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This page provides complete NCERT exercise solutions for CBSE Class 11 Economics Chapter 6 — Measures of Dispersion. Review the Key Concepts and formulas first, then attempt each numerical. Step-by-step working is shown for all calculations. Verify your answers against these solutions and ensure you understand the method for each measure of dispersion.
Key Concepts (Quick Recap)
- Dispersion: the extent to which values in a distribution vary from the average
- Absolute Measures: Range, Quartile Deviation, Mean Deviation, Standard Deviation, Variance
- Relative Measures: Coefficient of Range, Coefficient of QD, Coefficient of MD, Coefficient of Variation (CV)
- Range = L − S (Largest − Smallest); Coefficient of Range = (L − S)/(L + S)
- Quartile Deviation (QD) = (Q₃ − Q₁)/2; Coefficient of QD = (Q₃ − Q₁)/(Q₃ + Q₁)
- Mean Deviation (MD) = Σ|xᵢ − A|/n (A = mean, median, or mode)
- For series: MD about mean = Σf|xᵢ − x̄|/Σf
- Variance (σ²) = Σ(xᵢ − x̄)²/n (for population); s² = Σ(xᵢ − x̄)²/(n−1) (for sample)
- Standard Deviation (σ) = √Variance = √(Σ(xᵢ − x̄)²/n)
- Shortcut: σ = √(Σxᵢ²/n − x̄²)
- For grouped: σ = √(Σfᵢ(xᵢ − x̄)²/Σfᵢ)
- Step-deviation: σ = h × √(Σfu²/Σf − (Σfu/Σf)²)
- Coefficient of Variation (CV) = (σ/x̄) × 100 — lower CV means more consistency
- Combined SD: σ₁₂ = √[(n₁σ₁² + n₂σ₂² + n₁d₁² + n₂d₂²)/(n₁ + n₂)] where d₁ = x̄₁ − x̄₁₂
Intext Questions
Section 6.1 — Range and Quartile Deviation
Section 6.2 — Mean Deviation
| x | x − x̄ | \ | x − x̄\ | |
|---|---|---|---|---|
| 4 | -4 | 4 | ||
| 6 | -2 | 2 | ||
| 8 | 0 | 0 | ||
| 10 | 2 | 2 | ||
| 12 | 4 | 4 |
| MD = Σ | x − x̄ | /n = (4+2+0+2+4)/5 = 12/5 = 2.4 |
|---|
Chapter-End Exercise
Question 1
Calculate the range and coefficient of range from the following data:
| Class | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|
| f | 5 | 8 | 12 | 7 | 3 |
Range = Upper limit of highest class − Lower limit of lowest class = 60 − 10 = 50
Coefficient of Range = (60 − 10)/(60 + 10) = 50/70 = 0.714 or 71.43%
Question 2
Calculate the quartile deviation and its coefficient from the following:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| f | 6 | 10 | 15 | 10 | 9 |
| Class | f | cf |
|---|---|---|
| 0-10 | 6 | 6 |
| 10-20 | 10 | 16 |
| 20-30 | 15 | 31 |
| 30-40 | 10 | 41 |
| 40-50 | 9 | 50 |
N = 50
Q₁: Q₁ class: cf ≥ N/4 = 12.5. cf = 16 ≥ 12.5, so Q₁ class = 10-20.
Q₁ = 10 + [(12.5 − 6)/10] × 10 = 10 + (6.5/10) × 10 = 10 + 6.5 = 16.5
Q₃: Q₃ class: cf ≥ 3N/4 = 37.5. cf = 41 ≥ 37.5, so Q₃ class = 30-40.
Q₃ = 30 + [(37.5 − 31)/10] × 10 = 30 + (6.5/10) × 10 = 30 + 6.5 = 36.5
QD = (Q₃ − Q₁)/2 = (36.5 − 16.5)/2 = 20/2 = 10
Coefficient of QD = (Q₃ − Q₁)/(Q₃ + Q₁) = 20/53 = 0.377
Question 3
Calculate the mean deviation about the mean for the following distribution:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| f | 5 | 8 | 15 | 10 | 2 |
Step 1: Calculate mean (x̄)
| Class | x (mid-point) | f | fx |
|---|---|---|---|
| 0-10 | 5 | 5 | 25 |
| 10-20 | 15 | 8 | 120 |
| 20-30 | 25 | 15 | 375 |
| 30-40 | 35 | 10 | 350 |
| 40-50 | 45 | 2 | 90 |
| Total | 40 | 960 |
x̄ = 960/40 = 24
Step 2: Calculate mean deviation
| Class | x | f | \ | x − x̄\ | f\ | x − x̄\ | ||
|---|---|---|---|---|---|---|---|---|
| 0-10 | 5 | 5 | 19 | 95 | ||||
| 10-20 | 15 | 8 | 9 | 72 | ||||
| 20-30 | 25 | 15 | 1 | 15 | ||||
| 30-40 | 35 | 10 | 11 | 110 | ||||
| 40-50 | 45 | 2 | 21 | 42 | ||||
| Total | 40 | 334 |
| MD = Σf | x − x̄ | /Σf = 334/40 = 8.35 |
|---|
Question 4
Calculate the standard deviation from the following individual data:
10, 12, 14, 16, 18, 20, 22
Answer (Direct Method):
x̄ = (10+12+14+16+18+20+22)/7 = 112/7 = 16
| x | x − x̄ | (x − x̄)² |
|---|---|---|
| 10 | -6 | 36 |
| 12 | -4 | 16 |
| 14 | -2 | 4 |
| 16 | 0 | 0 |
| 18 | 2 | 4 |
| 20 | 4 | 16 |
| 22 | 6 | 36 |
Σ(x − x̄)² = 112
σ = √(Σ(x − x̄)²/n) = √(112/7) = √16 = 4
Shortcut Method: σ = √(Σx²/n − x̄²)
Σx² = 100 + 144 + 196 + 256 + 324 + 400 + 484 = 1904
σ = √(1904/7 − 256) = √(272 − 256) = √16 = 4
Question 5
Calculate the standard deviation from the following frequency distribution:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| f | 4 | 8 | 12 | 10 | 6 |
Answer (Step-Deviation Method): Let A = 25, h = 10
| Class | x | f | u = (x−25)/10 | fu | fu² |
|---|---|---|---|---|---|
| 0-10 | 5 | 4 | -2 | -8 | 16 |
| 10-20 | 15 | 8 | -1 | -8 | 8 |
| 20-30 | 25 | 12 | 0 | 0 | 0 |
| 30-40 | 35 | 10 | 1 | 10 | 10 |
| 40-50 | 45 | 6 | 2 | 12 | 24 |
| Total | 40 | 6 | 58 |
σ = h × √(Σfu²/Σf − (Σfu/Σf)²)
= 10 × √(58/40 − (6/40)²)
= 10 × √(1.45 − 0.0225)
= 10 × √1.4275
= 10 × 1.1948
= 11.95
Question 6
Calculate the variance and standard deviation:
| x | 5 | 10 | 15 | 20 | 25 |
|---|---|---|---|---|---|
| f | 3 | 5 | 8 | 4 | 2 |
| x | f | fx | x² | fx² |
|---|---|---|---|---|
| 5 | 3 | 15 | 25 | 75 |
| 10 | 5 | 50 | 100 | 500 |
| 15 | 8 | 120 | 225 | 1800 |
| 20 | 4 | 80 | 400 | 1600 |
| 25 | 2 | 50 | 625 | 1250 |
| Total | 22 | 315 | 5225 |
x̄ = Σfx/Σf = 315/22 = 14.32
Variance = (Σfx²/Σf) − x̄² = (5225/22) − (14.32)² = 237.5 − 205.06 = 32.44
σ = √32.44 = 5.70
Question 7
Compare the variability of two batsmen using coefficient of variation:
Batsman A: Mean = 50, SD = 10 Batsman B: Mean = 60, SD = 12
CV(A) = (σ/x̄) × 100 = (10/50) × 100 = 20%
CV(B) = (12/60) × 100 = 20%
Both batsmen have the same coefficient of variation (20%), which means they have equal consistency or equal relative variability, even though Batsman B has a higher average and higher absolute SD.
Question 8
Two factories A and B have the following data. Which is more consistent?
| Factory A | Factory B | |
|---|---|---|
| Mean wage | Rs 500 | Rs 400 |
| SD | Rs 50 | Rs 30 |
CV(A) = (50/500) × 100 = 10%
CV(B) = (30/400) × 100 = 7.5%
Factory B has a lower coefficient of variation (7.5% < 10%), so Factory B is more consistent in wages.
Question 9
Calculate the combined standard deviation:
Group 1: n₁ = 30, x̄₁ = 50, σ₁ = 8 Group 2: n₂ = 20, x̄₂ = 60, σ₂ = 10
Step 1: Combined mean
x̄₁₂ = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂) = (30×50 + 20×60)/50 = (1500 + 1200)/50 = 2700/50 = 54
Step 2: Combined SD
d₁ = x̄₁ − x̄₁₂ = 50 − 54 = −4; d₁² = 16 d₂ = x̄₂ − x̄₁₂ = 60 − 54 = 6; d₂² = 36
σ₁₂ = √[(n₁σ₁² + n₂σ₂² + n₁d₁² + n₂d₂²)/(n₁ + n₂)]
= √[(30×64 + 20×100 + 30×16 + 20×36)/50]
= √[(1920 + 2000 + 480 + 720)/50]
= √[5120/50]
= √102.4
= 10.12
Question 10
Distinguish between absolute and relative measures of dispersion.
| Basis | Absolute Measures | Relative Measures |
|---|---|---|
| Expression | In same units as data | Unit-free (ratio or percentage) |
| Examples | Range, SD, MD, QD | Coefficient of Range, CV, Coefficient of QD |
| Purpose | Measure variability in the same unit | Compare variability across different data sets |
| Use | For a single distribution | For comparing two or more distributions |
| Affected by | Units of measurement | Not affected by units |
Question 11
What are the merits and demerits of standard deviation?
Merits: 1. Based on all observations. 2. Rigidly defined by a mathematical formula. 3. Least affected by sampling fluctuations. 4. Suitable for algebraic treatment — combined SD can be calculated. 5. Used as the basis for other statistical measures (skewness, correlation). 6. Most widely used measure of dispersion.
Demerits: 1. Difficult to understand and calculate compared to range or MD. 2. Affected by extreme values (since deviations are squared). 3. Cannot be calculated for open-ended classes. 4. Requires mean as the reference point.
Question 12
What are the merits and demerits of mean deviation?
Merits: 1. Simple to understand and calculate. 2. Based on all observations. 3. Less affected by extreme values than SD. 4. Can be calculated about mean, median, or mode.
Demerits: 1. Ignores algebraic signs — not suitable for further mathematical treatment. 2. Not rigidly defined (can be calculated about different averages). 3. Cannot be used for combined calculations. 4. Not as popular as standard deviation.
Question 13
Explain the concept of Lorenz Curve.
- The line of equal distribution (45° line) represents perfect equality.
- The farther the Lorenz Curve is from the 45° line, the greater the inequality.
- The area between the 45° line and the Lorenz Curve measures the degree of inequality.
- It is widely used to study income and wealth distribution.
Question 14
Which measure of dispersion is the best and why?
1. It is based on all observations. 2. It has a precise mathematical definition. 3. It is least affected by sampling fluctuations. 4. It is suitable for further algebraic treatment. 5. It forms the basis for advanced statistical techniques like correlation and regression. 6. It allows comparison through the coefficient of variation.
However, for quick estimates or when simplicity is needed, range or QD may be preferred.
Question 15
Calculate the mean deviation about the median for: 15, 20, 25, 30, 35
| x | \ | x − Md\ | |
|---|---|---|---|
| 15 | 10 | ||
| 20 | 5 | ||
| 25 | 0 | ||
| 30 | 5 | ||
| 35 | 10 |
MD about median = (10 + 5 + 0 + 5 + 10)/5 = 30/5 = 6
Internal Links
- [Chapter 6 Guide — Measures of Dispersion](/school/cbse/class/11/economics/cbse-measures-of-dispersion)
- [All Class 11 Economics Solutions](/school/cbse/class/11/economics/solutions)
- [Class 11 Economics Subject Guide](/school/cbse/class/11/economics)