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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
Draw a frequency polygon from the information given in the following table.
| Age of blood donar (Years) | No. of blood donars |
| Less than 20 | 0 |
| Less than 25 | 30 |
| 75Less than 30 | 75 |
| 165Less than 35 | 127 |
| Less than 40 | 165 |
| Less than 45 | 185 |
| Less than 50 | 197 |
Draw a frequency polygon for the following grouped frequency distribution table.
|
Age of the donor
(Yrs.) |
20 - 24 | 25 - 29 | 30 - 34 | 35 - 39 | 40 - 44 | 45 - 49 |
| No. of blood doners | 38 | 46 | 35 | 24 | 15 | 12 |
The following table shows the average rainfall in 150 towns. Show the information by a frequency polygon.
|
Average rainfall (cm)
|
0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 |
| No. of towns | 14 | 12 | 36 | 48 | 40 |
The maximum bowing speed (km/hour) or 33 players at a cricket coaching centre is given below:
| Bowling Speed(km/hour) | 85-100 | 100-115 | 115-130 | 130-145 |
| Numbers of players | 9 | 11 | 8 | 5 |
find the modal bowling speed of players.
The distribution of heights (in cm) of 100 people is given below. Construct a histogram and the frequency polygon imposed on it.
| Height (in cm) |
125 − 135 | 136 − 146 | 147 − 157 | 158 − 168 | 169 − 179 | 180 − 190 | 191 − 201 |
| Frequency | 12 | 22 | 18 | 24 | 15 | 7 | 2 |
In the following frequency distribution, the heights of 90 students are recorded and the frequency polygon is drawn.
| Heights | 30 – 35 | 35 – 40 | 40 – 45 | 45 – 50 | 50 – 55 | 55 – 60 |
| No. of Students | 6 | 12 | 20 | 18 | 24 | 10 |
For the classes 18-19, 19-20, 20-21, 21-22, 22-23, and 23-24 with frequencies 4, 13, 15, 19, 17, and 6, the coordinate pair for the class 20-21 is:
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?
When plotting a frequency polygon, which variable is taken on the y-axis?
