For the calculation of mode of the following frequency distribution,$f_{1} = \dots \dots \dots \dots \dots \dots$
Class $0-100$ $100-200$ $200-300$ $300-400$ $400-500$ $500-600$ Frequency $12$ $18$ $27$ $20$ $17$ $6$
| Class | $0-100$ | $100-200$ | $200-300$ | $300-400$ | $400-500$ | $500-600$ |
| Frequency | $12$ | $18$ | $27$ | $20$ | $17$ | $6$ |
- A$18$
- B$20$
- C$27$
- D$17$
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For a given frequency distribution,$\bar{x}=20, \Sigma f_{i} u_{i}=-50, n=100$ and $c=10$. Then,assumed mean $A = \ldots \ldots \ldots \ldots$
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View SolutionThe mean of $15$ observations is $25$. Later on,it was found that one observation was taken by mistake as $20$ instead of $50$. Then,the correct mean is:
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Marks $30-35$ $35-40$ $40-45$ $45-50$ $50-55$ $55-60$ $60-65$ Frequency $14$ $16$ $18$ $23$ $18$ $8$ $3$
Determine the median percentage of marks.
| Marks | $30-35$ | $35-40$ | $40-45$ | $45-50$ | $50-55$ | $55-60$ | $60-65$ |
| Frequency | $14$ | $16$ | $18$ | $23$ | $18$ | $8$ | $3$ |
Determine the median percentage of marks.
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View SolutionWill the median class and modal class of grouped data always be different? Justify your answer.
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View SolutionFor the calculations of mode of a grouped data,the maximum frequency of a class is denoted by $\ldots \ldots \ldots . .$
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