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NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 PDF

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 has been provided here. Students can refer to these solutions before their examination for better understanding.
authorImageNeha Tanna21 Jan, 2025
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NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2: NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 focus on advanced concepts of Statistics, including calculation and interpretation of measures like the mean, median, and mode of grouped data.

The exercise helps students understand and solve problems related to cumulative frequency distribution and graphical representation through ogives (cumulative frequency curves). Detailed solutions provide step-by-step methods, making it easier for students to grasp concepts and improve problem-solving skills. These solutions are aligned with the latest NCERT curriculum and are essential for building a strong foundation in Statistics, a crucial topic for board exams and practical applications in real-life scenarios.

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 Overview

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 provide a comprehensive understanding of Statistics, focusing on key concepts like the mean, median, and mode of grouped data, cumulative frequency, and ogives. These solutions are crucial for mastering data interpretation, a vital skill for academics and real-life applications. By offering step-by-step explanations, they simplify complex calculations, enhancing problem-solving abilities. Statistics has broad applications in various fields like economics, social sciences, and research, making this exercise essential. Preparing with these solutions ensures a solid foundation for board exams and develops analytical skills important for higher studies and future careers.

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 PDF Download

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 focus on essential topics in Statistics , such as calculating the mean, median, and mode for grouped data, along with cumulative frequency and graphical representation. These solutions are designed to simplify complex concepts with step-by-step explanations, aiding students in mastering the topic effectively. They are aligned with the latest NCERT guidelines to help students excel in exams. Below, we have provided a PDF for easy reference and practice.

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 Statistics

Below is the NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 Statistics -

1. The following table shows the ages of the patients admitted to a hospital during a year:

Age (in years) 5-15 15-25 25-35 35-45 45-55 55-65
Number of patients 6 11 21 23 14 5

Find the mode and the mean of the data given above. Compare and interpret the two

measures of central tendency.

Solution: To find out the modal class, let us the consider the class interval with high frequency. Here, the greatest frequency = 23, so the modal class = 35 – 45, Lower limit of modal class = l = 35, class width (h) = 10, f m = 23, f 1 = 21 and f 2 = 14 The formula to find the mode is Mode = l + [(f m – f 1 )/ (2f m – f 1 – f 2 )] × h Substitute the values in the formula, we get Mode = 35+[(23-21)/(46-21-14)]×10 = 35 + (20/11) = 35 + 1.8 = 36.8 years So the mode of the given data = 36.8 years Calculation of Mean: First find the midpoint using the formula, x i = (upper limit +lower limit)/2
Class Interval Frequency (f i ) Mid-point (x i ) f i x i
5-15 6 10 60
15-25 11 20 220
25-35 21 30 630
35-45 23 40 920
45-55 14 50 700
55-65 5 60 300
Sum f i = 80 Sum f i x i = 2830
The mean formula is Mean = x̄ = ∑f i x i /∑f i = 2830/80 = 35.375 years Therefore, the mean of the given data = 35.375 years

2. The following data gives the information on the observed lifetimes (in hours) of 225

electrical components:

Lifetime (in hours) 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 10 35 52 61 38 29

Determine the modal lifetimes of the components.

Solution: From the given data the modal class is 60–80. Lower limit of modal class = l = 60, The frequencies are: f m = 61, f 1 = 52, f 2 = 38 and h = 20 The formula to find the mode is Mode = l+ [(f m – f 1 )/(2f m – f 1 – f 2 )] × h Substitute the values in the formula, we get Mode = 60 + [(61 – 52)/ (122 – 52 – 38)] × 20 Mode = 60 + [(9 × 20)/32] Mode = 60 + (45/8) = 60 + 5.625 Therefore, modal lifetime of the components = 65.625 hours.

3. The following data gives the distribution of total monthly household expenditure of 200

families of a village. Find the modal monthly expenditure of the families. Also, find the

mean monthly expenditure:

Expenditure (in Rs.) Number of families
1000-1500 24
1500-2000 40
2000-2500 33
2500-3000 28
3000-3500 30
3500-4000 22
4000-4500 16
4500-5000 7
Solution: Given data: Modal class = 1500-2000, l = 1500, Frequencies: f m = 40 f 1 = 24, f 2 = 33 and h = 500 Mode formula: Mode = l + [(f m – f 1 )/ (2f m – f 1 – f 2 )] × h Substitute the values in the formula, we get Mode = 1500 + [(40 – 24)/ (80 – 24 – 33)] × 500 Mode = 1500 + [(16 × 500)/23] Mode = 1500 + (8000/23) = 1500 + 347.83 Therefore, modal monthly expenditure of the families = Rupees 1847.83 Calculation for mean: First find the midpoint using the formula, x i =(upper limit +lower limit)/2 Let us assume a mean, (a) be 2750.
Class Interval f i x i d i = x i – a u i = d i /h f i u i
1000-1500 24 1250 -1500 -3 -72
1500-2000 40 1750 -1000 -2 -80
2000-2500 33 2250 -500 -1 -33
2500-3000 28 2750 = a 0 0 0
3000-3500 30 3250 500 1 30
3500-4000 22 3750 1000 2 44
4000-4500 16 4250 1500 3 48
4500-5000 7 4750 2000 4 28
f i = 200 f i u i = -35
The formula to calculate the mean, Mean = x̄ = a +(∑f i u i /∑f i ) × h Substitute the values in the given formula = 2750 + (-35/200) × 500 = 2750 – 87.50 = 2662.50 So, the mean monthly expenditure of the families = Rs. 2662.50

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures

No of students per teacher Number of states / U.T
15-20 3
20-25 8
25-30 9
30-35 10
35-40 3
40-45 0
45-50 0
50-55 2
Solution: Given data: Modal class = 30 – 35, l = 30, Class width (h) = 5, f m = 10, f 1 = 9 and f 2 = 3 Mode Formula: Mode = l + [(f m – f 1 )/ (2f m – f 1 – f 2 )] × h Substitute the values in the given formula Mode = 30 + [(10 – 9)/ (20 – 9 – 3)] × 5 = 30 + (5/8) = 30 + 0.625 = 30.625 Therefore, the mode of the given data = 30.625 Calculation of mean: Find the midpoint using the formula, x i =(upper limit +lower limit)/2
Class Interval Frequency (f i ) Mid-point (x i ) f i x i
15-20 3 17.5 52.5
20-25 8 22.5 180.0
25-30 9 27.5 247.5
30-35 10 32.5 325.0
35-40 3 37.5 112.5
40-45 0 42.5 0
45-50 0 47.5 0
50-55 2 52.5 105.0
Sum f i = 35 Sum f i x i = 1022.5
Mean = x̄ = ∑f i x i /∑f i = 1022.5/35 = 29.2 (approx) Therefore, mean = 29.2

5. The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.

Run Scored Number of Batsman
3000-4000 4
4000-5000 18
5000-6000 9
6000-7000 7
7000-8000 6
8000-9000 3
9000-10000 1
10000-11000 1

Find the mode of the data.

Solution: Given data: Modal class = 4000 – 5000, l = 4000, class width (h) = 1000, f m = 18, f 1 = 4 and f 2 = 9 Mode Formula: Mode = l + [(f m – f 1 )/ (2f m – f 1 – f 2 )] × h Substitute the values Mode = 4000 + [(18 – 4)/ (36 – 4 – 9)] × 1000 = 4000 + (14000/23) = 4000 + 608.695 = 4608.695 = 4608.7 (approximately) Thus, the mode of the given data is 4608.7 runs.

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data:

Number of cars Frequency
0-10 7
10-20 14
20-30 13
30-40 12
40-50 20
50-60 11
60-70 15
70-80 8
Solution: Given Data: Modal class = 40 – 50, l = 40, Class width (h) = 10, f m = 20, f 1 = 12 and f 2 = 11 Mode = l + [(f m – f 1 )/(2f m – f 1 – f 2 )] × h Substitute the values Mode = 40 + [(20 – 12)/ (40 – 12 – 11)] × 10 = 40 + (80/17) = 40 + 4.7 = 44.7 Thus, the mode of the given data is 44.7 cars.

Benefits of Using NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2

Concept Clarity : Step-by-step explanations simplify complex topics like mean, median, mode, and cumulative frequency, making them easier to understand.

Exam Preparation : Solutions align with the NCERT syllabus, ensuring students are well-prepared for board exams.

Time Management : Solved examples teach efficient problem-solving techniques, helping students solve questions faster during exams.

Strong Foundation : Builds a solid understanding of Statistics, a topic with applications in higher studies and real-life scenarios.

Self-Learning : The solutions provide clear guidance, enabling students to learn independently.

Error-Free Answers : Ensures accurate and reliable solutions for better learning outcomes.

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 FAQs

What is good knowledge of statistics?

Statistical knowledge helps to choose the proper method of collecting the data and employ those samples in the correct analysis process in order to effectively produce the results. In short, statistics is a crucial process which helps to make the decision based on the data.

What skill is statistics?

Statistics is a skill that involves collecting, analyzing, interpreting, and presenting data. Having a good grasp on statistics means that you can take large sets of numbers, see the patterns and trends, and explain them in a way that makes sense to others.

What are the 4 stages of statistics?

The four phases of a statistical investigation are posing a question, collect data, analyze the data, and interpret the results.

What are the two main types of statistics?

There are two kinds of Statistics, which are descriptive Statistics and inferential Statistics. In descriptive Statistics, the Data or Collection Data are described in a summarized way, whereas in inferential Statistics, we make use of it in order to explain the descriptive kind.
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