NCERT Solutions for Exercise 14.3 Class 9 Maths Chapter 14 - Statistics

# NCERT Solutions for Exercise 14.3 Class 9 Maths Chapter 14 - Statistics

Edited By Vishal kumar | Updated on Oct 18, 2023 09:11 AM IST

## NCERT Solutions for Class 9 Maths Chapter 14 - Statistics Exercise 14.3- Download Free PDF

NCERT Solutions for exercise 14.3 Class 9 Maths chapter 14 Statistics exercise 14.3 is the important exercise of NCERT solutions for Class 9 Mathematics. We have provided 9th class maths exercise 14.3 answers to them below, which will be quite important when performing assignments. As from the former exercise is concerned, we know that Statistics can be called a data analysis technique in an area of mathematics that comes with collecting and analyzing the amounts of numerical data at one or more places. Showing mains to this exercise 14.3 Class 9 Maths is an exercise followed by exercise 14.2 of the chapter which consists of numerical problems based on graphs of collection of data.

In this NCERT book exercise 14.3, we will be studying concepts of Histogram which we talk about in the previous exercise during the chapter study The Class 9 Maths chapter 14 exercise 14.3 basically works on important topics like Histogram, bar graph and frequency of polygon over graphs as mentioned with terms like frequency, maximum and minimum values which cover around each and every question of the exercise. The NCERT syllabus Class 9 Maths chapter 14 exercise 14.3 covers Initial topics with examples called in terms of numerical solved. Most of the real-life questions will be discussed in NCERT solutions for Class 9 Maths Chapter 14 exercise 14.3. Along with Class 9 Maths Chapter 1 exercise 14.3, the following exercises are also present.

**According to the CBSE Syllabus for the academic year 2023-24, this chapter has been renumbered as Chapter 12.

## Download PDF of NCERT Solutions for Class 9 Maths Chapter 14 – Statistics Exercise 14.3

Access Statistics Class 9 Chapter 14 Exercise: 14.3

 Serial Number Causes Female fatality rate (%) 1. Reproductive health conditions 31.8 2. Neuropsychiatric conditions 25.4 3. Injuries 12.4 4. Cardiovascular conditions 4.3 5. Respiratory conditions 4.1 6. Other causes 22.0

Represent the information given above graphically

The graphical representation of the given data is as follows

 Serial Number Causes Female fatality rate (%) 1. Reproductive health conditions 31.8 2. Neuropsychiatric conditions 25.4 3. Injuries 12.4 4. Cardiovascular conditions 4.3 5. Respiratory conditions 4.1 6. Other causes 22.0

Which condition is the major cause of women’s ill health and death worldwide?

From the graph we can see reproductive health conditions is the major cause of women’s ill health and death worldwide. The female fatality rate is 31.8% due to reproductive health conditions.

 Serial Number Causes Female fatality rate (%) 1. Reproductive health conditions 31.8 2. Neuropsychiatric conditions 25.4 3. Injuries 12.4 4. Cardiovascular conditions 4.3 5. Respiratory conditions 4.1 6. Other causes 22.0

Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause

Due to poor financial conditions and failure of the government to provide necessary healthcare condition to women, reproductive health conditions is the major cause of ill health and death of women worldwide.

 Section Number of girls per thousand boys Schedule Caste (SC) 940 Schedule Tribe (ST) 970 Non SC/ST 920 Backward districts 950 Non-backward districts 920 Rural 930 Urban 910

Represent the information above by a bar graph.

The graphical representation of the given information is as follows

 Section Number of girls per thousand boys Schedule Caste (SC) 940 Schedule Tribe (ST) 970 Non SC/ST 920 Backward districts 950 Non-backward districts 920 Rural 930 Urban 910

In the classroom discuss what conclusions can be arrived at from the graph

From the graph, we can see that the number of girls per thousand boys is the least in urban society and the highest in the Scheduled Tribes.

910 in case of urban society and 970 in that of Scheduled Tribes.

 Political Party A B C D E F Seats Won 75 55 37 29 10 37

Draw a bar graph to represent the polling results.

The representation of the given data in the form of a bar graph is as follows.

 Political Party A B C D E F Seats Won 75 55 37 29 10 37

Which political party won the maximum number of seats?

Party A has won the maximum number of seats. Party A has won 75 seats.

 Length (in mm) Number of leaves 118-126 3 127-135 5 136-144 9 145-153 12 154-162 5 163-171 4 172-180 2

Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]

As we can see from the given table that the data is discontinous and the difference between the upper limit of a class and the lower limit of the next class is 1 and therefore we change both of them by a value 1/2.

e.g 127 - 135 would become 126.5 - 235.5

The modified table therefore is

The representation of the above data through a histogram is as follows

 Length (in mm) Number of leaves 118-126 3 127-135 5 136-144 9 145-153 12 154-162 5 163-171 4 172-180 2

Is there any other suitable graphical representation for the same data?

A frequency polygon could be another suitable graphical representation for the same data.

 Length (in mm) Number of leaves 118-126 3 127-135 5 136-144 9 145-153 12 154-162 5 163-171 4 172-180 2

Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

No it is certainly not correct to conclude that the maximum number of leaves are 153 mm long because the given data does not tell us about the exact length of the leaves. It only tells us about the range in which their lengths lie. We can only conclude that the maximum number of leaves (12) have their lengths in the region 145 - 153.

 Life time (in hours) Number of lamps 300-400 14 400-500 56 500-600 60 600-700 86 700-800 74 800-900 62 900-1000 48

Represent the given information with the help of a histogram.

The representation of the given information in the form of a histogram is as follows.

 Life time (in hours) Number of lamps 300-400 14 400-500 56 500-600 60 600-700 86 700-800 74 800-900 62 900-1000 48

How many lamps have a lifetime of more than 700 hours?

Lamps having life time in the range 700 - 800 = 74

Lamps having life time in the range 800 - 900 = 62

Lamps having life time in the range 900 - 1000 = 48

Lamps having a life time of more than 700 hours = 74 + 62 + 48 = 184.

 Section A Section B Marks Frequency Marks Frequency 0-10 3 0-10 5 10-20 9 10-20 19 20-30 17 20-30 15 30-40 12 30-40 10 40-50 9 40-50 1

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections

To make the frequency polygon we first modify the table as follows

$\\Class\ marks= \frac{Upper\ limit\ of\ class\ interval+Lower\ limit\ of\ class\ interval}{2}$

To make the frequency polygon we mark the marks on the x-axis and the number of students on the y-axis.

The representation of the given information in the form of frequency polygon is as follows.

From the frequency polygon we can see that the performance of section A is better.

 Number of balls Team A Team B 1-6 2 5 7-12 1 6 13-18 8 2 19-24 9 10 25-30 4 5 31-36 5 6 37-42 6 3 43-48 10 4 49-54 6 8 55-60 2 10

Represent the data of both the teams on the same graph by frequency polygons. [ Hint : First make the class intervals continuous.]

The given data is not continous we therefore modify the limits of the class intervals as well to make the class intervals continous.

To make the frequency polygon we first modify the table as follows

$\\Class\ marks= \frac{Upper\ limit\ of\ class\ interval+Lower\ limit\ of\ class\ interval}{2}$

To make the frequency polygon we mark the number of balls on the x-axis and the runs scored on the y-axis.

The representation of the given information in the form of frequency polygon is as follows.

 Age (in years) Number of children 1-2 5 2-3 3 3-5 6 5-7 12 7-10 9 10-15 10 15-17 4

Draw a histogram to represent the data above.

Since the class sizes vary to make the histogram we have to calculate the weighted frequency for each rectangle as per its width

$\\Weighted\ frequency=\frac{Minimum\ class\ size}{Class\ size\ of\ the\ interval}\times Frequency\\ Length\ of\ the\ rectangle=\frac{Minimum\ width}{Width\ of\ the\ rectangle}\times Frequency$

Minimum class size = 2 - 1 = 1

The modified table showing the weighted frequency as per the size of the class intervals is as follows.

The histogram representing the information given in the above table is as follows.

 Number of letters Number of surnames 1-4 6 4-6 30 6-8 44 8-12 16 12-20 4

Since the class sizes vary to make the histogram we have to calculate the weighted frequency for each rectangle as per its width

$\\Weighted\ frequency=\frac{Minimum\ class\ size}{Class\ size\ of\ the\ interval}\times Frequency\\ Length\ of\ the\ rectangle=\frac{Minimum\ width}{Width\ of\ the\ rectangle}\times Frequency$

Minimum class size = 6 - 4 = 2

The modified table showing the weighted frequency as per the size of the class intervals is as follows.

The histogram representing the information given in the above table is as follows.

 Number of letters Number of surnames 1-4 6 4-6 30 6-8 44 8-12 16 12-20 4

Write the class interval in which the maximum number of surnames lie.

The class interval in which the maximum number of surnames lie is 6 - 8

The weighted frequency of this class interval (taking 2 as the minimum class size) is 44.

## More About NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.3

There are around 8-10 questions that cover all the practical and numerical aspects of exercise and chapter too as mentioned in Exercise 14.3 Class 9 Maths. The initial questions will give an intro idea of data analysis through graphs for grouped or ungrouped data that can be collected from day-to-day life. As the problem, the serial number increases the level of questions remains constant in NCERT solutions for class 9 maths Chapter 14 exercise 14.3. In the lasts of class 9 maths Chapter 14 exercise, 14.3 gives the best questions covered in of all the topics and examples discussed in the chapter. The Class 9th Maths Chapter 14 exercise 14.3 gives attributes to class interval theory over problems.

Also Read| Statistics Class 9 Notes

## Benefits of NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.3

• Taking other benefits apart from above mentioned of exercise 14.3 Class 9 Maths and the NCERT solutions for ex 14.3 class 9 into account is that students can hands-on practice over concepts which they learnt just again the exercise, Hence would be easy for them to Learn.

• The concept of bar graph and frequency polygons will also be discussed in exercise 14.3 class 9 maths

• Most of the students who can solve every question of this exercise 14.3 Class 9 Maths will be able to enjoy statistics happily from initial to another level that is given in class 9 maths ex 14.3.

• For students in Class 9, in the final exams, they might get some short answers or long answer questions which are the numerical type that is there in Class 9 Maths Chapter 14 exercise 14.3.

## Key Features of Class 9 Maths Chapter 14 Exercise 14.3 Solution

1. Comprehensive Coverage: 9th class maths exercise 14.3 answers covers various topics related to statistics, including data representation, frequency distribution, and graphical representation, offering a comprehensive understanding of the subject.

2. Expertly Crafted Solutions: The class 9 ex 14.3 solutions are meticulously designed by subject matter experts to provide clear, step-by-step explanations, making statistical concepts easy to grasp.

3. CBSE Syllabus Alignment: The class 9 maths ex 14.3 solutions are in line with the CBSE syllabus, making them suitable for homework, assignments, and exam preparation.

4. Clarity and Detail: The exercise 14.3 class 9 maths solutions offer in-depth explanations that help students build a strong foundation in statistics, ensuring conceptual clarity.

5. PDF Format: Students can access the solutions in PDF format, allowing for convenient offline use.

Also see-

## NCERT Solutions of Class 10 Subject Wise

1. Define Histogram?

Ans: a diagram made up of rectangles with areas proportional to the frequency of a variable and widths equal to the class interval

2. What do you mean Bar graph?

Ans: A bar graph is a chart or graphical presentation of facts, quantities, or numbers that practices that relate to or strips. Bar graphs are used to compare and contrast different types of information by simply comparing quantities, frequencies, or other measurements.

3. Can we generate a relation between frequency and histogram?

Ans:  A frequency histogram is a sort of bar graph that displays the frequency or the number of times, an outcome happens in a data collection. It has a title, an x-axis, a y-axis, and vertical bars to graphically portray the data...

4. What are Frequency Polygons?

Ans:   A frequency polygon is a graphical depiction of a distribution. The market looks are used to comprehend the form of a distribution.

5. What is the difference between a histogram and frequency Polygons?

Ans: The frequency histogram is analogous to a column graph, even though there are no gaps between the columns. A frequency polygon is a type of line graph that is used in statistics. These graphs can be created separately or in combination. These graphs may be created utilizing data from a frequency distribution table.

6. Can we use Bar Graphs for Grouped data?

Ans: Yes we can use bar graphs for grouped data.

7. Which graph is used for the representation of two or more sectional tables?

Ans: We generally use frequency polygon for these types of problems of joining or comparison

8. What is the use of the range of data?

Ans: Range of data gives the numerical value of (Max value  of data – Min value of data) which

Will give an idea of variations in data given in port.

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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

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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) Option 2) Option 3) Option 4)

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

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, 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