Frank Solutions Class 9 Maths Chapter 23 Graphical Representation of Statistical Data provides students 100% answers in a lucid manner. Students who have doubts relying on the covered concepts, are advised to follow Frank Solutions and clear their doubts quickly. Practising Frank Solutions on a regular basis enhances their skills, which are essential to boost exam preparation. Students can access Frank Solutions Class 9 Maths Chapter 23 Graphical Representation of Statistical Data PDF, from the links provided below.
Chapter 23 Graphical Representation of Statistical Data has problems which gives information of data using graphs. The way of analyzing numerical data is known as graphical representation. Referring to Frank Solutions helps students to speed up their problem solving skills, which is an important aspect from an exam point of view.
Frank Solutions for Class 9 Maths Chapter 23 Graphical Representation of Statistical Data Download PDF
Access Frank Solutions for Class 9 Maths Chapter 23 Graphical Representation of Statistical Data
1. Harmeet earns Rs 50,000 per month. He budget for his salary as per the following table:
|
Expenses |
Accommodation |
Food |
Clothing |
Travel |
Miscellaneous |
Savings |
|
Amount (Rs) |
12000 |
9000 |
2500 |
7500 |
4000 |
15000 |
Draw a bar graph for the above data.
Solution:
The bar graph for the above data is as follows:
2. The birth rate per thousand of the following states over a certain period is given below:
|
States |
Punjab |
Haryana |
U.P. |
Gujarat |
Rajasthan |
Jammu and Kashmir |
|
Birth Rate (per thousand) |
22.9 |
21.8 |
19.5 |
21.1 |
23.9 |
18.3 |
Draw a bar graph for the above data
Solution:
The bar graph for the above data is shown below
3. Fadil, a class IX student, scored marks in different subjects (each out of total 100) during his annual examination as given below
|
Subject |
Maths |
English |
Science |
Social Studies |
Hindi |
Physical Education |
|
Mark (out of 100) |
75 |
80 |
77 |
78 |
67 |
89 |
Draw horizontal bar graph for the above data.
Solution:
The horizontal bar graph for the above data is as follows:
4. The number of students in different sections of class IX of a certain school is given in the following table.
|
Section |
IX – A |
IX – B |
IX – C |
IX – D |
IX – E |
|
Number of students |
48 |
40 |
50 |
45 |
38 |
Draw horizontal bar graph for the above data.
Solution:
The horizontal bar graph for the above data is given below
5. The number of students (boys and girls) of class IX participating in different activities during their annual day function is given below:
|
Activities |
Dance |
Speech |
Singing |
Quiz |
Drama |
Anchoring |
|
Boys |
12 |
5 |
4 |
4 |
10 |
2 |
|
Girls |
10 |
8 |
6 |
3 |
9 |
1 |
Draw a double bar graph for the above data.
Solution:
The double bar graph for the above data is shown below
6. Draw a histogram for the following frequency distribution:
|
Train fare |
0 – 50 |
50 – 100 |
100 – 150 |
150 – 200 |
200 – 250 |
250 – 300 |
|
No. of travellers |
25 |
40 |
36 |
20 |
17 |
12 |
Solution:
This is an exclusive frequency distribution. We represent the class limits on the x-axis on a suitable scale and the frequencies on the y-axis on a suitable scale. Taking class intervals as bases and the corresponding frequencies as heights, we construct rectangles to obtain a histogram of the given frequency distribution.
The histogram for the above frequency distribution is shown below
7. Draw a histogram for the following frequency table:
|
Class Interval |
5 – 9 |
10 – 14 |
15 – 19 |
20 – 24 |
25 – 29 |
30 – 34 |
|
Frequency |
5 |
9 |
12 |
10 |
16 |
12 |
Solution:
We see that the class intervals are in an inclusive manner. First, we need to convert them into exclusive manner.
|
Class interval |
Frequency |
|
4.5 – 9.5 |
5 |
|
9.5 – 14.5 |
9 |
|
14.5 – 19.5 |
12 |
|
19.5 – 24.5 |
10 |
|
24.5 – 29.5 |
16 |
|
29.5 – 34.5 |
12 |
We take the true class limits on the x-axis on a suitable scale and the frequencies on the y-axis on a suitable scale. Taking class intervals as bases and the corresponding frequencies as heights, we construct rectangles to obtain a histogram of the given frequency distribution.
Here, as the class limits do not start from 0, we put a kink between 0 and the true lower boundary of the first class.
The histogram for the given frequency table is shown below
8. Draw a histogram for the following cumulative frequency table:
|
Marks |
Less than 10 |
Less than 20 |
Less than 30 |
Less than 40 |
Less than 50 |
Less than 60 |
|
Number of student |
7 |
18 |
30 |
45 |
55 |
60 |
Solution:
|
Marks |
0 – 10 |
10 – 20 |
20 – 30 |
30 – 40 |
40 – 50 |
50 – 60 |
|
Number of students |
7 |
11 |
12 |
15 |
10 |
5 |
The histogram for the cumulative frequency table is shown below
9. Draw a histogram for the following cumulative frequency table:
|
Class interval |
0 – 10 |
10 – 20 |
20 – 30 |
30 – 40 |
40 – 50 |
|
Cumulative Frequency |
6 |
10 |
18 |
32 |
40 |
Solution:
First convert the cumulative frequency table to an exclusive frequency distribution table.
|
Class interval |
0 – 10 |
10 – 20 |
20 – 30 |
30 – 40 |
40 – 50 |
|
Cumulative Frequency |
6 |
4 |
8 |
14 |
8 |
We take the class limits on the x-axis and the frequencies on the y-axis on suitable scales. We draw rectangles with the class intervals as bases and the corresponding frequencies as heights. The histogram for the given cumulative frequency table is shown below
10. Draw a histogram and a frequency polygon for the following data:
|
Marks |
0 – 20 |
20 – 40 |
40 – 60 |
60 – 80 |
80 – 100 |
|
Number of students |
12 |
18 |
30 |
25 |
15 |
Solution:
We represent the class limits on the x-axis and the frequencies on the y-axis on a suitable scale. Taking class intervals as bases and the corresponding frequencies as heights, we construct rectangles to obtain a histogram of the given frequency distribution.
Now,
Take the mid-points of the upper horizontal side of each rectangle. Join the mid-points of two imaginary class intervals, one on either side of the histogram, by line segments one after the other.
The histogram and a frequency polygon for a given data is as follows:
11. Draw a histogram and a frequency polygon for the following data:
|
Wages |
150 – 200 |
200 – 250 |
250 – 300 |
300 – 350 |
350 – 400 |
400 – 450 |
|
No. of workers |
25 |
40 |
35 |
28 |
30 |
22 |
Solution:
We represent the class limits on the x-axis and the frequencies on the y-axis on a suitable scale. Taking class intervals as bases and the corresponding frequencies as heights, we construct rectangles to obtain a histogram of the given frequency distribution.
Now,
Take the mid-points of the upper horizontal side of each rectangle. Join the mid-points of two imaginary class intervals, one on either side of the histogram, by line segments one after the other.
Here, as the class limits do not start from 0, we put a kink between 0 and the lower
boundary of the first class.
The histogram and a frequency polygon of the given data is as follows
12. Draw a frequency polygon for the following data:
|
Expenses |
100 – 150 |
150 – 200 |
200 – 250 |
250 – 300 |
300 – 350 |
350 – 400 |
|
No. of families |
22 |
37 |
26 |
18 |
10 |
5 |
Solution:
We take the class limits on the x-axis and the frequencies on the y-axis on suitable scales.
Now,
Find the class marks of all the class intervals. Locate the points (x1, y1) on the graph, where x1 denotes the class mark and y1 denotes the corresponding frequency. Join all the points plotted above with straight line segments. Join the first point and the last point to the points representing class marks of the class intervals before the first class interval and after the last class interval of the given frequency distribution.
Here, as the class limits do not start from 0, we put a kink between 0 and the lower boundary of the first class.
Frequency polygon for the given data is shown below
13. Draw a frequency polygon for the following data:
|
Class |
15 – 20 |
20 – 25 |
25 – 30 |
30 – 35 |
35 – 40 |
40 – 45 |
|
Frequency |
5 |
12 |
15 |
26 |
18 |
7 |
Solution:
We take the class limits on the x-axis and the frequencies on the y-axis on suitable scales.
Now,
Find the class marks of all the class intervals. Locate the points (x1, y1) on the graph, where x1 denotes the class mark and y1 denotes the corresponding frequency. Join all the points plotted above with straight line segments. Join the first point and the last point to the points representing class marks of the class intervals before the first class interval and after the last class interval of the given frequency distribution
Here, as the class limits do not start from 0, we put a kink between 0 and the lower boundary of the first class
Frequency polygon for the given data is shown below
14. Draw a frequency polygon for the following data:
|
Marks |
5 – 9 |
10 – 14 |
15 – 19 |
20 – 24 |
25 – 29 |
30 – 34 |
|
No. of students |
7 |
11 |
15 |
22 |
18 |
5 |
Solution:
We see that the class intervals are in an inclusive manner. We first need to convert them into exclusive manner.
|
Marks |
No. of students |
|
4.5 – 9.5 |
7 |
|
9.5 – 14.5 |
11 |
|
14.5 – 19.5 |
15 |
|
19.5 – 24.5 |
22 |
|
24.5 – 29.5 |
18 |
|
29.5 – 34.5 |
5 |
We take the class limits on the x-axis and the frequencies on the y-axis on suitable scales.
Now,
Find the class marks of all the class intervals. Locate the points (x1, y1) on the graph, where x1 denotes the class mark and y1 denotes the corresponding frequency. Join all the points plotted above with straight line segments. Join the first point and the last point to the points representing class marks of the class intervals before the first class interval and after the last class interval of the given frequency distribution.
Here, as the class limits do not start from 0, we put a kink between 0 and the lower boundary of the first class.
15. Read the following bar graph and answer the following questions:
a. What information is given by the graph?
b. Which state is the largest producer of wheat?
c. Which state is the largest producer of sugar?
d. Which state has total production of wheat and sugar as its maximum?
e. Which state has the total production of wheat and sugar minimum?
Solution:
a. The given graph gives information about production of wheat and sugar in five different states (U.P, Bihar, W.B, M.P, Punjab)
b. The largest producer of wheat is Punjab
c. The largest producer of sugar is U.P.
d. The state which has total production of wheat and sugar as its maximum is U.P.
e. The state which has total production of wheat and sugar minimum is W.B.
