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NCERT Solutions for Class 9 Maths Chapter 12 Statistics

NCERT Solutions for Class 9 Maths Chapter 12 Statistics

Edited By Komal Miglani | Updated on Apr 15, 2025 01:30 AM IST

Statistics is one of the important topics in the Class 9 NCERT syllabus. Statistics is a powerful branch of mathematics that helps us systematically collect, organize, and interpret data, allowing us to analyze large amounts of information in a clear and insightful manner. For instance, we will use statistics to know the average marks of a class and the maximum average age of college students. The main parts of statistics include data collection, data presentation (like tables and graphs), and data analysis (like finding the mean, median, and mode). It is beneficial in everyday life and plenty of fields. These NCERT solutions are created by the expert team at Careers360, keeping in mind the latest CBSE syllabus. In statistics class 9, the numbers or facts are collected for a purpose, and the collection is called data.

This Story also Contains
  1. Statistics Class 9 Questions And Answers PDF Free Download
  2. Statistics Class 9 Solutions - Important Formulae
  3. Key Features
  4. NCERT Solutions for class 9 Mathematics: Chapter-wise
  5. NCERT solutions for class 9 - Subject Wise
NCERT Solutions for Class 9 Maths Chapter 12 Statistics
NCERT Solutions for Class 9 Maths Chapter 12 Statistics

NCERT solutions for class 9 Maths chapter 12 Statistics covers the solutions for this particular topic in detail. This article on NCERT solutions for class 9 Maths Chapter 12 statistics offers clear and step-by-step solutions for the exercise problems in the NCERT Books for class 9 Maths. Students who need Statistics Class 9 solutions will find this article very useful. It covers all the important Class 9 Maths Chapter 12 question answers. NCERT solutions for class 9 maths and other subjects and classes can be downloaded from NCERT Solutions.

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Statistics Class 9 Questions And Answers PDF Free Download

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Statistics Class 9 Solutions - Important Formulae

Class Mark = (Lower Limit + Upper Limit) / 2

Measures of Central Tendency:

Mean (x̄): The mean is calculated as the sum of all observations divided by the total number of observations.

  • Mean (x̄) = Sum of all observations (∑xn) / Total Number of observations (N)

Median: The median is the middle value in a data set when the observations are arranged in ascending or descending order.

  • For an even number of observations, the median is the average of the two middlemost observations.

  • For an odd number of observations, the median is the value of the ((n+1)/2)-th observation.

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Mode: The mode is the observation that occurs most frequently or has the maximum frequency in the given data set.


Class 9 Maths Chapter 12 Question Answer: Exercise: 12.1
Total Questions: 9
Page number: 163-166

Q1 (i) A survey conducted by an organization for the cause of illness and death among women between the ages 15 - 44 (in years) worldwide, found the following figures (in o/o ):

Serial NumberCausesFemale fatality rate (%)
1.Reproductive health conditions31.8
2.Neuropsychiatric conditions25.4
3.Injuries12.4
4.Cardiovascular conditions4.3
5.Respiratory conditions4.1
6.Other causes22.0

Represent the information given above graphically

Answer:

The graphical representation of the given data is as follows

1640338415730

Q1 (ii) A survey conducted by an organization for the cause of illness and death among women between the ages 15 - 44 (in years) worldwide, found the following figures (in o/o ) :

Serial NumberCausesFemale fatality rate (%)
1.Reproductive health conditions31.8
2.Neuropsychiatric conditions25.4
3.Injuries12.4
4.Cardiovascular conditions4.3
5.Respiratory conditions4.1
6.Other causes22.0

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

Answer:

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

Q1 (iii) A survey conducted by an organization for the cause of illness and death among women between the ages 15 - 44 (in years) worldwide, found the following figures (in o/o ):

Serial NumberCausesFemale fatality rate (%)
1.Reproductive health conditions31.8
2.Neuropsychiatric conditions25.4
3.Injuries12.4
4.Cardiovascular conditions4.3
5.Respiratory conditions4.1
6.Other causes22.0

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

Answer:

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

Q2 (i) The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.

SectionNumber of girls per thousand boys
Schedule Caste (SC)940
Schedule Tribe (ST)970
Non SC/ST920
Backward districts950
Non-backward districts920
Rural930
Urban910

Represent the information above with a bar graph.

Answer:

The graphical representation of the given information is as follows.

1640338443736

Q2 (ii) The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.

SectionNumber of girls per thousand boys
Schedule Caste (SC)940
Schedule Tribe (ST)970
Non SC/ST920
Backward districts950
Non-backward districts920
Rural930
Urban910

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

Answer:

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

Q3 (i) Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

Political PartyABCDEF
Seats Won755537291037

Draw a bar graph to represent the polling results.

Answer:

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

1640338768092

Q3 (ii) Given below are the seats won by different political parties in the polling outcome of state assembly elections:

Political PartyABCDEF
Seats Won755537291037

Which political party won the maximum number of seats?

Answer:

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

Q4 (i) The length of 40 leaves of a plant is measured correctly by one millimeter, and the obtained data is represented in the following table:

Length (in mm)Number of leaves
118-1263
127-1355
136-1449
145-15312
154-1625
163-1714
172-1802

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

Answer:

As we can see from the given table the data is discontinuous 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

1640338796583

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

1640338809916

Q4 (ii) The length of 40 leaves of a plant is measured correctly by one millimeter, and the obtained data is represented in the following table:

Length (in mm)Number of leaves
118-1263
127-1355
136-1449
145-15312
154-1625
163-1714
172-1802

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

Answer:

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

Q4 (iii) The length of 40 leaves of a plant is measured correctly by one millimeter, and the obtained data is represented in the following table:

Length (in mm)Number of leaves
118-1263
127-1355
136-1449
145-15312
154-1625
163-1714
172-1802

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

Answer:

No, it is certainly not correct to conclude that the maximum number of leaves is 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.

Q5 (i) The following table gives the lifetimes of 400 neon lamps:

Lifetime (in hours)Number of lamps
300-40014
400-50056
500-60060
600-70086
700-80074
800-90062
900-100048

Represent the given information with the help of a histogram.

Answer:

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

1640338837985

Q5 (ii) The following table gives the lifetimes of 400 neon lamps:

LiLifetimein hours)Number of lamps
300-40014
400-50056
500-60060
600-70086
700-80074
800-90062
900-100048

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

Answer:

Lamps having a lifetime in the range of 700 - 800 = 74

Lamps having a lifetime in the range of 800 - 900 = 62

Lamps having a lifetime in the range 900 - 1000 = 48

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

Q6 The following table gives the distribution of students in two sections according to the marks obtained by them:

Section A
Section B
MarksFrequencyMarksFrequency
0-1030-105
10-20910-2019
20-301720-3015
30-401230-4010
40-50940-501

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

Answer:

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

Class marks=Upper limit of class interval+Lower limit of class interval2

1640338876105

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 a frequency polygon is as follows.

1640338889361

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

Q7 The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Number of ballsTeam ATeam B
1-625
7-1216
13-1882
19-24910
25-3045
31-3656
37-4263
43-48104
49-5468
55-60210

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

Answer:

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

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

Class marks=Upper limit of class interval+Lower limit of class interval2

1640338984649

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 a frequency polygon is as follows.

1640339009651

Q8 A random survey of the number of children of various age groups playing in a park was found as follows:

Age (in years)Number of children
1-25
2-33
3-56
5-712
7-109
10-1510
15-174

Draw a histogram to represent the data above.

Answer:

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=Minimum class sizeClass size of the interval×FrequencyLength of the rectangle=Minimum widthWidth of the rectangle×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.

1640339044515

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

1640339059729

Q9 (i) 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

Number of lettersNumber of surnames
1-46
4-630
6-844
8-1216
12-204

Answer:

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=Minimum class sizeClass size of the interval×FrequencyLength of the rectangle=Minimum widthWidth of the rectangle×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.

1640339087970

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

1640339102697

Q9 (ii) 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

Number of lettersNumber of surnames
1-46
4-630
6-844
8-1216
12-204

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

Answer:

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.

Key Features

Class 9 Maths Chapter 12 questions answer offers several key features that make them an excellent resource for students. Some of the important features of these solutions include:

Comprehensive coverage: NCERT Solutions for Maths Chapter 12 class 9 covers all the important topics related to Statistics, such as measures of central tendency, dispersion, and graphical representation of data.

Easy to understand: The statistics class 9 solutions are written clearly and concisely, making them easy for students to understand. They are presented in a step-by-step manner, which helps students to grasp the concepts better.

Helpful tips and tricks: Class 9 maths provides helpful tips and tricks to solve problems more efficiently, saving time and effort.

NCERT Solutions for class 9 Mathematics: Chapter-wise

NCERT solutions for class 9 - Subject Wise

Here are the subject-wise links for the NCERT solutions of class 9:

Also, check NCERT Books and NCERT Syllabus here:

Given below are some useful links for NCERT books and the NCERT syllabus for class 9:

Frequently Asked Questions (FAQs)

1. What are the important topics in class 9 maths statistics ?

Mean, median, mode, variance, and standard deviation are the important topics of this NCERT Book chapter. having command in the concepts of statistics will help students not only in exams but in understanding data in daily life and professional life as well. for ease students can study statistics class 9 pdf both online and offline mode.

2. How does the NCERT solutions are helpful ?

NCERT maths chapter 14 class 9 solutions are helpful for the students if they stuck while solving NCERT problems. Also, these solutions are provided in a very detailed manner which will give them conceptual clarity. practicing these NCERT solutions provide you indepth understanding of concepts that leads to confidence in exan and ultimately you will score more marks in the exam.

3. Where can I find the complete solutions of NCERT for class 9 maths ?

Here you will get the detailed NCERT solutions for class 9 maths by clicking on the link. Practicing these NCERT class 9 maths chapter 14 solutions are important for exam and indepth understanding of the concepts

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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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