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प्रश्न
The marks obtained by 50 students in Mathematics are given below.
(i) Make a frequency distribution table taking a class size of 10 marks
(ii) Draw a histogram and a frequency polygon.
| 52 | 33 | 56 | 52 | 44 | 59 | 47 | 61 | 49 | 61 |
| 47 | 52 | 67 | 39 | 89 | 57 | 64 | 58 | 63 | 65 |
| 32 | 64 | 50 | 54 | 42 | 48 | 22 | 37 | 59 | 63 |
| 36 | 35 | 48 | 48 | 55 | 62 | 74 | 43 | 41 | 51 |
| 08 | 71 | 30 | 18 | 43 | 28 | 20 | 40 | 58 | 49 |
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उत्तर
Maximum marks obtained = 89
Minimum marks obtained = 08
Range = Maximum marks – Minimim marks
= 89 – 08
= 81
Taking the class size = 10, then
Number of possible intervals = `"Range"/"Class size"`
= `81/10`
= 8.1
= 9
| Class Interval | Tally marks | Frequency |
| 0 − 10 | I | 1 |
| 10 − 20 | I | 1 |
| 20 − 30 | III | 3 |
| 30 − 40 | IIII III | 8 |
| 40 − 50 | IIII IIII III | 13 |
| 50 − 60 | IIII IIII II | 12 |
| 60 − 70 | IIII IIII | 9 |
| 70 − 80 | II | 2 |
| 80 − 90 | I | 1 |
| Total | 50 | 50 |
Now we have the continuous frequency table.
| Class Intervals | 0 − 10 | 10 − 20 | 20 − 30 | 30 − 40 | 40 − 50 | 50 − 60 | 60 − 70 | 70 − 80 | 80 − 90 |
| Frequency | 1 | 1 | 3 | 8 | 13 | 12 | 9 | 2 | 1 |
We will draw the histogram taking class interval in x-axis and frequency in y-axis as follows.

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संबंधित प्रश्न
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| 28-30 | 4 |
| 30-32 | 8 |
| 32-34 | 22 |
| 34-36 | 12 |
| 36-38 | 6 |
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| Maximum Speed (km/h) | Frequency |
| 10 – 20 | 5 |
| 20 – 30 | 5 |
| 30 – 40 | 10 |
| 40 – 50 | 8 |
| 50 – 60 | 0 |
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