1A 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.
(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.
Answer & Explanation:
(i) The data can be represented using a bar graph. The causes are plotted on the x-axis and the female fatality rate (%) on the y-axis. Bars of uniform width with equal spacing are drawn with heights proportional to the percentages.
(ii) Reproductive health conditions is the major cause with a fatality rate of 31.8%.
(iii) This part requires discussion with a teacher and is not answered in the textbook. (Based on concepts from Section 12.1 (A) Bar Graphs).
2The 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.
Answer & Explanation:
(i) The data can be represented using a bar graph. The sections are plotted on the x-axis and the number of girls per thousand boys on the y-axis. Bars of uniform width with equal spacing are drawn with heights proportional to the given numbers.
(ii) The graph shows that the number of girls per thousand boys is highest for Scheduled Tribes (ST) and lowest for Urban areas. This indicates a variation in the child sex ratio across different sections of society. (Based on concepts from Section 12.1 (A) Bar Graphs).
3Given 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?
Answer & Explanation:
(i) The data can be represented using a bar graph. The political parties are plotted on the x-axis and the number of seats won on the y-axis. Bars of uniform width with equal spacing are drawn with heights proportional to the seats won.
(ii) Political Party A won the maximum number of seats (75). (Based on concepts from Section 12.1 (A) Bar Graphs).
4The length of 40 leaves of a plant are measured correct to one millimetre, 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?
Answer & Explanation:
(i) The given class intervals are not continuous (e.g., 126 to 127 has a gap). To make them continuous, we adjust them by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class. The new intervals become 117.5-126.5, 126.5-135.5, ..., 171.5-180.5. A histogram can then be drawn with these continuous class intervals on the x-axis and the number of leaves on the y-axis. Rectangles of width equal to the class size (9 mm) and height proportional to the frequency are drawn.
(ii) Yes, a frequency polygon is another suitable graphical representation for this continuous data.
(iii) No, it is not correct. The data shows that the maximum number of leaves (12) fall in the length range of 145 mm to 153 mm. It does not mean that all these 12 leaves are exactly 153 mm long. (Based on concepts from Section 12.1 (B) Histogram).
5The following table gives the life times of 400 neon lamps:
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
(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours?
Answer & Explanation:
(i) The data has continuous class intervals of uniform width (100 hours). A histogram can be drawn with life time on the x-axis and the number of lamps on the y-axis. Rectangles of width 100 units and height proportional to the number of lamps are drawn for each class interval.
(ii) Lamps with a lifetime of more than 700 hours are in the intervals 700-800, 800-900, and 900-1000. Total number = 74 + 62 + 48 = 184 lamps. (Based on concepts from Section 12.1 (B) Histogram).
6The following table gives the distribution of students of two sections according to the marks obtained by them:
Marks
Frequency (Section A)
Frequency (Section B)
0-10
3
5
10-20
9
19
20-30
17
15
30-40
12
10
40-50
9
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.
Answer & Explanation:
To draw the frequency polygons:
1. Find the class-marks for each interval: 5, 15, 25, 35, 45.
2. Plot the points for Section A: (5,3), (15,9), (25,17), (35,12), (45,9).
3. Plot the points for Section B: (5,5), (15,19), (25,15), (35,10), (45,1).
4. Add points for class-marks before 0-10 (i.e., -5) and after 40-50 (i.e., 55) with frequency 0 for both sections.
5. Join the points for each section with line segments.
Comparison: The frequency polygon for Section B shows a higher concentration of students in the 10-20 marks range, while Section A has a more even distribution with a peak in the 20-30 range. Section A has more students in the higher marks ranges (30-50) compared to Section B. Overall, Section A appears to have performed better in the higher marks brackets. (Based on concepts from Section 12.1 (C) Frequency Polygon).
7The 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. [Hint: First make the class intervals continuous.]
Answer & Explanation:
The given class intervals are continuous (e.g., 1-6, 7-12...). However, for a frequency polygon, we need the class-marks.
1. Class-marks are: 3.5, 9.5, 15.5, 21.5, 27.5, 33.5, 39.5, 45.5, 51.5, 57.5.
2. Plot the points for Team A using these class-marks and their runs.
3. Plot the points for Team B similarly.
4. Add points for class-marks before 1-6 (i.e., -2.5) and after 55-60 (i.e., 63.5) with frequency 0 for both teams.
5. Join the points for each team with line segments to get two frequency polygons on the same graph. (Based on concepts from Section 12.1 (C) Frequency Polygon).
8A random survey of the number of children of various age groups playing in a 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.
Answer & Explanation:
The class intervals have varying widths. To draw a correct histogram, we must adjust the lengths of the rectangles so that their areas are proportional to the frequencies.
1. The minimum class size is 1 (from intervals 1-2 and 2-3).
2. Calculate the length of each rectangle as (Frequency / Class width) * Minimum class size.
Age (in years)
Frequency
Class width
Length of rectangle
1-2
5
1
(5/1)*1 = 5
2-3
3
1
(3/1)*1 = 3
3-5
6
2
(6/2)*1 = 3
5-7
12
2
(12/2)*1 = 6
7-10
9
3
(9/3)*1 = 3
10-15
10
5
(10/5)*1 = 2
15-17
4
2
(4/2)*1 = 2
3. Draw the histogram using these adjusted lengths. (Based on concepts from Section 12.1 (B) Histogram with varying widths, Example 3).
91100 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
1-4
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.
Answer & Explanation:
The class intervals have varying widths. To draw a correct histogram, we must adjust the lengths of the rectangles.
1. The minimum class size is 2 (from intervals 4-6 and 6-8).
2. Calculate the length of each rectangle as (Frequency / Class width) * Minimum class size.
Number of letters
Frequency
Class width
Length of rectangle
1-4
6
3
(6/3)*2 = 4
4-6
30
2
(30/2)*2 = 30
6-8
44
2
(44/2)*2 = 44
8-12
16
4
(16/4)*2 = 8
12-20
4
8
(4/8)*2 = 1
3. Draw the histogram using these adjusted lengths.
(ii) The maximum number of surnames (44) lie in the class interval 6-8. (Based on concepts from Section 12.1 (B) Histogram with varying widths, Example 3).