NCERT Solutions for Exercise 14.2 Class 10 Maths Chapter 14 - Statistics

# NCERT Solutions for Exercise 14.2 Class 10 Maths Chapter 14 - Statistics

Edited By Ramraj Saini | Updated on Nov 27, 2023 06:03 PM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.2

NCERT Solutions for Exercise 14.2 Class 10 Maths Chapter 14 Statistics are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 Maths ex 14.2 introduces us to the concept of finding the mode of grouped data. In Mathematics NCERT syllabus Class 9 we have learnt about the mode. Mode is the most occurring value amongst all the observations and till ninth, we know how to find the model of the ungrouped data that is a collection of discrete values. NCERT solutions for Class 10 Maths chapter 14 exercise 14.2 includes various types of problems which covers the entire topic of finding mode in the grouped data.

Exercise 14.2 Class 10 Maths also gives us information about what is multi modal data, what is modal class and how to find the mode within the modal class. 10th class Maths exercise 14.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Statistics Class 10 Chapter 14 Exercise: 14.2

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

The class having maximum frequency is the modal class.

The maximum frequency is 23 and hence the modal class = 35-45

Lower limit (l) of modal class = 35, class size (h) = 10

Frequency ( $f_1$ ) of the modal class = 23, frequency ( $f_0$ ) of class preceding the modal class = 21, frequency ( $f_2$ ) of class succeeding the modal class = 14.

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 35 + \left(\frac{23-21}{2(23)-21-14} \right).10 \\ \\ = 35 + \frac{2}{11}.10$

$= 36.8$

Now,

 Age Number of patients $f_i$ Classmark $x_i$ $f_ix_i$ 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 $\sum f_ix_i$ =2830

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{2830}{80} = 35.37$
The maximum number of patients are in the age group of 36.8, whereas the average age of all the patients is 35.37.

Determine the modal lifetimes of the components.

The class having maximum frequency is the modal class.

The maximum frequency is 61 and hence the modal class = 60-80

Lower limit (l) of modal class = 60, class size (h) = 20

Frequency ( $f_1$ ) of the modal class = 61 frequency ( $f_0$ ) of class preceding the modal class = 52, frequency ( $f_2$ ) of class succeeding the modal class = 38.

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 60 + \left(\frac{61-52}{2(61)-52-38} \right).20 \\ \\ = 60 + \frac{9}{32}.20$

$= 65.62$

Thus, the modal lifetime of 225 electrical components is 65.62 hours

The class having maximum frequency is the modal class.

The maximum frequency is 40 and hence the modal class = 1500-2000

Lower limit (l) of modal class = 1500, class size (h) = 500

Frequency ( $f_1$ ) of the modal class = 40 frequency ( $f_0$ ) of class preceding the modal class = 24, frequency ( $f_2$ ) of class succeeding the modal class = 33.

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 1500 + \left(\frac{40-24}{2(40)-24-33} \right).500 \\ \\ = 1500 + \frac{16}{23}.500$

$= 1847.82$

Thus, the Mode of the data is Rs. 1847.82

Now,

Let the assumed mean be a = 2750 and h = 500

 Expenditure Number of families $f_i$ Classmark $x_i$ $d_i = x_i -a$ $u_i = \frac{d_i}{h}$ $f_iu_i$ 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 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 $\sum f_i$ =200 $\sum f_ix_i$ = -35

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 2750 + \frac{-35}{200}\times500 = 2750 -87.5 = 2662.50$

Thus, the Mean monthly expenditure is Rs. 2662.50

The class having maximum frequency is the modal class.

The maximum frequency is 10 and hence the modal class = 30-35

Lower limit (l) of modal class = 30, class size (h) = 5

Frequency ( $f_1$ ) of the modal class = 10 frequency ( $f_0$ ) of class preceding the modal class = 9, frequency ( $f_2$ ) of class succeeding the modal class = 3

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 30 + \left(\frac{10-9}{2(10)-9-3} \right).5 \\ \\ = 30 + \frac{1}{8}.5$

$= 30.625$

Thus, Mode of the data is 30.625

Now,

Let the assumed mean be a = 32.5 and h = 5

 Class Number of states $f_i$ Classmark $x_i$ $d_i = x_i -a$ $u_i = \frac{d_i}{h}$ $f_iu_i$ 15-20 3 17.5 -15 -3 -9 20-25 8 22.5 -10 -2 -16 25-30 9 27.5 -5 -1 -9 30-35 10 32.5 0 0 0 35-40 3 37.5 5 1 3 40-45 0 42.5 10 2 0 45-50 0 47.5 15 3 0 50-55 2 52.5 20 4 8 $\sum f_i$ =35 $\sum f_ix_i$ = -23

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 32.5 + \frac{-23}{35}\times5= 29.22$

Thus, the Mean of the data is 29.22

Find the mode of the data.

The class having maximum frequency is the modal class.

The maximum frequency is 18 and hence the modal class = 4000-5000

Lower limit (l) of modal class = 4000, class size (h) = 1000

Frequency ( $f_1$ ) of the modal class = 18 frequency ( $f_0$ ) of class preceding the modal class = 4, frequency ( $f_2$ ) of class succeeding the modal class = 9

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ =4000 + \left(\frac{18-4}{2(18)-4-9} \right).1000 \\ \\ = 4000 + \frac{14}{23}.1000$

$= 4608.70$

Thus, Mode of the data is 4608.70

The class having maximum frequency is the modal class.

The maximum frequency is 20 and hence the modal class = 40-50

Lower limit (l) of modal class = 40, class size (h) = 10

Frequency ( $f_1$ ) of the modal class = 20 frequency ( $f_0$ ) of class preceding the modal class = 12, frequency ( $f_2$ ) of class succeeding the modal class = 11

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ =40+ \left(\frac{20-12}{2(20)-12-11} \right).10 \\ \\ = 40 + \frac{8}{17}.10$

$= 44.70$

Thus, Mode of the data is 44.70

## More About NCERT Solutions for Class 10 Maths Exercise 14.2

Class 10 Maths chapter 14 exercise 14.2 includes all the question of finding mode of the grouped data but in some way all the questions are different from each question because each question is represented with real life example problem such as ages of patients, life of electric component, expenditure of family, ratio of teacher and student, runs scored by batsman, counting passing cars, etc.

Along with this NCERT solutions for Class 10 Maths exercise 14.2 includes question in which we have to compare two central tendencies mode and mean. Exercise 14.2 Class 10 Maths covers all types of questions that can be formed in mode of grouped data and how we can use them in real life and compared with other central tendencies. Moreover, there are a few examples before NCERT Solutions for Class 10 Maths exercise 14.2 that have questions that are based on finding the mode of the ungrouped data.

Also Read| Statistics Class 10 Notes

## Benefits of NCERT Solutions for Class 10 Maths Exercise 14.2

• Class 10 Maths chapter 14 exercise 14.2 broadly covers all kinds of questions that can be formed on a central tendency mode especially on grouped data
• NCERT Class 10 Maths chapter 14 exercise 14.2, will be helpful in JEE Main as statistics is a major part of syllabus
• NCERT Class 10 Maths chapter 14 exercise 14.2, will be base for the class 11 chapter 15 statistics which is extension for this chapter
• Statistics is also an important part of data science which is a growing field in the world of internet

## Key Features For Class 10 Maths Chapter 14 Exercise 14.2

• Step-by-Step Solutions: Clear and detailed explanations for each problem in ex 14.2 class 10, ensuring students understand the methodology.
• Conceptual Understanding: Explanation of statistical concepts involved in the exercises, aiding students in grasping the underlying principles.
• Variety of Problems: In class 10 ex 14.2, diverse types of problems, offering a broad spectrum of challenges to reinforce understanding.
• Visual Aids, Diagrams, and Examples: Use of visual representations, diagrams, and real-life examples to simplify complex concepts and aid visual learners.
• Application-Based Problems: In class 10 maths ex 14.2, application-oriented problems that relate statistics to real-world scenarios, fostering practical understanding and problem-solving skills.

Also, See

## NCERT Solutions for Class 10 Subject Wise

1. What do you mean by central tendency ?

This ex 14.2 class 10 discuss about the concept of Central tendency refers to the centre value of all observations.

2. How can we measure central tendency of a data as covered in NCERT solutions for Class 10 Maths exercise 14.2?

These concepts are discussed ex 14.2 class 10 comprehensively. With the help of mode , median and mean we can measure central tendency of a data. Practice this exercise to command these concepts.

3. What do you mean by multimodal data ?

These concepts are discussed in class 10 maths ex 14.2 . The data which have more than one value which are repeated in same frequency are called multimodal data. Practice the problems to command these concepts.

4. What do you mean by modal class ?

These concepts are discussed in 10th class maths exercise 14.2 answers. Modal class is the class which have maximum frequency or we can say that the class which has highest occurring value in the data. Practice these problems to command the concepts.

5. Explain the variables covered in the formula of finding mode in grouped data according to NCERT book Class 10 Maths exercise 14.2 .
• l =  modal class's lowest limit

• h = denotes the length of the class interval (assuming all class sizes to be equal)

• f ₁ =represents the frequency of the modal class

• f ₀ =represents the frequency of the class preceding the modal class

• f ₂ =  represents the frequency of the class following the modal class.

6. How many questions are there in Class 10 Maths exercise 14.2 ?

There are six questions in class 10 ex 14.2. Practice these problems to command the concepts.

7. How many solved examples are there before Exercise 14.2 Class 10 Maths that are based on the mode of grouped data?

There are three problems based on the nature of root that are solved before the Class 10 Mathematics chapter 14 exercise 14.2.

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### Questions related to CBSE Class 10th

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According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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