In the formula $M = l + \frac{(\frac{n}{2} - cf)}{f} \times h$ for the median,$f = \ldots \ldots \ldots$
- Acumulative frequency of the class preceding the median class
- Bfrequency of the median class
- Clower limit of the median class
- Dclass size
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Similar Questions
Calculate the mean of the scores of $20$ students in a mathematics test:
Marks $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ Number of students $2$ $4$ $7$ $6$ $1$
| Marks | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ |
| Number of students | $2$ | $4$ | $7$ | $6$ | $1$ |
Medium
View SolutionThe weight of coffee in $70$ packets is shown in the following table:
Weight (in $g$) Number of packets $200-201$ $12$ $201-202$ $26$ $202-203$ $20$ $203-204$ $9$ $204-205$ $2$ $205-206$ $1$
Determine the modal weight (in $g$).
| Weight (in $g$) | Number of packets |
| $200-201$ | $12$ |
| $201-202$ | $26$ |
| $202-203$ | $20$ |
| $203-204$ | $9$ |
| $204-205$ | $2$ |
| $205-206$ | $1$ |
Determine the modal weight (in $g$).
Easy
View SolutionIs it true to say that the mean,mode,and median of grouped data will always be different? Justify your answer.
Easy
View SolutionFor some given data,if $Z = 95$ and $\bar{x} = 98,$ then $M = \ldots \ldots \ldots$
Easy
View SolutionFind the mode of the following frequency distribution:
Class $0-15$ $15-30$ $30-45$ $45-60$ $60-75$ $75-90$ $90-105$ Frequency $4$ $5$ $23$ $45$ $66$ $42$ $15$
| Class | $0-15$ | $15-30$ | $30-45$ | $45-60$ | $60-75$ | $75-90$ | $90-105$ |
| Frequency | $4$ | $5$ | $23$ | $45$ | $66$ | $42$ | $15$ |
Medium
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