The formula to find the mode of a grouped data is $\ldots \ldots \ldots . . .$
- A$Z=l+\frac{(\frac{n}{2}-cf)}{f} \times c$
- B$Z=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h$
- C$Z=3M - 2\bar{x}$
- D$Z= A +\frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h$
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Similar Questions
Find the mode of the following frequency distribution:
Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$ Frequency $8$ $9$ $26$ $23$ $20$ $14$
| Class | $0-20$ | $20-40$ | $40-60$ | $60-80$ | $80-100$ | $100-120$ |
| Frequency | $8$ | $9$ | $26$ | $23$ | $20$ | $14$ |
Medium
View SolutionThe wickets taken by a bowler in $10$ cricket matches are as follows:
$2, 6, 4, 5, 0, 2, 1, 3, 2, 3$
Find the mode of the data.
$2, 6, 4, 5, 0, 2, 1, 3, 2, 3$
Find the mode of the data.
Easy
View SolutionFind the mode of the following frequency distribution:
Class $0-100$ $100-200$ $200-300$ $300-400$ $400-500$ $500-600$ Frequency $7$ $21$ $37$ $13$ $12$ $10$
| Class | $0-100$ | $100-200$ | $200-300$ | $300-400$ | $400-500$ | $500-600$ |
|---|---|---|---|---|---|---|
| Frequency | $7$ | $21$ | $37$ | $13$ | $12$ | $10$ |
Medium
View SolutionIn a frequency distribution,the modal class is $70-85$ with frequency $25$. The frequencies of the classes succeeding and preceding the modal class are $20$ and $8$ respectively. Then,the values of $f_{0}, f_{1}$ and $f_{2}$ are respectively.
Medium
View SolutionIn the formula $Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$ for the mode,$l = \ldots \ldots \ldots$
Easy
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