Statistics is the branch of mathematics concerned with collecting, organising, analysing, and interpreting data. In Class 11, we extend our earlier knowledge of data representation to study measures of dispersion — how spread out a dataset is. While measures of central tendency (mean, median, mode) tell us the centre, measures of dispersion describe the spread.
Measures of Dispersion:
- 1.Range: The difference between the largest and smallest values.
- 2.Range = Maximum value - Minimum value
- Mean Deviation: Average of the absolute deviations from a central value (mean or median).
- Mean Deviation about Mean: MD(mean) = (1/n) × sum |xi - xbar|
- Mean Deviation about Median: MD(median) = (1/n) × sum |xi - M|
- Variance and Standard Deviation:
- Variance (sigma2) = (1/n) × sum (xi - xbar)2
- Standard Deviation (sigma) = sqrt[Variance]
- Alternative formula: sigma2 = (1/n) × sum xi2 - (xbar)2
- 1.Coefficient of Variation (CV): A relative measure used to compare two datasets with different units or means.
- 2.CV = (sigma / xbar) × 100
- For Grouped Data:
- Mean: xbar = sum(fi xi) / sum(fi)
- Variance: sigma2 = [sum(fi xi2) / sum(fi)] - xbar2
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Worked Examples
Find the range of the data: 12, 7, 19, 3, 25, 8.
Maximum = 25, Minimum = 3. Range = 25 - 3 = 22
Find the mean deviation about the mean for: 4, 7, 8, 9, 10, 12.
Mean = (4+7+8+9+10+12)/6 = 50/6 ≈ 8.33
|xi - mean|: 4.33, 1.33, 0.33, 0.67, 1.67, 3.67. Sum = 12. MD = 12/6 = 2
Find the variance of: 2, 4, 6, 8, 10.
Mean = 30/5 = 6. Deviations: -4, -2, 0, 2, 4. Squared: 16, 4, 0, 4, 16. Sum = 40.
Variance = 40/5 = 8. Standard deviation = √(8) = 2sqrt(2).
Compare the datasets A (mean = 50, SD = 5) and B (mean = 10, SD = 2) using CV.
CV(A) = (5/50) × 100 = 10%. CV(B) = (2/10) × 100 = 20%.
Dataset B is more variable (higher CV).
For data with n = 10, sum(xi) = 100, sum(xi2) = 1100. Find variance.
xbar = 100/10 = 10. Variance = 1100/10 - 100 = 110 - 100 = 10
The runs scored by 8 cricket players are: 50, 65, 70, 60, 45, 55, 80, 75. Find SD.
Mean = 500/8 = 62.5. Deviations squared: 156.25, 6.25, 56.25, 6.25, 306.25, 56.25, 306.25, 156.25. Sum = 1050.
Variance = 1050/8 = 131.25. SD = √(131.25) ≈ 11.46.
For a frequency distribution, if N = 20, sum(fi xi) = 400, sum(fi xi2) = 9200, find variance.
Mean = 400/20 = 20. Variance = 9200/20 - 400 = 460 - 400 = 60
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Common mistakes
- Variance is the average of squared deviations; standard deviation is the square root of variance. Never confuse the two.
- Mean deviation uses absolute values of deviations; variance uses squares. The absolute value sign cannot be replaced by squaring in MD.
- The coefficient of variation is expressed as a percentage. Always multiply by 100.
Summary
Measures of dispersion quantify the spread of data. Standard deviation and variance are the most important. The coefficient of variation allows comparison across datasets with different scales.