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NCERT Solutions For Class 9 Maths Chapter 12 Statistics

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Find Complete Statistics Class 9 Questions And Answers With Detailed Solutions

NCERT Solutions for Maths Class 9 Chapter 12 Statistics is one of the most important chapters that carries good marks, so learning this chapter correctly using Chapter 12 Statistics Class 9 NCERT Solutions is necessary. Class 9 Maths Chapter 14 Statistics is thus highly important from the examinationโ€™s perspective. The chapter covers data from the statistical aspect, grouping data, presenting data in a specific manner, the idea of primary and secondary data, and developing grouped frequency distribution tables.

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Get ready to unlock patterns, compare data sets, and better understand the world around you! Class 9 Maths Ch 12 PDF Solutions is provided below. The NCERT solutions provide step-by-step explanations, making it easier to grasp complex topics. By practising Class 9 Maths Ch 12 PDF, students can confidently strengthen their analytical skills and prepare for exams.


Access Exercise wise NCERT Solutions for Class 9 Maths Chapter 12

Current Syllabus Exercises of Class 9 Maths Chapter 12

NCERT Solutions of Class 9 Maths Statistics Exercise 12.1

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NCERT Solutions For Class 9 Maths Chapter 12 Statistics
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STATISTICS in One Shot (Complete Chapter) CBSE Class 9 Math Chapter 14 [Term 1 Exam] NCERT Vedantu
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Find Complete Statistics Class 9 Questions And Answers With Detailed Solutions

NCERT Solutions for Class 9 Maths Chapter 12, "Statistics," deals with the concepts related to the collection, organization, analysis, interpretation, and presentation of data. The chapter consists of the following exercises:


Exercise 12.1: Exercise 12.1 covers the graphical representation of data using histograms and frequency polygons. Students learn to construct histograms by dividing data into intervals and plotting frequencies as bars, which visually display data distribution. Frequency polygons involve plotting midpoints of intervals and connecting them with lines, aiding in data comparison and trend analysis. This exercise enhances students' skills in data visualization and interpretation.

Access NCERT Solutions for Class 9 Maths Chapter 12 โ€“ Statistics

Exercise  12.1

1. A survey conducted by an organisation for the cause of illness and death among the women between the ages \[15 - 44\] (in years) worldwide, found the following figures (in %)


S.No.

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


i. Represent the information given above graphically.

Ans: The graph of the information presented above can be produced as follows by depicting causes on the x-axis and family fatality rate on the y-axis, and selecting an acceptable scale (1 unit = 5% for the y axis).

A bar graph on the x-axis and family fatality rate on the y-axis


All the rectangle bars are of the same width and have equal spacing between them.


ii. Which condition is the major cause of womenโ€™s ill health and death worldwide?

Ans: Reproductive health issues are the leading cause of women's illness and mortality globally, affecting 31.8% of women.


iii. 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

Ans: The factors are as follows:

a. Lack of medical facilities

b. Lack of correct knowledge of treatment


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

Section 

Number of Girls Per Thousand Boys

Scheduled caste (SC)

Scheduled tribe (ST)

Non SC/ST

Backward districts

Non โ€“ backward districts

Rural

Urban 

940

970

920

950

920

930

910


i. Represent the information above by a bar graph.

Ans: The graph of the information presented above may be built by choosing an appropriate scale (1 unit = 100 girls for the y-axis) and representing section (variable) on the x-axis and number of girls per thousand boys on the y-axis.

A bar graph


Here, all the rectangle bars are of the same length and have equal spacing in between them.


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

Ans: The largest number of females per thousand boys (i.e., 970) is found in ST, while the lowest number of girls per thousand boys (i.e., 910) is found in urban areas.

In addition, the number of females per thousand boys is higher in rural regions than in cities, in backward districts than in non-backward districts, and in SC and ST districts than in non-SC/ST districts.


3. Given below are the seats won by different political parties in the polling outcome of a state assembly elections:


Political party

A

B

C

D

E

F

Seats won

75

55

37

29

10

37


i. Draw a bar graph to represent the polling results.

Ans:

A bar graph to represent the polling results


Here, all the rectangle bars are of the same length and have equal spacing in between them.

ii. Which political party won the maximum number of seats?

Ans: From the above graph it is clear that Political party โ€˜Aโ€™ won the maximum number of seats.


4. The length of\[40\] leaves of a plant are measured correct to one millimeter, and the obtained data is represented in the following table:


Length (in mm)

Number of Leaves

117.5-126.5

3

126.5-135.5

5

135.5-144.5

9

144.5-153.5

12

135.5-162.5

5

162.5-171.5

4

171.5-180.5

2


i. Draw a histogram to represent the given data.

Ans: The length of leaves is represented in a discontinuous class interval with a difference of \[1\] between them, as can be seen. To make the class intervals continuous, \[\dfrac{1}{2} = 0.5\] must be added to each upper class limit and \[0.5\] must be subtracted from the lower class limits.


Length (in mm)

Number of Leaves

117.5-126.5

3

126.5-135.5

5

135.5-144.5

9

144.5-153.5

12

135.5-162.5

5

162.5-171.5

4

171.5-180.5

2


Histogram built using the length of leaves on the x-axis and the number of leaves on the y-axis


The above histogram may be built using the length of leaves on the x-axis and the number of leaves on the y-axis.

On the y-axis, one unit symbolises two leaves.

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

Ans: Frequency polygon is another good graphical representation of this data.


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

Ans: No, because the maximum number of leaves (i.e.\[12\]) has a length of \[144.5{\text{mm}}\] to \[153.5{\text{mm}}\] It is not necessary for all of them to be \[153{\text{mm}}\]long.


5. The following table gives the life times of neon lamps: 

\[\boxed{\begin{array}{*{20}{c}}  {{\text{ Length (in hours) }}}&{{\text{ Number of lamps}}} \\   {300 - 400}&{14} \\   {400 - 500}&{56} \\   {500 - 600}&{60} \\   {600 - 700}&{86} \\   {700 - 800}&{74} \\   {800 - 900}&{62} \\   {900 - 1000}&{48} \\   {}&{} \end{array}}\]

i. Represent the given information with the help of a histogram.

Ans: The histogram of the given data may be produced by plotting the life duration (in hours) of neon lamps on the x-axis and the number of lamps on the y-axis. Here,1

The histogram of the given data by plotting the life duration (in hours) of neon lamps on the x-axis and the number of lamps on the y-axis


Here, 1 unit on the y-axis represents 10 lamps.


ii. How many lamps have a lifetime of more than \[700\] hours?

Ans: It may be deduced that the number of neon lamps with a lifetime more than \[700\]is equal to the sum of the numbers of neon lamps with lifetimes of \[700,800\]and \[900\]. As a result, there are \[184\] neon bulbs with a lifetime of more than \[700\] hours \[(74 + 62 + 48 = 184)\].


6. The following table gives the distribution of students of two sections according to the mark obtained by them:

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.

Ans: We can find the class marks of the given class intervals by using the following formula.

\[{\text{Class mark  = }}\dfrac{{{\text{Upper class limit  +  Lower class limit}}}}{2}\]

Section A

Section B

Marks

Class marks

Frequency 

Marks 

Class Marks

Frequency 

0-10

5

3

0-10

5

5

10-20

15

9

10-20

15

19

20-30

25

17

20-30

25

15

30-40

35

12

30-40

35

10

40-50

45

9

40-50

45

1


The frequency polygon can be constructed as follows, with class markings on the x-axis and frequency on the y-axis, and an appropriate scale \[(1{\text{ unit  =  3 for the y - axis}})\].


The frequency polygon


It can be observed that the performance of students of section โ€˜Aโ€™ is better than the students of section โ€˜Bโ€™ in terms of good marks.


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

\[\boxed{\begin{array}{*{20}{c}} {{\text{ }}\underline {{\text{Number of balls}}} {\text{ }}}&{{\text{ }}\underline {{\text{Class mark}}} {\text{ }}}&{{\text{ }}\underline {{\text{Team A}}} {\text{ }}}&{{\text{ }}\underline {{\text{Team B}}} {\text{ }}} \\   {0.5 - 6.5}&{3.5}&2&5 \\   {6.5 - 12.5}&{9.5}&1&6 \\   {12.5 - 18.5}&{15.5}&8&2 \\   {18.5 - 24.5}&{21.5}&9&{10} \\   {24.5 - 30.5}&{27.5}&4&5 \\  {30.5 - 36.5}&{33.5}&5&6 \\   {36.5 - 42.5}&{39.5}&6&3 \\  {42.5 - 48.5}&{45.5}&{10}&4 \\   {48.5 - 54.5}&{51.5}&6&8 \\  {54.5 - 60.5}&{57.5}&2&{10} \\   {}&{}&{}&{} \end{array}}\]

Represent the data of both the teams on the same graph by frequency polygons. 

(Hint: First make the class intervals continuous.)

Ans: A frequency polygon can be created by plotting class grades on the x-axis and running times on the y-axis.

A frequency polygon can be created by plotting class grades on the x-axis and running times on the y-axis


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

\[\boxed{\begin{array}{*{0}{c}}  {\underline {{\text{Age (in years)}}} }&{\underline {{\text{Number of children }}} }      \\   {{\text{1 - 2}}}&{\text{5}} \\   {{\text{2 - 3}}}&{\text{3}} \\   {{\text{3 - 5}}}&{\text{6}} \\ {{\text{5 - 7}}}&{{\text{12}}} \\  {{\text{7 - 10}}}&{\text{9}} \\   {{\text{10 - 15}}}&{{\text{10}}} \\   {{\text{15 - 17}}}&{\text{4}}    \end{array}}\]

Draw a histogram to represent the data above.

Ans:

Age (in years)

Frequency (Number of Children)

Width of Class

Length of Rectangle

1-2

5

1

\[\dfrac{{5 \times 1}}{1} = 5\]

2-3

3

1

\[\dfrac{{3 \times 1}}{1} = 3\]

3-5

6

2

\[\dfrac{{6 \times 1}}{2} = 3\]

5-7

12

2

\[\dfrac{{12 \times 1}}{2} = 6\]

7-10

9

3

\[\dfrac{{9 \times 1}}{3} = 3\]

10-15

10

5

\[\dfrac{{10 \times 1}}{5} = 2\]

15-17

4

2

\[\dfrac{{4 \times 1}}{2} = 2\]


A histogram


9. \[100\] 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 Letters

Number of Surnames

1-4

4-6

6-8

8-12

12-20

6

30

44

16

4


i. Draw a histogram to depict the given information.

Ans:


Number of Letters

Frequency (Number of Surnames)

Width of Class

Length of Rectangle

1-4

6

3

\[\dfrac{{6 \times 2}}{3} = 4\]

4-6

30

2

\[\dfrac{{30 \times 2}}{2} = 30\]

6-8

44

2

\[\dfrac{{44 \times 2}}{2} = 44\]

8-12

16

4

\[\dfrac{{16 \times 2}}{4} = 8\]

12-20

4

8

\[\dfrac{{4 \times 2}}{8} = 1\]


The histogram can be generated using the number of letters on the x-axis and the fraction of the number of surnames per 2 letters interval on the y-axis, as well as an acceptable scale (1 unit = 4 students for the y axis).

the number of letters on the x-axis and the fraction of the number of surnames per 2 letters interval on the y-axis, as well as an acceptable scale


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

Ans: The maximum number of surnames in the class interval is 6-8 since it contains 44 surnames, which is the maximum for this data.


Overview of Deleted Syllabus for CBSE Class 9 Maths Statistics

Chapter

Dropped Topics

Statistics

12.1 Introduction

12.2 Collection of data

12.3 Presentation of data

12.5 Measure of central tendency

12.6 Summary


Class 9 Maths Chapter 12: Exercises Breakdown

Exercise

Number of Questions

Exercise 12.1

9 Questions & Solutions


Conclusion

Maths NCERT Class 9 Statistics Chapter has equipped you with valuable tools to transform raw data into meaningful insights. You've explored various methods for data collection and delved into the power of visual representations like bar graphs and histograms. By organizing data into frequency distributions, you've learned to summarize large datasets effectively. Furthermore, you've grasped the significance of central tendency measures (mean, median, mode) in understanding the "typical" value within a data set. In previous years exams, around 2-3 questions have been asked from Class 9 Maths Ch Statistics.


Other Study Material for CBSE Class 9 Maths Chapter 12


Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Study Materials for CBSE Class 9 Maths

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FAQs on NCERT Solutions For Class 9 Maths Chapter 12 Statistics

1. Is a histogram the same as a bar graph in statistics?

No, a histogram is not the same as a bar graph, although both use rectangular bars to display data. The key difference lies in the type of data they represent and the presence of gaps between the bars.

This confusion often arises because they look visually similar. However, a histogram is used to represent the frequency distribution of continuous data, like the heights of students. Its bars are drawn adjacent to each other without any gaps. In contrast, a bar graph is used for discrete or categorical data, such as the number of students who prefer different fruits, and always has gaps between the bars to show that the categories are separate.

A quick check is to look for gaps: if there are no gaps between the bars, it is a histogram representing a continuous range of data.

Therefore, a histogram shows distribution over intervals, while a bar graph compares distinct categories.

2. Is the range a measure of central tendency like the mean?

No, the range is a measure of spread or dispersion, not central tendency. It tells you how spread out the data is by showing the difference between the highest and lowest values, unlike the mean, which identifies the central point of the data.

3. Do I have to pay to get the statistics class 9 questions and answers PDF?

No, you do not have to pay for the solutions. Reputable educational platforms like Vedantu provide a Free PDF download of the statistics class 9 ncert solutions, making them accessible for all students to use for practice and revision without any cost.

4. Do NCERT Solutions for Class 9 Maths Chapter 12 only provide the final answers?

NCERT Solutions provide complete, step-by-step explanations for every problem, not just the final answer. Many students mistakenly think these are just answer keys, but they are designed as learning tools that guide you through the entire problem-solving process from start to finish.

For example, when asked to construct a frequency distribution table, the solution will show how to determine the range, decide on class intervals, and use tally marks to accurately count frequencies. This detailed approach is crucial for understanding the logic behind the final table, not just copying it.

Following each step helps you pinpoint exactly where you might be making a mistake, whether in calculation or in applying a statistical rule.

This method ensures you learn the 'how' and 'why' behind each answer.

5. Can a dataset have more than one mode?

Yes, a dataset can absolutely have more than one mode. If two or more values appear with the same highest frequency, the dataset is called bimodal (two modes) or multimodal (more than two modes). A dataset can also have no mode if all values appear only once.

6. Are the class 9 maths chapter 12 ncert solutions only useful for checking homework?

No, the class 9 maths chapter 12 ncert solutions are valuable tools for complete exam preparation, not just for checking homework. The common myth is that their only purpose is to verify answers after you've finished an assignment.

In reality, these solutions are structured to build conceptual clarity. They explain the correct application of formulas for mean and mode and demonstrate the proper method for drawing histograms and frequency polygons, which are common formats in `statistics questions and answers cbse` exams. By studying these methods, you learn how to present your answers in a way that scores full marks.

Think of them as a guide that shows you the standard, expected method for solving problems, helping you align your practice with exam expectations.

Using them for regular practice builds confidence for the final exams.

7. Is primary data always better to use than secondary data?

No, primary data is not always better than secondary data; the best choice depends on the specific investigation's needs. Students often assume that firsthand data (primary) is more reliable, but collecting it can be expensive, time-consuming, and sometimes impossible.

For instance, if you need to study the population growth of India over the last 50 years, collecting this data yourself is not feasible. You would rely on secondary data from a reliable source like the Census of India. However, if you want to know the favourite sport among students in your class, collecting primary data through a direct survey is more efficient and appropriate.

The key is to evaluate the requirements of your study, including its scope, the time available, and the budget, to decide which data type is suitable.

8. Does the NCERT Statistics Class 9 PDF only contain solutions for the main exercise?

No, a good Statistics Class 9 PDF with solutions typically covers more than just the final exercise problems. The misconception is that it's only a key for end-of-chapter questions like the `Ex 12.1 Class 9 Statistics solutions`.

In fact, detailed solution PDFs also include explanations for the in-text questions and the solved examples presented within the chapter. This ensures you understand concepts as they are introduced, before you attempt the main exercises. For example, it will break down the steps for creating a grouped frequency table from raw data shown in a chapter example.

This comprehensive coverage allows for a much smoother learning curve, as you can practice and verify your understanding at every stage of the chapter.

It helps build a strong foundation before you even reach the final exercise.

9. Is it enough to just memorize the steps from the ncert solution class 9 maths chapter 12?

No, simply memorizing steps is not an effective learning strategy. The purpose of the ncert solution class 9 maths chapter 12 is to help you understand the logic behind each step, so you can apply the same concepts to solve different, unseen problems.

10. Is the class mark just the average of any two numbers in a class interval?

No, the class mark is specifically the average of the upper and lower class limits of a single interval. It is not the average of any random numbers. The formula is fixed: Class Mark = (Upper Class Limit + Lower Class Limit) / 2. It represents the precise midpoint of that class.