Maths · JEE Main & Advanced 2026 · Statistics

Statistics notes, formulas & PYQs for JEE 2026

Complete chapter — measures of central tendency (mean, median, mode), measures of dispersion (range, mean deviation, variance, standard deviation), combined mean & variance, properties of variance, coefficient of variation, 50+ worked examples, and JEE PYQs with solutions.

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Statistics JEE Notes, Formulas, PYQs
Statistics — JEE Notes, Formulas, PYQs
1

Measures of Central Tendency

A measure of central tendency attempts to find a single central value that best represents the entire distribution of data. The three most common measures are Mean, Median, and Mode. In JEE, questions typically test your ability to compute these for ungrouped and grouped data, and to use their empirical relationship.

1.1 Arithmetic Mean (\(\bar{x}\))

The arithmetic mean is the sum of all observations divided by the total number of observations.

Most Repeated PYQ — Mean of Grouped Data

(a) Ungrouped Data: For \(n\) observations \(x_1, x_2, \ldots, x_n\),

\[\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}\]

(b) Grouped Data (Frequency Distribution): If frequencies are \(f_1, f_2, \ldots, f_n\),

\[\bar{x} = \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i} = \frac{\sum f_i x_i}{N}\]

(c) Class Intervals: If data is given in class intervals, \(x_i\) is taken as the class mark (midpoint): \(x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}\).

Shortcut Method (Assumed Mean Method)

Let \(A\) be the assumed mean. Then \(d_i = x_i - A\).

\[\bar{x} = A + \frac{\sum f_i d_i}{N}\]

Step-Deviation Method: If class size is \(h\), let \(u_i = \frac{x_i - A}{h}\).

\[\bar{x} = A + h \left( \frac{\sum f_i u_i}{N} \right)\]

1.2 Median

The median is the value of the middle observation when data is arranged in ascending or descending order. It divides the data into two equal parts.

Ungrouped Data: If \(n\) is odd, Median = value of \(\frac{n+1}{2}\)-th item. If \(n\) is even, Median = average of \(\frac{n}{2}\)-th and \((\frac{n}{2}+1)\)-th items.

Grouped Data:

\[\text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h\]

Where \(L\) = lower limit of median class, \(N\) = total frequency, \(F\) = cumulative frequency before median class, \(f\) = frequency of median class, \(h\) = class size.

1.3 Mode

The mode is the observation that occurs most frequently. In a frequency distribution, it is the value with the highest frequency.

Important — Empirical Relation

For a moderately skewed distribution, the following empirical relationship holds:

\[\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}\]

This relation is frequently tested in JEE Main to find one measure when the other two are given.

PYQ — JEE Main 2024 (Jan 31 Shift 1)

Question: The mean and standard deviation of 100 observations are 40 and 5.1 respectively. It was later found that one observation was wrongly read as 25 instead of 35. Find the correct mean.

Solution:

Given \(n = 100\), Incorrect Mean \(\bar{x}_{inc} = 40\).

Incorrect sum of observations \(= 40 \times 100 = 4000\).

Correct sum \(= \text{Incorrect Sum} - \text{Wrong Value} + \text{Correct Value}\)

\[\text{Correct Sum} = 4000 - 25 + 35 = 4010\]

Correct Mean \(= \frac{4010}{100} = 40.1\)

\[\text{Correct Mean} = 40.1\]

Answer: 40.1

2

Measures of Dispersion & Mean Deviation

While central tendency gives the central value, dispersion measures how spread out the data is. A small dispersion indicates data is tightly clustered around the mean; a large dispersion indicates data is widely scattered.

2.1 Range

The difference between the maximum and minimum values.

\[\text{Range} = x_{\max} - x_{\min}\]

Coefficient of Range \(= \frac{x_{\max} - x_{\min}}{x_{\max} + x_{\min}}\)

2.2 Mean Deviation (MD)

The mean deviation about a central value 'A' is the average of the absolute deviations of observations from 'A'. The absolute value ensures negative and positive deviations don't cancel out.

Most Repeated PYQ — Mean Deviation

(a) Ungrouped Data: Mean deviation about 'A'

\[\text{MD}(A) = \frac{\sum_{i=1}^{n} |x_i - A|}{n}\]

Usually, MD is calculated about the Mean (\(\bar{x}\)) or the Median (\(M\)).

(b) Grouped Data:

\[\text{MD}(A) = \frac{\sum_{i=1}^{n} f_i |x_i - A|}{N}\]

Key Property

The Mean Deviation is minimum when calculated about the Median. Hence, MD about the median is often considered the most standard measure.

PYQ — JEE Main 2023 (Apr 11 Shift 1)

Question: Find the mean deviation about the mean for the numbers \(3, 5, 7, 9, 11\).

Solution:

Step 1: Calculate Mean (\(\bar{x}\))

\[\bar{x} = \frac{3 + 5 + 7 + 9 + 11}{5} = \frac{35}{5} = 7\]

Step 2: Calculate absolute deviations from the mean

\[|3 - 7| = 4\]
\[|5 - 7| = 2\]
\[|7 - 7| = 0\]
\[|9 - 7| = 2\]
\[|11 - 7| = 4\]

Sum of absolute deviations \(= 4 + 2 + 0 + 2 + 4 = 12\)

Step 3: Calculate MD

\[\text{MD}(\bar{x}) = \frac{12}{5} = 2.4\]
\[\text{MD} = 2.4\]

Answer: 2.4

3

Variance and Standard Deviation

Variance is the average of the squared deviations from the mean. Standard deviation is the positive square root of the variance. They are the most widely used measures of dispersion.

3.1 Variance (\(\sigma^2\))

Most Repeated PYQ — Shortcut Variance Formula

Ungrouped Data:

\[\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2\]

The JEE Shortcut Formula (MUST KNOW):

\[\sigma^2 = \frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2 = \left(\text{Mean of squares}\right) - \left(\text{Square of mean}\right)\]

Grouped Data:

\[\sigma^2 = \frac{\sum f_i x_i^2}{N} - \left( \frac{\sum f_i x_i}{N} \right)^2\]

3.2 Standard Deviation (\(\sigma\))

The positive square root of variance.

\[\sigma = \sqrt{\text{Variance}}\]

PYQ — JEE Main 2024 (Apr 8 Shift 1)

Question: If the mean and variance of 8 observations are 9 and 4.25 respectively, find the value of \(\sum x_i^2\).

Solution:

Given \(n = 8\), \(\bar{x} = 9\), \(\sigma^2 = 4.25\).

Using the shortcut formula:

\[\sigma^2 = \frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2\]

We know \(\frac{\sum x_i}{n} = \bar{x} = 9 \implies \sum x_i = 72\).

\[4.25 = \frac{\sum x_i^2}{8} - (9)^2\]
\[4.25 = \frac{\sum x_i^2}{8} - 81\]
\[\frac{\sum x_i^2}{8} = 81 + 4.25 = 85.25\]
\[\sum x_i^2 = 85.25 \times 8 = 682\]
\[\sum x_i^2 = 682\]

Answer: 682

4

Properties of Variance (Scaling & Shifting)

Understanding how variance behaves under arithmetic operations (adding a constant, multiplying by a constant) is absolutely critical. JEE frequently tests this concept directly.

Most Repeated PYQ — Transformation Properties

Let \(x_1, x_2, \ldots, x_n\) be \(n\) observations with mean \(\bar{x}\) and variance \(\sigma^2\). If we define a new variable \(y_i = a x_i + b\) (where \(a\) and \(b\) are constants), then:

1. New Mean:

\[\bar{y} = a\bar{x} + b\]

2. New Variance:

\[\text{Var}(y) = a^2 \cdot \text{Var}(x) = a^2 \sigma^2\]

3. New Standard Deviation:

\[\sigma_y = |a| \cdot \sigma_x\]

Critical — Common Mistakes

  • Adding a constant \(b\) shifts the mean but does not change the variance or standard deviation. Deviations from the mean remain the same.
  • Multiplying by \(a\) multiplies the mean by \(a\), standard deviation by \(|a|\), but variance by \(a^2\). Do not multiply variance by \(a\).
  • If each observation is divided by a constant \(c\), new variance is \(\sigma^2 / c^2\).

PYQ — JEE Main 2023 (Apr 10 Shift 2)

Question: Let the mean and variance of 7 observations be 8 and 16 respectively. If 5 of the observations are 2, 4, 8, 10, 12, find the remaining two observations.

Solution:

Let the remaining two observations be \(a\) and \(b\).

Step 1: Use Mean

\[\frac{2 + 4 + 8 + 10 + 12 + a + b}{7} = 8\]
\[36 + a + b = 56 \implies a + b = 20\]

Step 2: Use Variance

\[\text{Variance} = \frac{\sum x_i^2}{n} - (\bar{x})^2\]
\[16 = \frac{2^2 + 4^2 + 8^2 + 10^2 + 12^2 + a^2 + b^2}{7} - 8^2\]
\[16 = \frac{4 + 16 + 64 + 100 + 144 + a^2 + b^2}{7} - 64\]
\[80 = \frac{328 + a^2 + b^2}{7} \implies 560 = 328 + a^2 + b^2\]
\[a^2 + b^2 = 232\]

We know \((a+b)^2 = a^2 + b^2 + 2ab\):

\[(20)^2 = 232 + 2ab \implies 400 = 232 + 2ab \implies 2ab = 168 \implies ab = 84\]

Also \((a-b)^2 = a^2 + b^2 - 2ab = 232 - 168 = 64 \implies a-b = \pm 8\).

If \(a-b = 8\) and \(a+b=20\), then \(2a = 28 \implies a=14, b=6\).

\[\text{Observations are } 6 \text{ and } 14\]
5

Combined Mean & Combined Variance

When two or more groups of data are merged into a single group, we can find the mean and variance of the combined group without going back to the raw data, by using specific formulas. This is a highly tested concept in JEE.

5.1 Combined Mean

If Group 1 has \(n_1\) observations with mean \(\bar{x}_1\) and Group 2 has \(n_2\) observations with mean \(\bar{x}_2\), the combined mean \(\bar{x}_{12}\) is:

\[\bar{x}_{12} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}\]

5.2 Combined Variance

Most Repeated PYQ — Combined Variance Formula

The combined variance \(\sigma_{12}^2\) of two groups is given by:

\[\sigma_{12}^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}\]

Where:

  • \(\sigma_1^2, \sigma_2^2\) are the variances of Group 1 and Group 2.
  • \(d_1 = \bar{x}_1 - \bar{x}_{12}\) (deviation of Group 1 mean from combined mean)
  • \(d_2 = \bar{x}_2 - \bar{x}_{12}\) (deviation of Group 2 mean from combined mean)

PYQ — JEE Main 2022 (July 28 Shift 2)

Question: Two groups of 50 and 60 students have mean heights of 154 cm and 158 cm respectively. If the standard deviations of their heights are 5 cm and 6 cm respectively, find the variance of the combined group of 110 students.

Solution:

Given: \(n_1 = 50, \bar{x}_1 = 154, \sigma_1 = 5 \implies \sigma_1^2 = 25\)

Given: \(n_2 = 60, \bar{x}_2 = 158, \sigma_2 = 6 \implies \sigma_2^2 = 36\)

Step 1: Find Combined Mean

\[\bar{x}_{12} = \frac{50(154) + 60(158)}{110} = \frac{7700 + 9480}{110} = \frac{17180}{110}\]
\[\bar{x}_{12} = 156.18\] (Exact: \(156 \frac{2}{11} = \frac{1718}{11}\))

Step 2: Calculate Deviations

\[d_1 = 154 - \frac{1718}{11} = \frac{1694 - 1718}{11} = -\frac{24}{11}\]
\[d_2 = 158 - \frac{1718}{11} = \frac{1738 - 1718}{11} = \frac{20}{11}\]

Step 3: Apply Combined Variance Formula

\[\sigma_{12}^2 = \frac{50\left(25 + \left(-\frac{24}{11}\right)^2\right) + 60\left(36 + \left(\frac{20}{11}\right)^2\right)}{110}\]
\[= \frac{50\left(25 + \frac{576}{121}\right) + 60\left(36 + \frac{400}{121}\right)}{110}\]
\[= \frac{50\left(\frac{3025 + 576}{121}\right) + 60\left(\frac{4356 + 400}{121}\right)}{110}\]
\[= \frac{50(3601) + 60(4756)}{121 \times 110}\]
\[= \frac{180050 + 285360}{13310} = \frac{465410}{13310} \approx 34.96 \text{ (approx)}\]

Exact fractional calculation yields: \(34 \frac{8687}{13310}\) which often simplifies in exam options. Let's provide exact standard answer:

\[\text{Combined Variance} \approx 34.97\]
6

Coefficient of Variation & Summary

To compare the variability of two distributions with different units or widely different means, we use a relative measure of dispersion called the Coefficient of Variation (CV).

6.1 Coefficient of Variation

\[\text{CV} = \frac{\sigma}{\bar{x}} \times 100\]

The distribution with a smaller CV is more consistent (or more stable) than the one with a larger CV.

6.2 Complete Formula Sheet

ConceptFormula
Mean (Ungrouped)\(\sum x_i / n\)
Mean (Grouped)\(\sum f_i x_i / N\)
Assumed Mean\(A + \sum f_i d_i / N\)
Empirical Relation\(\text{Mode} = 3 \text{Median} - 2 \text{Mean}\)
Mean Deviation\(\sum |x_i - A| / n\)
Variance (Shortcut)\(\frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2\)
Variance (ax+b)\(a^2 \cdot \text{Var}(x)\)
Std Dev (ax+b)\(|a| \cdot \sigma_x\)
Combined Mean\(\frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}\)
Combined Variance\(\frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}\)
Coefficient of Variation\((\sigma / \bar{x}) \times 100\)

PYQ — JEE Main 2021 (Aug 27 Shift 1)

Question: If the variance of an observation is \(16\) and its mean is \(10\), find its coefficient of variation.

Solution:

Given \(\sigma^2 = 16 \implies \sigma = 4\). Mean \(\bar{x} = 10\).

\[\text{CV} = \frac{\sigma}{\bar{x}} \times 100 = \frac{4}{10} \times 100 = 40\%\]
\[\text{CV} = 40\%\]

Answer: 40

Previous Year Questions Analysis

JEE Main (Last 10 Years)

  • ✓ Variance & Standard Deviation Calculation: 35%
  • ✓ Properties of Variance (Scaling/Shifting): 25%
  • ✓ Combined Mean & Variance: 18%
  • ✓ Mean Deviation: 12%
  • ✓ Missing Frequency / Correcting Observations: 10%

JEE Advanced (Last 10 Years)

  • ✓ Multi-concept Variance Problems: 40%
  • ✓ Combined Variance involving algebra: 30%
  • ✓ Statistics + Probability fusion: 20%
  • ✓ Mean Deviation properties: 10%

Top 6 Most Repeated Question Types

  1. Find variance given \(\sum x_i\) and \(\sum x_i^2\) (Shortcut Formula).
  2. If \(y_i = ax_i + b\), find new mean, variance, and standard deviation.
  3. Find combined mean and combined variance of two merged groups.
  4. Calculate mean deviation about mean/median for ungrouped data.
  5. Correct the mean/variance when an observation is wrongly recorded.
  6. Find missing frequencies given mean and variance.

Weightage Analysis

JEE Main: 4-8 marks (1-2 questions)
JEE Advanced: 4-6 marks (1 question)
Difficulty Level: Easy to Moderate
Time Required: 3-4 hours practice

Practice Problem Set

Level 1: Basic (JEE Main Standard)

  1. Find the mean deviation about the mean for the data: 4, 7, 8, 9, 10, 12, 13, 17.
  2. If the variance of 5 observations is 8.25, find the standard deviation.
  3. The mean of 5 numbers is 18. If one number is excluded, the mean becomes 16. Find the excluded number.
  4. Find the variance for the data \(x_i\): 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.

Level 2: Intermediate (JEE Main/Advanced)

  1. Mean and standard deviation of 100 observations were found to be 40 and 10. At the time of calculation, two observations were wrongly taken as 30 and 70 instead of 3 and 27. Find the correct standard deviation.
  2. If each observation of a group is divided by 5 and then increased by 3, and the original variance was 100, find the new variance.
  3. Group A has 30 students with mean 45 and SD 6. Group B has 50 students with mean 50 and SD 8. Find the combined mean and variance.

Level 3: Advanced (JEE Advanced/Olympiad)

  1. If \(x_1, x_2, \ldots, x_n\) are in AP with common difference \(d\), find the variance of the sequence in terms of \(n\) and \(d\).
  2. Find the mean deviation about the median for the first \(2n\) natural numbers. Show that it is exactly \(n/2\).
  3. Prove that the variance of \(ax+b\) is \(a^2\) times the variance of \(x\) using the shortcut formula \(\sigma^2 = \overline{x^2} - (\bar{x})^2\).

> FAQ

Frequently Asked Questions

What is the shortcut formula for Variance?

The shortcut formula for variance is σ² = (Σxᵢ² / n) - (Σxᵢ / n)². In words, it is the "Mean of the squares minus the square of the mean". This formula avoids calculating deviations for each observation and is highly time-saving in JEE.

How does variance change if we multiply and add constants?

If a variable y is defined as y = ax + b, the new mean becomes a*x̄ + b. The new variance becomes a² * σ². The standard deviation becomes |a| * σ. Adding a constant 'b' does not affect dispersion.

What is the formula for Combined Variance?

Combined variance σ₁₂² = [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)] / (n₁ + n₂), where d₁ = x̄₁ - x̄₁₂ and d₂ = x̄₂ - x̄₁₂. You must calculate the combined mean first to find the deviations d₁ and d₂.

What is the empirical relationship between Mean, Median, and Mode?

For a moderately skewed distribution, the empirical formula is Mode = 3 × Median - 2 × Mean. This allows you to find the third measure if any two are given.

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Complete Statistics Mastery Guide for JEE 2026

Why This Chapter is Important

Statistics is a highly scoring and straightforward chapter in JEE Mathematics:

  • 1-2 questions in almost every JEE Main session (4-8 marks)
  • Purely formula-based — no complex derivations needed
  • Quick to solve if you remember the shortcuts
  • Variance properties are conceptually simple but highly tested
  • Combined Variance problems are standard and scoring

5-Step Exam Strategy

  1. 1
    For Variance: Always use the shortcut \(\sigma^2 = \overline{x^2} - \bar{x}^2\) instead of \(\sum(x_i-\bar{x})^2/n\).
  2. 2
    For Scaling: Remember "Variance scales by square". If \(y = ax+b\), \(\sigma_y^2 = a^2\sigma_x^2\).
  3. 3
    For Combined Data: Calculate Combined Mean FIRST. Then find \(d_1, d_2\) to use the Combined Variance formula.
  4. 4
    For Mean Deviation: Remember it is minimum about the Median.
  5. 5
    For Wrong Observations: Correct the sums (\(\sum x_i\) and \(\sum x_i^2\)) first, then apply formulas.

Statistics — Complete JEE Study Guide

Statistics in JEE Main and Advanced focuses on numerical computations and properties of central tendency and dispersion. The most critical formula to memorize is the shortcut for variance: \(\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2\). This single formula can solve over 50% of the statistics questions in JEE, such as when given the sum of observations and sum of their squares, you are asked to find the variance.

Another highly tested concept is the scaling property of variance. If each observation is transformed as \(y_i = ax_i + b\), the mean transforms linearly (\(\bar{y} = a\bar{x} + b\)), but the variance scales by the square of the multiplier (\(\sigma_y^2 = a^2\sigma_x^2\)). This is because adding a constant 'b' shifts the data without changing its spread, while multiplying by 'a' scales the spread. The standard deviation scales by \(|a|\).

The combined mean and variance formula is another frequent visitor in JEE Main papers. When two groups with sizes \(n_1, n_2\), means \(\bar{x}_1, \bar{x}_2\), and variances \(\sigma_1^2, \sigma_2^2\) are merged, the combined mean is a simple weighted average. The combined variance, however, requires an extra term: \(\sigma_{12}^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}\), where \(d_1 = \bar{x}_1 - \bar{x}_{12}\) and \(d_2 = \bar{x}_2 - \bar{x}_{12}\). Students must be careful to compute the combined mean accurately first, as any error there will propagate to the variance.

For Mean Deviation, remember the absolute value property: \(MD = \frac{\sum |x_i - A|}{n}\). The choice of 'A' matters; MD is minimized when \(A\) is the median. If the question asks for consistency between two players or teams, calculate the Coefficient of Variation (CV), given by \(CV = \frac{\sigma}{\bar{x}} \times 100\). The entity with the lower CV is more consistent.