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प्रश्न
Following table shows a frequency distribution for the speed of cars passing through at a particular spot on a high way:
| Class interval (km/h) | Frequency |
| 30 – 40 | 3 |
| 40 – 50 | 6 |
| 50 – 60 | 25 |
| 60 – 70 | 65 |
| 70 – 80 | 50 |
| 80 – 90 | 28 |
| 90 – 100 | 14 |
Draw the frequency polygon representing the above data without drawing the histogram.
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उत्तर
First we obtain in the class marks (mid-marks) of the given table as:
Class-marks = `("Lower limit" + "Upper limit")/2`
Since, the new table is shown below:
| Class interval | Class marks | Frequency |
| 30 – 40 | 35 | 3 |
| 40 – 50 | 45 | 6 |
| 50 – 60 | 55 | 25 |
| 60 – 70 | 65 | 65 |
| 70 – 80 | 75 | 50 |
| 80 – 90 | 85 | 28 |
| 90 – 100 | 95 | 14 |
Now, let’s draw a frequency polygon by plotting the class marks along the x-axis and the frequency along y-axis.
Also, plotting all the points as B(35, 3), C(45, 6), (D(55, 25), E(65, 65), F(75, 50), G(85, 28) and H(95, 14).
Then join all these point line segment, shown below:
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