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 %):
(i). Represent the information given above graphically.
(ii). Which condition is the major cause of women's ill health and death worldwide?
(i). Represent the information above by a bar graph.
(i). Draw a bar graph to represent the polling results.
(ii). Which political party won the maximum number of seats?
(i). Draw a histogram to represent the given data.
(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?
(i). Represent the given information with the help of a histogram.
(ii) How many lamps have a lifetime of more than 700 hours?
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.
Represent the data of both the teams on the same graph by frequency polygons.
Now by taking class marks on x axis and runs scored on y axis we can construct frequency polygon as following -
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 (%)
|
1.
2.
3.
4.
5.
6.
|
Reproductive health conditions
Neuropsychiatric conditions
Injuries
Cardiovascular conditions
Respiratory conditions
Other causes
|
31.8
25.4
12.4
4.3
4.1
22.0
|
(i). Represent the information given above graphically.
(ii). Which condition is the major cause of women's ill health and death worldwide?
Solution
(i) By representing causes on x axis and family fatality rate on y axis and choosing an appropriate scale (1 unit = 5% for y axis) we can draw the graph of information given above, as following
All the rectangle bars are of same width and having equal spacing between them.
(ii) Reproductive health condition is the major cause of women's ill health and death worldwide as 31.8% of women are affected by it.
(ii) Reproductive health condition is the major cause of women's ill health and death worldwide as 31.8% of women are affected by it.
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)
Scheduled Tribe (ST)
Non SC/ST
Backward districts
Non-backward districts
Rural
Urban
|
940
970
920
950
920
930
910
|
(i). Represent the information above by a bar graph.
Solution
(i). By representing section (variable) on x axis and number of girls per thousand boys on y axis we can draw the graph of information given as above and choosing an appropriate scale (1 unit = 100 girls for y axis)
Here all the rectangle bars are of same width and have equal spacing in between them.
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). By taking polling results on x axis and seats won as y axis and choosing an appropriate scale (1 unit = 10 seats for y axis) we can draw the required graph of above information as below -
Here rectangle bars are of same width and have equal spacing in between them.
(ii). We may find that political party 'A' won maximum number of seats.
(ii). We may find that political party 'A' won maximum number of seats.
Question 4. The length of 40 leaves of a plant are measured correct to one millimeter, and the obtained data is represented in the following table:
Length (in mm)
|
Number of leaves
|
118 - 126
127 - 135
136 - 144
145 - 153
154 - 162
163 - 171
172 - 180
|
3
5
9
12
5
4
2
|
(i). Draw a histogram to represent the given data.
(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). Length of leaves are represented in a discontinuous class intervals having a difference of 1 in between them. So we have to add to each upper class limit and also have to subtract 0.5 from the lower class limits so as to make our class intervals continuous.
Now taking length of leaves on x axis and number of leaves on y axis we can draw the histogram of this information as below -
Length (in mm)
|
Number of leaves
|
117.5 - 126.5
|
3
|
126.5 - 135.5
|
5
|
135.5 - 144.5
|
9
|
144.5 - 153.5
|
12
|
153.5 - 162.5
|
5
|
162.5 - 171.5
|
4
|
171.5 - 180.5
|
2
|
Now taking length of leaves on x axis and number of leaves on y axis we can draw the histogram of this information as below -
Here 1 unit on y axis represents 2 leaves.
(ii). Other suitable graphical representation of this data could be frequency polygon.
(iii). No as maximum number of leaves (i.e. 12) have their length in between of 144.5 mm and 153.5 mm. It is not necessary that all have their lengths as 153 mm.
(ii). Other suitable graphical representation of this data could be frequency polygon.
(iii). No as maximum number of leaves (i.e. 12) have their length in between of 144.5 mm and 153.5 mm. It is not necessary that all have their lengths as 153 mm.
Question 5. The following table gives the life times of neon lamps:
Length (in hours)
|
Number of lamps
|
300 - 400
400 - 500
500 - 600
600 - 700
700 - 800
800 - 900
900 - 1000
|
14
56
60
86
74
62
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). By taking life time (in hours) of neon lamps on x axis and number of lamps on y axis we can draw the histogram of the given information as below -
Here 1 unit on y axis represents 10 lamps.
(ii). Number of neon lamps having their lifetime more than 700 are sum of number of neon lamps having their lifetime as 700 - 800, 800 - 900, and 900 - 1000.
So number of neon lamps having their lifetime more than 700 hours is 184. (74 + 62 + 48 = 184)
(ii). Number of neon lamps having their lifetime more than 700 are sum of number of neon lamps having their lifetime as 700 - 800, 800 - 900, and 900 - 1000.
So number of neon lamps having their lifetime more than 700 hours is 184. (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
10 - 20
20 - 30
30 - 40
40 - 50
|
3
9
17
12
9
|
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
|
5
19
15
10
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 can find class marks of given class intervals by using formula -
Class mark
Class mark
Section A
|
Section B
| ||||
Marks
|
Class marks
|
Frequency
|
Marks
|
Class marks
|
Frequency
|
0 - 10
|
5
|
3
|
0 - 10
|
5
|
5
|
10 - 20
|
15
|
9
|
10 - 20
|
15
|
19
|
20 - 30
|
25
|
17
|
20 - 30
|
25
|
15
|
30 - 40
|
35
|
12
|
30 - 40
|
35
|
10
|
40 - 50
|
45
|
9
|
40 - 50
|
45
|
1
|
Now taking class marks on x axis and frequency on y axis and choosing an appropriate scale (1 unit = 3 for y axis) we can draw frequency polygon as below -
From the graph we can see performance of students of section 'A' is better than the students of section 'B' as for good marks.
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
7 - 12
13 - 18
19 - 24
25 - 30
31 - 36
37 - 42
43 - 48
49 - 54
55 - 60
|
2
1
8
9
4
5
6
10
6
2
|
5
6
2
10
5
6
3
4
8
10
|
Represent the data of both the teams on the same graph by frequency polygons.
Solution
We observe that given data is not having its class intervals continuous. There is a gap of 1 in between of them. So we have to add 
= 0.5 to upper class limits and subtract 0.5 from
= 0.5 to upper class limits and subtract 0.5 from
lower class limits.
Also class mark of each interval can be found by using formula -
Also class mark of each interval can be found by using formula -
Class mark 
Now continuous data with class mark of each class interval can be represented as following -
Number of balls
|
Class mark
|
Team A
|
Team B
|
0.5 - 6.5
|
3.5
|
2
|
5
|
6.5 - 12.5
|
9.5
|
1
|
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 by taking class marks on x axis and runs scored on y axis we can construct frequency polygon as following -
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
2 - 3
3 - 5
5 - 7
7 - 10
10 - 15
15 - 17
|
5
3
6
12
9
10
4
|
Draw a histogram to represent the data above.
Solution
Here data is having class intervals of varying width. We may find proportion of children per 1 year interval as following -
Age (in years)
|
Frequency (Number of children)
|
Width of class
|
Length of rectangle
|
1 - 2
|
5
|
1
|  |
2 - 3
|
3
|
1
|  |
3 - 5
|
6
|
2
|  |
5 - 7
|
12
|
2
|  |
7 - 10
|
9
|
3
|  |
10 - 15
|
10
|
5
|  |
15 - 17
|
4
|
2
|  |
Now taking age of children on x axis and proportion of children per 1 year interval on y axis we may draw histogram as 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:
Number of letters
|
Number of surnames
|
1 - 4
4 - 6
6 - 8
8 - 12
12 - 20
|
6
30
44
16
4
|
(i). Draw a histogram to depict the given information.
(ii). Write the class interval in which the maximum number of surnames lie.
(ii). Write the class interval in which the maximum number of surnames lie.
Solution
(i). Given data is having class intervals of varying width. We need to compute the adjusted frequency
Now by taking number of letters on x axis and proportion of number of surnames per 2 letters interval on y axis and choosing an appropriate scale (1 unit = 4 students for y axis) we will construct the histogram as below
Number of letters
|
Frequency (Number of surnames)
|
Width of class
|
Length of rectangle
|
1 - 4
|
6
|
3
|  |
4 - 6
|
30
|
2
|  |
6 - 8
|
44
|
2
|  |
8 -12
|
16
|
4
|  |
12 - 20
|
4
|
8
|  |
Now by taking number of letters on x axis and proportion of number of surnames per 2 letters interval on y axis and choosing an appropriate scale (1 unit = 4 students for y axis) we will construct the histogram as below
(ii). The class interval in which the maximum number of surname lie is
6 - 8 as there are 44 number of surnames in it i.e. maximum for this data.
6 - 8 as there are 44 number of surnames in it i.e. maximum for this data.
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