Form the frequency distribution table from the following cumulative frequency data:
Marks (out of $90$) Number of candidates More than or equal to $80$ $4$ More than or equal to $70$ $6$ More than or equal to $60$ $11$ More than or equal to $50$ $17$ More than or equal to $40$ $23$ More than or equal to $30$ $27$ More than or equal to $20$ $30$ More than or equal to $10$ $32$ More than or equal to $0$ $34$
| Marks (out of $90$) | Number of candidates |
| More than or equal to $80$ | $4$ |
| More than or equal to $70$ | $6$ |
| More than or equal to $60$ | $11$ |
| More than or equal to $50$ | $17$ |
| More than or equal to $40$ | $23$ |
| More than or equal to $30$ | $27$ |
| More than or equal to $20$ | $30$ |
| More than or equal to $10$ | $32$ |
| More than or equal to $0$ | $34$ |
(N/A) To form the frequency distribution table,we subtract the cumulative frequencies of consecutive intervals to find the frequency of each specific class interval.
Given that the total number of candidates is $34$ (as seen in the 'More than or equal to $0$' category),we calculate the frequency for each class interval as follows:
Given that the total number of candidates is $34$ (as seen in the 'More than or equal to $0$' category),we calculate the frequency for each class interval as follows:
| Class interval | Frequency |
| $0-10$ | $34-32 = 2$ |
| $10-20$ | $32-30 = 2$ |
| $20-30$ | $30-27 = 3$ |
| $30-40$ | $27-23 = 4$ |
| $40-50$ | $23-17 = 6$ |
| $50-60$ | $17-11 = 6$ |
| $60-70$ | $11-6 = 5$ |
| $70-80$ | $6-4 = 2$ |
| $80-90$ | $4$ |
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The median class in the following frequency distribution is ..........
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