Measures of Dispersion: ISC Class 11 Economics Notes

What Measures of Dispersion is about

Weightage: Measures of Dispersion is part of the unit Statistics of the ISC Class XI Economics syllabus. CISCE gives no Class XI unit weightage for Economics: Paper I (Theory) carries 80 marks, and the school sets the paper.Source: CISCE ISC Economics syllabus, examination year 2028, Class XI paper pattern

Key concepts

Dispersion: meaning, purpose and types of measures

Dispersion is the extent to which the values of a series differ from their average (or from one another). It shows how representative the average is: low dispersion means values are close to the average.

Purposes: to judge the reliability of the average; to compare the variability (consistency) of two or more series; to control variability (quality control, price stability); as a basis for further statistics (correlation, sampling).

Absolute measures are in the units of the data (₹, kg, marks): range, quartile deviation, mean deviation, standard deviation. They cannot compare series in different units or with very different averages.

Relative measures (coefficients) are pure numbers (ratios or percentages): coefficient of range, coefficient of QD, coefficient of MD, coefficient of variation. They allow comparisons.

Qualities of a good measure (as for averages): rigidly defined, easy, based on all values, not unduly affected by extremes, capable of algebraic treatment.

Worked example

Why can't the standard deviations of heights (in cm) and weights (in kg) of a class be compared directly? [1 mark]

Try it first, then show the solutionSolution

1. SDs are absolute measures in different units; a relative measure (coefficient of variation) is needed.

Answer: Because absolute measures carry the units of the data; use the coefficient of variation to compare.

Range

Range = Largest value (L) − Smallest value (S).

Coefficient of range = (L − S) ÷ (L + S).

For grouped data, L = upper limit of the highest class and S = lower limit of the lowest class.

Example: 7, 11, 13, 15, 19, 22, 25, 28, 30, 34, 40: range = 40 − 7 = 33; coefficient = 33 ÷ 47 = 0.70.

Merits: simplest; quick idea of spread; used in quality control and in daily price (high-low) and temperature reports.

Demerits: depends only on two extreme values; ignores all others; greatly affected by one unusual value; cannot be found for open-ended classes.

Worked example

Suppose the daily highest and lowest prices of a share during a week were ₹482 and ₹436. Find the range and its coefficient. [1 mark]

Try it first, then show the solutionSolution

1. Range = 482 − 436 = ₹46; coefficient = 46 ÷ 918 = 0.05.

Answer: Range ₹46; coefficient 0.05.

Quartile deviation (semi-interquartile range)

Quartiles divide ordered data into four equal parts: Q1 (25% below it), Q2 = median, Q3 (75% below it).

Interquartile range = Q3 − Q1. Quartile deviation (QD) = (Q3 − Q1) ÷ 2.

Coefficient of QD = (Q3 − Q1) ÷ (Q3 + Q1).

Positions: individual and discrete series: Q1 = size of the $\frac{N+1}{4}$th item; Q3 = size of the $\frac{3(N+1)}{4}$th item. Continuous series: Q1 class contains the $\frac{N}{4}$th item, and

$Q_1=L_1+\frac{\frac{N}{4}-c.f.}{f}\times i$, $Q_3=L_1+\frac{\frac{3N}{4}-c.f.}{f}\times i$

Example (individual): 7, 11, 13, 15, 19, 22, 25, 28, 30, 34, 40 (N = 11). Q1 = 3rd item = 13; Q3 = 9th item = 30. QD = (30 − 13) ÷ 2 = 8.5; coefficient = 17 ÷ 43 = 0.40.

Example (wages of 50 workers): N ÷ 4 = 12.5 → class 200-300 (c.f. 5, f 8): Q1 = 200 + (12.5 − 5) ÷ 8 × 100 = 293.75. 3N ÷ 4 = 37.5 → class 400-500 (c.f. 28, f 12): Q3 = 400 + (37.5 − 28) ÷ 12 × 100 = 479.17. QD = ₹92.71.

Merits: not affected by extremes; usable with open-ended classes. Demerits: ignores the lowest and highest 25% of values; not suited to algebraic treatment.

Worked example

Using Q1 = ₹293.75 and Q3 = ₹479.17 for the wage data, find the coefficient of quartile deviation. [1 mark]

Try it first, then show the solutionSolution

1. Coefficient = (479.17 − 293.75) ÷ (479.17 + 293.75) = 185.42 ÷ 772.92 = 0.24.

Answer: 0.24.

Mean (average) deviation

Mean deviation (MD) is the arithmetic mean of the absolute deviations of the values from an average (the median or the mean), ignoring signs.

• Individual: $MD=\frac{\sum |D|}{N}$
• Discrete and continuous: $MD=\frac{\sum f|D|}{\sum f}$ (D from class marks in a continuous series)

Coefficient of MD = MD ÷ the average from which it was calculated.

MD is least when taken from the median, so the median is often preferred.

Example: 7, 11, 13, 15, 19, 22, 25, 28, 30, 34, 40; median = 22. |D| = 15, 11, 9, 7, 3, 0, 3, 6, 8, 12, 18; sum = 92; MD = 92 ÷ 11 = 8.36; coefficient = 8.36 ÷ 22 = 0.38.

Wage data: deviations of class marks from the median ₹380: Σf|D| = 5,580; MD = 5,580 ÷ 50 = ₹111.60.

Merits: based on all values; easy to understand; less affected by extremes than SD. Demerits: ignoring signs is mathematically awkward, so it is unsuitable for further algebraic treatment.

Worked example

Find the mean deviation from the mean of 4, 8, 10, 14, 19. [2 marks]

Try it first, then show the solutionSolution

1. Mean = 55 ÷ 5 = 11; |D| = 7, 3, 1, 3, 8; Σ|D| = 22.
2. MD = 22 ÷ 5 = 4.4; coefficient = 4.4 ÷ 11 = 0.4.

Answer: MD = 4.4; coefficient of MD = 0.4.

Standard deviation

Standard deviation (σ) is the square root of the mean of the squared deviations from the arithmetic mean. It is the most widely used measure of dispersion (Karl Pearson, 1893). Its square, σ², is the variance.

Individual series:

• Actual mean method: $\sigma =\sqrt{\frac{\sum x^2}{N}}$ where $x=X-\bar{X}$.
• Assumed mean method: $\sigma =\sqrt{\frac{\sum d^2}{N}-{\left(\frac{\sum d}{N}\right)}^2}$ where $d=X-A$.

Discrete and continuous series (step deviation): $\sigma =\sqrt{\frac{\sum f{d}^{\prime 2}}{N}-{\left(\frac{\sum f{d}^{\prime }}{N}\right)}^2}\times i$

Example (individual): 12, 15, 18, 21, 24: mean 18; x = −6, −3, 0, 3, 6; Σx² = 90; σ = √(90 ÷ 5) = √18 = 4.24.

Example (wages): A = 350, i = 100; d' = −2, −1, 0, 1, 2, 3; f = 5, 8, 15, 12, 6, 4; Σfd' = 18; Σfd'² = 20 + 8 + 0 + 12 + 24 + 36 = 100.

σ = √(100 ÷ 50 − (18 ÷ 50)²) × 100 = √(2 − 0.1296) × 100 = √1.8704 × 100 = ₹136.76.

Properties: not affected by change of origin (adding a constant to every value); affected by change of scale (multiplying every value by k multiplies σ by k).

Merits: rigidly defined; based on all values; suitable for algebraic treatment; least affected by sampling fluctuations. Demerits: harder to compute; gives more weight to extreme values (because deviations are squared).

Worked example

Suppose the number of children in 30 households is: X 1, 2, 3, 4, 5 with f 3, 7, 12, 6, 2. Find the mean and standard deviation. [2 marks]

Try it first, then show the solutionSolution

1. ΣfX = 3 + 14 + 36 + 24 + 10 = 87; mean = 87 ÷ 30 = 2.9.
2. ΣfX² = 3 + 28 + 108 + 96 + 50 = 285; σ = √(285 ÷ 30 − 2.9²) = √(9.5 − 8.41) = √1.09 = 1.04.

Answer: Mean 2.9 children; σ ≈ 1.04.

Coefficient of variation

The coefficient of variation (CV) is the standard deviation expressed as a percentage of the mean:

$CV=\frac{\sigma }{\bar{X}}\times 100$

It is the relative measure based on SD, used to compare the variability, consistency, stability or uniformity of two or more series, even with different units or averages.

Rule: the series with the lower CV is more consistent (more stable, more uniform); the higher CV is more variable.

Example: two batsmen's scores in 8 matches:

• Batsman P: 42, 18, 65, 30, 55, 10, 72, 28 → mean 40, σ = 20.93, CV = 52.34%.
• Batsman Q: 38, 45, 40, 36, 44, 39, 42, 36 → mean 40, σ = 3.20, CV = 8.00%.

Both average 40, but Q is far more consistent (lower CV). P is the more 'risky' selection.

Wage data: CV = 136.76 ÷ 386 × 100 = 35.43%.

Uses in economics: comparing price stability of commodities, inequality of incomes across regions, variability of crop yields, risk of investments.

Worked example

Suppose Firm A's monthly sales have mean ₹80 lakh and SD ₹12 lakh; Firm B's have mean ₹50 lakh and SD ₹9 lakh. Which firm's sales are more stable? [2 marks]

Try it first, then show the solutionSolution

1. CV(A) = 12 ÷ 80 × 100 = 15%; CV(B) = 9 ÷ 50 × 100 = 18%.
2. Lower CV means more stable: Firm A, although its SD is larger.

Answer: Firm A (CV 15% vs 18%): more stable, despite a larger absolute SD.

Measures of Dispersion in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• Dispersion: Range, Quartile Deviation and Mean Deviation
• Standard Deviation, Variance and Coefficient of Variation

Measures of Dispersion: key facts

Range and its coefficient

Range = L − S; Coefficient = (L − S) ÷ (L + S)

Quartile deviation

QD = (Q3 − Q1) ÷ 2; Coefficient of QD = (Q3 − Q1) ÷ (Q3 + Q1)

Q1 at (N + 1)/4 (individual) or N/4 (continuous).

Mean deviation

$MD=\frac{\sum f|D|}{\sum f}$; Coefficient = MD ÷ average used

Standard deviation (step deviation)

$\sigma =\sqrt{\frac{\sum f{d}^{\prime 2}}{N}-{\left(\frac{\sum f{d}^{\prime }}{N}\right)}^2}\times i$

Standard deviation (actual mean)

$\sigma =\sqrt{\frac{\sum x^2}{N}}$, where $x=X-\bar{X}$

Coefficient of variation

$CV=\frac{\sigma }{\bar{X}}\times 100$

Lower CV = more consistent.

Measures of Dispersion: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Measures of Dispersion

The question patterns to expect, and the method that works for each.

Tricks to remember Measures of Dispersion

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Absolute has units; relative has none

  Every absolute measure has a 'coefficient' twin for comparisons.

  Use it when: Choosing a measure.

• QD: half the middle half

  Q3 − Q1 covers the middle 50%; QD is half of it.

  Use it when: Quartile deviation.

• SD: square, mean, root

  Deviations from the mean → square → average → square root.

  Use it when: Standard deviation.

• Low CV, high consistency

  CV is the 'wobble per unit of average'.

  Use it when: Comparing series.

Measures of Dispersion: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section C style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Measures of Dispersion

The slips examiners see most often, and what to write instead.

• ✗Mistake: Comparing series by SD when means differ

  SD is absolute; a larger mean naturally allows a larger SD.

  ✓Do this: Compare CVs.

  Usually costs: 1-2 marks.

• ✗Mistake: Forgetting to multiply by i in step deviation

  d' is in class-width units; the result must be scaled back.

  ✓Do this: Write '× i' in the formula before substituting.

  Usually costs: 1 mark.

• ✗Mistake: Keeping signs in mean deviation

  Signed deviations from the mean sum to zero; MD uses absolute values.

  ✓Do this: Write |D| and drop minus signs.

  Usually costs: 1 mark.

• ✗Mistake: Using N/4 for quartiles of an individual series

  Individual and discrete series use (N + 1)/4; continuous series use N/4.

  ✓Do this: Check the type of series first.

  Usually costs: 1 mark.

Practice questions on Measures of Dispersion

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The coefficient of variation is calculated as:
   (a) mean ÷ SD × 100
   (b) SD ÷ mean × 100
   (c) QD ÷ median
   (d) range ÷ mean

   Hint

   SD relative to mean.

   Answer

   (b) SD ÷ mean × 100.

2. Q2 · 1 mark · Very short answerExam level

   Find the range and coefficient of range of: 45, 62, 38, 70, 55.

   Hint

   L − S; (L − S) ÷ (L + S).

   Answer

   Range = 70 − 38 = 32; coefficient = 32 ÷ 108 = 0.30.

3. Q3 · 2 marks · Short answerExam level

   Distinguish between absolute and relative measures of dispersion.

   Hint

   Units.

   Answer

   Absolute measures (range, QD, MD, SD) are in the units of the data and cannot compare series with different units or averages. Relative measures (coefficients, CV) are pure numbers and allow such comparisons.

4. Q4 · 2 marks · Short answerHarder exam level

   Find the standard deviation of 8, 12, 15, 20, 25 by the actual mean method.

   Hint

   Mean = 16.

   Answer

   Mean = 80 ÷ 5 = 16; x = −8, −4, −1, 4, 9; Σx² = 64 + 16 + 1 + 16 + 81 = 178; σ = √35.6 = 5.97.

5. Q5 · 2 marks · Short answerHarder exam level

   Find the mean deviation from the median of: 3, 9, 6, 12, 15, 8, 10.

   Hint

   Median is the 4th item.

   Answer

   Ordered: 3, 6, 8, 9, 10, 12, 15; median = 9; |D| = 6, 3, 1, 0, 1, 3, 6; Σ = 20; MD = 20 ÷ 7 = 2.86; coefficient = 2.86 ÷ 9 = 0.32.

6. Q6 · 4 marks · Short answerHarder exam level

   Calculate the quartile deviation and its coefficient for the marks of 40 students: 0-10: 3, 10-20: 7, 20-30: 12, 30-40: 10, 40-50: 8.

   Hint

   N/4 = 10, 3N/4 = 30.

   Answer

   c.f. 3, 10, 22, 32, 40. Q1: 10th item, class 10-20: Q1 = 10 + (10 − 3) ÷ 7 × 10 = 20. Q3: 30th item, class 30-40: Q3 = 30 + (30 − 22) ÷ 10 × 10 = 38. QD = 9; coefficient = 18 ÷ 58 = 0.31.

7. Q7 · 4 marks · Short answerHOTS / competency

   Calculate the standard deviation of the marks of 40 students: 0-10: 3, 10-20: 7, 20-30: 12, 30-40: 10, 40-50: 8 (mean 28.25), and the coefficient of variation.

   Hint

   Step deviation with A = 25, i = 10.

   Answer

   d' = −2, −1, 0, 1, 2; fd' = −6, −7, 0, 10, 16 (Σ = 13); fd'² = 12, 7, 0, 10, 32 (Σ = 61). σ = √(61 ÷ 40 − (13 ÷ 40)²) × 10 = √(1.525 − 0.105625) × 10 = √1.419375 × 10 = 11.91. CV = 11.91 ÷ 28.25 × 100 = 42.17%.

8. Q8 · 2 marks · Short answerHOTS / competency

   Suppose the mean and SD of a group of 25 workers' wages are ₹400 and ₹50. If each wage is raised by 10% and then a flat ₹20 bonus is added, find the new mean, SD and CV.

   Hint

   Scale affects SD; origin does not.

   Answer

   Mean: 400 × 1.1 + 20 = ₹460. SD: 50 × 1.1 = ₹55 (the bonus does not change it). CV = 55 ÷ 460 × 100 = 11.96%.

9. Q9 · 2 marks · Short answerHarder exam level

   Suppose two factories' daily outputs have: Factory A: mean 500 units, SD 45; Factory B: mean 320 units, SD 32. Which is more consistent? Also explain why comparing SDs alone would mislead.

   Hint

   CV.

   Answer

   CV(A) = 45 ÷ 500 × 100 = 9%; CV(B) = 32 ÷ 320 × 100 = 10%. A is more consistent. Comparing SDs (45 vs 32) would wrongly suggest B is steadier, because A's output is much larger on average.

10. Q10 · 6 marks · Long answerHOTS / competency

    'Standard deviation is the best measure of dispersion.' Discuss with reference to its merits and demerits compared with range, QD and MD.

    Hint

    Rigid, all values, algebra; but extremes.

    Answer

    Merits of SD: rigidly defined; based on all observations (unlike range and QD); uses squared deviations, so signs are handled correctly (unlike MD, which ignores signs); capable of algebraic treatment (combined SD, variance analysis); least affected by sampling fluctuations; basis of CV, correlation and further statistics. Demerits: harder to calculate; gives extra weight to extreme values because deviations are squared; cannot be calculated for open-ended classes. Range is quick but uses only two values; QD resists extremes but ignores half the data; MD uses all values but is mathematically awkward. Conclusion: SD is generally the best for further analysis and comparison (through CV), but QD or MD may be preferred for skewed data with extreme values or open-ended classes.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Coefficient of range?Show answer

   (L − S) ÷ (L + S).

2. 2.Quartile deviation?Show answer

   (Q3 − Q1) ÷ 2.

3. 3.MD is least from which average?Show answer

   The median.

4. 4.Effect of adding a constant on SD?Show answer

   None (independent of change of origin).

5. 5.CV formula?Show answer

   σ ÷ mean × 100.

6. 6.Lower CV means?Show answer

   More consistent or stable series.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Dispersion = scatter around the average; shows how representative it is.
• ☐Range L − S; coefficient (L − S) ÷ (L + S).
• ☐QD = (Q3 − Q1) ÷ 2; coefficient (Q3 − Q1) ÷ (Q3 + Q1).
• ☐MD = Σf|D| ÷ N; coefficient MD ÷ average.
• ☐σ = √(Σfd'²/N − (Σfd'/N)²) × i; unaffected by origin, affected by scale.
• ☐CV = σ ÷ mean × 100; lower CV = more consistent.

Measures of Dispersion: frequently asked questions