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Measures of Central Tendency — Mean, Median & Mode

Master Mean Median and Mode for JEE Main JEE Advanced and Board Exams. Covers arithmetic mean for ungrouped grouped and discrete data, three methods of computing grouped mean, median formula for continuous data, mode formula, empirical relation Mode equals 3 Median minus 2 Mean, combined mean, weighted mean with fully verified solved examples.

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When we collect data — whether exam scores, rainfall figures, or stock prices — we need a single number that best represents the entire dataset. Measures of central tendency are statistical tools that locate the "centre" of a distribution. The three most important measures are the mean (arithmetic average), the median (middle value), and the mode (most frequent value). Each captures a different aspect of the data's centre and is appropriate in different situations. For JEE (Main & Advanced), this topic appears directly in statistics questions and also as a prerequisite for measures of dispersion. Expect 1–2 questions per paper on mean, median, and their properties.

1. Arithmetic Mean JEE Main & Advanced

The arithmetic mean (or simply mean) $\bar{x}$ of a dataset is the sum of all observations divided by the number of observations.

Mean for Ungrouped Data

For $n$ observations $x_{1}, x_{2}, \dots, x_{n}$ :

$\bar{x} = \frac{x_{1} + x_{2} + \dots + x_{n}}{n} = \frac{\sum_{i = 1}^{n} x_{i}}{n} = \frac{Σ x_{i}}{n}$

Mean for Discrete Frequency Distribution

For values $x_{1}, x_{2}, \dots, x_{k}$ with frequencies $f_{1}, f_{2}, \dots, f_{k}$ (where $Σ f_{i} = N$ ):

$\bar{x} = \frac{Σ f_{i} x_{i}}{Σ f_{i}} = \frac{Σ f_{i} x_{i}}{N}$

Mean for Continuous (Grouped) Data — Three Methods

Method 1 — Direct Method:

$\bar{x} = \frac{Σ f_{i} m_{i}}{N}$

where $m_{i}$ is the midpoint of the $i$ th class interval: $m_{i} = \frac{lower limit + upper limit}{2}$ .

Method 2 — Assumed Mean (Short-cut) Method:

$\bar{x} = A + \frac{Σ f_{i} d_{i}}{N}$ where $d_{i} = m_{i} - A$

$A$ is the assumed mean (usually the midpoint of the class with the highest frequency). This method reduces arithmetic when numbers are large.

Method 3 — Step Deviation Method:

$\bar{x} = A + \frac{Σ f_{i} u_{i}}{N} \times h$ where $u_{i} = \frac{d_{i}}{h} = \frac{m_{i} - A}{h}$

$h$ = class width (common class size). Simplifies further when all class widths are equal.

Properties of Arithmetic Mean

Property	Statement
Sum of deviations	$Σ (x_{i} - \bar{x}) = 0$ — deviations from mean always sum to zero
Shift by constant	If each $x_{i}$ is replaced by $x_{i} + k$ , new mean $= \bar{x} + k$
Scale by constant	If each $x_{i}$ is replaced by $k x_{i}$ , new mean $= k \bar{x}$
Minimises sum of squared deviations	$Σ (x_{i} - a)^{2}$ is minimum when $a = \bar{x}$ (least squares property)
Combined mean	For two groups with means ${\bar{x}}_{1}$ , ${\bar{x}}_{2}$ and sizes $n_{1}$ , $n_{2}$ : $\bar{x} = \frac{n_{1} {\bar{x}}_{1} + n_{2} {\bar{x}}_{2}}{n_{1} + n_{2}}$
Effect of wrong entry	Corrected mean $= \bar{x} + \frac{correct value - wrong value}{n}$

Weighted Mean

When observations have different weights $w_{i}$ :

${\bar{x}}_{w} = \frac{Σ w_{i} x_{i}}{Σ w_{i}}$

2. Median JEE Main & Advanced

The median is the middle value of a dataset when arranged in ascending or descending order. It divides the distribution into two equal halves.

Median for Ungrouped Data

Arrange data in ascending order. Then:

If $n$ is odd: Median $= (\frac{n + 1}{2})$ th observation.
If $n$ is even: Median $= \frac{(\frac{n}{2}) th + (\frac{n}{2} + 1) th observation}{2}$ .

Median for Grouped (Continuous) Data

$Median = l + \frac{\frac{N}{2} - c f}{f} \times h$

where:

$l$ = lower boundary of the median class
$N$ = total frequency ( $Σ f_{i}$ )
$c f$ = cumulative frequency of the class preceding the median class
$f$ = frequency of the median class
$h$ = class width (height of the median class)
Median class = the class whose cumulative frequency first exceeds $N / 2$

Properties of Median

Median minimises $Σ | x_{i} - a |$ (sum of absolute deviations) — the least absolute deviations property.
Not affected by extreme values (outliers) — unlike the mean.
Suitable for skewed distributions and ordinal data.
For a symmetric distribution: Mean = Median = Mode.

3. Mode JEE Main & Advanced

The mode is the observation (or class) that occurs with the highest frequency in a dataset.

Mode for Ungrouped Data

Simply the value that appears most often. A dataset can be:

Unimodal: one mode
Bimodal: two modes
Multimodal: more than two modes
No mode: all values appear equally often

Mode for Grouped Data

The modal class is the class with the highest frequency. Mode is found by:

$Mode = l + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times h$

where:

$l$ = lower boundary of the modal class
$f_{1}$ = frequency of the modal class (highest frequency)
$f_{0}$ = frequency of the class preceding the modal class
$f_{2}$ = frequency of the class following the modal class
$h$ = class width

Properties of Mode

Easiest to compute — no calculation needed for ungrouped data.
Not affected by extreme values.
May not exist or may not be unique.
Useful for qualitative (categorical) data.

4. Empirical Relationship Between Mean, Median and Mode JEE Main & Advanced

For a moderately skewed (asymmetric) distribution, Karl Pearson established the empirical relationship:

$Mode = 3 Median - 2 Mean$

This is a very frequently tested result — it allows computing one measure when the other two are known.

Distribution Type	Relationship	Shape
Symmetric	Mean = Median = Mode	Bell-shaped; no skew
Positively skewed (right-skewed)	Mean $>$ Median $>$ Mode	Long tail on the right
Negatively skewed (left-skewed)	Mean $<$ Median $<$ Mode	Long tail on the left

5. Geometric Mean and Harmonic Mean CBSE Boards

Geometric Mean (GM) of $n$ positive observations:

$G M = (x_{1} \cdot x_{2} \dots x_{n})^{1 / n}$

Used for averaging ratios, growth rates, and index numbers.

Harmonic Mean (HM) of $n$ observations:

$H M = \frac{n}{\frac{1}{x_{1}} + \frac{1}{x_{2}} + \dots + \frac{1}{x_{n}}} = \frac{n}{Σ (1 / x_{i})}$

Used for averaging rates (speed, frequency).

Inequality: AM $\geq$ GM $\geq$ HM

For positive numbers, the arithmetic mean is always $\geq$ geometric mean $\geq$ harmonic mean, with equality iff all observations are equal.

6. Choosing the Right Measure of Central Tendency

Situation	Best Measure	Reason
Symmetric distribution, no outliers	Mean	Uses all data; most efficient
Skewed distribution or outliers present	Median	Not distorted by extreme values
Categorical (qualitative) data	Mode	Mean/median not applicable for non-numeric data
Most common value needed	Mode	Directly gives the most frequent value
Averaging growth rates, ratios	Geometric Mean	Appropriate for multiplicative relationships
Averaging speeds (same distance)	Harmonic Mean	Gives correct average speed

Mean, Median, Mode — Quick Computation Framework

Task	Key Formula / Step
Mean of grouped data (large numbers)	Use step deviation: $\bar{x} = A + \frac{Σ f_{i} u_{i}}{N} \times h$ ; choose $A$ as middle class midpoint
Find median class	Compute cumulative frequencies; median class = first class where $c f \geq N / 2$
Find mode class	Modal class = class with highest frequency $f_{1}$ ; note $f_{0}$ (before) and $f_{2}$ (after)
Given mean and median, find mode	Mode $= 3 Median - 2 Mean$ (empirical formula — extremely fast)
Combined mean of two groups	$\bar{x} = \frac{n_{1} {\bar{x}}_{1} + n_{2} {\bar{x}}_{2}}{n_{1} + n_{2}}$ ; never average the means directly
Correcting mean after wrong entry	Corrected $Σ x = n \bar{x} - wrong + correct$ ; new mean $=$ corrected $Σ x \div n$

JEE Power Tips — Avoid These Common Mistakes

Empirical formula: Mode = 3 Median − 2 Mean. This is Karl Pearson's empirical relation for moderately skewed distributions. It is tested directly in JEE — memorise it exactly. Wrong version: Mode = 2 Median − 3 Mean is a very common error.
For grouped data, use class midpoints $m_{i}$ , not class limits. The midpoint of a class $[10, 20)$ is $15$ , not $10$ or $20$ . Using boundaries instead of midpoints gives a wrong mean.
Median class: cumulative frequency first exceeds $N / 2$ , not equals $N / 2$ . If $N / 2 = 25$ and cumulative frequencies are $20, 25, 35, \dots$ : if $c f = 25$ exactly, the median class is the next one (where $c f$ first strictly exceeds $N / 2$ ). The formula uses $c f$ of the class before the median class.
In the mode formula, $f_{0}$ and $f_{2}$ are frequencies of adjacent classes, not cumulative frequencies. Using cumulative frequencies here is a common error that gives a completely wrong answer.
Combined mean is a weighted average, not a simple average of means. If group 1 has mean $10$ ( $n_{1} = 100$ ) and group 2 has mean $20$ ( $n_{2} = 10$ ), the combined mean is $\frac{100 (10) + 10 (20)}{110} = \frac{1200}{110} \approx 10.9$ , not $\frac{10 + 20}{2} = 15$ .
Mean is affected by outliers; median is not. In a dataset ${1, 2, 3, 4, 100}$ : Mean $= 22$ (distorted by 100), Median $= 3$ (unaffected). This distinction is a standard MCQ.
$Σ (x_{i} - \bar{x}) = 0$ always — the sum of deviations from the mean is always zero. If you are asked to verify this or use it to find a missing value, this identity is the key.

Practice Questions (JEE / Board Level)

Q1: The mean of 5 observations is 8. If one observation 6 is replaced by 14, the new mean is:

A) 8.8
B) 9.6
C) 10
D) 9

Answer: B) 9.6.

Explanation:

First, find the original sum of the observations:
Original sum = $5 \times 8 = 40$ .

Adjust the sum by subtracting the old observation and adding the new one:
New sum = $40 - 6 + 14 = 48$ .

Calculate the new mean:
New mean = $\frac{48}{5} = 9.6$ .

Q2: The median of the data {3, 7, 2, 9, 5, 11, 4} is:

A) 4
B) 5
C) 7
D) 9

Answer: B) 5.

Explanation:

First, arrange the data in ascending order: 2, 3, 4, 5, 7, 9, 11.

The number of observations is $n = 7$ (which is odd).
The position of the median is the $(\frac{n + 1}{2})$ th observation:
$(\frac{7 + 1}{2}) = \frac{8}{2} = 4 th observation$ .

The 4th observation in the sorted list is 5.

Q3: If the mean and median of a moderately skewed distribution are 12 and 10 respectively, then the mode is:

A) 4
B) 6
C) 8
D) 14

Answer: B) 6.

Explanation:

For a moderately skewed distribution, the relationship between mean, median, and mode is given by Karl Pearson's empirical formula:
$Mode = 3 (Median) - 2 (Mean)$

Substitute the given values:
$Mode = 3 (10) - 2 (12)$
$Mode = 30 - 24 = 6$ .

Q4: The mean of two groups of sizes 40 and 60 are 30 and 50 respectively. The combined mean is:

A) 40
B) 42
C) 44
D) 45

Answer: B) 42.

Explanation:

The formula for the combined mean ( $\bar{x}$ ) of two groups is:
$\bar{x} = \frac{n_{1} {\bar{x}}_{1} + n_{2} {\bar{x}}_{2}}{n_{1} + n_{2}}$

Substitute the given values ( $n_{1} = 40, {\bar{x}}_{1} = 30$ and $n_{2} = 60, {\bar{x}}_{2} = 50$ ):
$\bar{x} = \frac{40 (30) + 60 (50)}{40 + 60}$
$\bar{x} = \frac{1200 + 3000}{100}$
$\bar{x} = \frac{4200}{100} = 42$ .

Q5: Consider the following frequency distribution:

Class	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	15	12	10

The modal class is:

A) 10–20
B) 20–30
C) 30–40
D) 40–50

Answer: B) 20–30.

Explanation:

The modal class is defined as the class interval that has the highest frequency. Looking at the table, the frequencies are 5, 8, 15, 12, and 10. The highest frequency is 15, which corresponds to the class interval 20–30.

Q6: Using the data from Q5, the mode of the distribution is:

A) 25
B) 27
C) 28
D) 30

Answer: B) 27.

Explanation:

From the previous question, the modal class is 20–30.
Lower limit of modal class ( $l$ ) = 20
Class size ( $h$ ) = 10
Frequency of modal class ( $f_{1}$ ) = 15
Frequency of preceding class ( $f_{0}$ ) = 8
Frequency of succeeding class ( $f_{2}$ ) = 12

Use the mode formula for grouped data:
$Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h$

Substitute the values:
$Mode = 20 + (\frac{15 - 8}{2 (15) - 8 - 12}) \times 10$
$Mode = 20 + (\frac{7}{30 - 20}) \times 10$
$Mode = 20 + (\frac{7}{10}) \times 10$
$Mode = 20 + 7 = 27$ .

Q7 (JEE type): The mean of $n$ observations is $\bar{x}$ . If the first observation is increased by 1, the second by 2, the third by 3, and so on, the new mean is:

A) $\bar{x} + n$
B) $\bar{x} + \frac{n}{2}$
C) $\bar{x} + \frac{n + 1}{2}$
D) $\bar{x} + \frac{n - 1}{2}$

Answer: C) $\bar{x} + \frac{n + 1}{2}$ .

Explanation:

The original sum of the $n$ observations is $n \bar{x}$ .

The total increase added to the observations is the sum of the first $n$ natural numbers:
$Total increase = 1 + 2 + 3 + \dots + n$
$= \frac{n (n + 1)}{2}$ .

The new sum of the observations is:
$New sum = n \bar{x} + \frac{n (n + 1)}{2}$ .

To find the new mean, divide the new sum by the number of observations ( $n$ ):
$New mean = \frac{n \bar{x} + \frac{n (n + 1)}{2}}{n}$
$New mean = \bar{x} + \frac{n + 1}{2}$ .

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Measures of Central Tendency — Mean, Median & Mode

1. Arithmetic Mean JEE Main & Advanced

Mean for Ungrouped Data

Mean for Discrete Frequency Distribution

Mean for Continuous (Grouped) Data — Three Methods

Properties of Arithmetic Mean

Weighted Mean

2. Median JEE Main & Advanced

Median for Ungrouped Data

Median for Grouped (Continuous) Data

Properties of Median

3. Mode JEE Main & Advanced

Mode for Ungrouped Data

Mode for Grouped Data

Properties of Mode

4. Empirical Relationship Between Mean, Median and Mode JEE Main & Advanced

5. Geometric Mean and Harmonic Mean CBSE Boards

Inequality: AM $\geq$ GM $\geq$ HM

6. Choosing the Right Measure of Central Tendency

Mean, Median, Mode — Quick Computation Framework

Practice Questions (JEE / Board Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

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Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Measures of Central Tendency — Mean, Median & Mode

1. Arithmetic Mean JEE Main & Advanced

Mean for Ungrouped Data

Mean for Discrete Frequency Distribution

Mean for Continuous (Grouped) Data — Three Methods

Properties of Arithmetic Mean

Weighted Mean

2. Median JEE Main & Advanced

Median for Ungrouped Data

Median for Grouped (Continuous) Data

Properties of Median

3. Mode JEE Main & Advanced

Mode for Ungrouped Data

Mode for Grouped Data

Properties of Mode

4. Empirical Relationship Between Mean, Median and Mode JEE Main & Advanced

5. Geometric Mean and Harmonic Mean CBSE Boards

Inequality: AM ≥ GM ≥ HM

6. Choosing the Right Measure of Central Tendency

Mean, Median, Mode — Quick Computation Framework

Practice Questions (JEE / Board Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

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We Value Your Privacy

Inequality: AM $\geq$ GM $\geq$ HM