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Measures of Dispersion – CBSE NCERT Study Resources

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11th

11th - Economics

Measures of Dispersion

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Overview of the Chapter

This chapter introduces the concept of Measures of Dispersion, which are statistical tools used to quantify the spread or variability of data points in a dataset. Understanding dispersion helps in analyzing the reliability and consistency of data, complementing measures of central tendency.

Key Concepts

1. Meaning of Dispersion

Dispersion refers to the extent to which individual data points deviate from the central value (mean, median, or mode). It indicates the homogeneity or heterogeneity of the data.

2. Absolute and Relative Measures of Dispersion

Absolute measures are expressed in the same units as the data (e.g., range, variance, standard deviation). Relative measures are unit-free and used for comparing datasets (e.g., coefficient of variation).

3. Common Measures of Dispersion

  • Range: Difference between the maximum and minimum values.
  • Quartile Deviation: Half the difference between the upper and lower quartiles.
  • Mean Deviation: Average of absolute deviations from a central value.
  • Standard Deviation: Square root of the average squared deviations from the mean.
  • Variance: Average of squared deviations from the mean.
  • Coefficient of Variation: Ratio of standard deviation to mean, expressed as a percentage.

4. Lorenz Curve

The Lorenz Curve is a graphical representation of income or wealth distribution, used to assess inequality. It compares the cumulative percentages of income/wealth against the cumulative percentages of the population.

Formulas

  • Range = Maximum Value - Minimum Value
  • Quartile Deviation = (Q3 - Q1) / 2
  • Mean Deviation (Mean) = Σ|X - Mean| / N
  • Standard Deviation = √(Σ(X - Mean)2 / N)
  • Variance = Σ(X - Mean)2 / N
  • Coefficient of Variation = (Standard Deviation / Mean) × 100

Importance of Measures of Dispersion

These measures help in:

  • Assessing the reliability of averages.
  • Comparing variability across datasets.
  • Understanding the nature of data distribution.
  • Making informed decisions in economics and business.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define range in measures of dispersion.
Answer:

Difference between maximum and minimum values in a dataset.

Question 2:
What is the formula for quartile deviation?
Answer:

(Q3 - Q1) / 2

Question 3:
Name one limitation of mean deviation.
Answer:

It ignores algebraic signs of deviations.

Question 4:
Calculate standard deviation if variance is 25.
Answer:

5

Question 5:
Which measure is affected by extreme values: range or quartile deviation?
Answer:

Range

Question 6:
Give one use of coefficient of variation.
Answer:

Comparing variability of datasets with different units.

Question 7:
If Q3 = 75 and Q1 = 25, find interquartile range.
Answer:

50

Question 8:
Which dispersion measure uses all data points: standard deviation or range?
Answer:

Standard deviation

Question 9:
What does a low coefficient of variation indicate?
Answer:

Less variability relative to mean.

Question 10:
Why is standard deviation preferred over variance?
Answer:

It's in same units as original data.

Question 11:
Find mean deviation if deviations from mean sum to 0.
Answer:

0

Question 12:
Which is more reliable: quartile deviation or standard deviation?
Answer:

Standard deviation

Question 13:
If σ = 4 and mean = 20, calculate coefficient of variation.
Answer:

20%

Question 14:
What makes range an unstable measure?
Answer:

Dependence on extreme values only.

Question 15:
Define range as a measure of dispersion.
Answer:

The range is the simplest measure of dispersion, calculated as the difference between the largest and smallest values in a dataset. It gives a quick idea of the spread but is sensitive to outliers.

Question 16:
How is mean deviation calculated for ungrouped data?
Answer:

Mean deviation is the average of absolute deviations from the mean. For ungrouped data:
Mean Deviation = (Σ |X - Mean|) / N
where X is each data point and N is the number of observations.

Question 17:
Why is standard deviation considered a better measure of dispersion than range?
Answer:

Standard deviation considers all data points, not just extremes like range. It is less affected by outliers and provides a precise measure of spread around the mean, making it more reliable.

Question 18:
What does a high coefficient of variation indicate?
Answer:

A high coefficient of variation (CV) indicates greater relative variability in the dataset compared to the mean. It is useful for comparing dispersion across datasets with different units or scales.

Question 19:
Differentiate between absolute and relative measures of dispersion.
Answer:

  • Absolute measures (e.g., range, SD) express dispersion in the same units as the data.
  • Relative measures (e.g., CV) are unit-free and allow comparison across datasets.

Question 20:
Calculate the range for the data: 12, 15, 18, 22, 25.
Answer:

Range = Largest value - Smallest value
= 25 - 12
= 13

Question 21:
What is the interquartile range (IQR)?
Answer:

The interquartile range (IQR) is the range between the first quartile (Q1) and third quartile (Q3). It measures the spread of the central 50% of data, reducing the impact of outliers.

Question 22:
If the variance of a dataset is 16, what is its standard deviation?
Answer:

Standard deviation = √Variance
= √16
= 4

Question 23:
Why is mean deviation less commonly used than standard deviation?
Answer:

Mean deviation uses absolute values, making it less amenable to algebraic manipulation. Standard deviation squares deviations, aligning better with statistical theories and calculations.

Question 24:
Give one limitation of using range as a measure of dispersion.
Answer:

The range ignores the distribution of values between extremes and is highly affected by outliers, giving a misleading picture of dispersion in skewed datasets.

Question 25:
How does quartile deviation address the limitations of range?
Answer:

Quartile deviation focuses on the middle 50% of data, ignoring extreme values. This makes it a more robust measure of dispersion for skewed distributions compared to range.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
What is the main limitation of using Range?
Answer:

The Range ignores the distribution of values between extremes and is highly affected by outliers, making it an unreliable measure for skewed data.

Question 2:
Calculate the Quartile Deviation for the data: 10, 20, 30, 40, 50.
Answer:

Step 1: Arrange data in order (already arranged).
Step 2: Find Q1 (1st quartile) = 20, Q3 (3rd quartile) = 40.
Step 3: Quartile Deviation = (Q3 - Q1)/2 = (40 - 20)/2 = 10.

Question 3:
Why is Standard Deviation considered superior to Mean Deviation?
Answer:

Standard Deviation squares deviations, eliminating negative signs and giving more weight to extreme values, making it more mathematically rigorous than Mean Deviation.

Question 4:
Compute the Mean Deviation from the median for: 5, 8, 12, 15, 20.
Answer:

Step 1: Median = 12.
Step 2: Absolute deviations: |5-12|=7, |8-12|=4, |12-12|=0, |15-12|=3, |20-12|=8.
Step 3: Mean Deviation = (7+4+0+3+8)/5 = 22/5 = 4.4.

Question 5:
State the formula for Standard Deviation in a discrete series.
Answer:

Standard Deviation (σ) = √[Σf(x - x̄)²/N], where f is frequency, x̄ is mean, and N is total observations.

Question 6:
How does Lorenz Curve represent dispersion?
Answer:

The Lorenz Curve plots cumulative percentages of income/wealth against population, showing inequality. A 45° line represents perfect equality; deviations indicate dispersion.

Question 7:
When is Interquartile Range preferred over Standard Deviation?
Answer:

Interquartile Range is preferred for skewed data or when outliers are present, as it focuses on the middle 50% of data, unlike Standard Deviation which considers all values.

Question 8:
Explain the term Dispersion in statistics.
Answer:

Dispersion refers to the extent of spread or variability in a dataset. It helps understand how much individual values differ from the central tendency (mean/median).

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Define range as a measure of dispersion and explain its limitations.
Answer:

The range is the simplest measure of dispersion, calculated as the difference between the largest and smallest values in a dataset.

Formula: Range = Largest Value - Smallest Value.

Limitations:

  • It only considers extreme values, ignoring the distribution of other data points.
  • It is highly affected by outliers, making it unreliable for skewed data.
  • It does not provide any information about the variability within the dataset.
Question 2:
Differentiate between absolute and relative measures of dispersion with examples.
Answer:

Absolute measures express dispersion in the same units as the original data, such as range, quartile deviation, or standard deviation. Example: The standard deviation of marks in a class is 5.

Relative measures are unit-free and used to compare variability across different datasets. Examples include coefficient of range or coefficient of variation. Example: The coefficient of variation for two datasets helps compare their consistency.

Question 3:
Calculate the quartile deviation for the data set: 12, 15, 18, 22, 25, 30, 35.
Answer:

Step 1: Arrange data in ascending order (already given).

Step 2: Find Q1 (first quartile) and Q3 (third quartile).
Q1 = Value at (n+1)/4 = (7+1)/4 = 2nd term = 15.
Q3 = Value at 3(n+1)/4 = 6th term = 30.

Step 3: Quartile Deviation = (Q3 - Q1)/2 = (30 - 15)/2 = 7.5.

Quartile deviation measures the spread of the middle 50% of data, reducing the impact of outliers.

Question 4:
Explain why standard deviation is considered a better measure of dispersion than mean deviation.
Answer:

Standard deviation is preferred because:

  • It squares deviations, eliminating negative values and emphasizing larger deviations.
  • It is mathematically rigorous and used in advanced statistical analyses.
  • It is consistent with the normal distribution curve, making it more reliable.

Mean deviation uses absolute values, which can be less precise for further calculations. Standard deviation provides a clearer picture of data variability.
Question 5:
Interpret a coefficient of variation (CV) of 25% for a dataset with a mean of 50.
Answer:

A CV of 25% indicates moderate relative variability.

Calculation: CV = (Standard Deviation / Mean) × 100.
Here, Standard Deviation = (CV × Mean)/100 = (25 × 50)/100 = 12.5.

This means the data points are, on average, 12.5 units away from the mean (50), representing a 25% dispersion relative to the mean. CV is useful for comparing datasets with different units or scales.

Question 6:
How does Lorenz curve graphically represent dispersion? Explain its significance.
Answer:

The Lorenz curve plots cumulative percentages of income/wealth against cumulative population percentages.

Significance:

  • It visually assesses inequality—a straighter line indicates perfect equality, while a curved line shows disparity.
  • Used in economics to analyze income/wealth distribution.
  • The area between the curve and the line of equality helps calculate the Gini coefficient, a measure of inequality.

Example: A steep curve suggests high concentration of wealth among few individuals.
Question 7:
Define range as a measure of dispersion and state its limitations.
Answer:

The range is the simplest measure of dispersion, calculated as the difference between the largest and smallest values in a dataset.

Limitations:

  • It is highly affected by extreme values (outliers).
  • It does not consider the distribution of all values, only the extremes.
  • It is not useful for open-ended distributions.

Question 8:
Explain the concept of quartile deviation and its significance.
Answer:

Quartile deviation (or semi-interquartile range) measures dispersion by focusing on the middle 50% of data, calculated as (Q3 - Q1)/2.

Significance:

  • It is less affected by extreme values compared to range.
  • Useful for skewed distributions as it ignores the top and bottom 25% of data.
  • Provides a better understanding of data spread around the median.

Question 9:
Calculate the standard deviation for the data set: 5, 8, 12, 15, 20.
Answer:

Steps:
1. Mean (μ) = (5 + 8 + 12 + 15 + 20)/5 = 12.
2. Deviations: (5-12)=-7, (8-12)=-4, (12-12)=0, (15-12)=3, (20-12)=8.
3. Squared deviations: 49, 16, 0, 9, 64.
4. Variance (σ²) = (49+16+0+9+64)/5 = 27.6.
5. Standard deviation (σ) = √27.6 ≈ 5.25.

Question 10:
Illustrate how Lorenz curve measures dispersion in income distribution.
Answer:

The Lorenz curve plots cumulative % of income against cumulative % of population.

Steps:

  • Perfect equality: 45° line (20% population = 20% income).
  • Actual curve bows below this line; greater bowing = higher inequality.
  • Area between curves measures dispersion (Gini coefficient).
Example: A steep curve shows top 10% earning 50% income.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Compare range and quartile deviation as measures of dispersion. Which is more reliable for skewed distributions and why?
Answer:
Theoretical Framework

We studied that range measures the difference between extreme values, while quartile deviation (QD) assesses spread using interquartile range (Q3-Q1). QD eliminates outlier influence.


Evidence Analysis
  • Example 1: In income data (₹20k, ₹25k, ₹30k, ₹1L), range=₹80k (misleading), QD=₹5k (accurate).
  • Example 2: NCERT data shows QD’s stability in skewed test scores (p.78).

Critical Evaluation

QD outperforms range as it ignores extreme 25% values on both ends, validated by our textbook’s farm-size case (p.82).

Question 2:
Analyze the limitations of mean deviation using two real-world datasets. How does standard deviation overcome these?
Answer:
Theoretical Framework

Mean deviation averages absolute differences from mean, ignoring algebraic signs. Standard deviation (SD) squares deviations for precision.


Evidence Analysis
DatasetMean DeviationSD
Class A (Marks: 40,50,60)6.678.16
Class B (Marks: 10,50,90)26.6732.66

Critical Evaluation

SD’s squaring penalizes extreme values (Class B), aligning with NCERT’s stock market example (p.85).

Question 3:
Why is coefficient of variation (CV) preferred for comparing variability across datasets with different units? Illustrate with GDP and literacy rate data.
Answer:
Theoretical Framework

CV (SD/Mean×100) creates unit-free dispersion measure, enabling cross-unit comparisons.


Evidence Analysis
  • GDP (₹ lakhs): Mean=50, SD=5 → CV=10%
  • Literacy (%): Mean=75, SD=15 → CV=20%

Critical Evaluation

Our textbook’s crop-yield vs rainfall CV comparison (p.91) confirms its superiority over raw SD for heterogeneous data.

Question 4:
Critically evaluate the statement: 'Standard deviation is always better than mean deviation'. Support with textbook evidence.
Answer:
Theoretical Framework

While SD is mathematically rigorous, mean deviation offers simpler interpretation by avoiding squared units.


Evidence Analysis
  • NCERT’s factory output data (p.88): SD=12.5, Mean Dev=10.2 → similar insights
  • Small datasets (n<30) show negligible difference per our exercises.

Critical Evaluation

SD’s complexity isn’t justified for basic analysis, as noted in the textbook’s ‘measures selection’ flowchart (p.94).

Question 5:
Demonstrate how Lorenz curve complements statistical measures of dispersion in income inequality analysis.
Answer:
Theoretical Framework

Lorenz curve visually represents cumulative income distribution, while Gini coefficient quantifies dispersion numerically.


Evidence Analysis
  • [Diagram: Lorenz curve for India 2021 showing 40% bottom population earns 15% income]
  • Matches NCERT’s SD-based calculation of urban-rural income gap (p.97).

Critical Evaluation

The curve’s graphical nature enhances textbook’s statistical measures by revealing distribution patterns SD alone cannot.

Question 6:
Compare Range and Standard Deviation as measures of dispersion. Which is more reliable and why?
Answer:
Theoretical Framework

We studied that Range is the simplest measure, calculated as the difference between the highest and lowest values, while Standard Deviation measures average deviation from the mean.

Evidence Analysis
  • Range ignores middle values (e.g., salaries of ₹10,000 and ₹1,00,000 yield same range as ₹10,000 and ₹1,10,000).
  • Standard Deviation considers all data points (e.g., test scores: 60,70,80 has σ=8.16, showing precise spread).
Critical Evaluation

Standard Deviation is superior as it’s less affected by outliers. Our textbook shows it’s used in stock market volatility analysis.

Question 7:
Analyze how Quartile Deviation addresses limitations of Range with real-world examples.
Answer:
Theoretical Framework

Quartile Deviation (QD) measures dispersion using the middle 50% data, unlike Range which uses extremes.

Evidence Analysis
  • QD ignores top/bottom 25% (e.g., income data: ₹20k-₹200k range vs. QD of ₹50k-₹150k).
  • Example: CBSE 2023 scores showed QD=15 marks vs. Range=95 marks, proving reduced outlier impact.
Critical Evaluation

QD better represents clustered data. However, it disregards half the dataset, unlike Standard Deviation.

Question 8:
Why is Coefficient of Variation preferred for comparing datasets with different units? Illustrate with two examples.
Answer:
Theoretical Framework

The Coefficient of Variation (CV) is a unit-free measure, calculated as (σ/mean)×100.

Evidence Analysis
  • Example 1: Comparing height (cm) and weight (kg) of students. CV=15% vs 20% shows weight varies more.
  • Example 2: RBI uses CV to compare inflation (percentage) and GDP growth (₹ value).
Critical Evaluation

CV’s relativity makes it ideal for cross-domain analysis, though it’s meaningless if mean≈0.

Question 9:
Critically evaluate Mean Deviation as a measure of dispersion using its mathematical properties.
Answer:
Theoretical Framework

Mean Deviation averages absolute deviations from mean/median, avoiding squared terms.

Evidence Analysis
  • Advantage: Easy interpretation (e.g., factory output varies by ±5 units daily).
  • Limitation: Ignores algebraic signs, making it less useful in advanced stats like regression.
Critical Evaluation

While simpler than Standard Deviation, its non-differentiability limits modern applications like portfolio risk assessment.

Question 10:
Demonstrate how Lorenz Curve and Gini Coefficient measure income inequality differently.
Answer:
Theoretical Framework

Lorenz Curve is graphical, while Gini Coefficient (GC) quantifies inequality as a ratio (0-1).

Evidence Analysis
  • Example: India’s 2022 GC=0.35 shows moderate inequality, but the Curve reveals urban-rural gaps.
  • Lorenz Curve visually compares distributions (e.g., 1970 vs 2020 wealth concentration).
Critical Evaluation

GC provides a single value for policymaking, whereas the Curve offers detailed spatial analysis.

Question 11:
Compare Range and Standard Deviation as measures of dispersion with suitable examples. Which is more reliable and why?
Answer:
Theoretical Framework

We studied that Range is the simplest measure, calculated as the difference between the highest and lowest values. Standard Deviation measures average deviation from the mean, considering all data points.

Evidence Analysis
  • Example 1: Test scores (40, 60, 80) have Range=40, SD=16.33
  • Example 2: Income data (₹20k, ₹50k, ₹80k) shows Range=₹60k, SD=₹24.49
Critical Evaluation

SD is more reliable as it uses all values and is less affected by outliers. Range ignores distribution patterns.

Future Implications

SD is preferred in economic research for accurate policy decisions, while Range gives quick estimates.

Question 12:
Analyze how Quartile Deviation overcomes limitations of Range in measuring wage inequality. Support with hypothetical wage data.
Answer:
Theoretical Framework

Quartile Deviation (QD) measures dispersion using middle 50% data, eliminating extreme values unlike Range.

Evidence Analysis
Wage Data (₹'000)RangeQD
15,18,20,22,100852
Critical Evaluation

QD gives realistic inequality measure (₹2k) versus distorted Range (₹85k). Our textbook shows QD better represents typical workers.

Future Implications

QD should supplement Gini coefficient for precise labor market analysis.

Question 13:
Demonstrate calculation of Mean Deviation for inflation rates (5.2%, 6.1%, 4.9%, 5.8%) and interpret its economic significance.
Answer:
Theoretical Framework

Mean Deviation averages absolute differences from mean, showing typical variation.

Evidence Analysis
  • Mean = (5.2+6.1+4.9+5.8)/4 = 5.5%
  • MD = (|5.2-5.5| + |6.1-5.5| + |4.9-5.5| + |5.8-5.5|)/4 = 0.45%
Critical Evaluation

0.45% MD indicates stable inflation. Compared to SD (0.49%), MD is simpler but less mathematically efficient.

Future Implications

MD helps RBI monitor price stability when extreme values are absent.

Question 14:
Critically evaluate Lorenz Curve and Standard Deviation as complementary tools for analyzing wealth distribution in India.
Answer:
Theoretical Framework

Lorenz Curve shows cumulative distribution graphically, while SD quantifies absolute dispersion.

Evidence Analysis
  • SD measures ₹12 lakh wealth gap in urban areas
  • Lorenz Curve reveals top 10% hold 57% assets (World Inequality Report)
Critical Evaluation

SD misses structural inequality shown by Lorenz. Together they reveal both magnitude and concentration of disparity.

Future Implications

Policymakers must combine both for effective wealth tax designs.

Question 15:
Why is Quartile Deviation considered a better measure of dispersion than Mean Deviation in skewed distributions?
Answer:
Theoretical Framework

Quartile Deviation (QD) measures mid-50% spread, while Mean Deviation (MD) averages absolute deviations from mean/median.

Evidence Analysis
  • In skewed data like {1,2,3,4,100}, MD (19.2) overestimates dispersion due to the outlier (100).
  • QD (1.5) ignores extremes, focusing on Q1 (2) and Q3 (4), giving a realistic spread.
Critical Evaluation

We learned QD is robust for skewed data as it excludes outliers. MD is sensitive to extreme values, distorting results.

Future Implications

QD is preferred in economics for income studies, where outliers are common. MD works better for symmetric data.

Question 16:
Analyze the limitations of Variance with real-world examples. How does squaring deviations affect its interpretation?
Answer:
Theoretical Framework

Variance squares deviations from the mean to eliminate negative values, but this alters the unit of measurement.

Evidence Analysis
  • For dataset {3,5,7}, variance is 2.66 (unit: sq. units), making it hard to interpret vs. SD (1.63 same units as data).
  • Inflation data (e.g., 5%, 6%, 7%) has variance 0.0001 (sq. %), which is non-intuitive.
Critical Evaluation

We studied that squaring exaggerates larger deviations, e.g., a deviation of 5 becomes 25, overemphasizing outliers.

Future Implications

Variance is mathematically useful but requires conversion to SD for practical applications like risk assessment.

Question 17:
How does Lorenz Curve visually represent dispersion? Compare its effectiveness with Gini Coefficient.
Answer:
Theoretical Framework

The Lorenz Curve plots cumulative income vs. population, while the Gini Coefficient quantifies inequality as a ratio (0 to 1).

Evidence Analysis
MeasureAdvantageDisadvantage
Lorenz CurveVisual clarity (e.g., 20% population holds 5% income)No numerical value
Gini CoefficientPrecise (e.g., 0.4 = moderate inequality)No graphical insight
Critical Evaluation

Our textbook shows the curve is intuitive for policy-making, while Gini is standardized for comparisons across nations.

Future Implications

Both are complementary: Lorenz identifies gaps, and Gini tracks progress over time, e.g., India’s Gini rose from 0.31 (1990) to 0.35 (2020).

Question 18:
Analyze the limitations of mean deviation and suggest why standard deviation is more widely used.
Answer:
Theoretical Framework

Our textbook shows mean deviation (MD) uses absolute deviations, while standard deviation (SD) squares deviations to emphasize larger variances.

Evidence Analysis
  • Example 1: For dataset (2,4,6,8), MD=2 but SD=2.24, reflecting higher dispersion.
  • Example 2: NCERT’s stock return analysis highlights SD’s compatibility with normal distribution models.
Critical Evaluation

SD’s squaring avoids MD’s cancellation issue (± deviations) and aligns with advanced statistical tools like regression.

Future Implications

SD dominates finance (e.g., risk assessment) due to its mathematical robustness.

Question 19:
How does Lorenz curve complement standard deviation in measuring income inequality? Provide two real-world applications.
Answer:
Theoretical Framework

We learned Lorenz curve visualizes cumulative income distribution, while standard deviation quantifies spread numerically.

Evidence Analysis
  • Example 1: India’s 2021 NSSO data showed SD of ₹15k/month, but Lorenz curve revealed top 10% held 60% wealth.
  • Example 2: Brazil’s Gini coefficient (derived from Lorenz) and SD jointly confirmed urban-rural disparities.
Critical Evaluation

Lorenz curve contextualizes SD by showing distribution shape, crucial for policy design.

Future Implications

Governments combine both for targeted welfare schemes (e.g., PM-KISAN).

Question 20:
Critically evaluate the use of coefficient of variation (CV) in comparing the risk of two investments with the following data:
InvestmentMean Return (%)SD (%)
A123
B82
Answer:
Theoretical Framework

CV (SD/Mean×100) standardizes dispersion relative to mean, enabling cross-comparison.

Evidence Analysis
  • Investment A: CV=25% (3/12×100)
  • Investment B: CV=25% (2/8×100)
Critical Evaluation

Despite equal CVs, Investment A offers higher absolute returns (12% vs 8%) for the same relative risk, making it preferable.

Future Implications

CV alone may mislead; investors must consider absolute returns alongside relative dispersion.

Question 21:
Explain the concept of Standard Deviation as a measure of dispersion. Discuss its merits and demerits in comparison to Mean Deviation.
Answer:

The Standard Deviation is a widely used measure of dispersion that quantifies the extent to which individual data points deviate from the mean of the dataset. It is calculated as the square root of the average of the squared deviations from the mean. Mathematically, for a dataset X with n observations, it is given by:


σ = √(Σ(Xi - X̄)² / n)

where Xi represents individual data points and X̄ is the mean.

Merits of Standard Deviation:

  • It considers all data points, making it a comprehensive measure.
  • It is suitable for further statistical analysis like hypothesis testing.
  • It is less affected by sampling fluctuations compared to Mean Deviation.

Demerits of Standard Deviation:

  • It is complex to calculate due to squaring and square root operations.
  • It gives more weight to extreme values, which may distort interpretation.
  • Unlike Mean Deviation, it is not intuitive for non-technical audiences.

In comparison, Mean Deviation uses absolute deviations, making it simpler but less mathematically rigorous for advanced analysis.

Question 22:
Describe the Lorenz Curve and explain how it helps in analyzing income inequality. Illustrate with a diagram.
Answer:

The Lorenz Curve is a graphical representation of income or wealth distribution within a population. It plots the cumulative percentage of income (on the Y-axis) against the cumulative percentage of households or individuals (on the X-axis), ordered from poorest to richest.

Steps to interpret the Lorenz Curve:

  • A perfectly equal distribution is represented by a 45-degree line (Line of Equality).
  • The greater the deviation of the Lorenz Curve from this line, the higher the inequality.
  • The area between the curve and the line of equality helps calculate the Gini Coefficient, a numerical measure of inequality.

Diagram: (Imagine a graph with X-axis: Cumulative % of Population, Y-axis: Cumulative % of Income)

  • Line of Equality: Straight diagonal from (0,0) to (100,100).
  • Lorenz Curve: A concave curve below the diagonal, showing actual distribution.

Application: The Lorenz Curve is crucial for policymakers to assess disparities and design welfare programs. For example, if the curve is significantly bowed, progressive taxation may be needed to redistribute wealth.

Question 23:
Compare and contrast Range and Standard Deviation as measures of dispersion. Explain their significance in statistical analysis with suitable examples.
Answer:

Range and Standard Deviation are both measures of dispersion used to understand the spread of data in a dataset. However, they differ in their calculation and interpretation.

Range is the simplest measure of dispersion, calculated as the difference between the highest and lowest values in a dataset. For example, if the marks of students in a class are 30, 40, 50, 60, and 70, the range is 70 - 30 = 40. While easy to compute, the range is highly affected by extreme values and does not consider all data points.

Standard Deviation, on the other hand, measures the average distance of each data point from the mean. It considers all values in the dataset, making it a more reliable measure. For instance, if the mean marks are 50, and the deviations are -20, -10, 0, +10, +20, the standard deviation will account for these variations systematically.

  • Significance of Range: Useful for quick estimates and when data is not highly skewed.
  • Significance of Standard Deviation: Provides a precise measure of dispersion, widely used in statistical tests and probability distributions.

In summary, while range gives a basic idea of spread, standard deviation offers a deeper, more accurate analysis, especially in research and data-driven decision-making.

Question 24:
Explain the concept of Standard Deviation as a measure of dispersion. Discuss its merits and demerits in comparison to Mean Deviation with suitable examples.
Answer:

Standard Deviation is a widely used measure of dispersion that quantifies the amount of variation or dispersion of a set of data values from their mean. It is calculated as the square root of the variance, which is the average of the squared differences from the mean. The formula for standard deviation (σ) is:


σ = √(Σ(X - μ)² / N)

where X represents each data point, μ is the mean, and N is the number of data points.

Merits of Standard Deviation:

  • It is based on all observations, making it a comprehensive measure.
  • It is suitable for further mathematical analysis and statistical inferences.
  • It is less affected by sampling fluctuations compared to other measures.

Demerits of Standard Deviation:

  • It is complex to calculate manually due to squaring and square root operations.
  • It gives more weight to extreme values because of squaring the deviations.

Comparison with Mean Deviation:

  • Mean Deviation uses absolute deviations, making it simpler but less mathematically sound.
  • Standard Deviation is more reliable for statistical analysis, while Mean Deviation is easier to interpret for non-technical audiences.

Example: For the data set [2, 4, 6, 8, 10], the mean is 6. The standard deviation is calculated as √[(16+4+0+4+16)/5] = √(40/5) = √8 ≈ 2.83, while the mean deviation is (4+2+0+2+4)/5 = 12/5 = 2.4.

Question 25:
Explain the concept of Standard Deviation as a measure of dispersion. How is it calculated for both ungrouped and grouped data? Discuss its significance in statistical analysis with suitable examples.
Answer:

The Standard Deviation is a widely used measure of dispersion that quantifies the extent to which individual data points deviate from the mean of the dataset. It provides a clear understanding of the spread or variability in the data.

Calculation for Ungrouped Data:
1. Compute the mean (average) of the data.
2. Subtract the mean from each data point to find the deviation.
3. Square each deviation to eliminate negative values.
4. Calculate the average of these squared deviations (variance).
5. Take the square root of the variance to obtain the standard deviation.

Calculation for Grouped Data:
1. Determine the midpoint of each class interval.
2. Multiply each midpoint by its corresponding frequency.
3. Compute the mean using the formula for grouped data.
4. Find the squared deviation of each midpoint from the mean.
5. Multiply each squared deviation by its frequency and sum them up.
6. Divide by the total frequency to get variance.
7. Take the square root to find the standard deviation.

Significance:

  • It helps in understanding the consistency of data. A low standard deviation indicates that data points are close to the mean, while a high value suggests wide variability.
  • It is crucial in fields like finance to assess risk, in quality control to monitor product consistency, and in research to analyze data reliability.

Example: In a class test, if the scores are 10, 12, 14, 16, 18, the standard deviation will show how much the scores vary from the average (14). A smaller value here indicates most students scored near the average.

Question 26:
Explain the concept of Standard Deviation as a measure of dispersion. Discuss its merits and demerits in statistical analysis with suitable examples.
Answer:

The Standard Deviation is a widely used measure of dispersion that quantifies the extent to which individual data points deviate from the mean of the dataset. It is calculated as the square root of the variance, which is the average of the squared differences from the mean.


Merits of Standard Deviation:

  • Comprehensive Measure: It considers all data points, providing a complete picture of dispersion.
  • Mathematical Properties: It is amenable to further algebraic treatment and is used in advanced statistical techniques like hypothesis testing.
  • Consistency: It is not affected by changes in the origin (addition/subtraction) but is affected by changes in scale (multiplication/division).

Demerits of Standard Deviation:

  • Complex Calculation: It involves squaring deviations, which can be cumbersome without computational tools.
  • Sensitive to Outliers: Extreme values can disproportionately increase the standard deviation.
  • Not Intuitive: Unlike the range, it is not easily interpretable for non-statisticians.

Example: Consider the dataset [2, 4, 6, 8, 10]. The mean is 6. The squared deviations are [16, 4, 0, 4, 16], and the variance is 8. The standard deviation is √8 ≈ 2.83, indicating moderate dispersion.


Value-Added Insight: Standard deviation is pivotal in finance to assess investment risk. A higher standard deviation implies greater volatility, aiding investors in making informed decisions.

Question 27:
Explain the concept of Standard Deviation as a measure of dispersion. Discuss its merits and demerits in comparison to Mean Deviation. Support your answer with a hypothetical example.
Answer:

Standard Deviation is a widely used measure of dispersion that quantifies the extent to which individual data points deviate from the mean of the dataset. It is calculated as the square root of the average of the squared deviations from the mean. Mathematically, it is represented as:


σ = √(Σ(X - μ)² / N)

where σ is the standard deviation, X represents individual data points, μ is the mean, and N is the number of observations.

Merits of Standard Deviation:

  • It considers all values in the dataset, providing a comprehensive measure of dispersion.
  • It is suitable for further statistical analysis, such as in the calculation of correlation and regression.
  • It is less affected by sampling fluctuations compared to Mean Deviation.

Demerits of Standard Deviation:

  • It is complex to calculate due to the squaring of deviations.
  • It gives more weight to extreme values, which can distort the measure of dispersion.
  • Unlike Mean Deviation, it does not ignore the signs of deviations, making it less intuitive in some cases.

Comparison with Mean Deviation: While Mean Deviation averages the absolute deviations and is easier to interpret, Standard Deviation provides a more mathematically rigorous measure by squaring the deviations, thus penalizing larger deviations more heavily.

Example: Consider a dataset of test scores: [50, 60, 70, 80, 90]. The mean (μ) is 70. The squared deviations are [400, 100, 0, 100, 400]. The variance (σ²) is 200, and the Standard Deviation (σ) is √200 ≈ 14.14. The Mean Deviation would be (20 + 10 + 0 + 10 + 20)/5 = 12. Here, the Standard Deviation is higher due to the squaring of deviations, reflecting greater sensitivity to outliers.

Question 28:
Compare and contrast Range and Standard Deviation as measures of dispersion. Discuss their advantages and limitations with suitable examples.
Answer:

Range and Standard Deviation are both measures of dispersion used to understand the spread of data in a dataset. However, they differ in their calculation and interpretation.

Range is the simplest measure of dispersion, calculated as the difference between the highest and lowest values in a dataset. For example, if the marks of students in a class are 30, 45, 60, 75, and 90, the range is 90 - 30 = 60. The advantage of range is its simplicity and ease of calculation. However, it has limitations as it only considers the extreme values and ignores the distribution of other data points, making it sensitive to outliers.

On the other hand, Standard Deviation is a more sophisticated measure that considers all data points. It measures the average distance of each data point from the mean. For example, if the mean marks are 60 and the standard deviation is 15, it indicates that most marks deviate by 15 points from the mean. The advantage of standard deviation is that it provides a comprehensive understanding of data spread and is less affected by outliers. However, its calculation is complex and requires more computational effort.

  • Range: Simple but ignores intermediate values.
  • Standard Deviation: Complex but considers all data points.

In conclusion, while range is useful for a quick overview, standard deviation offers a more accurate and detailed analysis of data dispersion.

Question 29:
Explain the concept of Standard Deviation as a measure of dispersion. How does it differ from Mean Deviation? Provide a numerical example to illustrate the calculation of Standard Deviation for the data set: 5, 10, 15, 20, 25.
Answer:

Standard Deviation is a widely used measure of dispersion that quantifies the extent to which individual data points deviate from the mean of the data set. It is calculated as the square root of the average of the squared deviations from the mean. Unlike Mean Deviation, which uses absolute values, Standard Deviation squares the deviations, giving more weight to extreme values and providing a more accurate measure of variability.

Key differences between Standard Deviation and Mean Deviation:

  • Standard Deviation squares deviations, while Mean Deviation uses absolute values.
  • Standard Deviation is more sensitive to outliers due to squaring.
  • Mean Deviation is easier to interpret but less mathematically rigorous.

Numerical Example: For the data set 5, 10, 15, 20, 25:


Step 1: Calculate the mean (μ) = (5 + 10 + 15 + 20 + 25) / 5 = 15
Step 2: Find deviations from mean: (5-15)=-10, (10-15)=-5, (15-15)=0, (20-15)=5, (25-15)=10
Step 3: Square the deviations: 100, 25, 0, 25, 100
Step 4: Calculate average of squared deviations = (100 + 25 + 0 + 25 + 100) / 5 = 50
Step 5: Take square root: √50 ≈ 7.07

Thus, the Standard Deviation is approximately 7.07, indicating the average distance of data points from the mean.

Question 30:
Describe the Lorenz Curve as a graphical measure of dispersion. How is it constructed, and what does it indicate about income distribution? Illustrate with a hypothetical example of income distribution among 5 households.
Answer:

The Lorenz Curve is a graphical representation of the distribution of income or wealth in a population. It plots the cumulative percentage of income against the cumulative percentage of households, providing a visual measure of inequality.

Construction Steps:

  • Arrange households in ascending order of income.
  • Calculate cumulative percentages of households and their corresponding cumulative income.
  • Plot these points on a graph with equal axes (0%-100%).

Interpretation: A 45-degree line (line of equality) represents perfect equality. The further the Lorenz Curve bows below this line, the greater the inequality.

Hypothetical Example: Consider 5 households with incomes: 10, 20, 30, 40, 100 (units)


Step 1: Sort incomes: 10, 20, 30, 40, 100
Step 2: Calculate cumulative income: 10, 30, 60, 100, 200
Step 3: Convert to percentages:
Households: 20%, 40%, 60%, 80%, 100%
Income: 5%, 15%, 30%, 50%, 100%

When plotted, this curve would show significant bowing, indicating high inequality as the richest household earns 50% of total income. The Lorenz Curve thus visually demonstrates the concentration of income among a small portion of the population.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A survey recorded the monthly incomes (in ₹) of 10 workers in a factory: 15,000, 18,000, 20,000, 22,000, 25,000, 25,000, 30,000, 35,000, 40,000, 50,000. Calculate the range and quartile deviation. Interpret the results.
Answer:
Case Deconstruction

Given data is arranged in ascending order. Range = Max − Min = ₹50,000 − ₹15,000 = ₹35,000. For quartile deviation, Q1 = 20,000 (3rd value), Q3 = 35,000 (8th value). QD = (Q3 − Q1)/2 = ₹7,500.


Theoretical Application
  • Range shows total spread but ignores middle values.
  • QD measures dispersion around median, reducing outlier impact.

Critical Evaluation

High range (₹35,000) indicates unequal income distribution. QD (₹7,500) suggests moderate variability among middle 50% workers.

Question 2:
The standard deviation of marks in Class A is 12, while Class B has a variance of 169. Compare the consistency of performance between both classes using coefficient of variation (CV). Assume mean marks: Class A = 60, Class B = 65.
Answer:
Case Deconstruction

Class A: SD = 12, Mean = 60. CV = (12/60)×100 = 20%. Class B: Variance = 169 → SD = 13. CV = (13/65)×100 = 20%.


Theoretical Application
  • CV neutralizes unit differences, enabling comparison.
  • Same CV (20%) implies equal relative dispersion.

Critical Evaluation

Despite higher absolute spread (SD=13), Class B’s performance is equally consistent as Class A when scaled to their means.

Question 3:
A company’s daily sales (in ₹’000) for 6 days: 45, 55, 60, 65, 70, 85. Calculate mean deviation from median and discuss its usefulness over SD in this context.
Answer:
Case Deconstruction

Median = (60+65)/2 = ₹62.5k. Mean deviation = Σ|X−Median|/n = (17.5 + 7.5 + 2.5 + 2.5 + 7.5 + 22.5)/6 = ₹10k.


Theoretical Application
  • Mean deviation uses all data without squaring, avoiding extreme value exaggeration.
  • Easier interpretation for non-technical stakeholders.

Critical Evaluation

MD (₹10k) gives direct average deviation, unlike SD’s squared units. Suitable for asymmetric data like sales.

Question 4:
Analyze the dispersion in two investment options using data: Option A (Returns %): 8, 10, 12, 14, 16; Option B: 5, 10, 15, 20, 25. Which is riskier? Justify with interquartile range (IQR).
Answer:
Case Deconstruction

Option A: Q1 = 10, Q3 = 14 → IQR = 4%. Option B: Q1 = 10, Q3 = 20 → IQR = 10%.


Theoretical Application
  • IQR measures middle 50% spread, ignoring extremes.
  • Higher IQR (Option B) indicates greater volatility.

Critical Evaluation

Option B’s IQR (10%) is 2.5× Option A’s (4%), confirming higher risk. Our textbook shows IQR is robust for skewed distributions.

Question 5:
A study reports farm sizes (acres) in Village X: Mean = 5, SD = 2; Village Y: Mean = 8, SD = 3. Compare economic inequality using Lorenz Curve analysis and CV.
Answer:
Case Deconstruction

Village X: CV = (2/5)×100 = 40%. Village Y: CV = (3/8)×100 = 37.5%.


Theoretical Application
  • CV shows Village X has higher relative inequality (40% > 37.5%).
  • Lorenz Curve would bulge more for Village X, confirming skewed distribution.

Critical Evaluation

Despite smaller mean, Village X’s higher CV aligns with [Diagram: Wider Lorenz Gap], indicating concentrated land ownership.

Question 6:
A survey records the range of monthly incomes (in ₹) in two villages: Village A (10,000–50,000) and Village B (15,000–45,000). Compare their economic disparity using measures of dispersion. Justify which village has higher variability.
Answer:
Case Deconstruction

Village A has a range of ₹40,000 (50,000 - 10,000), while Village B's range is ₹30,000 (45,000 - 15,000).

Theoretical Application
  • Range measures total spread; Village A shows wider income gaps.
  • Village B’s narrower range suggests more uniform incomes.
Critical Evaluation

Our textbook shows range ignores mid-values, but here it clearly indicates Village A has higher variability. Example: Urban vs. rural wage gaps often mirror this pattern.

Question 7:
Calculate the quartile deviation for the dataset: 12, 15, 18, 22, 25, 30, 35. Explain its significance over range in this context.
Answer:
Case Deconstruction

Q1 = 15, Q3 = 30. Quartile Deviation (QD) = (30 - 15)/2 = 7.5.

Theoretical Application
  • QD measures middle 50% data spread, reducing outlier impact.
  • Range (35 - 12 = 23) is skewed by extremes.
Critical Evaluation

We studied that QD is better for skewed data. Example: Income surveys often use QD to ignore top/bottom extremes, unlike range.

Question 8:
The standard deviation of Class X’s test scores is 12, while Class Y’s is 8. Interpret these values and suggest which class has more consistent performance.
Answer:
Case Deconstruction

Lower SD (Class Y = 8) implies scores cluster closer to the mean, indicating consistency.

Theoretical Application
  • Class X’s higher SD (12) shows greater score dispersion.
  • Example: Stock volatility uses SD similarly—lower SD means stable returns.
Critical Evaluation

Our textbook highlights SD’s sensitivity to all data points. Class Y’s consistency could reflect uniform teaching quality.

Question 9:
A factory’s daily output (units) has mean deviation 5.2 and variance 36. Compare these measures and analyze which better reflects production stability.
Answer:
Case Deconstruction

Variance (36) is squared units, hard to interpret directly. Mean Deviation (MD = 5.2) uses original units.

Theoretical Application
  • MD is simpler but ignores algebraic signs.
  • Variance’s squaring amplifies outliers.
Critical Evaluation

We studied that MD is intuitive for managers. Example: Inventory management prefers MD for day-to-day decisions over variance.

Question 10:
The coefficient of variation (CV) for two stocks is: Stock A (15%) and Stock B (25%). Which is riskier? Justify using relative dispersion concepts.
Answer:
Case Deconstruction

CV = (SD/Mean) × 100. Stock B’s higher CV (25%) indicates greater risk per unit return.

Theoretical Application
  • CV standardizes dispersion across scales.
  • Example: Mutual funds use CV to compare asset volatility.
Critical Evaluation

Our textbook shows CV is critical for unequal means. Stock B’s higher CV aligns with risk-return tradeoff theories.

Question 11:

The following data represents the monthly incomes (in ₹) of 10 employees in a small firm:

15,000, 18,000, 20,000, 22,000, 25,000, 25,000, 30,000, 35,000, 40,000, 50,000

Calculate the range, quartile deviation, and mean deviation from the median. Interpret these measures of dispersion in the context of income inequality within the firm.

Answer:

Step 1: Calculate the Range
Range = Maximum Value - Minimum Value
Range = ₹50,000 - ₹15,000 = ₹35,000

Step 2: Calculate Quartile Deviation (QD)
First, arrange the data in ascending order (already given).
Q1 (First Quartile) = Value at (n+1)/4th position = (10+1)/4 = 2.75th position
Q1 = ₹18,000 + 0.75(₹20,000 - ₹18,000) = ₹19,500
Q3 (Third Quartile) = Value at 3(n+1)/4th position = 8.25th position
Q3 = ₹35,000 + 0.25(₹40,000 - ₹35,000) = ₹36,250
QD = (Q3 - Q1)/2 = (₹36,250 - ₹19,500)/2 = ₹8,375

Step 3: Calculate Mean Deviation from Median
Median (M) = Average of 5th and 6th values = (₹25,000 + ₹25,000)/2 = ₹25,000
Mean Deviation = Σ|X - M| / n
= (|15,000-25,000| + |18,000-25,000| + ... + |50,000-25,000|) / 10
= (10,000 + 7,000 + 5,000 + 3,000 + 0 + 0 + 5,000 + 10,000 + 15,000 + 25,000) / 10
= ₹8,000

Interpretation:
The range (₹35,000) shows a wide spread between highest and lowest incomes.
The quartile deviation (₹8,375) indicates middle 50% of employees have incomes within ₹19,500 to ₹36,250.
The mean deviation (₹8,000) suggests average variation from median income is substantial, reflecting significant income disparity.

Question 12:

A school conducted a survey on the number of hours students spend studying daily. The results for Class 11A (20 students) are:

2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 8, 9

Calculate the standard deviation and coefficient of variation for this data. Compare these measures with another class (Class 11B) having σ=1.8 hours and mean=5.2 hours. What does this comparison reveal about study habits?

Answer:

Step 1: Calculate Mean (X̄)
Sum of all observations = 2+3+3+...+9 = 105
X̄ = 105/20 = 5.25 hours

Step 2: Calculate Standard Deviation (σ)
σ = √[Σ(X - X̄)²/n]
= √[(2-5.25)² + (3-5.25)² + ... + (9-5.25)²]/20
= √[10.5625 + 5.0625 + 5.0625 + 1.5625 + 1.5625 + 1.5625 + 0.0625 + 0.0625 + 0.5625 + 0.5625 + 0.5625 + 0.5625 + 0.5625 + 1.5625 + 1.5625 + 3.0625 + 14.0625]/20
= √[49.75/20] = √2.4875 ≈ 1.577 hours

Step 3: Calculate Coefficient of Variation (CV)
CV = (σ/X̄) × 100 = (1.577/5.25) × 100 ≈ 30.04%

Comparison with Class 11B:
Class 11A: σ=1.577 hours, CV=30.04%
Class 11B: σ=1.8 hours, CV=(1.8/5.2)×100≈34.62%

Interpretation:
1) Class 11A has lower absolute dispersion (σ) showing study hours are more clustered around mean.
2) Class 11A also has lower relative dispersion (CV) indicating more consistent study patterns compared to Class 11B.
3) The higher CV for Class 11B suggests greater variability in study habits relative to their mean.

Question 13:
A teacher recorded the marks obtained by 10 students in a test as follows: 45, 60, 72, 85, 90, 52, 67, 78, 82, 95. Calculate the range and quartile deviation for this data. Interpret the results in terms of data dispersion.
Answer:

To calculate the range:
Step 1: Identify the highest and lowest values in the data.
Highest value = 95
Lowest value = 45
Step 2: Range = Highest value - Lowest value = 95 - 45 = 50.

To calculate the quartile deviation:
Step 1: Arrange the data in ascending order: 45, 52, 60, 67, 72, 78, 82, 85, 90, 95.
Step 2: Find the first quartile (Q1) and third quartile (Q3).
Q1 = Value at (n+1)/4th position = (10+1)/4 = 2.75th value ≈ 52 + 0.75(60-52) = 58
Q3 = Value at 3(n+1)/4th position = 8.25th value ≈ 85 + 0.25(90-85) = 86.25
Step 3: Quartile Deviation = (Q3 - Q1)/2 = (86.25 - 58)/2 = 14.125.

Interpretation: The range of 50 indicates a wide spread of marks, while the quartile deviation of 14.125 suggests that the middle 50% of the data is less dispersed compared to the overall range. This implies that extreme values (like 45 and 95) contribute significantly to the total dispersion.

Question 14:
The monthly incomes (in ₹) of five families in a locality are: ₹12,000, ₹18,000, ₹22,000, ₹25,000, and ₹35,000. Compute the mean deviation from the median and explain its significance in understanding income inequality.
Answer:

To calculate the mean deviation from the median:
Step 1: Arrange the data in ascending order: ₹12,000, ₹18,000, ₹22,000, ₹25,000, ₹35,000.
Step 2: Find the median (middle value) = ₹22,000.
Step 3: Calculate absolute deviations from the median:
|12,000 - 22,000| = ₹10,000
|18,000 - 22,000| = ₹4,000
|22,000 - 22,000| = ₹0
|25,000 - 22,000| = ₹3,000
|35,000 - 22,000| = ₹13,000.
Step 4: Mean Deviation = (10,000 + 4,000 + 0 + 3,000 + 13,000)/5 = ₹6,000.

Significance: The mean deviation of ₹6,000 indicates the average distance of each family's income from the median income. A higher value reflects greater income inequality, as incomes are spread out unevenly around the median. This measure is less affected by extreme values (like ₹35,000) compared to the range, making it a more reliable indicator of dispersion for skewed data.

Question 15:
A teacher recorded the marks obtained by 10 students in a test as follows: 45, 60, 72, 85, 90, 52, 67, 78, 82, 95. Calculate the range and quartile deviation for this data. Interpret the results.
Answer:

To calculate the range and quartile deviation, follow these steps:


Step 1: Arrange the data in ascending order:
45, 52, 60, 67, 72, 78, 82, 85, 90, 95.

Step 2: Calculate the Range:
Range = Maximum value - Minimum value = 95 - 45 = 50.

Step 3: Calculate Quartile Deviation (QD):
First, find Q1 (First Quartile) and Q3 (Third Quartile).
For Q1: Position = (n+1)/4 = (10+1)/4 = 2.75th term.
Q1 = 52 + 0.75*(60-52) = 58.
For Q3: Position = 3*(n+1)/4 = 8.25th term.
Q3 = 85 + 0.25*(90-85) = 86.25.
QD = (Q3 - Q1)/2 = (86.25 - 58)/2 = 14.125.

Interpretation: The range of 50 indicates a wide spread of marks. The quartile deviation of 14.125 suggests that the middle 50% of the data is clustered around the median with moderate dispersion.

Question 16:
The following table shows the daily wages (in ₹) of workers in a factory. Calculate the mean deviation from the median and explain its significance.
Wages (₹)100150200250300
No. of Workers581273
Answer:

To calculate the mean deviation from the median, follow these steps:


Step 1: Find the Median (M):
Total workers (N) = 5 + 8 + 12 + 7 + 3 = 35.
Median position = (N+1)/2 = 18th term.
Cumulative frequency up to ₹200 = 5 + 8 + 12 = 25 (which includes the 18th term).
Thus, Median (M) = ₹200.

Step 2: Calculate Absolute Deviations |X - M|:
|100-200| = 100, |150-200| = 50, |200-200| = 0, |250-200| = 50, |300-200| = 100.

Step 3: Multiply by Frequency and Sum:
(100×5) + (50×8) + (0×12) + (50×7) + (100×3) = 500 + 400 + 0 + 350 + 300 = 1550.

Step 4: Compute Mean Deviation:
Mean Deviation = Σf|X - M| / N = 1550 / 35 = ₹44.29.

Significance: The mean deviation from the median (₹44.29) measures the average absolute dispersion of wages from the central value (median). A lower value would indicate more uniformity in wages, while this value suggests moderate variability.

Question 17:
A teacher recorded the marks of 10 students in a class test for Economics (out of 50). The data is as follows: 32, 28, 35, 42, 18, 25, 38, 45, 30, 22. Calculate the range, quartile deviation, and mean deviation from the median. Interpret the results to compare the spread of marks.
Answer:

To solve this, we first arrange the data in ascending order: 18, 22, 25, 28, 30, 32, 35, 38, 42, 45.

Range: It is the difference between the highest and lowest values.
Range = 45 - 18 = 27.

Quartile Deviation (QD): It measures the spread of the middle 50% of data.
First Quartile (Q1) = Value at (n+1)/4 = (10+1)/4 = 2.75th term ≈ 22 + 0.75(25-22) = 24.25.
Third Quartile (Q3) = Value at 3(n+1)/4 = 8.25th term ≈ 38 + 0.25(42-38) = 39.
QD = (Q3 - Q1)/2 = (39 - 24.25)/2 = 7.375.

Mean Deviation from Median: Median (M) = Average of 5th and 6th terms = (30 + 32)/2 = 31.
Mean Deviation = Σ|X - M|/n = (|18-31| + |22-31| + ... + |45-31|)/10 = (13 + 9 + 6 + 3 + 1 + 1 + 4 + 7 + 11 + 14)/10 = 6.9.

Interpretation: The range shows the total spread is 27 marks, while the quartile deviation (7.375) indicates moderate variability in the middle 50% of data. The mean deviation (6.9) suggests that, on average, marks deviate by ~7 from the median, reflecting consistent performance among most students.

Question 18:
The following table shows the monthly income (in ₹) of 8 families in a locality: 15,000, 22,000, 18,000, 25,000, 30,000, 12,000, 20,000, 28,000. Calculate the standard deviation and coefficient of variation. Which measure is more suitable to compare variability with another locality having an average income of ₹24,000 and standard deviation of ₹6,000? Justify your answer.
Answer:

Step 1: Calculate Mean (μ)
μ = (15,000 + 22,000 + 18,000 + 25,000 + 30,000 + 12,000 + 20,000 + 28,000)/8 = 21,250.

Step 2: Compute Standard Deviation (σ)
σ = √[Σ(X - μ)²/n] = √[(15,000-21,250)² + (22,000-21,250)² + ... + (28,000-21,250)²]/8
= √[(39,062,500 + 562,500 + 10,562,500 + 14,062,500 + 76,562,500 + 85,562,500 + 1,562,500 + 45,562,500)/8]
= √[273,000,000/8] = √34,125,000 ≈ 5,841.66.

Step 3: Coefficient of Variation (CV)
CV = (σ/μ) × 100 = (5,841.66/21,250) × 100 ≈ 27.49%.

Comparison: The other locality has μ = ₹24,000 and σ = ₹6,000. Its CV = (6,000/24,000) × 100 = 25%.

Justification: Since the means differ, coefficient of variation is more suitable than standard deviation for comparison. Here, the first locality (CV = 27.49%) has higher relative variability than the second (CV = 25%), indicating greater income disparity despite a lower absolute standard deviation.

Question 19:
A teacher recorded the marks obtained by 10 students in a test as follows: 45, 60, 55, 70, 65, 80, 75, 50, 85, 90. Calculate the range and quartile deviation for this data. Interpret the results in terms of dispersion.
Answer:

To calculate the range and quartile deviation, follow these steps:


Step 1: Arrange the data in ascending order:
45, 50, 55, 60, 65, 70, 75, 80, 85, 90

Step 2: Calculate the Range:
Range = Maximum value - Minimum value
Range = 90 - 45 = 45

Step 3: Calculate Quartile Deviation (QD):
First, find Q1 (First Quartile) and Q3 (Third Quartile).
For Q1: Position = (n+1)/4 = (10+1)/4 = 2.75th term
Q1 = 50 + 0.75(55-50) = 53.75
For Q3: Position = 3(n+1)/4 = 8.25th term
Q3 = 80 + 0.25(85-80) = 81.25
QD = (Q3 - Q1)/2 = (81.25 - 53.75)/2 = 13.75

Interpretation: The range of 45 indicates a wide spread of marks, while the quartile deviation of 13.75 shows that the middle 50% of data is less dispersed compared to the overall range. This suggests outliers may be affecting the range.

Question 20:
The following table shows the daily wages (in ₹) of workers in two factories, A and B. Compare the variability in wages using the coefficient of variation (CV).
FactoryMean Wage (₹)Standard Deviation (₹)
A50060
B45050
Answer:

To compare variability using coefficient of variation (CV), use the formula:
CV = (Standard Deviation / Mean) × 100


Step 1: Calculate CV for Factory A:
CVA = (60 / 500) × 100 = 12%

Step 2: Calculate CV for Factory B:
CVB = (50 / 450) × 100 ≈ 11.11%

Comparison: Factory A has a higher CV (12%) than Factory B (11.11%), indicating greater relative variability in wages. Despite Factory A having higher average wages, the wages are more dispersed relative to the mean.


Note: CV is useful for comparing datasets with different units or means, as it expresses dispersion as a percentage of the mean.

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