute deviations for MD or squared deviations for variance.
Use frequencies throughout and preserve precision until the final step.
Check nonnegative dispersion, units and agreement with an alternative formula.
Connections that matter
Algebra derives shortcut formulas and transformation rules.
Probability later connects distributions with expected values and variance.
Economics and science use dispersion to judge consistency and uncertainty.
Common mistakes
Adding signed deviations without absolute values makes mean deviation collapse to zero.
Dividing by the number of distinct values ignores frequencies.
Reporting variance in original units overlooks its squared units.
Averaging group standard deviations ignores between-group differences.
Revision checklist
I can calculate range and mean deviation about a stated centre.
I can obtain variance and standard deviation by two methods.
I can process frequency and grouped data correctly.
I can use coding and transformation rules.
I can interpret variability and explain grouped-data limitations.