1. Chapter at a Glance
Two batsmen both average 50 runs. One scores 48, 50, 52, 49, 51; the other scores 0, 100, 5, 95, 50. Same average — very different consistency! The mean alone cannot tell them apart. We need measures of dispersion, which describe how spread out the data is around the centre. This chapter covers range, mean deviation, variance and standard deviation for ungrouped, discrete and grouped data — tools used in every field that handles data.
- Why dispersion matters; range.
- Mean deviation about the mean and about the median.
- Mean deviation for discrete and grouped (continuous) frequency distributions.
- Variance and standard deviation: definitions and shortcut formulas.
- Step-deviation method for grouped data.
- Enrichment: coefficient of variation for comparing consistency.
Note: all data in this chapter is practice data for learning.
2. Range
Range = maximum value − minimum value. It is easy to find but depends only on the two extreme values and ignores all others.
Batsman A (48–52): range 4. Batsman B (0–100): range 100. B is far less consistent.
3. Mean Deviation
Mean deviation about a central value a (mean or median):
We take absolute values because the plain deviations Σ(xᵢ − x̄) always add up to 0.
Find the mean deviation about the mean of 6, 7, 10, 12, 13, 4, 8, 12.
Solution: Mean = 72/8 = 9. |xᵢ − 9|: 3, 2, 1, 3, 4, 5, 1, 3 → sum 22. M.D. = 22/8 = 2.75.
Find the mean deviation about the median of 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.
Solution: Sorted: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21. Median (6th) = 9. |xᵢ − 9|: 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12 → 58. M.D. = 58/11 ≈ 5.27.
Grouped (continuous) data: use the class marks xᵢ as representative values. For the median, use Median = l + [(N/2 − cf)/f] × h.
Find the mean deviation about the mean:
| Class | fᵢ | xᵢ | fᵢxᵢ | |xᵢ − 45| | fᵢ|xᵢ − 45| |
|---|---|---|---|---|---|
| 10–20 | 2 | 15 | 30 | 30 | 60 |
| 20–30 | 3 | 25 | 75 | 20 | 60 |
| 30–40 | 8 | 35 | 280 | 10 | 80 |
| 40–50 | 14 | 45 | 630 | 0 | 0 |
| 50–60 | 8 | 55 | 440 | 10 | 80 |
| 60–70 | 3 | 65 | 195 | 20 | 60 |
| 70–80 | 2 | 75 | 150 | 30 | 60 |
| Total | 40 | 1800 | 400 |
Solution: Mean = 1800/40 = 45. M.D. = 400/40 = 10.
4. Variance and Standard Deviation
Instead of absolute values, we can square the deviations — this is easier to work with algebraically and gives more weight to large deviations.
- Variance σ² = Σ(xᵢ − x̄)² / n (ungrouped); Σfᵢ(xᵢ − x̄)² / N (frequency data).
- Standard deviation σ = √(variance) — in the same units as the data.
Shortcut: σ² = (Σxᵢ²)/n − x̄² — "mean of squares minus square of mean".
Step deviation: with yᵢ = (xᵢ − A)/h, σ = (h/N)√[NΣfᵢyᵢ² − (Σfᵢyᵢ)²].
Find the variance and SD of 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
Solution: Mean = 150/10 = 15. Deviations: −9, −7, −5, −3, −1, 1, 3, 5, 7, 9; squares sum = 2(81 + 49 + 25 + 9 + 1) = 330. Variance = 33; SD = √33 ≈ 5.74.
Find the mean and variance: xᵢ: 4, 8, 11, 17, 20, 24, 32 with fᵢ: 3, 5, 9, 5, 4, 3, 1.
Solution: N = 30, Σfx = 12 + 40 + 99 + 85 + 80 + 72 + 32 = 420 → mean = 14. Σf(x − 14)² = 3(100) + 5(36) + 9(9) + 5(9) + 4(36) + 3(100) + 1(324) = 300 + 180 + 81 + 45 + 144 + 300 + 324 = 1374. Variance = 1374/30 = 45.8; SD ≈ 6.77.
5. Enrichment: Comparing Consistency
To compare the variability of two series with different means, use the coefficient of variation: CV = (σ / x̄) × 100. The series with the smaller CV is more consistent.
Practice data: Team X has mean score 82, SD 7.3; Team Y has mean 48, SD 8.2. Which is more consistent?
Solution: CV(X) = 7.3/82 × 100 ≈ 8.9%; CV(Y) = 8.2/48 × 100 ≈ 17.1%. Team X is more consistent.
6. Solved Problems (Board Pattern)
The mean and variance of eight observations are 9 and 9.25. Six of them are 6, 7, 10, 12, 12 and 13. Find the other two.
Solution: Sum = 72 → x + y = 72 − 60 = 12. Σxᵢ² = 8(9.25 + 81) = 722; the six known squares add to 642 → x² + y² = 80. Then 2xy = 144 − 80 = 64 → xy = 32. So x, y are 4 and 8.
Each observation of a data set is multiplied by 3. How do the mean and SD change? What if 5 is added to each?
Solution: ×3: mean ×3, SD ×3 (variance ×9). +5: mean + 5, SD unchanged.
The mean and SD of 20 observations are 10 and 2. One observation 8 was wrongly read as 12. Find the correct mean and SD.
Solution: Wrong sum 200 → correct sum 196 → mean 9.8. Wrong Σx² = 20(4 + 100) = 2080 → correct = 2080 − 144 + 64 = 2000. Variance = 2000/20 − 9.8² = 100 − 96.04 = 3.96 → SD = √3.96 ≈ 1.99.
7. Master Formula Table
| Measure | Formula |
|---|---|
| Range | max − min |
| M.D. (ungrouped) | Σ|xᵢ − a|/n |
| M.D. (frequency) | Σfᵢ|xᵢ − a|/N |
| Variance | Σ(xᵢ − x̄)²/n = Σxᵢ²/n − x̄² |
| Variance (frequency) | Σfᵢ(xᵢ − x̄)²/N |
| SD (step deviation) | (h/N)√[NΣfy² − (Σfy)²] |
| Coefficient of variation | σ/x̄ × 100 |
| Change of origin/scale | y = (x − A)/h ⇒ σₓ = hσᵧ |
8. Master Summary Table
| Measure | Strength | Weakness |
|---|---|---|
| Range | Very easy | Uses only extremes |
| Mean deviation | Uses all values; intuitive | Absolute values awkward algebraically |
| Variance | Algebraically convenient | Squared units |
| Standard deviation | Same units as data; widely used | Affected by outliers |
| Coefficient of variation | Compares different series | Needs positive mean |
9. Mind Map
- max − min
- Crude
- About mean
- About median
- Squared deviations
- Shortcut, step-deviation
- CV = σ/x̄ × 100
- Smaller = consistent
10. Tips and Tricks
- Use the shortcut σ² = Σx²/n − x̄² when the mean is not a whole number.
- Step deviation keeps the arithmetic small for large class marks.
- Adding a constant shifts data but does not change its spread.
- Forgetting absolute values in mean deviation (the sum would be 0).
- Reporting variance when SD is asked (take the square root).
- Using class limits instead of class marks for grouped data.
11. Important One-liners
- Σ(xᵢ − x̄) = 0 always.
- Mean deviation about the median is the least among mean deviations about any point.
- SD is the square root of variance.
- Adding a constant to every value leaves SD unchanged.
- Multiplying every value by k multiplies SD by |k|.
12. Exam-pattern Questions with Detailed Answers
The range of 12, 7, 25, 3, 18 is:
- 22
- 25
- 15
- 13
Show answer
25 − 3 = 22.
Answer: (A)
If the variance of a data set is 16, what is its SD?
Show answer
√16 = 4.
Answer: 4
Find the mean deviation about the mean of 4, 7, 8, 9, 10, 12, 13, 17.
Show answer
Mean 10. |deviations|: 6, 3, 2, 1, 0, 2, 3, 7 → 24. M.D. = 3.
Answer: 3
Find the variance of the first 10 natural numbers.
Show answer
For 1…n: variance = (n² − 1)/12 = 99/12 = 8.25.
Answer: 8.25
Find the mean deviation about the median of 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
Show answer
Sorted: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18. Median = (13 + 14)/2 = 13.5. |dev|: 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5 = 28. M.D. = 28/12 ≈ 2.33.
Answer: ≈ 2.33
Find the SD of 5, 5, 5, 5.
Show answer
No spread: 0.
Answer: 0
The mean of 5 observations is 4.4 and their variance is 8.24. Three observations are 1, 2 and 6. Find the other two.
Show answer
Sum 22 → x + y = 13. Σx² = 5(8.24 + 19.36) = 138 → x² + y² = 138 − 41 = 97. (x + y)² = 169 → 2xy = 72 → xy = 36. x, y = 4 and 9.
Answer: 4 and 9
If each observation of a set with SD 5 is increased by 10, what is the new SD? If each is multiplied by −2?
Show answer
+10: SD stays 5. ×(−2): SD = 10.
Answer: 5 and 10
Find the mean and SD of the practice data: class 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5, 8, 15, 16, 6.
Show answer
A = 25, h = 10, y: −2, −1, 0, 1, 2. N = 50; Σfy = −10 − 8 + 0 + 16 + 12 = 10; Σfy² = 20 + 8 + 0 + 16 + 24 = 68. Mean = 25 + 10 × 10/50 = 27. σ = (10/50)√(50 × 68 − 100) = 0.2√3300 ≈ 11.49.
Answer: Mean 27, SD ≈ 11.49
The mean and SD of 100 observations were 40 and 5.1. Later it was found that 50 was misread as 40. Find the correct mean and SD.
Show answer
Sum 4000 → 4010; mean 40.1. Σx² = 100(26.01 + 1600) = 162601 → +2500 − 1600 = 163501. Variance = 1635.01 − 1608.01 = 27; SD = √27 ≈ 5.20.
Answer: Mean 40.1, SD ≈ 5.20
Practice data: two plants produce bolts with mean lengths 20 mm and 25 mm, and SDs 2 mm and 2.4 mm. Which plant is more consistent?
Show answer
CV₁ = 2/20 × 100 = 10%; CV₂ = 2.4/25 × 100 = 9.6%. Plant 2 is slightly more consistent.
Answer: Plant 2
Show that the variance of a, a + d, a + 2d, …, a + 2nd (2n + 1 terms) is n(n + 1)d²/3.
Show answer
Mean = a + nd. Deviations: −nd, …, 0, …, nd. Σ(dev)² = 2d²(1² + … + n²) = 2d² n(n + 1)(2n + 1)/6. Divide by 2n + 1: n(n + 1)d²/3. ∎
Answer: Proved
13. Quick Revision Summary
- Dispersion measures spread: range, mean deviation, variance, SD.
- M.D. = Σ|x − a|/n (a = mean or median); use class marks for grouped data.
- σ² = Σ(x − x̄)²/n = Σx²/n − x̄²; σ = √σ².
- Step deviation simplifies grouped calculations.
- Shift: SD unchanged; scale by k: SD × |k|; CV compares consistency.