CA Foundation Statistics Chapter 14 Unit 1: Measures of Central Tendency

Chapter Overview

Measures of Central Tendency in One Minute

Different averages answer different questions. The correct average depends on whether we want a general value, a middle position, the most common item, a growth rate or a rate such as speed.

AAMGeneral numerical average

MMedianMiddle positional value

MoModeMost frequent value

GGMGrowth and compound change

HHMRates and equal-distance speed

PPartition ValuesDivide ordered data into parts

Meaning and Need for an Average

Suppose the marks of 100 students are available. Reading all 100 values does not immediately tell us the general performance of the class. A single representative value helps us understand the overall level of the data.

Core Idea A measure of central tendency is one value used to represent a complete set of observations.

Why Do We Need an Average?

It summarises a large mass of data.
It helps compare two or more groups.
It gives a quick idea of the general level of observations.
It supports business, economic and statistical decisions.

From Data to Decision

1CollectObserve the data

2SummariseFind a suitable average

3CompareUse the result for decisions

Qualities of an Ideal Measure

Clearly Defined

Different persons should obtain the same answer from the same data.

Simple

It should be easy to understand and reasonably easy to calculate.

Uses the Data

It should use all observations wherever the nature of the measure permits.

Reliable

It should be stable, useful for comparison and suitable for further analysis.

No Single Average is Perfect Arithmetic Mean uses every value but is affected by extremes. Median resists extremes but does not use every value fully. The correct measure depends on the problem.

Which Average Should You Use?

Arithmetic Mean

Use when all numerical observations should contribute to one general average.

Memory rule: Ordinary average

Median

Use when position matters or extreme values and open-ended classes are present.

Memory rule: Middle position

Mode

Use when the most common size, choice, brand or value is required.

Memory rule: Most frequent

Use Geometric Mean When

Growth takes place over several periods.
Rates combine multiplicatively.
Compound returns or index numbers are involved.

Use Harmonic Mean When

Average speed is required over equal distances.
Rates or “per-unit” quantities are averaged.
The variable appears in the denominator.

Situation	Best Measure	Reason
General numerical data	Arithmetic Mean	Uses all observations and permits algebraic treatment.
Income data with a few very rich persons	Median	Extreme values do not pull the result heavily.
Most demanded shoe size	Mode	Identifies the value occurring most frequently.
Average annual compound growth	Geometric Mean	Correct for multiplicative growth.
Average speed for equal distances	Harmonic Mean	Correct for rates with equal quantities in the numerator.

Property Comparison: What Each Average Can and Cannot Do

ICAI frequently asks these properties directly. Learn the table as a decision sheet rather than trying to derive the answer during the examination.

Property	AM	Median	Mode	GM	HM
Rigidly or uniquely defined	Yes	Yes	Not always	Yes	Yes
Based on all observations	Yes	No	No	Yes	Yes
Affected by extreme values	Yes	No	Generally little	Yes	Yes, especially by small values
Suitable for open-ended classes	No	Yes	Often possible	No	No
Capable of algebraic treatment	Excellent	Limited	Limited	Yes	Yes
Can be located graphically	No	Yes, by ogive	Yes, by histogram	No	No
Main use	General average	Position and skewed data	Most common value	Growth and ratios	Rates and equal-distance speed

Direct ICAI Answers Most commonly used measure → AM. Open-end classification → Median. Measure not always uniquely defined → Mode. Average rates → GM and HM. Quartiles and Median graphically → Ogive.

Quick Recall General value → AM; Middle position → Median; Most common → Mode; Growth → GM; Rates → HM.

Arithmetic Mean

Arithmetic Mean is obtained by adding all observations and dividing by the number of observations.

Individual Data
x̄ = Σx / n

Symbol	Meaning
x̄	Arithmetic Mean of the observations.
Σ	“Sum of”. It tells us to add all the values that follow it.
x	Each individual observation or value.
Σx	Total of all individual observations.
n	Number of individual observations.

Frequency Distribution
x̄ = Σfx / Σf = Σfx / N

Symbol	Meaning
x̄	Arithmetic Mean of the distribution.
f	Frequency corresponding to a value or class.
x	Value in a discrete series or midpoint of a class in grouped data.
fx	Product of frequency and the corresponding value or class midpoint.
Σfx	Total of all the fx products.
Σf or N	Total frequency, meaning the total number of observations.

Individual Data

Fully Solved Illustration: Arithmetic Mean of Individual Data

Question: Daily wages of nine workers are ₹58, ₹62, ₹48, ₹53, ₹70, ₹52, ₹60, ₹84 and ₹75. Find the arithmetic mean wage.

Step 1: Write the formulax̄ = Σx ÷ n

Step 2: Find Σx58 + 62 + 48 + 53 + 70 + 52 + 60 + 84 + 75 = 562

Step 3: Count observationsn = 9 workers

Step 4: Substitutex̄ = 562 ÷ 9

Step 5: Calculatex̄ = ₹62.44

Interpretation: The nine workers earn an average daily wage of approximately ₹62.44.

Grouped Frequency Distribution

For grouped data, first calculate the midpoint of every class. Multiply the midpoint by the corresponding frequency and then divide the total of fx by the total frequency.

Weight (kg)	Frequency (f)	Midpoint (x)	fx
44–48	3	46	138
49–53	4	51	204
54–58	5	56	280
59–63	7	61	427
64–68	9	66	594
69–73	8	71	568
Total	36	—	2211

Fully Solved Illustration: Arithmetic Mean of Grouped Data

Step 1: Find class midpointsx = (Lower limit + Upper limit) ÷ 2

Step 2: Multiply f by xFor 44–48: 3 × 46 = 138; repeat for every class

Step 3: Find totalsΣf = 36 and Σfx = 2211

Step 4: Apply formulax̄ = Σfx ÷ Σf = 2211 ÷ 36

Step 5: Final answerx̄ = 61.42 kg

Interpretation: The estimated average weight of the 36 persons is 61.42 kg.

Step-Deviation Method

x̄ = A + (Σfd / N) × C, where d = (x − A) / C

Symbol	Meaning
x̄	Arithmetic Mean to be calculated.
A	Assumed Mean, normally a convenient central value or midpoint.
x	Value or class midpoint.
C	Common factor or common class width used to simplify deviations.
d	Step deviation, calculated as (x − A) / C.
f	Frequency corresponding to each value or class.
fd	Product of frequency and step deviation.
Σfd	Total of all fd values.
N	Total frequency, Σf.

1Choose ASelect a convenient assumed mean

2Find dCalculate (x − A) ÷ C

3ApplyUse A + (Σfd/N) × C

Fully Solved Illustration: Step-Deviation Method

Question: Find the mean for class midpoints 10, 20, 30, 40 and 50 with frequencies 2, 3, 5, 4 and 1.

x	f	d = (x − 30) ÷ 10	fd
10	2	−2	−4
20	3	−1	−3
30	5	0	0
40	4	1	4
50	1	2	2
Total	15	—	−1

Step 1: Choose valuesA = 30, C = 10, N = 15 and Σfd = −1

Step 2: Substitutex̄ = 30 + (−1 ÷ 15) × 10

Step 3: Calculate adjustment(−1 ÷ 15) × 10 = −0.67

Step 4: Final answerx̄ = 30 − 0.67 = 29.33

Common Mistake For grouped data, x means the class midpoint. Do not use the lower or upper class limit directly.

Properties of Arithmetic Mean and Combined Mean

Property	Meaning
Constant observations	If every observation equals k, the mean is also k.
Sum of deviations	Σ(x − x̄) = 0 and Σf(x − x̄) = 0.
Change of origin and scale	If y = a + bx, then ȳ = a + bx̄.
Combined mean	Means of two or more groups are combined using their group sizes.

Combined Mean for Two Groups
x̄ = (n₁x̄₁ + n₂x̄₂) / (n₁ + n₂)

Symbol	Meaning
x̄	Combined Mean of both groups.
n₁ and n₂	Number of observations in the first and second groups.
x̄₁ and x̄₂	Mean of the first and second groups.
n₁x̄₁ and n₂x̄₂	Total value represented by each group.
n₁ + n₂	Total number of observations in both groups together.

Combined Mean Example

40 female workers earn an average of ₹5,200 and 60 male workers earn an average of ₹6,800.

Female wage total40 × 5,200 = ₹2,08,000

Male wage total60 × 6,800 = ₹4,08,000

Combined Mean₹6,16,000 ÷ 100 = ₹6,160

Change of Origin and Scale: Direct Transformation Questions

ICAI often gives a relationship such as y = a + bx and asks for the Mean, Median or Mode of y. Do not recalculate the complete series.

Transformation Rules
Mean(y) = a + b Mean(x)
Median(y) = a + b Median(x)
Mode(y) = a + b Mode(x)

Symbol	Meaning
x	Original variable.
y	Transformed variable.
a	Constant added to every observation; change of origin.
b	Constant multiplying every observation; change of scale.

Fully Solved Illustration: Median under Transformation

Question: y = 2x − 3 and Median(x) = 20. Find Median(y).

Step 1: Identify a and ba = −3 and b = 2

Step 2: Apply the ruleMedian(y) = a + b Median(x)

Step 3: SubstituteMedian(y) = −3 + 2(20)

Final AnswerMedian(y) = 37

Fully Solved Illustration: Mean from an Equation

Question: 2u + v + 7 = 0 and Mean(u) = 10. Find Mean(v).

Step 1: Express vv = −2u − 7

Step 2: Transform the meanMean(v) = −2 Mean(u) − 7

Step 3: Substitute−2(10) − 7

Final AnswerMean(v) = −27

Exam Trap These simple linear rules apply directly to AM, Median and Mode. Do not automatically use the same additive rule for GM or HM.

Missing Frequency Problems

Total FrequencyUse Σf = N

Mean EquationUse Σfx = N × x̄

Unknown ValuesSolve the equations

Symbol	Meaning
Σf	Total of all known and unknown frequencies.
N	Total number of observations.
Σfx	Total of frequency multiplied by the corresponding value.
x̄	Given Arithmetic Mean.
Unknown frequency	The missing value is usually represented by a letter such as m, x or y.

Exam Strategy First form the total-frequency equation. Then use the given mean to form the second equation.

Fully Solved Illustration: Missing Frequency

Question: Values 10, 20, 30 and 40 have frequencies 3, 5, m and 2. If the mean is 25, find m.

x	f	fx
10	3	30
20	5	100
30	m	30m
40	2	80
Total	10 + m	210 + 30m

Step 1: Use mean formula25 = (210 + 30m) ÷ (10 + m)

Step 2: Cross multiply250 + 25m = 210 + 30m

Step 3: Rearrange40 = 5m

Step 4: Final answerm = 8

Median

Median is the middle-most value after arranging observations in ascending or descending order. It is a positional average.

Odd Number of Observations

Median position = (n + 1) / 2 th item.

Example: For 7 observations, median is the 4th item.

Even Number of Observations

Median is the average of the n/2 th and (n/2 + 1) th items.

Example: For 8 observations, average the 4th and 5th items.

Symbol	Meaning in Individual-Series Median
n	Total number of observations after arranging them in order.
(n + 1) / 2	Position of the median when the number of observations is odd.
n / 2 and (n / 2) + 1	The two middle positions when the number of observations is even.
Median	For even observations, the average of the values at the two middle positions.

Odd Number Example

Marks: 72, 85, 56, 80, 65, 52, 68

Ordered data: 52, 56, 65, 68, 72, 80, 85

Median68

Even Number Example

Ordered wages: 56, 82, 82, 96, 100, 106, 110, 120

Middle values96 and 100

Median(96 + 100) ÷ 2 = ₹98

Grouped Frequency Distribution

Median = L + [(N/2 − c.f.) / f] × C

Symbol	Meaning
L	Lower class boundary of the median class.
N	Total frequency, Σf.
N/2	Position of the median observation.
c.f.	Cumulative frequency of the class immediately preceding the median class.
f	Frequency of the median class.
C	Class width or class size of the median class.

How to Read the Formula Start from the lower boundary of the median class and move proportionately inside that class according to the position of the median observation.

1Find N/2Locate the middle position

2Identify ClassFirst cumulative frequency above N/2

3Apply FormulaUse L, c.f., f and C

Fully Solved Illustration: Median of Grouped Data

Question: Find the median of the following distribution.

Class	f	Cumulative f
0–10	5	5
10–20	9	14
20–30	12	26
30–40	8	34
40–50	6	40

Step 1: Find N and N/2N = 40; N/2 = 20

Step 2: Locate median classThe first cumulative frequency above 20 is 26; therefore 20–30 is the median class

Step 3: Identify formula valuesL = 20, c.f. = 14, f = 12 and C = 10

Step 4: SubstituteMedian = 20 + [(20 − 14) ÷ 12] × 10

Step 5: CalculateMedian = 20 + 5 = 25

Interpretation: Half the observations lie below approximately 25 and half lie above it.

Do Not Confuse The median class is not necessarily the class with the highest frequency. It is the class containing the N/2th item.

Properties of Median

If y = a + bx, then Median of y = a + b × Median of x.
The sum of absolute deviations Σ|x − A| is minimum when A is the median.
Median is suitable for open-ended and highly skewed distributions.

Partition Values

Partition values divide an ordered distribution into equal parts.

Quartiles

Divide the distribution into four equal parts.

Q₁, Q₂ and Q₃

Deciles

Divide the distribution into ten equal parts.

D₁ to D₉

Percentiles

Divide the distribution into one hundred equal parts.

P₁ to P₉₉

Important Identity Q₂ = Median = D₅ = P₅₀

Unclassified Data

Position = (n + 1)p th item

Symbol	Meaning
n	Total number of ordered observations.
p	Required proportion of the distribution.
k	Number of the required quartile, decile or percentile.
p = k/4	Used for quartiles, such as Q₁ where p = 1/4.
p = k/10	Used for deciles, such as D₇ where p = 7/10.
p = k/100	Used for percentiles, such as P₆₀ where p = 60/100.

Use p = 1/4, 2/4 or 3/4 for quartiles; p = k/10 for deciles; and p = k/100 for percentiles.

Fully Solved Illustration: Quartile in Individual Data

Question: Find Q₁ for the ordered values 5, 8, 12, 15, 18, 21, 25, 30, 34 and 40.

Step 1: Number of observationsn = 10

Step 2: Find Q₁ position(n + 1) ÷ 4 = 11 ÷ 4 = 2.75th item

Step 3: Identify neighbouring values2nd item = 8 and 3rd item = 12

Step 4: InterpolateQ₁ = 8 + 0.75(12 − 8)

Step 5: Final answerQ₁ = 8 + 3 = 11

Graphical Determination of Median and Quartiles

Median, quartiles, deciles and percentiles are positional values and can be located from an ogive, that is, a cumulative-frequency curve.

1Find PositionN/2 for Median, N/4 for Q₁, 3N/4 for Q₃

2Mark on Y-axisLocate the required cumulative frequency

3Read X-axisMove to the curve and then down to obtain the value

Direct Recall Quartiles are determined graphically using an ogive, not a histogram, frequency polygon or pie chart.

Grouped Data

Partition Value = L + [(Np − c.f.) / f] × C

Symbol	Meaning
L	Lower class boundary of the class containing the required partition value.
N	Total frequency, Σf.
p	Required proportion, such as 1/4 for Q₁, 7/10 for D₇ or 60/100 for P₆₀.
Np	Position of the required quartile, decile or percentile.
c.f.	Cumulative frequency of the class immediately preceding the relevant class.
f	Frequency of the class containing the partition value.
C	Class width or class size.

Fully Solved Illustration: First Quartile of Grouped Data

Using the grouped distribution in the median example, find Q₁.

Step 1: Find positionN/4 = 40/4 = 10th item

Step 2: Locate Q₁ classThe 10th item lies in 10–20 because cumulative frequency rises from 5 to 14

Step 3: Identify valuesL = 10, c.f. = 5, f = 9 and C = 10

Step 4: SubstituteQ₁ = 10 + [(10 − 5) ÷ 9] × 10

Step 5: Final answerQ₁ = 10 + 5.56 = 15.56

Open-End Classification Median and partition values can still be used because they depend mainly on position. Arithmetic Mean may not be suitable when the first or last class is open-ended.

Mode

Mode is the value that occurs most frequently. It represents the most common or most popular value.

Mode is Useful For

Most demanded shoe or garment size.
Most popular brand or model.
Most common wage, price or choice.

Possible Forms

Unimodal: one mode.
Bimodal: two modes.
Multimodal: more than two modes.
No mode: equal frequencies.

Simple Example

5, 3, 8, 9, 5, 6

Mode5

Grouped Frequency Distribution

Mode = L + [(f₀ − f₋₁) / (2f₀ − f₋₁ − f₁)] × C

Symbol	Meaning
L	Lower class boundary of the modal class.
f₀	Frequency of the modal class, which is normally the highest frequency.
f₋₁	Frequency of the class immediately preceding the modal class.
f₁	Frequency of the class immediately succeeding the modal class.
C	Class width or class size of the modal class.

The modal class is the class with the highest frequency. The formula estimates the position of the mode within that class.

Fully Solved Illustration: Mode of Grouped Data

Question: Frequencies for classes 0–10, 10–20, 20–30, 30–40 and 40–50 are 4, 7, 12, 9 and 3. Find the mode.

Step 1: Identify modal class20–30 has the highest frequency 12

Step 2: Identify valuesL = 20, f₀ = 12, f₋₁ = 7, f₁ = 9 and C = 10

Step 3: SubstituteMode = 20 + [(12 − 7) ÷ (24 − 7 − 9)] × 10

Step 4: SimplifyMode = 20 + (5 ÷ 8) × 10

Step 5: Final answerMode = 26.25

Empirical Relationship

Mean − Mode = 3(Mean − Median)
Mode = 3 Median − 2 Mean

Term	Meaning
Mean	Arithmetic Mean of the distribution.
Median	Middle positional value of the distribution.
Mode	Most frequently occurring or most typical value.
3	Empirical coefficient used in the approximate relationship for a moderately skewed distribution.

Use with Care The empirical relationship is only an approximation for a moderately skewed distribution. It is not a universal identity.

Missing Frequencies Using Median or Mode

When frequencies are missing, first use the total-frequency condition. Then use the given Median or Mode formula to form another equation.

Illustration: Missing Frequencies when Median is Given

A distribution has total frequency 100. Two frequencies are missing. The Median is 32.

Step 1Use Σf = 100 to form the first equation.

Step 2Since Median = 32 lies in 30–40, identify L, c.f., f and C.

Step 3Substitute in Median = L + [(N/2 − c.f.)/f] × C.

Step 4Solve the Median equation together with the total-frequency equation.

Exam point: Do not guess the missing frequencies. The two equations must be solved simultaneously.

Illustration: Missing Frequency when Mode is Given

If the modal value is given, identify the modal class first. The unknown frequency may appear as f₀, f₋₁ or f₁.

Step 1Identify modal class from the given Mode value.

Step 2Write L, C, f₀, f₋₁ and f₁.

Step 3Substitute in Mode = L + [(f₀ − f₋₁)/(2f₀ − f₋₁ − f₁)] × C.

Step 4Solve for the unknown frequency and then calculate the required Mean or Median.

Geometric Mean

For n positive observations, Geometric Mean is the nth root of their product.

Individual Data: G = (x₁ × x₂ × ... × xₙ)^1/n

Symbol	Meaning
G	Geometric Mean.
x₁, x₂, …, xₙ	Positive individual observations.
n	Number of observations.
1/n	Indicates that the nth root of the product is to be taken.

Frequency Data: log G = Σf log x / N

Symbol	Meaning
G	Geometric Mean.
log G	Logarithm of the Geometric Mean.
x	Positive value or class midpoint.
log x	Logarithm of each value or class midpoint.
f	Frequency corresponding to each value.
Σf log x	Total of frequency multiplied by the logarithm of the corresponding value.
N	Total frequency, Σf.
Antilog	Reverse logarithm used to obtain G after calculating log G.

Fully Solved Illustration: Geometric Mean of Individual Data

Question: Find the GM of 3, 6 and 12.

Step 1: Write formulaG = (x₁ × x₂ × x₃)^1/3

Step 2: SubstituteG = (3 × 6 × 12)^1/3

Step 3: Multiply3 × 6 × 12 = 216

Step 4: Take cube rootG = 216^1/3 = 6

Fully Solved Illustration: Average Compound Growth Rate

Sales grow by 10%, 20% and 5% in three successive years. Find the average annual compound growth rate.

Step 1: Convert rates into relatives1.10, 1.20 and 1.05

Step 2: Multiply relatives1.10 × 1.20 × 1.05 = 1.386

Step 3: Take cube rootGM relative = 1.386^1/3 ≈ 1.115

Step 4: Convert back to percentage(1.115 − 1) × 100 ≈ 11.5%

The compound average is approximately 11.5% per year, not the simple average of 11.67%.

Use GM For

Compound growth rates.
Average investment returns.
Index numbers and ratios.

Remember

All observations should be positive.
GM is suitable for multiplicative change.
Logarithms simplify its calculation.

Harmonic Mean

Harmonic Mean is the reciprocal of the arithmetic mean of the reciprocals.

Individual Data: H = n / Σ(1/x)

Symbol	Meaning
H	Harmonic Mean.
n	Number of individual observations.
x	Each positive observation.
1/x	Reciprocal of each observation.
Σ(1/x)	Total of the reciprocals of all observations.

Frequency Data: H = N / Σ(f/x)

Symbol	Meaning
H	Harmonic Mean.
N	Total frequency, Σf.
f	Frequency corresponding to each value.
x	Value or class midpoint.
f/x	Frequency divided by the corresponding value.
Σ(f/x)	Total of all f/x values.

Fully Solved Illustration: Harmonic Mean of Individual Data

Question: Find the HM of 4, 6 and 10.

Step 1: Write formulaH = n ÷ Σ(1/x)

Step 2: Find reciprocals1/4, 1/6 and 1/10

Step 3: Add reciprocals1/4 + 1/6 + 1/10 = 31/60

Step 4: SubstituteH = 3 ÷ (31/60)

Step 5: Final answerH = 180/31 = 5.81 approximately

Fully Solved Illustration: Average Speed over Equal Distances

A vehicle covers 120 km at 40 km/h and another 120 km at 60 km/h.

Step 1: Total distance120 + 120 = 240 km

Step 2: Time for first part120 ÷ 40 = 3 hours

Step 3: Time for second part120 ÷ 60 = 2 hours

Step 4: Total time3 + 2 = 5 hours

Step 5: Average speed240 ÷ 5 = 48 km/h

This is also obtained by HM: 2 ÷ (1/40 + 1/60) = 48 km/h.

Equal Distances

Use Harmonic Mean for average speed.

Example: Equal kilometres at 40 km/h and 60 km/h.

Equal Times

Use Arithmetic Mean for average speed.

Example: One hour at 40 km/h and one hour at 60 km/h.

Concept Check Equal distances → HM. Equal times → AM.

Relationship between AM, GM and HM

AM ≥ GM ≥ HM

Symbol	Meaning
AM	Arithmetic Mean.
GM	Geometric Mean.
HM	Harmonic Mean.
≥	Greater than or equal to.

For the same set of positive observations, AM cannot be smaller than GM, and GM cannot be smaller than HM. Equality occurs only when all observations are equal.

Fully Solved Illustration: Verify AM ≥ GM ≥ HM

For the observations 6, 8, 12 and 36:

Arithmetic Mean(6 + 8 + 12 + 36) ÷ 4 = 62 ÷ 4 = 15.5

Geometric Mean(6 × 8 × 12 × 36)^1/4 = 20736^1/4 = 12

Harmonic Mean4 ÷ (1/6 + 1/8 + 1/12 + 1/36) = 4 ÷ (29/72) = 9.93

Therefore, 15.5 ≥ 12 ≥ 9.93.

Special Result for Two Positive Numbers

AM × HM = GM²

Symbol	Meaning
AM	Arithmetic Mean of exactly two positive observations.
GM	Geometric Mean of the same two observations.
HM	Harmonic Mean of the same two observations.
GM²	Square of the Geometric Mean.

Important Limitation AM × HM = GM² holds for exactly two positive observations. It does not generally hold for more than two observations.

Weighted Averages

When observations do not have equal importance, a weight is attached to each observation.

Weighted AM = Σwx / Σw

Symbol	Meaning
x	Observation or value.
w	Weight or relative importance attached to the observation.
wx	Product of the weight and the observation.
Σwx	Total of all weighted values.
Σw	Total of all weights.

Weighted GM = Antilog [Σw log x / Σw]

Symbol	Meaning
x	Positive observation or growth relative.
w	Weight attached to each observation.
log x	Logarithm of each observation.
Σw log x	Total of weight multiplied by the logarithm of each observation.
Σw	Total of all weights.
Antilog	Reverse logarithm used to convert the result back to the original scale.

Weighted HM = Σw / Σ(w/x)

Symbol	Meaning
x	Positive observation, rate or price.
w	Weight attached to each observation.
w/x	Weight divided by the corresponding observation.
Σ(w/x)	Total of all w/x values.
Σw	Total of all weights.

Fully Solved Illustration: Weighted Arithmetic Mean

A student scores 70, 80 and 90 in components carrying weights of 20%, 30% and 50%.

Score x	Weight w	wx
70	20	1,400
80	30	2,400
90	50	4,500
Total	100	8,300

Step 1: Apply formulaWeighted AM = Σwx ÷ Σw

Step 2: Substitute8,300 ÷ 100

Step 3: Final answerWeighted mean = 83 marks

Fully Solved Illustration: Weighted Harmonic Mean

A buyer purchases 20 units at ₹10 per unit and 30 units at ₹15 per unit. Find the weighted harmonic mean price where quantities are the weights.

Step 1: Write formulaWeighted HM = Σw ÷ Σ(w/x)

Step 2: Substitute(20 + 30) ÷ [(20/10) + (30/15)]

Step 3: Simplify denominator2 + 2 = 4

Step 4: Final answerWeighted HM = 50 ÷ 4 = ₹12.50

Meaning of Weight A larger weight gives the corresponding observation greater influence on the final average.

Quick Comparative Review

Measure	Main Strength	Main Weakness	Best Use
AM	Uses all observations and is algebraically useful	Affected by extreme values	General numerical analysis
Median	Resists extremes and works with open ends	Does not use every value fully	Income and skewed data
Mode	Shows the most common value	May be absent or multiple	Popular size, brand or choice
GM	Correct for compound growth	Cannot ordinarily use zero or negative values	Growth rates and index numbers
HM	Correct for rates and ratios	Highly affected by very small values	Speed and per-unit rates

Worked Examples and Exam Traps

Example 1: Mode from Mean and Median

Mean = 55.60 and Median = 52.40.

FormulaMode = 3 Median − 2 Mean

Calculation3(52.40) − 2(55.60) = 46

Example 2: Two Numbers from AM and GM

For two positive numbers, AM = 5 and GM = 4.

Suma + b = 10

Productab = 16

Numbers8 and 2

Example 3: Average Speed

A vehicle covers equal distances at 40 km/h and 60 km/h.

Correct AverageHarmonic Mean

Average Speed2 ÷ (1/40 + 1/60) = 48 km/h

Question-Type Coverage Checklist

The following table converts the exercise pattern into direct revision instructions. A student should be able to answer each line without deriving the rule during the examination.

Question Type	Direct Rule to Recall
Meaning of central tendency	It measures the central location, not dispersion or scatterness.
Grouped AM assumption	All observations in a class are represented by the class midpoint.
Sampling fluctuation	AM is affected by sampling fluctuations; the statement “not affected” is wrong.
Open-end classification	Median is generally the best measure.
Extreme observations	Median is not materially affected; AM is affected.
Even number of observations	Median is the simple average of the two middle ordered values.
Uniqueness	Mode may not be uniquely defined.
Average rates	GM and HM are considered; HM is especially relevant for rates with equal quantities.
Profits and losses	GM cannot ordinarily be used when values include negatives.
Graphical quartiles	Use an ogive.
Linear transformation	Apply y = a + bx directly to Mean, Median and Mode.
AM = GM	All positive observations are equal, therefore HM is also equal.

Solved Numerical Drill

1. Find Two Numbers from AM and GM

AM = 6.5 and GM = 6 for two positive numbers.

Suma + b = 2 × 6.5 = 13

Productab = GM² = 36

Equationt² − 13t + 36 = 0

Factorise(t − 9)(t − 4) = 0

AnswerThe numbers are 9 and 4.

2. Combined Mean and Group Proportion

Unskilled workers earn ₹10,000 on average, skilled workers earn ₹15,000, and the combined mean is ₹12,000. Find the percentage of skilled workers.

Assume proportionsSkilled = p; Unskilled = 1 − p

Weighted equation15,000p + 10,000(1 − p) = 12,000

Solve5,000p = 2,000, so p = 0.40

Answer40% workers are skilled.

3. Combined Harmonic Mean

Two groups contain 15 and 13 observations and have HMs of 75 and 65 respectively.

Combined HM = (n₁ + n₂) ÷ [(n₁/H₁) + (n₂/H₂)]

Substitute28 ÷ [(15/75) + (13/65)]

Simplify denominator0.20 + 0.20 = 0.40

AnswerCombined HM = 28 ÷ 0.40 = 70

4. Equal-Distance Average Speed

An aircraft travels from A to B at 500 km/h and returns over the same distance at 700 km/h.

Average speed = 2ab ÷ (a + b)

Substitute2(500)(700) ÷ (500 + 700)

Calculate700,000 ÷ 1,200

Answer583.33 km/h

Minimum-Thinking Rule First classify the question: property, position, frequency, growth, rate, transformation or combined group. Then select the ready-made rule from the notes instead of rebuilding the concept from scratch.

Common Exam Traps

Trap	Correct Idea
Mean is unaffected by extremes	Wrong. Mean is strongly affected by extreme values.
Median class has the highest frequency	Wrong. It contains the N/2th item.
Mode always exists	Wrong. It may be absent or more than one.
GM can ordinarily use negative values	Wrong in the usual CA Foundation treatment.
Average speed is always AM	Use HM when equal distances are covered.
AM × HM = GM² for every data set	It holds only for two positive observations.

MCQ Shortcut Identify what the question is really asking: ordinary level, middle position, most common value, compound growth or rate. That usually reveals the correct average.

Revision Mind Map

Measures of Central Tendency — One Page Recall

First identify the nature of the data and the purpose of the average. Then choose AM, Median, Mode, GM or HM.

1. Meaning and Choice

One representative value summarises the data.
AM for general numerical use.
Median for position, open ends and skewed data.
Mode for the most common value.
GM for growth and HM for rates.

2. Arithmetic Mean

x̄ = Σx/n or Σfx/N.
Uses every observation.
Σ(x − x̄) = 0.
Suitable for algebraic treatment.
Affected by extreme values.

Formula Recall: Combined Mean = Total of all groups ÷ Total number of observations.

3. Median

Middle positional value.
Arrange data before locating it.
Grouped data uses N/2.
Works with open-ended classes.
Minimises absolute deviations.

4. Partition Values

Quartiles divide into 4 parts.
Deciles divide into 10 parts.
Percentiles divide into 100 parts.
Q₂ = D₅ = P₅₀ = Median.

5. Mode

Most frequently occurring value.
Modal class has the highest frequency.
Mode ≈ 3 Median − 2 Mean.
May be absent or multiple.

6. Geometric Mean

Used for compound and multiplicative change.
Useful for growth rates and index numbers.
Requires positive observations.
Logarithms simplify calculation.

7. Harmonic Mean

Reciprocal of the AM of reciprocals.
Used for rates and per-unit quantities.
Equal distances → HM for average speed.
Equal times → AM for average speed.

8. Relationship and Weighted Averages

For positive data: AM ≥ GM ≥ HM.
Equality when all observations are equal.
For two positive values: AM × HM = GM².
Weighted AM = Σwx/Σw.
Larger weight means greater influence.

Final Memory: Ordinary values → AM; Middle position → Median; Most common → Mode; Growth → GM; Rates → HM.

Measures Of Central Tendency – Chapter Notes

Meaning and Need for an Average

Why Do We Need an Average?

From Data to Decision

Qualities of an Ideal Measure

Clearly Defined

Simple

Uses the Data

Reliable

Which Average Should You Use?

Arithmetic Mean

Median

Mode

Use Geometric Mean When

Use Harmonic Mean When

Property Comparison: What Each Average Can and Cannot Do

Arithmetic Mean

Individual Data

Fully Solved Illustration: Arithmetic Mean of Individual Data

Grouped Frequency Distribution

Fully Solved Illustration: Arithmetic Mean of Grouped Data

Step-Deviation Method

Fully Solved Illustration: Step-Deviation Method

Properties of Arithmetic Mean and Combined Mean

Combined Mean Example

Change of Origin and Scale: Direct Transformation Questions

Fully Solved Illustration: Median under Transformation

Fully Solved Illustration: Mean from an Equation

Missing Frequency Problems

Fully Solved Illustration: Missing Frequency

Median

Odd Number of Observations

Even Number of Observations

Odd Number Example

Even Number Example

Grouped Frequency Distribution

Fully Solved Illustration: Median of Grouped Data

Properties of Median

Partition Values

Quartiles

Deciles

Percentiles

Unclassified Data

Fully Solved Illustration: Quartile in Individual Data

Graphical Determination of Median and Quartiles

Grouped Data

Fully Solved Illustration: First Quartile of Grouped Data

Mode

Mode is Useful For

Possible Forms

Simple Example

Grouped Frequency Distribution

Fully Solved Illustration: Mode of Grouped Data

Empirical Relationship

Missing Frequencies Using Median or Mode

Illustration: Missing Frequencies when Median is Given

Illustration: Missing Frequency when Mode is Given

Geometric Mean

Fully Solved Illustration: Geometric Mean of Individual Data

Fully Solved Illustration: Average Compound Growth Rate

Use GM For

Remember

Harmonic Mean

Fully Solved Illustration: Harmonic Mean of Individual Data

Fully Solved Illustration: Average Speed over Equal Distances

Equal Distances

Equal Times

Relationship between AM, GM and HM

Fully Solved Illustration: Verify AM ≥ GM ≥ HM

Special Result for Two Positive Numbers

Weighted Averages

Fully Solved Illustration: Weighted Arithmetic Mean

Fully Solved Illustration: Weighted Harmonic Mean

Quick Comparative Review

Worked Examples and Exam Traps

Example 1: Mode from Mean and Median

Example 2: Two Numbers from AM and GM

Example 3: Average Speed

Question-Type Coverage Checklist

Solved Numerical Drill