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प्रश्न
Find the median of the following frequency distribution:
| Class: | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 |
| Frequency: | 6 | 8 | 5 | 9 | 7 |
उत्तर
| Class Interval | Frequency | Cumulative frequency (cf) |
| 0 – 20 | 6 | 6 |
| 20 – 40 | 8 | 14 |
| 40 – 60 | 5 | 19 |
| 60 – 80 | 9 | 28 |
| 80 – 100 | 7 | 35 |
| N = `sumf_i` = 35 |
Here, N = 35
⇒ `"N"/2 = 35/2`
So, Median class is 40 – 60
Lower limit of median class (l) = 40
Class width (h) = 20
Frequency of median class (f) = 5
Precceding cf of median class (`"C"_f`) = 14
∴ Median = `l + ((N/2 - "C"_f)/f) xx h`
= `40 + ((35/2 - 14)/5) xx 20`
= 40 + 7 × 2
= 54
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संबंधित प्रश्न
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| 118 − 126 | 3 |
| 127 – 135 | 5 |
| 136 − 144 | 9 |
| 145 – 153 | 12 |
| 154 – 162 | 5 |
| 163 – 171 | 4 |
| 172 – 180 | 2 |
Find the median length of the leaves.
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| More than 150 | 0 |
| More than 140 | 12 |
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| More than 110 | 105 |
| More than 100 | 124 |
| More than 90 | 141 |
| More than 80 | 150 |
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| Marks | Number of students |
| More than or equal to 0 | 100 |
| More than or equal to 10 | 93 |
| More than or equal to 20 | 88 |
| More than or equal to 30 | 70 |
| More than or equal to 40 | 59 |
| More than or equal to 50 | 42 |
| More than or equal to 60 | 34 |
| More than or equal to 70 | 20 |
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