1. Review: Measures of Central Tendency & Dispersion

Concept:
Measures of central tendency summarize data around a center (average performance). Measures of dispersion describe the spread/variability of data (consistency). Together, they help analyze performance, compare datasets, and interpret reliability.

Applications:
Education, statistics, research, business analytics, quality control.

2. Central Tendency

1.Mean (Arithmetic Average) – Sum of all values divided by number of observations.
2.Median – Middle value when data is arranged in ascending order.
3.Mode – Most frequent value.

Example:
Marks of 2 students (8 subjects):

Observation:

Higher average → better achievement → Student B.

3. Dispersion (Variability)

1.Purpose: Measure how spread out data values are from the center.
2.Types:
1.Absolute Measures: Same units as data (Range, Quartile Deviation, Mean Deviation, Standard Deviation)
2.Relative Measures: Unit-free coefficients (Coefficient of Variation) → useful for comparing series with different units.

Key Notes:

Low dispersion → consistent/homogeneous data.

High dispersion → scattered/variable data.

Example: Range
Student Range
A 46–65 → 19
B 25–85 → 60 → more scattered

4. Quartile Deviation (QD)

1.Definition: Measures spread using middle 50% of data (ignores extremes).
2.Formulas:
QD = (Q3 - Q1)/2
Coefficient = (Q3 - Q1)/(Q3 + Q1)

3.Quartile Positions (Continuous Series):
1.Q1 = N/4 → 25th percentile
2.Q2 = Median = N/2 → 50th percentile
3.Q3 = 3N/4 → 75th percentile

4.Quartile Formula:
Q = l + ((Position - Cf)/f) × i

Where:
l = lower limit of quartile class
Cf = cumulative frequency before class
f = frequency of quartile class
i = class width

Example:
Q1 = 43.5, Q3 = 64.375 → QD = (64.375 − 43.5)/2 = 10.437
Coefficient = 20.875 / 107.875 ≈ 0.193

5. Mean Deviation (MD)

1.Definition: Average of absolute deviations from mean or median.
2.Formulas:
MD = Σ|x - Mean| / N
Coefficient = MD / Mean

3.Steps to Compute:
1.Find Mean/Median.
2.Calculate absolute deviations |x − central value|.
3.Take average (weighted if frequency exists).

Example:
Marks: 40,50,65,60,58,62,55,46 → Mean = 54.5
Σ|deviations| = 80 → MD = 10
Coefficient = 10/54.5 ≈ 0.183

6. Standard Deviation (SD)

1.Definition: Most reliable measure; square root of average squared deviations from mean. Uses all data and sensitive to extremes.
2.Formulas:
σ = √(Σ(x - Mean)² / N)

3.Alternate Methods:
1.Direct Method: σ = √(Σx²/N - Mean²)
2.Assumed Mean Method: d = x − A → σ = √(Σfd²/N - (Σfd/N)²)
3.Step Deviation Method: d' = d/i → σ = √(Σfd'²/N - (Σfd'/N)²) × i

4.Coefficient of Variation (CV):
CV = σ / Mean × 100%

Low CV → more consistent

High CV → more variable

Example:
Marks → σ ≈ 16.39, CV = 16.39 / 54.5 × 100 ≈ 30%

7. Key Concepts Summary

1.Concept Key Points
2.Central Tendency Mean, Median, Mode → center of data
3.Dispersion Range, QD, MD, SD → spread of data
4.Absolute Measures Same units as data → Range, QD, MD, SD
5.elative Measures Unit-free → Coefficient of Variation, Coefficient of QD/MD
6.Best Measure for Consistency SD → uses all data, stable, algebraic properties
7.Consistency vs Variability Low SD/CV → uniform, high SD/CV → scattered

8. Important Solved Examples – Measures of Central Tendency & Dispersion

Example 1: Total and Average Marks

Question:
Marks of Student A: 40, 50, 65, 60, 58, 62, 55, 46
Marks of Student B: 50, 65, 80, 35, 55, 70, 85, 25

Find the total and average marks for each student.

Solution:

Answer: Total A = 436, Average A = 54.5; Total B = 465, Average B = 58.125

Example 2: Identify More Scattered

Question:
Determine which student’s marks are more scattered using the range.

Solution:

Conclusion: B is more scattered.

Example 3: Better Achievement

Question:
Compare students’ average marks to see who performed better overall.

Solution:

Conclusion: B has better overall performance.

Example 4: Quartile Deviation (QD)

Question:
Given continuous series marks with Q1 = 43.5, Q3 = 64.375, find Quartile Deviation (QD) and its coefficient.

Solution:

QD = (Q3 − Q1)/2 = (64.375 − 43.5)/2 = 10.437

Coefficient of QD = (Q3 − Q1)/(Q3 + Q1) = 20.875 / 107.875 ≈ 0.193

Answer: QD = 10.437, Coefficient ≈ 0.193

Example 5: Find Quartiles from Frequency Table

Question:
Marks frequency table:

Find Q1 and Q3.

Solution:

Cumulative Frequency Table:

Total N = 14

Q1 position = N/4 = 14/4 = 3.5 → lies in 40–49

Q3 position = 3N/4 = 10.5 → lies in 60–69

Formula: Q = l + [(position − Cf)/f] × i

Q1 = 40 + [(3.5 − 2)/3] × 10 = 45

Q3 = 60 + [(10.5 − 9)/3] × 10 = 65

Answer: Q1 = 45, Q3 = 65

Example 6: Mean Deviation from Mean

Question:
Find Mean Deviation (MD) and coefficient for marks: 40, 50, 65, 60, 58, 62, 55, 46

Solution:

Mean = (ΣMarks)/N = 436/8 = 54.5

MD = Σ|deviation| / N = 82/8 = 10.25

Coefficient = MD / Mean ≈ 10.25 / 54.5 ≈ 0.188

Answer: MD ≈ 10.25, Coefficient ≈ 0.188

Example 7: Standard Deviation (Direct Method)

Question:
Compute standard deviation (σ) and coefficient of variation (CV%) for marks: 40, 50, 65, 60, 58, 62, 55, 46

Solution:

Σm² = 25384

Mean = 54.5

σ = √(Σm²/N − Mean²) = √(25384/8 − 54.5²) ≈ 16.39

CV = σ / Mean × 100 ≈ 16.39 / 54.5 × 100 ≈ 30%

Answer: σ ≈ 16.39, CV ≈ 30%

Example 8: Comparison Using Coefficient of Variation

Question:
Two firms have CVs: A = 1.53%, B = 1.73%. Which firm is more consistent?

Solution:

Conclusion: Lower CV → more consistency. Firm A is more uniform.

Example 9: Mean Deviation from Median

Question:
Find Mean Deviation (MD) from median for marks: 40, 50, 65, 60, 58, 62, 55, 46

Solution:

Median (Q2) = 56.5

MD = Σ|m−Median| / N = 54 / 8 = 6.75

Coefficient = 6.75 / 56.5 ≈ 0.119

Answer: MD ≈ 6.75, Coefficient ≈ 0.119

Example 10: Absolute vs Relative Measures

Question:
Compare absolute and relative measures of dispersion for the given data.

Solution:

Conclusion: CV is unit-free, allowing comparison across different series.

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