
The runs scored by the 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.
Number of balls Team A Team B 1-16 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
| Number of balls | Team A | Team B |
| 1-16 | 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 |
Answer
551.7k+ views
Hint: First of all we have to know what frequency polygons are, a frequency polygon is a graphical form of representation of data. The relevance of presentation of data in the pictorial or graphical form is immense. Frequency polygon, it is used to depict the shape of the data and to depict trends. It is usually drawn with the help of a histogram but can be drawn without it as well. So basically the frequency polygon is a graphical device for understanding the shapes of distributions.
Complete step-by-step solution:
Here it is given the problem that there are 2 teams named A and B, and their scored runs in the first 60 balls in a cricket match.
Let us learn the step by step process of drawing a frequency polygon:
First we have to calculate the class mark of each class interval. Where the formula for the class mark is given by:
$ \Rightarrow $Class mark = (Upper limit + Lower limit) / 2
The class marks are also known as the mid-value of each class.
Now mark the class marks for each class interval on the horizontal axis.
Plotting the runs of each team on the vertical axis.
Corresponding to each class mark, plot the runs of both the teams as given.
The height always depicts the runs scored by the teams.
We have to make sure that the runs scored by the teams are plotted against the class mark but not the upper or lower limit of the class interval.
Now join all the plotted points using a line segment. The curve obtained here will be kinked, this is called the frequency polygon graph.
Hence the resulting curve is called the frequency polygon.
Now plotting the frequency polygon here, as given below:
Note: Frequency polygons are a good choice for displaying cumulative frequency distributions. Also note that we can also draw a frequency polygon with the use of a histogram as well, by joining the midpoints of the bars.
Complete step-by-step solution:
Here it is given the problem that there are 2 teams named A and B, and their scored runs in the first 60 balls in a cricket match.
Let us learn the step by step process of drawing a frequency polygon:
First we have to calculate the class mark of each class interval. Where the formula for the class mark is given by:
$ \Rightarrow $Class mark = (Upper limit + Lower limit) / 2
The class marks are also known as the mid-value of each class.
Now mark the class marks for each class interval on the horizontal axis.
Plotting the runs of each team on the vertical axis.
Corresponding to each class mark, plot the runs of both the teams as given.
The height always depicts the runs scored by the teams.
We have to make sure that the runs scored by the teams are plotted against the class mark but not the upper or lower limit of the class interval.
Now join all the plotted points using a line segment. The curve obtained here will be kinked, this is called the frequency polygon graph.
Hence the resulting curve is called the frequency polygon.
Now plotting the frequency polygon here, as given below:
Note: Frequency polygons are a good choice for displaying cumulative frequency distributions. Also note that we can also draw a frequency polygon with the use of a histogram as well, by joining the midpoints of the bars.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

