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NCERT Solutions For Class 9 Maths Chapter 12 Statistics - 2025-26

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Statistics - Exercise-wise Questions and Answers For Class 9 Maths - Free PDF Download

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 examination’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 - 2025-26
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Exercises Under NCERT Solutions for Class 9 Maths Chapter 12 Statistics

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 – 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 women’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 – 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 ‘A’ 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 ‘A’ is better than the students of section ‘B’ 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 - 2025-26

1. How can I find reliable, step-by-step NCERT Solutions for Class 9 Maths Chapter 12, Statistics?

Vedantu provides comprehensive NCERT Solutions for Class 9 Maths Chapter 12, created by expert teachers. These solutions strictly follow the CBSE 2025-26 syllabus and marking guidelines. Each answer offers a detailed, step-by-step method to help you master the correct procedures for solving problems related to data presentation and measures of central tendency.

2. What is the correct method for calculating the mean of ungrouped data as per the NCERT Class 9 textbook?

To find the mean of ungrouped data in NCERT exercises, you should follow this exact procedure:
1. Add up all the observations or data points given in the problem. This sum is denoted by Σxᵢ.
2. Count the total number of observations, which is denoted by 'n'.
3. Divide the sum of observations by the total number of observations. The formula is: Mean (x̄) = (Σxᵢ) / n. Following these steps ensures you get the correct answer as per the CBSE pattern.

3. How do you solve questions asking to construct a histogram for data with continuous class intervals in Chapter 12?

To correctly construct a histogram for continuous data as per the NCERT solutions, you must follow these steps:
1. Represent the class intervals on the horizontal axis (x-axis) and the frequencies on the vertical axis (y-axis).
2. Choose an appropriate scale for both axes to fit the data.
3. For each class interval, draw a rectangle with the interval as the base and the corresponding frequency as its height.
4. Ensure there are no gaps between consecutive rectangles, as this visually represents the continuous nature of the data.

4. What is the crucial first step if an NCERT problem provides discontinuous class intervals for constructing a histogram?

If the class intervals in a problem are discontinuous (e.g., 1-5, 6-10), you must first make them continuous before drawing a histogram. This is a critical step for an accurate solution.
1. Find the gap between the upper limit of one class and the lower limit of the next (e.g., 6 - 5 = 1).
2. Calculate half of this gap (1 ÷ 2 = 0.5). This is the adjustment factor.
3. Subtract the adjustment factor from all lower limits and add it to all upper limits to create new, continuous class intervals (e.g., 0.5-5.5, 5.5-10.5). You can now proceed with drawing the histogram.

5. How do you find the median and mode for an ungrouped dataset in Class 9 Maths Chapter 12 exercises?

The correct method for finding the median and mode of ungrouped data is as follows:

  • For the Median: First, arrange all the data points in ascending order. If the number of observations (n) is odd, the median is the ((n+1)/2)th term. If n is even, the median is the average of the (n/2)th and ((n/2)+1)th terms.
  • For the Mode: The mode is the observation that appears most frequently in the dataset. It is found by simply identifying the data point with the highest frequency.

6. What is the key difference when solving a question that requires a bar graph versus one that needs a histogram?

The correct method depends on the type of data given in the NCERT problem.

  • A bar graph is the correct solution for representing discrete or non-continuous data (e.g., favourite sports, months of birth). There must be gaps between the bars to show they represent separate, distinct categories.
  • A histogram is the correct solution for representing continuous data presented in class intervals (e.g., height, weight, marks). There should be no gaps between the bars, indicating a continuous data range.

7. What is the step-by-step process for solving a question that asks to draw a frequency polygon?

To correctly draw a frequency polygon as per the NCERT Chapter 12 method, follow these steps:
1. Calculate the class mark for each class interval using the formula: Class Mark = (Upper Limit + Lower Limit) / 2.
2. Plot these class marks on the x-axis and their corresponding frequencies on the y-axis.
3. Join the plotted points sequentially using straight line segments.
4. To complete the polygon, join the first and last points to the x-axis by plotting points for two imaginary classes (one before the first and one after the last) with a frequency of zero.

8. Why is it important to master the step-by-step solutions for NCERT Class 9 Statistics for future classes?

Mastering the solutions in Class 9 Statistics is crucial as it builds the foundation for more advanced topics in Class 10 Maths and beyond. The methods for data presentation (histograms, frequency polygons) and calculating measures of central tendency for ungrouped data are essential prerequisites for understanding grouped data calculations, cumulative frequency, and probability, which are key topics in higher grades and competitive exams.