NCERT Solutions for Exercise 14.3 Class 10 Maths Chapter 14

NCERT Solutions for Exercise 14.3 Class 10 Maths Chapter 14 - Statistics

Edited By Ramraj Saini | Updated on Dec 02, 2023 10:37 AM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3

NCERT Solutions for Exercise 14.3 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 14.3 introduces us to the concept of finding the median. In ninth grade, we learned about the median of grouped data. The median is a measure of central tendency that represents the value of the data's middle observation. The median depends on the number of observations. The value of the median is different if the number of observations is odd and different if the number of observations is even. This exercise class 10 ex 14.3 feature a variety of questions that cover the complete topic of determining the median in grouped data.

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NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3
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Statistics C lass 10 Chapter 14 Statistics Exercise: 14.3
More About NCERT Solutions for Class 10 Maths Exercise 14.3
Benefits of NCERT Solutions for Class 10 Maths Exercise 14.3
Key Features For Class 10 Maths Chapter 14 Exercise 14.3
NCERT Solutions for Class 10 Subject Wise

NCERT Solutions for Exercise 14.3 Class 10 Maths Chapter 14 - Statistics

Exercise 14.3 Class 10 Maths also gives us information about what is cumulative frequency column, median class and also we will know about the relation between mean, median and mode. 10th class Maths exercise 14.3 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 C lass 10 Chapter 14 Statistics Exercise: 14.3

Q1 The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

1640762287658

Answer:

Let the assumed mean be a = 130 and h = 20

Class	Number of consumers $f_{i}$	Cumulative Frequency	Classmark $x_{i}$	$d_{i} = x_{i} - a$	$u_{i} = \frac{d_{i}}{h}$	$f_{i} u_{i}$
65-85	4	4	70	-60	-3	-12
85-105	5	9	90	-40	-2	-10
105-125	13	22	110	-20	-1	-13
125-145	20	42	130	0	0	0
145-165	14	56	150	20	1	14
165-185	8	64	170	40	2	16
185-205	4	68	190	60	3	12
		$\sum f_{i} = N$ = 68				$\sum f_{i} x_{i}$ = 7

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MEDIAN:
$N = 68 ⟹ \frac{N}{2} = 34$
$∴$ Median class = 125-145; Cumulative Frequency = 42; Lower limit, l = 125;

c.f. = 22; f = 20; h = 20
$M e d i a n = l + (\frac{\frac{n}{2} - c . f}{f}) . W$
$= 125 + (\frac{34 - 22}{20}) .20 = 125 + 12$

$= 137$

Thus, the median of the data is 137

MODE:

The class having maximum frequency is the modal class.
The maximum frequency is 20 and hence the modal class = 125-145
Lower limit (l) of modal class = 125, class size (h) = 20
Frequency ( $f_{1}$ ) of the modal class = 20; frequency ( $f_{0}$ ) of class preceding the modal class = 13, frequency ( $f_{2}$ ) of class succeeding the modal class = 14.

$M o d e = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) . h$

$= 125 + (\frac{20 - 13}{2 (20) - 13 - 14}) .20 = 125 + \frac{7}{13} .20$

$= 135.76$

Thus, Mode of the data is 135.76

MEAN:

Mean,
$\overset{―}{x} = a + \frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h$
$= 130 + \frac{7}{68} \times 20 = 137.05$

Thus, the Mean of the data is 137.05

Q2 If the median of the distribution given below is 28.5, find the values of x and y.

1640762346146

Answer:

Class	Number of consumers $f_{i}$	Cumulative Frequency
0-10	5	5
10-20	x	5+x
20-30	20	25+x
30-40	15	40+x
40-50	y	40+x+y
50-60	5	45+x+y
	$\sum f_{i} = N$ = 60

$N = 60 ⟹ \frac{N}{2} = 30$

Now,
Given median = 28.5 which lies in the class 20-30

Therefore, Median class = 20-30
Frequency corresponding to median class, f = 20
Cumulative frequency of the class preceding the median class, c.f. = 5 + x
Lower limit, l = 20; Class height, h = 10

$M e d i a n = l + (\frac{\frac{n}{2} - c . f}{f}) . W$
$⟹ 28.5 = 20 + (\frac{30 - 5 - x}{20}) .10 ⟹ 8.5 = \frac{25 - x}{2} ⟹ 25 - x = 8.5 (2) ⟹ x = 25 - 17 = 8$

Also,

$60 = 45 + x + y ⟹ x + y = 60 - 45 = 15 ⟹ y = 15 - x = 15 - 8 (∵ x = 8) ⟹ y = 7$

Therefore, the required values are: x=8 and y=7

Q3 A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

1640762364185

Answer:

Class	Frequency $f_{i}$	Cumulative Frequency
15-20	2	2
20-25	4	6
25-30	18	24
30-35	21	45
35-40	33	78
40-45	11	89
45-50	3	92
50-55	6	98
55-60	2	100

$N = 100 ⟹ \frac{N}{2} = 50$
Therefore, Median class = 35-45
Frequency corresponding to median class, f = 21
Cumulative frequency of the class preceding the median class, c.f. = 24
Lower limit, l = 35; Class height, h = 10

$M e d i a n = l + (\frac{\frac{n}{2} - c . f}{f}) . W$
$= 35 + (\frac{50 - 45}{33}) .5$

$= 35.75$

Thus, the median age is 35.75 years.

Q4 The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table :

1640762380533

Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to
117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

Answer:

The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes.

Class	Frequency $f_{i}$	Cumulative Frequency
117.5-126.5	3	3
126.5-135.5	5	8
135.5-144.5	9	17
144.5-153.5	12	29
153.5-162.5	5	34
162.5-171.5	4	38
171.5-180.5	2	40

$\dpi 100 N = 40 ⟹ \frac{N}{2} = 20$
Therefore, Median class = 144.5-153.5

Lower limit, l = 144.5; Class height, h = 9
Frequency corresponding to median class, f = 12
Cumulative frequency of the class preceding the median class, c.f. = 17

$M e d i a n = l + (\frac{\frac{n}{2} - c . f}{f}) . W$
$= 144.5 + (\frac{20 - 17}{12}) .9$

$= 146.75$

Thus, the median length of the leaves is 146.75 mm

Q5 The following table gives the distribution of the lifetime of 400 neon lamps :
1640762410074

Find the median lifetime of a lamp.

Answer:

Class	Frequency $f_{i}$	Cumulative Frequency
1500-2000	14	14
2000-2500	56	70
2500-3000	60	130
3000-3500	86	216
3500-4000	74	290
4000-4500	62	352
4500-5000	48	400

$\dpi 100 N = 400 ⟹ \frac{N}{2} = 200$
Therefore, Median class = 3000-3500

Lower limit, l = 3000; Class height, h = 500
Frequency corresponding to median class, f = 86
Cumulative frequency of the class preceding the median class, c.f. = 130

$M e d i a n = l + (\frac{\frac{n}{2} - c . f}{f}) . W$
$= 3000 + (\frac{200 - 130}{86}) .500 = 3000 + 406.97$

$= 3406.97$

Thus, the median lifetime of a lamp is 3406.97 hours

$= 146.75$

Thus, the median length of the leaves is 146.75 mm

Q6 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

1640762454336

Determine the median number of letters in the surnames. Find the mean number of
letters in the surnames? Also, find the modal size of the surnames.

Answer:

Class	Number of surnames $f_{i}$	Cumulative Frequency	Classmark $x_{i}$	$f_{i} x_{i}$
1-4	6	6	2.5	15
4-7	30	36	5.5	165
7-10	40	76	8.5	340
10-13	16	92	11.5	184
13-16	4	96	14.5	51
16-19	4	100	17.5	70
		$\sum f_{i} = N$ = 100		$\sum f_{i} x_{i}$ = 825

MEDIAN:
$N = 100 ⟹ \frac{N}{2} = 50$
$∴$ Median class = 7-10; Lower limit, l = 7;

Cumulative frequency of preceding class, c.f. = 36; f = 40; h = 3
$M e d i a n = l + (\frac{\frac{n}{2} - c . f}{f}) . W$
$= 7 + (\frac{50 - 36}{40}) .3$

$= 8.05$

Thus, the median of the data is 8.05

MODE:

The class having maximum frequency is the modal class.
The maximum frequency is 40 and hence the modal class = 7-10
Lower limit (l) of modal class = 7, class size (h) = 3
Frequency ( $f_{1}$ ) of the modal class = 40; frequency ( $f_{0}$ ) of class preceding the modal class = 30, frequency ( $f_{2}$ ) of class succeeding the modal class = 16

$M o d e = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) . h$

$= 7 + (\frac{40 - 30}{2 (40) - 30 - 16}) .3 = 125 + \frac{10}{34} .3$

$= 7.88$

Thus, Mode of the data is 7.88

MEAN:

Mean,
$\overset{―}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$
$= \frac{825}{100} = 8.25$

Thus, the Mean of the data is 8.25

Q7 The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

1640762471389

Answer:

Class	Number of students $f_{i}$	Cumulative Frequency
40-45	2	2
45-50	3	5
50-55	8	13
55-60	6	19
60-65	6	25
65-70	3	28
70-75	2	30

MEDIAN:
$N = 30 ⟹ \frac{N}{2} = 15$
$∴$ Median class = 55-60; Lower limit, l = 55;

Cumulative frequency of preceding class, c.f. = 13; f = 6; h = 5
$M e d i a n = l + (\frac{\frac{n}{2} - c . f}{f}) . W$
$= 55 + (\frac{15 - 13}{6}) .5 = 55 + \frac{2}{6} .5$

$= 56.67$

Thus, the median weight of the student is 56.67 kg

Benefits of NCERT Solutions for Class 10 Maths Exercise 14.3

Chapter 14 exercise 14.3 in Class 10 Maths presents a wide range of issues that might be based on a central tendency median, particularly on grouped data.
NCERT Statistics is a big element of the NCERT curriculum for Class 10 Mathematics chapter 14 exercise 14.3. It will be useful in JEE Main (joint admission exam) as statistics is a major part of the syllabus.
Class 10 Mathematics chapter 14 exercise 14.3 will serve as a foundation for Class 11 statistics chapter 15 (which is a continuation of this chapter).
The median can be used as a measure of location when one attaches reduced importance to extreme values

Key Features For Class 10 Maths Chapter 14 Exercise 14.3

Step-by-Step Solutions: Clear and detailed explanations for each problem in Ex 14.3 class 10, ensuring students understand the methodology.
Conceptual Understanding: Explanation of statistical concepts involved in the exercises, aiding students in grasping the underlying principles are discussed in class 10 maths ex 14.3.
Variety of Problems: In class 10 ex 14.3, 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: In10th class maths exercise 14.3 answers, 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. How does the median depend on the number of observations?

Median gives the value of the observation which is at the center so it is dependent on either the observation is odd number or even number. Practice ex 14.3 class 10 to command these concepts.

2. What is the formula of median of odd observation ?

The concept related to discusses in class 10 ex 14.3. Formula of median of odd observation is n/2th observation. Students can practice ex 14.3 class 10 to get deeper understanding of concepts.

3. What is the formula of median of even observation?

The concepts related to median is discussed in class 10 maths ex 14.3. Formula of median of even observation is (n/2+ 1 )th observation. Also practice class 10 ex 14.3 which is discussed in this article to get deeper understanding of the concepts.

4. What is the relation between mean, medium and mode as mentioned in Class 10 Maths chapter 14 exercise 14.3?

3 Median = Mode + 2 Mean is the relation between mean, medium and mode as mentioned in Class 10 maths chapter 14 exercise 14.3.

5. What do you mean by Cumulative Frequency Table ?

These concepts are discussed in 10th class maths exercise 14.3 answers. practice them to command the concepts. Cumulative Frequency Table is the cumulative frequency is calculated by adding each frequency from a frequency distribution table to the sum of its predecessors.

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i studied 1 to 10th CBSE board abroad and 11th 12th in karnataka state board . i am a domicile of karnataka and comes under minority, income under 5 lakhs. how neet councillng will allocate my seat

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

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4. Required Documents :

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Mitali Sharma 30 January,2025

show the previous year exams ?

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

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SWETCHCHA MISKA 10 July,2024

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

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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Manasvini Gupta 23 July,2024

i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

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Ashvadha 03 July,2024

i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

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.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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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NCERT Solutions for Exercise 14.3 Class 10 Maths Chapter 14 - Statistics

NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3

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Statistics C lass 10 Chapter 14 Statistics Exercise: 14.3

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Answer:

Answer:

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More About NCERT Solutions for Class 10 Maths Exercise 14.3

Benefits of NCERT Solutions for Class 10 Maths Exercise 14.3

Key Features For Class 10 Maths Chapter 14 Exercise 14.3

NCERT Solutions for Class 10 Subject Wise