Where are quartiles used?

Quartiles are used to understand the spread and distribution of data. They divide the data into four equal parts and are helpful in fields like economics and finance to analyze income distribution, stock performance, etc.

CBSE 2026-27 | Statistics for Economics

NCERT Solutions for Class 11 Statistics Chapter 5: Measures of Central Tendency

Q: What are the 3 main measures of central tendency?

The three main measures are the Mean (the arithmetic average), the Median (the middle value), and the Mode (the most frequent value). These are crucial for CBSE Class 11 Statistics Chapter 5.

Q: What is the formula for Mode in a continuous series?

The formula is Z = l₁ + ((f₁ - f₀) / (2f₁ - f₀ - f₂)) * h, where l₁ is the lower limit of the modal class, f₁ is its frequency, f₀ and f₂ are frequencies of the preceding and succeeding classes, and h is the class interval.

Q: When should I use Mean vs. Median?

Use Mean for symmetrical data without extreme values. Use Median when the data has extreme values (outliers) or for open-ended class distributions, as it gives a more representative value.

📚 Class 11 CBSE 📈 Statistics for Economics ⚡ 6–8 Marks 🔵 Easy to Moderate

Struggling with Mean, Median, and Mode? This ultimate guide simplifies CBSE Class 11 Statistics Chapter 5, Measures of Central Tendency. Master these core concepts for your 2026-27 board exams and competitive tests. Let's make statistics easy and scoring, helping you ace your exams with confidence!

📋Chapter at a Glance

📚

Chapter 5: Measures of Central Tendency – Quick Reference

Chapter Name	Measures of Central Tendency
Subject	Statistics (Economics)
Board / Class	CBSE Class 11
Target Year	2026-27
Important Topics	Arithmetic Mean, Median, Mode, Quartiles, Weighted Mean.
Difficulty Level	Easy to Moderate
Exam Weightage	6–8 Marks (Can be part of a higher-mark question)

📊Key Measures – Formulas to Memorise

➕;

Mean (Average)

$ \bar{X} = \frac{\sum X}{N} $

〰

Median (Middle)

$ M = (\frac{N+1}{2})^{th} $ item

📍

Mode (Frequent)

Most Occurring

🔗

Empirical Relation

Mode ≈ 3M - 2$ \bar{X} $

📌

Quartile 1 (Q₁)

$ (\frac{N+1}{4})^{th} $ item

📍

Quartile 3 (Q₃)

$ 3(\frac{N+1}{4})^{th} $ item

🎯Learning Objectives

Understand the meaning and purpose of a measure of central tendency.

Calculate Arithmetic Mean for individual, discrete, and continuous series.

Calculate Median and other positional values like Quartiles, Deciles, and Percentiles.

Determine the Mode for various types of statistical series.

Compare the different measures and understand their appropriate usage.

Calculate Weighted Arithmetic Mean.

💡Key Concepts, Definitions, & Formulas

Mastering these formulas is the first step to scoring full marks!

Central Tendency

A single value that describes a set of data by identifying the central position within that set. Also called a statistical average.

Arithmetic Mean (Mean)

The sum of all values divided by the number of values. Most common average.

$$ \bar{X} = \frac{\sum fX}{\sum f} $$

Median (M)

The **middle-most value** in a dataset when it's arranged in ascending or descending order. It is a positional average.

Mode (Z)

The value that appears **most frequently** in a dataset.

Highest Frequency

Quartiles (Q₁, Q₃)

Values that divide the data into four equal parts. Q₁ is the lower quartile, and Q₃ is the upper quartile.

Weighted Mean

An average where each value is assigned a weight according to its importance.

$$ \bar{X}_w = \frac{\sum WX}{\sum W} $$

🔗

Important Formulas for Continuous Series

1. Mean (Step-Deviation): $ \bar{X} = A + \left( \frac{\sum fd'}{\sum f} \right) \times h $
2. Median: $ M = l_1 + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h $
3. Mode: $ Z = l_1 + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h $

📝Full NCERT Solutions – Solved Exercise Questions

NCERT Q1: Calculate the Arithmetic Mean for the marks of 10 students: 45, 32, 37, 46, 39, 36, 41, 48, 36, 40 ▼

✅ Model Answer

This is an individual series. The formula for the Arithmetic Mean ($ \bar{X} $) is:

$$ \bar{X} = \frac{\sum X}{N} $$

Step 1: Sum all the observations (ΣX).
ΣX = 45 + 32 + 37 + 46 + 39 + 36 + 41 + 48 + 36 + 40 = 400
Step 2: Count the number of observations (N).
N = 10
Step 3: Apply the formula.
$ \bar{X} = \frac{400}{10} = 40 $

Therefore, the arithmetic mean of the marks is 40.

NCERT Q2: Find the Median from the data: 15, 21, 17, 25, 12, 19, 23 ▼

✅ Model Answer

This is an individual series.

Step 1: Arrange the data in ascending order.
12, 15, 17, 19, 21, 23, 25
Step 2: Find the number of observations (N).
N = 7 (an odd number)
Step 3: Use the formula for the Median (M).
$$ M = \text{Size of } \left( \frac{N+1}{2} \right)^{th} \text{ item} $$ $$ M = \text{Size of } \left( \frac{7+1}{2} \right)^{th} \text{ item} = 4^{th} \text{ item} $$
Step 4: Identify the 4th item in the arranged data.
The 4th item is 19.

Therefore, the Median of the data is 19.

NCERT Q3: Calculate the Mean for the following frequency distribution using the Direct Method. ▼

✅ Model Answer

This is a discrete series. The formula for Mean is: $$ \bar{X} = \frac{\sum fX}{\sum f} $$

Step 1: Create a calculation table.

Marks (X) No. of Students (f) fX
10 8 80
20 12 240
30 20 600
40 6 240
50 4 200
Total Σf = 50 ΣfX = 1360
Step 2: Find the sum of frequencies (Σf) and the sum of fX (ΣfX).
From the table, Σf = 50 and ΣfX = 1360.
Step 3: Apply the formula.
$$ \bar{X} = \frac{1360}{50} = 27.2 $$

Marks (X)	No. of Students (f)	fX
10	8	80
20	12	240
30	20	600
40	6	240
50	4	200
Total	Σf = 50	ΣfX = 1360

Therefore, the mean marks are 27.2.

NCERT Q4: Calculate the Median for the following continuous series data. ▼

✅ Model Answer

Step 1: Find the Median Class.
First, create a cumulative frequency (cf) column.

Class Interval Frequency (f) Cumulative Frequency (cf)
0-10 5 5
10-20 8 13
20-30 15 28
30-40 16 44
40-50 6 50
Total N = 50

Median Item = Size of $ (\frac{N}{2})^{th} $ item = $ \frac{50}{2} = 25^{th} $ item.
The 25th item lies in the cumulative frequency 28, so the Median Class is 20-30.
Step 2: Identify values for the Median formula.
$$ M = l_1 + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h $$
- $ l_1 $ (lower limit of median class) = 20
- $ \frac{N}{2} $ = 25
- cf (cumulative frequency of the preceding class) = 13
- f (frequency of median class) = 15
- h (class interval) = 10
Step 3: Substitute the values.
$$ M = 20 + \left( \frac{25 - 13}{15} \right) \times 10 $$ $$ M = 20 + \left( \frac{12}{15} \right) \times 10 $$ $$ M = 20 + 0.8 \times 10 = 20 + 8 = 28 $$

Class Interval	Frequency (f)	Cumulative Frequency (cf)
0-10	5	5
10-20	8	13
20-30	15	28
30-40	16	44
40-50	6	50
Total	N = 50

Therefore, the Median is 28.

NCERT Q5: Calculate the Mode from the data: Class [0-10, 10-20, ...], Freq [2, 5, 7, 5, 2] ▼

✅ Model Answer

Step 1: Identify the Modal Class.
The modal class is the class with the highest frequency. Here, the highest frequency is 7, which corresponds to the class interval 20-30.
Step 2: Identify values for the Mode formula.
$$ Z = l_1 + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h $$
- $ l_1 $ (lower limit of modal class) = 20
- $ f_1 $ (frequency of modal class) = 7
- $ f_0 $ (frequency of pre-modal class) = 5
- $ f_2 $ (frequency of post-modal class) = 5
- h (class interval) = 10
Step 3: Substitute the values.
$$ Z = 20 + \left( \frac{7 - 5}{2 \times 7 - 5 - 5} \right) \times 10 $$ $$ Z = 20 + \left( \frac{2}{14 - 10} \right) \times 10 $$ $$ Z = 20 + \left( \frac{2}{4} \right) \times 10 = 20 + 5 = 25 $$

Therefore, the Mode is 25.

❓Extra MCQs – Practice & Self-Test

💡

How to Use

Click an option to check if it's correct or wrong. The explanation will appear instantly.

Difficulty: Easy

Q1. The measure of central tendency that is most affected by extreme values is:

✅ Correct: (c) Arithmetic Mean. It uses all values in its calculation, so one very large or small value can pull the average up or down significantly.

Difficulty: Medium

Q2. In a moderately skewed distribution, if the Mean is 30 and the Mode is 24, the Median is:

✅ Correct: (b) 28. Use the empirical formula: Mode ≈ 3 Median - 2 Mean. So, 24 ≈ 3 Median - 2(30) => 24 ≈ 3 Median - 60 => 84 ≈ 3 Median => Median ≈ 28.

Difficulty: Easy

Q3. The value that divides the series into four equal parts is known as:

✅ Correct: (d) Quartile. 'Quart' means four. Quartiles Q₁, Q₂, and Q₃ divide the data into four equal segments.

Difficulty: Medium

Q4. The sum of deviations of items from the Arithmetic Mean is always:

✅ Correct: (b) 0. This is a fundamental property of the arithmetic mean. The sum of positive deviations exactly cancels out the sum of negative deviations.

Difficulty: Easy

Q5. Which of the following is a positional average?

✅ Correct: (c) Median. Its value depends on its position (the middle) in the dataset, not the magnitude of all the other values.

🚀Extra Board Exam Questions (2026-27)

📌 Short Answer Questions

Medium Q6: If the arithmetic mean of 5, 8, 10, 12, x, and 15 is 10, find the value of x. ▼

✅ Model Answer

We know Mean = (Sum of observations) / Number of observations.

1. **Set up the equation:**
$ 10 = \frac{5 + 8 + 10 + 12 + x + 15}{6} $

2. **Simplify the sum:**
$ 10 = \frac{50 + x}{6} $

3. **Solve for x:**
$ 10 \times 6 = 50 + x $
$ 60 = 50 + x $
$ x = 60 - 50 = 10 $

The value of x is 10.

Easy Q7: List two merits and two demerits of the Arithmetic Mean. ▼

✅ Model Answer

Merits:
1. It is easy to calculate and understand.
2. It is based on all observations in the dataset.
Demerits:
1. It is highly affected by extreme values (outliers).
2. It cannot be calculated for qualitative data or open-ended classes.

Medium Q8: When is the Median a more suitable average than the Mean? ▼

✅ Model Answer

The Median is more suitable than the Mean when:

The data has **extreme values (outliers)**. The median is not affected by them, while the mean is heavily skewed. For example, in salary data with a few billionaire CEOs.
The distribution has **open-ended classes**, as the exact values of the extremes are not needed to find the middle position.
The data is **qualitative and can be ranked** (ordinal data), such as "good", "better", "best".

Medium Q9: Find the first quartile (Q₁) for the data: 22, 26, 14, 30, 18, 11, 35, 41, 12, 32. ▼

✅ Model Answer

Step 1: Arrange data in ascending order. (N=10)
11, 12, 14, 18, 22, 26, 30, 32, 35, 41
Step 2: Find the position of Q₁.
$$ Q_1 = \text{Size of } \left( \frac{N+1}{4} \right)^{th} \text{ item} $$ $$ Q_1 = \text{Size of } \left( \frac{10+1}{4} \right)^{th} \text{ item} = 2.75^{th} \text{ item} $$
Step 3: Calculate the value.
This means Q₁ is the 2nd item plus 0.75 of the difference between the 3rd and 2nd items.
Q₁ = 2nd item + 0.75 * (3rd item - 2nd item)
Q₁ = 12 + 0.75 * (14 - 12) = 12 + 0.75 * 2 = 12 + 1.5 = 13.5

The first quartile (Q₁) is 13.5.

Easy Q10: What is a 'modal class' and how is it determined? ▼

✅ Model Answer

In a continuous frequency distribution, the **'modal class'** is the class interval that has the highest frequency.

It is determined by simply observing the frequency column and identifying the class corresponding to the largest frequency value.

📌 Long Answer & Case-Based Questions

Hard Q11: Calculate Mean, Median, and Mode for the data and comment on the result. ▼

✅ Model Answer

1. Mean Calculation: Approx. 29.5 (Using step-deviation is recommended).

2. Median Calculation:
N = 50, N/2 = 25. Median class is 25-35 (cf goes from 20 to 45).
l₁=25, cf=20, f=25, h=10.
M = 25 + ((25-20)/25) * 10 = 25 + (5/25)*10 = 25 + 2 = 29.

3. Mode Calculation:
Modal class is 25-35 (highest frequency 25).
l₁=25, f₁=25, f₀=15, f₂=10, h=10.
Z = 25 + ((25-15)/(2*25-15-10)) * 10 = 25 + (10/25)*10 = 25 + 4 = 29.

(Note: The provided answer in markdown had a slight error in mode calculation. I've corrected it. Let's re-calculate with the new value.)
Z = 25 + (10 / (50 - 25)) * 10 = 25 + (10/25)*10 = 25+4 = 29.
(Ah, the markdown's answer for Mode was 28.33. Let's check my calculation. f1=25, f0=15, f2=10. 2f1-f0-f2 = 50-15-10 = 25. (f1-f0) = 10. Z=25 + (10/25)*10 = 25+4 = 29. The markdown's answer was incorrect. My calculated answer is 29). Let's go with 29.

Final Results: Mean ≈ 29.5, Median = 29, Mode = 29.

Comment: Since the Mean, Median, and Mode are very close to each other (Mean ≈ Median ≈ Mode), the distribution is nearly symmetrical and not significantly skewed.

Case-Based Q12: A company manager wants to understand the average salary. Some executives have very high salaries. Which measure is appropriate and why? Calculate it. ▼

✅ Model Answer

a) Appropriate Measure and Reasoning:
The **Median** would be more appropriate. The presence of very high salaries for top executives act as extreme values (outliers). These outliers would disproportionately inflate the Arithmetic Mean, giving a misleadingly high average salary. The Median, being a positional average, is not affected by these extreme values and will provide a better representation of the 'typical' employee salary.

b) Median Calculation:

Step 1: Create a cumulative frequency table.
Salary (₹1000s) Employees (f) cf
20-30 20 20
30-40 25 45
40-50 30 75
50-60 15 90
60-70 10 100
Total N=100
Step 2: Find the median class.
N=100, N/2 = 50. The 50th item falls in the class 40-50.
Step 3: Apply the formula.
l₁=40, cf=45, f=30, h=10.
Median = 40 + ((50-45)/30) * 10 = 40 + (5/30)*10 = 40 + 1.67 = 41.67.

Salary (₹1000s)	Employees (f)	cf
20-30	20	20
30-40	25	45
40-50	30	75
50-60	15	90
60-70	10	100
Total	N=100

The median salary is ₹41.67 thousand or ₹41,670.

Medium Q14: Calculate the weighted arithmetic mean from the data for Rice, Wheat, Sugar, and Oil. ▼

✅ Model Answer

The formula for Weighted Mean is: $$ \bar{X}_w = \frac{\sum WX}{\sum W} $$

Step 1: Create a calculation table.
Item Price (X) Weight (W) WX
Rice 100 30 3000
Wheat 80 50 4000
Sugar 120 15 1800
Oil 200 5 1000
Total ΣW = 100 ΣWX = 9800
Step 2: Find ΣW and ΣWX.
ΣW = 100, ΣWX = 9800.
Step 3: Apply the formula.
Weighted Mean = 9800 / 100 = 98.

Item	Price (X)	Weight (W)	WX
Rice	100	30	3000
Wheat	80	50	4000
Sugar	120	15	1800
Oil	200	5	1000
Total		ΣW = 100	ΣWX = 9800

The weighted mean price is ₹98.

⚠Common Mistakes to Avoid

🧮

Calculation Errors

Simple mistakes in addition, multiplication, or division are very common. Double-check your calculations, especially in tables.

🔄

Formula Mix-up

Confusing the formula for Median with Mode, or using the wrong mean formula for a given series type. Create a formula sheet and practice.

🔎

Forgetting 'cf'

In the Median formula, students often use the cf of the median class itself. Remember to use the cumulative frequency of the class *preceding* it.

📐

Not Arranging Data

Forgetting to arrange data in ascending/descending order when calculating the median for an individual or discrete series. It's the first and most crucial step.

📚Exam Preparation Tips for 2026-27

📝

Master the Formulas

Create a formula sheet for Mean (all 3 methods), Median, Mode, and Quartiles. Revise it daily.

📋

Practice One of Each

Every day, solve one problem each for Mean, Median, and Mode from a continuous series. This will build speed and accuracy.

💡

Understand, Don't Memorize

Know *why* you are using a particular measure. Understand the difference between Mean and Median's suitability in different scenarios.

🖊

Presentation Matters

Write formulas clearly, draw proper tables, and highlight the final answer. This helps the examiner and reduces your chances of making silly mistakes.

🅾Frequently Asked Questions (FAQs)

What are the 3 main measures of central tendency?▼

The three main measures are the Mean (the arithmetic average), the Median (the middle value), and the Mode (the most frequent value). These are crucial for CBSE Class 11 Statistics Chapter 5.

What is the formula for Mode in a continuous series?▼

The formula is $ Z = l_1 + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h $, where `l₁` is the lower limit of the modal class, `f₁` is its frequency, `f₀` and `f₂` are frequencies of the preceding and succeeding classes, and `h` is the class interval.

When should I use Mean vs. Median?▼

Use Mean for symmetrical data without extreme values. Use Median when the data has extreme values (outliers) or for open-ended class distributions, as it gives a more representative value.

Is Chapter 5 Measures of Central Tendency difficult?▼

No, it is a foundational and high-scoring chapter. With consistent practice of the formulas and different types of problems, you can easily master it.

Master Measures of Central Tendency 📈

This chapter is the backbone of statistical analysis. Remember, the key to success is not just understanding but consistent practice. Solve all the NCERT questions, practice the important questions provided, and you'll find this chapter to be one of the most scoring in your exam.

⚡ Practice Chapter 5 MCQs Free

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