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प्रश्न
In a study of dental problem, the following data were obtained.
| Ages | 0 − 10 | 10 − 20 | 20 − 30 | 30 − 40 | 40 − 50 | 50 − 60 | 60 − 70 | 70 − 80 |
| No. of patients | 5 | 13 | 25 | 14 | 30 | 35 | 43 | 50 |
Represent the above data by a frequency polygon.
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उत्तर
Finding the midpoints of the class interval we get.
| Ages | Mid point (x) | No. of patients |
| 0 – 10 | 5 | 5 |
| 10 – 20 | 15 | 13 |
| 20 – 30 | 25 | 25 |
| 30 – 40 | 35 | 14 |
| 40 – 50 | 45 | 30 |
| 50 – 60 | 55 | 35 |
| 60 – 70 | 65 | 43 |
| 70 – 80 | 75 | 50 |
The points to be plotted are A(5, 5), B(15, 13), C(25, 25), D(35, 14), E(45, 30), G(65, 43), H(75, 50) to obtain the frequency polygon ZABCDEFGHI.
Where I imagined class between 80 and 90.
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संबंधित प्रश्न
In a handloom factory different workers take different periods of time to weave a saree. The number of workers and their required periods are given below. Present the information by a frequency polygon.
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8 - 10 | 10 - 12 | 12 - 14 | 14 - 16 | 16 - 18 | 18 - 20 |
| No. of workers | 5 | 16 | 30 | 40 | 35 | 14 |
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20 - 24 | 25 - 29 | 30 - 34 | 35 - 39 | 40 - 44 | 45 - 49 |
| No. of blood doners | 38 | 46 | 35 | 24 | 15 | 12 |
A _________ is a line graph for the graphical representation of the continuous frequency distribution
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| Heights | 30 – 35 | 35 – 40 | 40 – 45 | 45 – 50 | 50 – 55 | 55 – 60 |
| No. of Students | 6 | 12 | 20 | 18 | 24 | 10 |
A graph formed by plotting class marks against their corresponding frequencies and joining consecutive points with straight line segments is called a:
Each point plotted in a frequency polygon has coordinates:
Which formula gives the class mark of a class interval?
What is the class mark for the class interval 18-19?
Why are two neighbouring zero-frequency classes added when constructing a frequency polygon without a histogram?
For the class intervals 18-19 through 23-24, the frequency polygon is extended to touch the x-axis at which class marks?
