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This page provides complete NCERT exercise solutions for CBSE Class 11 Economics Chapter 5 — Measures of Central Tendency. 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 focus on understanding the method, not just the final result.
Key Concepts (Quick Recap)
- Central Tendency: a single value that represents the entire distribution
- Arithmetic Mean (x̄): sum of all values divided by number of observations
- Individual series: x̄ = Σx/n
- Discrete series: x̄ = Σfx/Σf
- Continuous series: x̄ = Σfx/Σf (x = mid-point of class)
- Shortcut method: x̄ = A + (Σfd/Σf) where d = x − A
- Step-deviation method: x̄ = A + (Σfu/Σf) × h where u = (x − A)/h
- Median (Md): the middle value when data is arranged in order
- Individual series (odd n): Md = (n+1)/2th value
- Individual series (even n): Md = average of n/2 and (n/2 + 1)th values
- Continuous series: Md = L + [(N/2 − cf)/f] × h
- Mode (Mo): the value that occurs most frequently
- Individual/discrete series: most frequent value
- Continuous series: Mo = L + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
- Empirical Relationship: Mo = 3Md − 2x̄ (for moderately skewed distributions)
- Weighted Mean: x̄w = Σwx/Σw
- Combined Mean: x̄₁₂ = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂)
- Partition Values: Quartiles (Q₁, Q₂, Q₃), Deciles (D₁...D₉), Percentiles (P₁...P₉₉)
Intext Questions
Section 5.1 — Arithmetic Mean
Σx = 6 + 8 + 10 + 12 + 14 = 50, n = 5
x̄ = 50/5 = 10
| x | d = x − A |
|---|---|
| 95 | -2 |
| 96 | -1 |
| 97 | 0 |
| 98 | 1 |
| 99 | 2 |
| 100 | 3 |
Σd = −2 − 1 + 0 + 1 + 2 + 3 = 3, n = 6
x̄ = A + Σd/n = 97 + 3/6 = 97 + 0.5 = 97.5
Section 5.2 — Median
Md = (n+1)/2 = (5+1)/2 = 3rd value = 7
Md = average of n/2th and (n/2 + 1)th values = average of 3rd and 4th values = (6 + 8)/2 = 7
Chapter-End Exercise
Question 1
Calculate the arithmetic mean from the following data:
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| No. of Students | 5 | 8 | 15 | 10 | 2 |
Answer (Direct Method):
| Marks (Class) | Mid-point (x) | Frequency (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 | Σf = 40 | Σfx = 960 |
x̄ = Σfx/Σf = 960/40 = 24 marks
Question 2
Calculate the mean using the step-deviation method:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| f | 4 | 8 | 12 | 10 | 6 |
| Class | Mid-point (x) | f | u = (x−25)/10 | fu |
|---|---|---|---|---|
| 0-10 | 5 | 4 | -2 | -8 |
| 10-20 | 15 | 8 | -1 | -8 |
| 20-30 | 25 | 12 | 0 | 0 |
| 30-40 | 35 | 10 | 1 | 10 |
| 40-50 | 45 | 6 | 2 | 12 |
| Total | Σf = 40 | Σfu = 6 |
x̄ = A + (Σfu/Σf) × h = 25 + (6/40) × 10 = 25 + 1.5 = 26.5
Question 3
Calculate the median from the following data:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| f | 6 | 9 | 15 | 12 | 8 |
| Class | f | Cumulative Frequency (cf) |
|---|---|---|
| 0-10 | 6 | 6 |
| 10-20 | 9 | 15 |
| 20-30 | 15 | 30 |
| 30-40 | 12 | 42 |
| 40-50 | 8 | 50 |
N = 50, N/2 = 25
Median class: the class where cf ≥ N/2 = 25. cf = 30 ≥ 25, so median class = 20-30.
L = 20, cf (of preceding class) = 15, f = 15, h = 10
Md = L + [(N/2 − cf)/f] × h = 20 + [(25 − 15)/15] × 10 = 20 + (10/15) × 10 = 20 + 6.67 = 26.67
Question 4
Calculate the mode from the following data:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| f | 5 | 8 | 20 | 12 | 7 |
L = 20, f₁ = 20 (frequency of modal class), f₀ = 8 (frequency of preceding class), f₂ = 12 (frequency of succeeding class), h = 10
Mo = L + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
Mo = 20 + [(20 − 8)/(2×20 − 8 − 12)] × 10
Mo = 20 + [12/(40 − 20)] × 10 = 20 + (12/20) × 10 = 20 + 6 = 26
Question 5
Calculate mean, median, and mode from the following data and verify the empirical relationship:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| f | 5 | 8 | 15 | 10 | 2 |
Mean (from Question 1): x̄ = 24
Median:
N = 40, N/2 = 20. cf = 13 < 20, next cf = 28 ≥ 20. Median class = 20-30.
Md = 20 + [(20 − 13)/15] × 10 = 20 + (7/15) × 10 = 20 + 4.67 = 24.67
Mode: Modal class = 20-30 (highest f = 15)
Mo = 20 + [(15 − 8)/(2×15 − 8 − 10)] × 10 = 20 + [7/(30 − 18)] × 10 = 20 + (7/12) × 10 = 20 + 5.83 = 25.83
Verification of empirical relationship: Mo = 3Md − 2x̄
RHS = 3(24.67) − 2(24) = 74.01 − 48 = 26.01
LHS (Mo) = 25.83 ≈ 26.01 (approximately equal, confirming the empirical relationship)
Question 6
The mean of 20 observations is 15. If one observation 20 is removed, find the new mean.
Total of 20 observations = 20 × 15 = 300
After removing 20: Total = 300 − 20 = 280, n = 19
New mean = 280/19 = 14.74
Question 7
The mean of 25 observations is 50. One observation 70 was misread as 20. Find the correct mean.
Incorrect total = 25 × 50 = 1250
Correct total = 1250 − 20 + 70 = 1300
Correct mean = 1300/25 = 52
Question 8
Calculate the weighted mean from the following data:
| Item | Price (x) | Weight (w) |
|---|---|---|
| A | 10 | 2 |
| B | 20 | 3 |
| C | 30 | 5 |
| D | 40 | 4 |
| Item | x | w | wx |
|---|---|---|---|
| A | 10 | 2 | 20 |
| B | 20 | 3 | 60 |
| C | 30 | 5 | 150 |
| D | 40 | 4 | 160 |
| Total | Σw = 14 | Σwx = 390 |
x̄w = Σwx/Σw = 390/14 = 27.86
Question 9
The mean of 100 items is 50 and the mean of 50 items is 30. Find the combined mean.
n₁ = 100, x̄₁ = 50, n₂ = 50, x̄₂ = 30
Combined mean = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂) = (100×50 + 50×30)/(100 + 50) = (5000 + 1500)/150 = 6500/150 = 43.33
Question 10
Find the missing frequency if the mean of the following distribution is 18:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| f | 5 | f₁ | 10 | f₂ | 5 |
Total frequency = 40.
| Class | x | f | fx |
|---|---|---|---|
| 0-10 | 5 | 5 | 25 |
| 10-20 | 15 | f₁ | 15f₁ |
| 20-30 | 25 | 10 | 250 |
| 30-40 | 35 | f₂ | 35f₂ |
| 40-50 | 45 | 5 | 225 |
Mean = Σfx/Σf = 18
(25 + 15f₁ + 250 + 35f₂ + 225)/40 = 18
500 + 15f₁ + 35f₂ = 720
15f₁ + 35f₂ = 220 ... (ii)
From (i): f₂ = 20 − f₁
Substituting: 15f₁ + 35(20 − f₁) = 220
15f₁ + 700 − 35f₁ = 220
−20f₁ = −480
f₁ = 24 — but this exceeds 20, so let me re-examine. Since f₁ + f₂ = 20 and both must be positive, let's verify: with f₁ + f₂ = 20, 15f₁ + 35f₂ = 220.
35(20 − f₁) + 15f₁ = 220 → 700 − 35f₁ + 15f₁ = 220 → 700 − 20f₁ = 220 → 20f₁ = 480 → f₁ = 24.
Since f₁ = 24 > 20, there may be a data inconsistency. However, solving both equations simultaneously: f₁ = 24, f₂ = −4. This indicates an error in the given data, but the method is correct.
Question 11
Calculate the median from the following discrete series:
| x | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| f | 4 | 6 | 10 | 8 | 2 |
| x | f | Cumulative Frequency |
|---|---|---|
| 10 | 4 | 4 |
| 20 | 6 | 10 |
| 30 | 10 | 20 |
| 40 | 8 | 28 |
| 50 | 2 | 30 |
The 15.5th value lies in the group where cf = 20, which corresponds to x = 30
Question 12
Calculate Q₁, Q₂, and Q₃ from the following:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| f | 4 | 8 | 12 | 10 | 6 |
Q₁: Q₁ class: where cf ≥ N/4 = 10. cf = 12 ≥ 10, so Q₁ class = 10-20.
Q₁ = L + [(N/4 − cf)/f] × h = 10 + [(10 − 4)/8] × 10 = 10 + (6/8) × 10 = 10 + 7.5 = 17.5
Q₂ (Median): cf ≥ N/2 = 20. cf = 24 ≥ 20, so Q₂ class = 20-30.
Q₂ = 20 + [(20 − 12)/12] × 10 = 20 + (8/12) × 10 = 20 + 6.67 = 26.67
Q₃: cf ≥ 3N/4 = 30. cf = 34 ≥ 30, so Q₃ class = 30-40.
Q₃ = 30 + [(30 − 24)/10] × 10 = 30 + (6/10) × 10 = 30 + 6 = 36
Question 13
What are the merits and demerits of arithmetic mean?
Merits: 1. Simple to understand and easy to calculate. 2. Based on all observations — representative of the entire data. 3. Rigorously defined by a mathematical formula. 4. Suitable for algebraic treatment — combined mean can be calculated. 5. Stable measure — does not vary much with sampling fluctuations.
Demerits: 1. Affected by extreme values — outliers distort the mean. 2. Cannot be calculated for open-ended classes. 3. Cannot be determined graphically. 4. May not be an actual value in the data (e.g., mean = 4.5 children). 5. Not suitable for qualitative data. 6. Gives misleading results for skewed distributions.
Question 14
What are the merits and demerits of median?
Merits: 1. Simple to understand and easy to calculate. 2. Not affected by extreme values. 3. Can be calculated for open-ended classes. 4. Can be determined graphically (using ogive). 5. Suitable for qualitative data that can be ranked. 6. Better measure for skewed distributions.
Demerits: 1. Not based on all observations — ignores extreme values. 2. Not suitable for algebraic treatment. 3. Affected by sampling fluctuations more than mean. 4. Requires data to be arranged in order (time-consuming for large data). 5. Not rigidly defined for grouped data (depends on interpolation formula).
Question 15
What are the merits and demerits of mode?
Merits: 1. Simple to understand — the most common value. 2. Not affected by extreme values. 3. Can be calculated for open-ended classes. 4. Represents the most typical value. 5. Can be determined graphically (histogram).
Demerits: 1. Not rigidly defined — can be ill-defined (no mode, multiple modes). 2. Not based on all observations. 3. Not suitable for algebraic treatment. 4. Affected by sampling fluctuations. 5. Difficult to calculate for grouped data (requires complex formula).
Question 16
When is the geometric mean preferred over the arithmetic mean?
1. Dealing with ratios, rates, or percentages (e.g., growth rates, index numbers). 2. Calculating average rate of change over time (compound growth). 3. Data has wide variation and arithmetic mean is distorted by extreme values. 4. Calculating average of index numbers (e.g., Fisher's index).
Question 17
The average marks of 30 boys is 60 and average marks of 20 girls is 70. Find the combined average.
n₁ = 30, x̄₁ = 60, n₂ = 20, x̄₂ = 70
Combined mean = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂) = (30×60 + 20×70)/(30 + 20) = (1800 + 1400)/50 = 3200/50 = 64
Question 18
Which measure of central tendency is the best and why?
- Mean is best for symmetric distributions and when all values matter.
- Median is best for skewed distributions and when extreme values exist.
- Mode is best when we need the most typical or common value.
However, the arithmetic mean is most widely used because it is based on all observations, has mathematical properties, and is suitable for further statistical analysis.
Question 19
Calculate mean, median, and mode for the following individual series:
12, 15, 11, 18, 13, 15, 10, 15, 14, 16
Mean: x̄ = Σx/n = (12+15+11+18+13+15+10+15+14+16)/10 = 139/10 = 13.9
Median: Arrange in order: 10, 11, 12, 13, 14, 15, 15, 15, 16, 18. n = 10 (even)
Md = average of 5th and 6th values = (14 + 15)/2 = 14.5
Mode: 15 appears 3 times (highest frequency). Mode = 15
Question 20
Explain the partition values — quartiles, deciles, and percentiles.
Quartiles divide data into 4 equal parts:
- Q₁ (first quartile) = value below which 25% of data lies. Position = (N+1)/4
- Q₂ (second quartile) = median. Position = (N+1)/2
- Q₃ (third quartile) = value below which 75% of data lies. Position = 3(N+1)/4
Deciles divide data into 10 equal parts:
- D₁ to D₉. Position of Dₖ = k(N+1)/10
Percentiles divide data into 100 equal parts:
- P₁ to P₉₉. Position of Pₖ = k(N+1)/100
For grouped data, the formulas use the same interpolation method as the median: Qₖ = L + [(kN/4 − cf)/f] × h
Internal Links
- [Chapter 5 Guide — Measures of Central Tendency](/school/cbse/class/11/economics/cbse-measures-of-central-tendency)
- [All Class 11 Economics Solutions](/school/cbse/class/11/economics/solutions)
- [Class 11 Economics Subject Guide](/school/cbse/class/11/economics)