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

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 :

(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.

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 :

(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.

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 :

(ii) Party A’ won the maximum number of seats i.e., 75.

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
[ ∵ ]

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 :

(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.

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 :

(ii) Number of lamps have a lifetime of more than 700 hours = 74 + 62 + 48 = 184

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 :

It is clear that from the polygon that the performance of section A is better in comparison of section B.

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 :

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)
=
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 :

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
4-6	30	2
6-8	44	2
8-12	16	4
12-20	4	8

So, the correct histogram with varying width is given below :

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

 

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