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
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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.

This Story also Contains
  1. NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.2
  2. Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 14 exercise 14.2
  3. Assess NCERT Solutions for Class 10 Maths chapter 14 exercise 14.2
  4. Statistics Class 10 Chapter 14 Exercise: 14.2
  5. More About NCERT Solutions for Class 10 Maths Exercise 14.2
  6. Benefits of NCERT Solutions for Class 10 Maths Exercise 14.2
  7. Key Features For Class 10 Maths Chapter 14 Exercise 14.2
  8. NCERT Solutions for Class 10 Subject Wise
NCERT Solutions for Exercise 14.2 Class 10 Maths Chapter 14 - Statistics
NCERT Solutions for Exercise 14.2 Class 10 Maths Chapter 14 - Statistics

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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Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 14 exercise 14.2

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Assess NCERT Solutions for Class 10 Maths chapter 14 exercise 14.2

Statistics Class 10 Chapter 14 Exercise: 14.2

Q1 The following table shows the ages of the patients admitted in a hospital during a year:

1640761737516

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

Answer:

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.

Q2 The following data gives information on the observed lifetimes (in hours) of 225 electrical components :

1640761774675

Determine the modal lifetimes of the components.

Answer:

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

Q3 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 :

1640761796831

Answer:

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

Q4 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.

1640761815121

Answer:

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

Q5 The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

1640762201359

Find the mode of the data.

Answer:

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

Q6 A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

1640762258431

Answer:

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

Frequently Asked Questions (FAQs)

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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Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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