CBSE Class 10 Mathematics Chapter 14 Statistics NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 10 PDF

This chapter provides comprehensive NCERT Solutions for Class 10 Mathematics, focusing on Statistics (Chapter 14). Students will learn to analyze data by calculating the mean, median, and mode for grouped data. The solutions cover various methods for finding the mean, including the direct method, assumed mean method, and step-deviation method, explaining when each is most appropriate. It also details how to find the median and mode from frequency distribution tables. These solutions are designed to help students understand the concepts thoroughly and build a strong foundation for statistical analysis, crucial for exam preparation and data interpretation.

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Quick info

Board	CBSE
Class	Class 10
Subject	Mathematics
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 14

Chapter summary

Chapter 14 on Statistics for Class 10 Maths NCERT Solutions covers the fundamental concepts of data analysis. It focuses on calculating measures of central tendency for grouped data: mean, median, and mode. The solutions provide step-by-step guidance on using different methods like the direct method, assumed mean method, and step-deviation method for finding the mean. It also explains how to determine the median and mode from frequency tables, equipping students with essential skills for interpreting statistical information.

Learning outcomes

Understand the concepts of mean, median, and mode for grouped data.
Apply the direct method to calculate the mean of a frequency distribution.
Utilize the assumed mean method for calculating the mean efficiently.
Implement the step-deviation method for finding the mean with large class marks.
Determine the median of a grouped frequency distribution.
Calculate the mode for a grouped frequency distribution.
Choose the most appropriate method for calculating the mean based on the data.

Topics covered

Paper topics

Statistics
Mean of grouped data
Direct Method for Mean
Assumed Mean Method for Mean
Step-Deviation Method for Mean
Class Mark
Class Interval
Frequency Distribution
Median of grouped data
Mode of grouped data

Important topics

Mean of grouped data (all methods)
Median of grouped data
Mode of grouped data
Choosing the appropriate method for calculating the mean
Understanding class intervals and class marks

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Questions and Solutions

Question 1

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants | Number of houses

0 - 2 | 2

2 - 4 | 3

4 - 6 | 1

6 - 8 | 5

8 - 10 | 6

10 - 12 | 2

12 - 14 | 3

Which method did you use for finding the mean, and why?

Solution:

To find the mean number of plants per house, we can use the direct method. First, we need to calculate the class mark (x_i) for each class interval using the formula:

x_i = \frac{\text{Upper limit} + \text{Lower limit}}{2}

Next, we calculate the product of each class mark (x_i) and its corresponding frequency (f_i), which represents the number of houses. Finally, we sum these products (Σf_i x_i) and the frequencies (Σf_i) to find the mean using the formula:

\bar{x} = \frac{\sum f_i x_i}{\sum f_i}

Let's construct the table:

Number of plants (Class Interval)	Number of houses (f_i)	Class Mark (x_i)	f_i x_i
0 - 2	2	$(0+2)/2 = 1$	$2 \times 1 = 2$
2 - 4	3	$(2+4)/2 = 3$	$3 \times 3 = 9$
4 - 6	1	$(4+6)/2 = 5$	$1 \times 5 = 5$
6 - 8	5	$(6+8)/2 = 7$	$5 \times 7 = 35$
8 - 10	6	$(8+10)/2 = 9$	$6 \times 9 = 54$
10 - 12	2	$(10+12)/2 = 11$	$2 \times 11 = 22$
12 - 14	3	$(12+14)/2 = 13$	$3 \times 13 = 39$
Total	$\sum f_i = 2 + 3 + 1 + 5 + 6 + 2 + 3 = 22$		$\sum f_i x_i = 2 + 9 + 5 + 35 + 54 + 22 + 39 = 166$

From the table, we have:

\sum f_i = 22

\sum f_i x_i = 166

Now, we calculate the mean:

\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{166}{22}

\bar{x} = \frac{83}{11} \approx 7.55

The mean number of plants per house is approximately 7.55.

Method Used: The Direct Method was used. This method is suitable here because the class marks (x_i) and the frequencies (f_i) are relatively small, making the calculation of f_i x_i straightforward and less prone to arithmetic errors.

Question 2

Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.

Daily wages (in Rs)

100 - 120 | 120 - 140 | 140 - 160 | 160 - 180 | 180 - 200

Number of workers | 12 | 14 | 10 | 8 | 6

Solution:

To find the mean daily wages, we first need to calculate the class mark (x_i) for each class interval. The formula for the class mark is:

x_i = \frac{\text{Upper limit} + \text{Lower limit}}{2}

The class size (h) for this data is uniform, $h = 120 - 100 = 20$ .

Since the class marks might be larger numbers, the Assumed Mean Method or the Step-Deviation Method would be appropriate. Let's use the Assumed Mean Method. We assume a mean (a) from the class marks, preferably the one with the highest frequency. Let's choose $a = 150$ (the class mark for 140-160).

We then calculate the deviation $d_i = x_i - a$ for each class.

The mean is then calculated using the formula:

\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}

Let's construct the table:

Daily Wages (Class Interval)	Number of Workers (f_i)	Class Mark (x_i)	Deviation (d_i = x_i - a) (a=150)	f_i d_i
100 - 120	12	$(100+120)/2 = 110$	$110 - 150 = -40$	$12 \times (-40) = -480$
120 - 140	14	$(120+140)/2 = 130$	$130 - 150 = -20$	$14 \times (-20) = -280$
140 - 160	10	$(140+160)/2 = 150$	$150 - 150 = 0$	$10 \times 0 = 0$
160 - 180	8	$(160+180)/2 = 170$	$170 - 150 = 20$	$8 \times 20 = 160$
180 - 200	6	$(180+200)/2 = 190$	$190 - 150 = 40$	$6 \times 40 = 240$
Total	$\sum f_i = 12 + 14 + 10 + 8 + 6 = 50$			$\sum f_i d_i = -480 - 280 + 0 + 160 + 240 = -360$

From the table, we have:

\sum f_i = 50

\sum f_i d_i = -360

Now, we calculate the mean using the Assumed Mean Method formula:

\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 150 + \frac{-360}{50}

\bar{x} = 150 - \frac{36}{5} = 150 - 7.2

\bar{x} = 142.8

Therefore, the mean daily wages of the workers of the factory is Rs 142.8.

Appropriate Method: The Assumed Mean Method was used because the class marks and deviations were manageable, simplifying the calculation compared to the direct method.

Common mistakes

Incorrectly calculating the class mark (midpoint) for each class interval.
Errors in summing the frequencies (Σf_i) or the products (Σf_i x_i, Σf_i u_i).
Using the wrong formula for mean, median, or mode.
Misidentifying the median class or modal class.
Arithmetic errors in calculations, especially with negative numbers or fractions.

Revision tips

Practice calculating the mean using all three methods (direct, assumed mean, step-deviation) to understand their applications.
Pay close attention to identifying the correct median class and applying the median formula accurately.
Ensure you correctly determine the modal class and use the mode formula precisely.
Review the formulas for class mark, mean, median, and mode before attempting problems.
Work through the examples and exercises systematically to build confidence.

Practice MCQs

Q1. What is the formula for calculating the class mark (x_i) of a class interval?

Q2. Which method is generally preferred for calculating the mean when class marks (x_i) and frequencies (f_i) are small?

Q3. In the assumed mean method for calculating the mean, what does 'd_i' represent?

Q4. What is the class size (h) in the context of grouped data?

Q5. The step-deviation method is particularly useful when:

Frequently asked questions

What is Chapter 14 of Class 10 Maths NCERT about?

Chapter 14 of Class 10 Maths NCERT Solutions focuses on Statistics, covering methods to calculate measures of central tendency like mean, median, and mode for grouped data.

What are the different methods to find the mean of grouped data?

The three main methods to find the mean of grouped data are the Direct Method, the Assumed Mean Method, and the Step-Deviation Method.

When should I use the Assumed Mean Method or Step-Deviation Method instead of the Direct Method?

These methods are preferred when the class marks (x_i) and frequencies (f_i) are large, as they simplify calculations. The Step-Deviation Method is especially useful when class sizes are uniform and deviations share a common factor.

How do these NCERT Solutions help in exam preparation?

These solutions provide clear, step-by-step explanations for each problem, helping students understand the concepts and methods thoroughly, which is essential for performing well in exams.

Are the formulas for median and mode covered in these solutions?

Yes, the solutions for Chapter 14 typically include explanations and applications of the formulas for calculating the median and mode of grouped frequency distributions.

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