The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:
Represent the data of both the teams on the same graph by frequency polygons. [Hint: First make the class intervals continuous.]
Solution:
- It can be observed from the given data that the class intervals of the given data are not continuous. There is a gap of ‘1’ unit between them. So, to make the class intervals continuous, 0.5 has to be added to every upper-class limit and 0.5 has to be subtracted from the lower-class limit:
- The number of balls class mark can also be found as shown below:
Class Mark = (Upper Limit + Lower Limit) / 2
The data table with continuous interval and with class mark is as below:
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 |
The frequency polygon for the above data can be constructed by
- Number of balls on x-axis using class mark values.
- Runs scored on y-axis with an approximate scale of “1 unit = 1 run” as the lowest run is 1 and the highest is10.
Video Solution:
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below: Represent the data of both the teams on the same graph by frequency polygons
NCERT Solutions for Class 9 Maths - Chapter 14 Exercise 14.3 Question 7:
Summary:
The data of runs scored by two teams A and B on the first 60 balls in a cricket match is given. We have made a frequency polygon to represent the data of both teams.