# NCERT Solutions for Class 9 Maths Chapter 14 Statistics

NCERT solutions for class 9 maths chapter 14 Statistics: Statistics is one of the important topics under class 9 NCERT syllabus. You must have seen information like the temperature of cities, cricket batsman or bowler ranking, results of particular exams either in news, tv, online sites or magazines. All the mentioned examples are represented through a table or graphs. CBSE NCERT solutions for class 9 maths chapter 14 Statistics is covering the solutions for this particular topic in detail. In statistics, the numbers or facts are collected for a purpose and the collection is called data. From these collected data you can understand and conclude some results. For example, if we have data of temperature of 40 cities for a particular day, we can infer that which city has a higher temperature, which city has lower temperature, what is the average temperature of 40 cities, which value of temperature is most repeated, what is the middle value of temperature if we arranged the temperature in either increasing or decreasing order etc. In this, you are arranging the collected data and then  analyze it and can infer certain things. In solutions of NCERT for class 9 maths chapter 14 Statistics the answers to the questions of mode, average, median, etc are covered. Statistics is the part of maths in which you learn to get some results based on the collected data. In this chapter, there are a total of 4 exercises consist of total of 35 questions. NCERT solutions for class 9 maths chapter 14 Statistics is covering every question is a step by step manner so that you do not lose a single mark in any question. Apart from this particular chapter NCERT solutions are available for other chapters as well.

## NCERT solutions for class 9 maths chapter 14 Statistics Excercise: 14.1

Five examples of data that we can collect in our daily life are

(i) Number of students in a class.

(ii) The number of books in a library.

(iii) Toys sold on a particular day at a shop.

(iv) People who voted for a particular candidate.

(v) Runs scored by a batsman on each ball in a particular evening.

(i) The number of students in a class.

(ii) The number of books in a library.

(iii) Toys sold on a particular day at a shop.

(iv) People who voted for a particular candidate.

(v) Runs scored by a batsman on each ball on a particular evening.

All of the data in Q.1 is primary data.

## NCERT solutions for class 9 maths chapter 14 Statistics Excercise: 14.2

A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,

A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.

Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?

The representation of the given data in the form of a frequency distribution table is as follows.

From the table we can see that O is the most common and AB is the rarest blood group.

5         3         10         20         25         11         13         7         12         31

19        10       12        17         18         11          32        17         16         2

7          9         7          8            3          5           12         15         18         3

12        14       2           9           6         15          15           7          6          12

Construct a grouped frequency distribution table with class size $5$ for the data given above taking the first interval as $0-5$ ($5$ not included). What main features do you observe from this tabular representation?

As the minimum and maximum distances of an engineer from his place of work is 2 and 32 respectively the class intervals with class size 5 would be the following.

0 - 5, 5 - 10, 10 - 15, 15 - 20, 20 - 25, 25 - 30, 30 - 35

The representation of the given data in the form of a grouped frequency distribution table is as follows

Frequencies of the class intervals 5 - 10 and 10 - 15 are maximum and equal to 11 each and frequencies of the class intervals 20 - 25 and 125 - 30 are minimum and equal to 1 each.

3.14159265358979323846264338327950288419716939937510

Make a frequency distribution of the digits from 0 to 9 after the decimal point.

The representation of the given data in the form of a frequency distribution table is as follows.

3.14159265358979323846264338327950288419716939937510

What are the most and the least frequently occurring digits?

The most frequently occurring digits are 3 and 9 with a frequency of 8.

## NCERT solutions for class 9 maths chapter 14 Statistics Excercise: 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 life time 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.

## NCERT solutions for class 9 maths chapter 14 Statistics Excercise: 14.4

2,         3,         4,         5,         0,         1,         3,         3,         4,         3

Find the mean, median and mode of these scores.

$Mean=\frac{Sum\ of\ observations}{Number \ of\ observations}$

Number of observations, n = 10

$\\\bar{X}=\frac{\sum_{i=1}^{n=10}x_{i}}{n}\\ \bar{X}=\frac{ 2+ 3+ 4+ 5+ 0+ 1+ 3+ 3+ 4+ 3}{10}\\ \bar{X}=\frac{28}{10}\\ \bar{X}=2.8$

Mean is 2.8

To find the median we have to arrange the given data in ascending order as follows:

0, 1, 2, 3, 3, 3, 3, 4, 4, 5

n = 10 (even)

$\\Median=\frac{(\frac{n}{2})^{th}\ term+(\frac{n}{2}+1)^{th}\ term}{2}\\ Median=\frac{(\frac{10}{2})^{th}\ term+(\frac{10}{2}+1)^{th}\ term}{2}\\ Median=\frac{(5)^{th}\ term+(6)^{th}\ term}{2}\\ Median=\frac{3+3}{2}\\ Median=3$

In the given data 3 occurs the maximum number of times (4)

Therefore, Mode = 3

41,     39,     48,     52,     46,     62,     54,     40,     96,     52,     98,     40,     42,     52,     60

Find the mean, median and mode of this data.

$Mean=\frac{Sum\ of\ observations}{Number \ of\ observations}$

Number of observations, n = 15

$\\\bar{X}=\frac{\sum_{i=1}^{n=15}x_{i}}{n}\\ \bar{X}=\frac{41+ 39+ 48+ 52+ 46+ 62+ 54+ 40+ 96+ 52+ 98+ 40+ 42+ 52+ 60}{15}\\ \bar{X}=\frac{822}{15}\\ \bar{X}=54.8$

Mean is 54.8

To find the median we have to arrange the given data in ascending order as follows:

39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98

n = 15 (odd)

$\\Median=(\frac{n+1}{2})^{th}\ term\\ Median=(\frac{15+1}{2})^{th}\ term\\ Median=8^{th}\ term\\ Median=52$

In the given data 52 occurs the maximum number of times ()

Therefore, Mode = 52

29,     32,     48,     50,     x,     x + 2,     72,     78,     84,     95

The given data is already in ascending order

Number of observations, n = 10 (even)

$\\Median=\frac{(\frac{n}{2})^{th}\ term+(\frac{n}{2}+1)^{th}\ term}{2}\\ Median=\frac{(\frac{10}{2})^{th}\ term+(\frac{10}{2}+1)^{th}\ term}{2}\\ Median=\frac{(5)^{th}\ term+(6)^{th}\ term}{2}\\ Median=\frac{(x)+(x+2)}{2}\\ Median=\frac{2x+2}{2}\\ Median=x+1$

x + 1 = 63

x = 62

In the given data 14 is occuring the maximum number of times (4)

Mode of the given data is therefore 14.

 Salary (in Rs) Number of workers 3000 16 4000 12 5000 10 6000 8 7000 6 8000 4 9000 3 10000 1 Total 60

 Salary ( in Rs)(xi) Number of workers(fi) fixi 3000 16 48000 4000 12 48000 5000 10 50000 6000 8 48000 7000 6 42000 8000 4 32000 9000 3 27000 10000 1 10000 Total $\sum_{i=1}^{8}f_{i}=60$ $\sum_{i=1}^{8}f_{i}x_{i}=305000$

The mean of the above data is given by

$\\\bar{X}=\frac{\sum_{i=1}^{8}f_{i}x_{i}}{\sum_{i=1}^{8}f_{i}}\\ \bar{X}=\frac{305000}{60}\\ \bar{X}=5083.33$

The mean salary of the workers working in the factory is Rs 5083.33

The mean is an appropriate measure of central tendency in case the observations are close to each other. An example of such a case is height of the students in a class.

## Statistics Excercise: 14.4

The mean is not an appropriate measure of central tendency in case the observations are not close to each other. An example of such a case is prices of the toys in a toy shop.

## NCERT solutions for class 9 maths chapter wise

 Chapter No. Chapter Name Chapter 1 NCERT solutions for class 9 maths chapter 1 Number Systems Chapter 2 CBSE NCERT solutions for class 9 maths chapter 2 Polynomials Chapter 3 Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry Chapter 4 NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables Chapter 5 CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry Chapter 6 Solutions of NCERT class 9 maths chapter 6 Lines And Angles Chapter 7 NCERT solutions for class 9 maths chapter 7 Triangles Chapter 8 CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals Chapter 9 Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles Chapter 10 NCERT solutions for class 9 maths chapter 10 Circles Chapter 11 CBSE NCERT solutions for class 9 maths chapter 11 Constructions Chapter 12 Solutions of NCERT class 9 maths chapter 12 Heron’s Formula Chapter 13 NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Chapter 14 CBSE NCERT solutions for class 9 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 9 maths chapter 15 Probability

## How to use NCERT solutions for class 9 maths chapter 14 Statistics?

• Go through some tables and understand that representation of data.

• Take a look through some examples to understand the calculation of mean, median, mode.

• Memorize the formulas to calculate mean, median, mode.

• Implement the application of all the concepts and formulas on the practice problems.

• While practicing you can use NCERT solutions for class 9 maths chapter 14 Statistics.

Keep working hard and happy learning!