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# NCERT Solutions for Class 9 Maths Chapter 12 Statistics

Last updated date: 20th Sep 2024
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## NCERT Solutions for Maths Class 9 Chapter 12 Statistics - 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 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.

Table of Content
1. NCERT Solutions for Maths Class 9 Chapter 12 Statistics - FREE PDF Download
2. Glance on Class 9 Maths Chapter 12 - Statistics
3. Access Exercise wise NCERT Solutions for Class 9 Maths Chapter 12
4. Exercises Under NCERT Solutions for Class 9 Maths Chapter 12 Statistics
5. Access NCERT Solutions for Class 9 Maths Chapter 12 โ Statistics
6. Overview of Deleted Syllabus for CBSE Class 9 Maths Statistics
7. Class 9 Maths Chapter 12: Exercises Breakdown
8. Other Study Material for CBSE Class 9 Maths Chapter 12
9. Chapter-Specific NCERT Solutions for Class 9 Maths
FAQs

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.

## Glance on Class 9 Maths Chapter 12 - Statistics

• Understanding the concept of data and its different types (numerical, categorical).

• Exploring methods for collecting data like surveys, experiments, and observations.

• Mastering how to represent data visually using Data Representation:

• Bar Graphs: Illustrating comparisons between categories.

• Histograms: Depicting the frequency distribution of numerical data.

• Organizing data into categories and calculating their frequencies (how often each category appears).

• Measures of Central Tendency:

• Mean (average): The sum of all values divided by the number of values.

• Median: The middle value when data is arranged in ascending or descending order.

• Mode: The most frequent value in the data set.

• There is only one exercise in Class 9th Maths Chapter 12 Statistics and 9 fully solved questions.

## 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
Competitive Exams after 12th Science
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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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## 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).

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/STBackward districtsNon โ backward districtsRuralUrban 940970920950920930910

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.

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:

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

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

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 $(1{\text{ unit = 3 for the y - axis}})$.

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.

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 $\dfrac{{4 \times 1}}{2} = 2$

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 1-44-66-88-1212-20 63044164

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).

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.

## 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.

## FAQs on NCERT Solutions for Class 9 Maths Chapter 12 Statistics

1. How can we find the median of grouped data?

From the class intervals and the corresponding frequency, write the cumulative frequency(cf) in the third column. On finding n/2, locate the median class whose cumulative frequency is greater than and nearest to n/2.

Now Median = l +ย  (((n/2) - cf) / f) h

Where
l = lower limit of the median class

n = no. of observations

cf = cumulative frequency of the class preceding the median class

f = frequency of the median class

h = class size ( where class sizes are equal)

2. What is the relationship between mean, median and mode of a data set?

In the case of any data set, the mean, median and mode are related by the following equation:

Mode = 3 Median - 2 Mean

Any problem on central tendency, that is mean, median and mode can be solved using this formula or a spin-off from the same.

3. Give five examples of data that you can collect from your day-to-day life.

Five examples from day-to-day life:

• Number of students in our class

• Number of fans in our school

• Electricity bills for our house for the last two years

• Election results obtained from television or newspapers

• Literacy rate figures obtained from the Educational Survey

4. What are the three measures of central tendency for ungrouped data?

The three measures of central tendency are Mean, Median and mode.

5. When are frequency polygons used?

Frequency polygons are used when data is very large and continuous.ย  It is generally used to compare two different sets of data of the same nature. For example, comparing the performance of two different sections of the same class.

6. What is the main focus of 12th Ch Statistics Class 9?

The 12th Ch Statistics Class 9 is centered on cumulative frequency distributions. It teaches students how to construct cumulative frequency tables and graphs, which help in understanding data trends and distributions more comprehensively.

7. How do cumulative frequency graphs help in data analysis in Chapter 12 statistics class 9?

Cumulative frequency graphs, or ogives, assist in visualizing the running total of frequencies. They are useful for determining the median, quartiles, and percentiles, making it easier to analyze the spread and concentration of data points.

8. What are the key skills students develop in class 9th statistics?

In class 9th statistics , students develop skills in creating cumulative frequency tables, drawing ogives, and interpreting these graphs. These skills are essential for summarizing large data sets and making informed statistical inferences.