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NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3.

  • Statistics Class 9 Ex 14.1
  • Statistics Class 9 Ex 14.2
  • Statistics Class 9 Ex 14.3
  • Statistics Class 9 Ex 14.4
Board CBSE
Textbook NCERT
Class Class 9
Subject Maths
Chapter Chapter 14
Chapter Name Statistics
Exercise  Ex 14.3
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3

Ex 14.3 Class 9 Maths Question 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
(in %)
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.
(ii) Which condition is the Igor 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) We draw the bar graph of this data in the following steps. Note that, the unit in the second column is percentage.

  1. We represent the causes (variable) on the horizontal axis choosing any scale, since width of the bar is not important but for clarity, we take equal widths for all bars and maintain equal gaps in between. Let one cause be represented by one unit
  2. We represent the female fatality rate (value) on the vertical axis. Here, we can choose the scale as 1 unit = 4%.
  3. To represent our first cause i.e., reproductive health conditions, we draw a rectangle bar with width 1 unit and height 31.8 units.
  4. Similarly, other heads are represented leaving a gap of 1 unit in between two consecutive bars.

Now, the graph is drawn in figure as given below :
NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3.1
(ii)
From graph, we observe that ‘reproductive health conditions’ is the major cause of women’s ill health and death world wide because it has maximum percentage among the causes e., 31.8%.
(iii) Two other factors which play a major role in the cause in (ii) above are injuries and other causes.

Ex 14.3 Class 9 Maths Question 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) 940
Scheduled Tribe (ST) 970
Non SC/ST 920
Backward districts 950
Non-backward districts 920
Rural 930
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) We draw the bar graph of this data, note that the unit in the second column is number of girls per thousand boys.

  1. We represent the section on the horizontal axis choosing any scale, since width of the bar is not important but for clarity, we take equal width for all bars and maintain equal gaps in between. Let one section be represented by one unit.
  2. We represent the number of girls per thousand boys on the vertical axis.
    Here, we can choose the scale as 1 unit = 100.

Now, the graph is drawn in figure as given below :
NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3.2

(ii) From graph, we observe that scheduled tribe (ST) number of girls is major section in different sections of Indian society, because it has maximum number of girls per thousand boys i.e., 970.

Ex 14.3 Class 9 Maths Question 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.
(ii) Which political party won the maximum number of seats?
Solution.
(i) We draw the bar graph of this data, note that the unit in the second column is seats won by political party.
We represent the political party on the horizontal axis choosing any scale, since width of the bar is not important but for clarity, we take equal widths for all bars and maintain equal gaps in between. Let one political party be represented by one unit.
We represent the seats won on the vertical axis. Here, we can choose the scale as 1 unit = 10.
Now, the graph is drawn as follows :
NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3.3
(ii)
Party A’ won the maximum number of seats i.e., 75.

Ex 14.3 Class 9 Maths Question 4.
The length of 40 leaves of a plant measured correct to one millimeter and the obtained data is represented in the following table

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

(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 are 153 mm long ? Why?
Solution.
(i) We know that the areas of the rectangles are proportional to the frequencies in a histogram. Here, the given frequency distribution is not continuous. For continuous distribution, we get first interval as
(118 – 0.5) – (126 + 0.5)= 117.5-126.5. The class width in this case is 9
[ ∵ \(\cfrac { 127-126 }{ 2 } =0.5 \) ]

So, we get the following modified table of given data :

Length (in mm) Frequency Width of the class
117.5-126.5 3 9
126.5-135.5 5 9
135.5-144.5 9 9
144.5-153.5 12 9
153.5-162.5 5 9
162.5-171.5 4 9
171.5-180.5 2 9

Now, we can draw the histogram for given data as follows :
NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3.4
(ii) Yes, other suitable graphical representation for the same data is frequency polygon.
(iii) No, because the maximum number of leaves have their lengths lying in the interval 145-153.

Ex 14.3 Class 9 Maths Question 5.
The following table gives the lifetimes of 400 neon lamps :

Lifetime (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

(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 for given table is as follows :
NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3.5
(ii)
Number of lamps have a lifetime of more than 700 hours = 74 + 62 + 48 = 184

Ex 14.3 Class 9 Maths Question 6.
The following table gives the distribution of students of two  sections according to the marks 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.
Solution.
We make modified table by given data as shown below :

Classes Class Marks Frequency (Section A) Frequency (Section B)
0-10 5 3 5
10-20 15 9 19
20-30 25 17 15
30-40 35 12 10
40-50 45 9 1

 Now, the required frequency polygons are as follows :
NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3.6
It is clear that from the polygon that the performance of section A is better in comparison of section B.

Ex 14.3 Class 9 Maths Question 7.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below :

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.Represent the data of
[Hint : First make the class intervals continuous]
Solution.
First, we make the class intervals continuous then modified table of given data is as shown below :

Number of balls Class marks Team A Team B
0.5-6.5 3.5 2 5
6.5-12.5 9.5 r 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

Now, frequency polygon for both teams are given below :
NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3.7

Ex 14.3 Class 9 Maths Question 8.
A random survey of the number of children of various age groups playing in park was found 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.
Solution.
Here, the widths of the rectangles are varying. So, we need to make certain modifications in the lengths of the rectangles, so that the areas are became proportional to the frequencies. The minimum class size is 1. Length of rectangle (Adjusted frequency)
= \(\cfrac { Minimum\quad class\quad size }{ Class\quad of\quad this\quad class } \times Frequency \)
Then modified table of given data is shown below :

Age

(in years)

Number of children (Frequency) Width of the class Length of the rectangle
1-2 5 1 1/1 x 5= 5
2-3 3 1 1/1 x 3 = 3
3-5 6 2 1/2 x 6 = 3
5-7 12 2 1/2 x 12 = 6
7-10 9 3 1/3 x 9 = 3
10-15 10 5 1/ 5 x 10 = 2
15-17 4 2 1 / 2 x 4 = 2

So, the histogram with varying width is given below :

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3.8

Ex 14.3 Class 9 Maths Question 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 :
Solution.

Number of Letters Number of Surnames
1-4 6
4-6 30
6-8 44
8-12 16
12-20 4

 (i) Draw a histogram to depict the given information.
(ii)
Write the class interval in which the maximum number of surnames lie.
Solution.
(i) We know that, the areas of the rectangles are proportional to the frequencies in a histogram. Here, the widths of the rectangles are varying. So, we need to make certain modifications in the lengths of the rectangles. So that the areas are again proportional to the frequencies.

  1. Select a class interval with the minimum class size. The minimum class size is 2.
  2. The lengths of the rectangles are then modified to be proportionate to the class size 2.
    Since, we have calculated these lengths for interval of 2 letters in each case, we may call these lengths as ‘proportion of surnames per 2 mark interval’.
    Here, we make a modified table by given data with minimum class size 2.
Number of letters Number of surnames Width of the class

Length of the rectangle

1-4 6 3 \(\cfrac { 6 }{ 3 } \times 2=4 \)
4-6 30 2 \(\cfrac { 30 }{ 2 } \times 2=30 \)
6-8 44 2 \(\cfrac { 44 }{ 2 } \times 2=44 \)
8-12 16 4 \(\cfrac { 16 }{ 4 } \times 2=8 \)
12-20 4 8 \(\cfrac { 4 }{ 8 } \times 2=1 \)

So, the correct histogram with varying width is given below :
NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3.9

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

 

We hope the NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3, drop a comment below and we will get back to you at the earliest.

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