The formula to find the mean of an ungrouped data is ........
- A$\bar{x}=\Sigma x_{i}$
- B$\bar{x}=\frac{\Sigma x_{i}}{n}$
- C$\bar{x}=\frac{\Sigma x_{i}}{2}$
- D$\bar{x}=n \cdot \Sigma x_{i}$
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Similar Questions
The mean of the following frequency distribution is $18$. Find the missing frequency $f$.
Class $11-13$ $13-15$ $15-17$ $17-19$ $19-21$ $21-23$ $23-25$ Frequency $3$ $6$ $9$ $13$ $f$ $5$ $4$
| Class | $11-13$ | $13-15$ | $15-17$ | $17-19$ | $19-21$ | $21-23$ | $23-25$ |
| Frequency | $3$ | $6$ | $9$ | $13$ | $f$ | $5$ | $4$ |
Medium
View SolutionThe median class in the following frequency distribution is ..........
Class $20-25$ $25-30$ $30-35$ $35-40$ $40-45$ $45-50$ $50-55$ Frequency $2$ $5$ $8$ $10$ $7$ $10$ $3$
| Class | $20-25$ | $25-30$ | $30-35$ | $35-40$ | $40-45$ | $45-50$ | $50-55$ |
| Frequency | $2$ | $5$ | $8$ | $10$ | $7$ | $10$ | $3$ |
Easy
View SolutionFind the mean of the following data:
Class $200-299$ $300-399$ $400-499$ $500-599$ $600-699$ $700-799$ $800-899$ Frequency $3$ $61$ $118$ $139$ $126$ $151$ $2$
| Class | $200-299$ | $300-399$ | $400-499$ | $500-599$ | $600-699$ | $700-799$ | $800-899$ |
| Frequency | $3$ | $61$ | $118$ | $139$ | $126$ | $151$ | $2$ |
Medium
View SolutionThe abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
Easy
View SolutionCumulative frequency distribution is used to calculate the $\ldots \ldots \ldots$ of a frequency distribution.
Easy
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