Q-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:
Table shows the length of number of leaves
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.
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):
Given frequency distribution is not continuous. So, first convert into frequency distribution.
So, Difference between lower limit of class and upper limit of preceding class:
Hence, added to each upper class limit and subtract from the lower class limits.
Obtain distribution is given as under:
Obtain distribution as per given data
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
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Solution (II): Other suitable graphical representation of this data is frequency polygon.
Solution (III): No, maximum number of leaves has in between 144.5 mm and 153.5 mm. It is not necessary that all leaves length as 153 mm.
Q-5 The following table gives the life times of 400 neon lamps:
Life times of neon lamps are shown in table
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 lifetime of more than 700 hours?
Solution (I): Life time of x-axis and the number of lamps on y-axis.
Histogram:
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Solution (II):
Lifetime more than 700 is the sum of the number of neon lamps having their lifetime as, and
So, neon lamp having more than 700 hour’s lifetime
Q-6 The following table gives the distribution of students of two sections according to the mark obtained by them:
Distribution of students according to marks
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:
The class marks of the given class intervals by using the formula.
Class interval (Marks)
Frequency polygons data table of Section A and B
Section A
Section B
Marks
Class marks
Frequency
Mark
Class marks
Frequency
5
3
5
5
15
9
15
19
25
17
25
15
35
12
35
10
45
9
45
1
Class marks on x-axis and frequency on y-axis.
Choose appropriate scale.
Below drawn the frequency polygons:
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