cl 9 math 14.3

Class 9 Mathematics 14.3 Solutions

Chapter Overview: 

Assam Eduverse resources complement the SEBA curriculum by providing detailed study materials, chapter-wise solutions, and interactive learning experiences to simplify these complex topics. The focus is on practical application, encouraging students to draw examples from real-life situations and perform qualitative analysis of data to select the most appropriate method of presentation. The curriculum emphasizes developing calculation speed and visualization skills through ample practice problems. 

Statistics is a vital branch of mathematics that deals with the collection, analysis, interpretation, and presentation of numerical data. The primary objective of the Class 9 chapter is to introduce students to fundamental statistical concepts and methods. Key topics include understanding different types of data (primary and secondary), organizing raw information into frequency distribution tables, and calculating essential measures like frequency, class intervals, and midpoints.
The chapter then transitions to the graphical representation of data, which aids in visual understanding and comparison. Students learn to construct and interpret bar graphs, histograms (for continuous data), and frequency polygons. These visual tools help in quickly identifying patterns, distributions, and trends within datasets, making complex data more accessible and understandable.
Finally, the curriculum covers “measures of central tendency” for ungrouped data. Students are taught how to calculate the mean (average), median (middle value in an ordered set), and mode (most frequent value). These measures provide a single, representative value for the entire dataset, allowing for effective summarization and comparison of different data sets. The chapter lays the groundwork for more advanced statistical studies in future grades.

Class 9 mathematics solution

Q1. 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.CausesFemale 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

(i) Represent the information given above graphically.

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

(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.

Solution:

(i) The information given in the question is represented below graphically.

Ncert solutions class 9 chapter 14-2

(ii) We can observe from the graph that reproductive health conditions are the major cause of women’s ill health and death worldwide.

(iii) Two factors responsible for the cause in (ii) are

  • Lack of proper care and understanding.
  • Lack of medical facilities.

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

S.No.SectionNumber of girls per thousand boys
1.Scheduled Caste (SC)  940
2.Scheduled Tribe (ST) 970
3.Non-SC/ST 920
4.Backward districts 950
5.Non-backward districts 920
6.Rural 930
7.Urban 910

(i) Represent the information above by a bar graph.

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

Solution:

(i) The information given in the question is represented below graphically.

Ncert solutions class 9 chapter 14-3

(ii) From the above graph, we can conclude that the maximum number of girls per thousand boys is present in section ST. We can also observe that the backward districts and rural areas have more girls per thousand boys than non-backward districts and urban areas.

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

Political party    A        B       C         D         E         F    
Seats won755537291037

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

(ii) Which political party won the maximum number of seats?

Solution:

(i) The bar graph representing the polling results is given below.

Ncert solutions class 9 chapter 14-4

(ii) From the bar graph, it is clear that Party A won the maximum number of seats.

Q4. The length of 40 leaves of a plant is measured correctly to one millimetre, and the obtained data is represented in the following table:

S.No.Length (in mm)Number of leaves 
1.118 – 1263
2.127 – 1355
3.136 – 1449
4.145 – 15312
5.154 – 1625
6.163 – 1714
7.172 – 1802

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

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

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

Solution:

(i) The data given in the question is represented in the discontinuous class interval. So, we have to make it in the continuous class interval. The difference is 1, so taking half of 1, we subtract ½ = 0.5 from the lower limit and add 0.5 to the upper limit. Then, the table becomes

S.No.Length (in mm)Number of leaves
1.117.5 – 126.53
2.126.5 – 135.55
3.135.5 – 144.59
4.144.5 – 153.512
5.153.5 – 162.55
6.162.5 – 171.54
7.171.5 – 180.52

Ncert solutions class 9 chapter 14-5

(ii) Yes, the data given in the question can also be represented by a frequency polygon.

(iii) No, we cannot conclude that the maximum number of leaves is 153 mm long because the maximum number of leaves are lying in-between the length of 144.5 – 153.5

Q5. The following table gives the lifetimes of 400 neon lamps.

Life Time (in hours)Number of Lamps
300 – 40014
400 – 50056
500 – 60060
600 – 70086
700 – 80074
800 – 90062
900 – 100048

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

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

Solution:

(i) The histogram representation of the given data is given below.

Ncert solutions class 9 chapter 14-6

(ii) The number of 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.

Ncert solutions class 9 chapter 14-7

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.

Solution:

The class-marks = (lower limit + upper limit)/2

For section A,

MarksClass-marksFrequency
0-1053
10-20159
20-302517
30-403512
40-50459

For section B,

MarksClass-marksFrequency
0-1055
10-201519
20-302515
30-403510
40-50451

Representing these data on a graph using two frequency polygon, we get

 

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

From the graph, we can conclude that the students of Section A performed better than Section B.

Q7The runs scored by two teams, A and B, on the first 60 balls in a cricket match are given below.

Ncert solutions class 9 chapter 14-9

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

[Hint: First, make the class intervals continuous.]

Solution:

The data given in the question is represented in the discontinuous class interval. So, we have to make it in the continuous class interval. The difference is 1, so taking half of 1, we subtract ½ = 0.5 = 0.5 from the lower limit and add 0.5 to the upper limit. Then, the table becomes

Number of BallsClass MarkTeam ATeam B
0.5-6.53.525
6.5-12.59.516
12.5-18.515.582
18.5-24.521.5910
24.5-30.527.545
30.5-36.533.556
36.5-42.539.563
42.5-48.545.5104
48.5-54.551.568
54.5-60.557.5210

The data of both teams are represented on the graph below by frequency polygons.

Ncert solutions class 9 chapter 14-10

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

Ncert solutions class 9 chapter 14-11

Draw a histogram to represent the data above.

Solution:

The width of the class intervals in the given data varies.

We know that,

The area of the rectangle is proportional to the frequencies in the histogram.

Thus, the proportion of children per year can be calculated as given in the table below.

Age

(in years)

Number of children (frequency)Width of classLength of rectangle
1-251(5/1)×1 = 5
2-331(3/1)×1 = 3
3-562(6/2)×1 = 3
5-7122(12/2)×1 = 6
7-1093(9/3)×1 = 3
10-15105(10/5)×1 = 2
15-1742(4/2)×1 = 2

Let x-axis = the age of children

y-axis = proportion of children per 1-year interval

Ncert solutions class 9 chapter 14-12Q9. 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:

Ncert solutions class 9 chapter 14-13

(i) Draw a histogram to depict the given information.

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

Solution:

(i) The width of the class intervals in the given data is varying.

We know that,

The area of the rectangle is proportional to the frequencies in the histogram.

Thus, the proportion of the number of surnames per 2 letters interval can be calculated as given in the table below.

Number of lettersNumber of surnamesWidth of classLength of rectangle
1-463(6/3)×2 = 4
4-6302(30/2)×2 = 30
6-8442(44/2)×2 = 44
8-12164(16/4)×2 = 8
12-2048(4/8)×2 = 1

Ncert solutions class 9 chapter 14-14

(ii) 6-8 is the class interval in which the maximum number of surnames lie.

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