For a frequency distribution,the mean deviation from the mean is computed by:
- A$M.D. = \frac{\sum d}{\sum f}$
- B$M.D. = \frac{\sum fd}{\sum f}$
- C$M.D. = \frac{\sum f|d|}{\sum f}$
- D$M.D. = \frac{\sum f}{\sum f|d|}$
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Similar Questions
Find the mean deviation about the mean for the given data:
$x_i$ $5$ $10$ $15$ $20$ $25$ $f_i$ $7$ $4$ $6$ $3$ $5$
| $x_i$ | $5$ | $10$ | $15$ | $20$ | $25$ |
| $f_i$ | $7$ | $4$ | $6$ | $3$ | $5$ |
Medium
View SolutionThe mean deviation from the mean of the data given below is
$\text{Marks}$ $10$ $15$ $20$ $25$ $30$ $\text{Number of students}$ $2$ $4$ $6$ $8$ $5$
| $\text{Marks}$ | $10$ | $15$ | $20$ | $25$ | $30$ |
| $\text{Number of students}$ | $2$ | $4$ | $6$ | $8$ | $5$ |
The mean deviation about the median for the observations $70, 42, 63, 34, 44, 54, 55, 46, 38, 48$ is:
Medium
View SolutionThe mean deviation about the mean for the following data is:
$\text{Marks obtained}$ $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $\text{Number of Boys}$ $6$ $8$ $10$ $4$ $2$
| $\text{Marks obtained}$ | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ |
| $\text{Number of Boys}$ | $6$ | $8$ | $10$ | $4$ | $2$ |
Calculate the mean deviation about the median age for the age distribution of $100$ persons given below:
Age (in years) $16-20$ $21-25$ $26-30$ $31-35$ $36-40$ $41-45$ $46-50$ $51-55$ Number of persons $5$ $6$ $12$ $14$ $26$ $12$ $16$ $9$
| Age (in years) | $16-20$ | $21-25$ | $26-30$ | $31-35$ | $36-40$ | $41-45$ | $46-50$ | $51-55$ |
| Number of persons | $5$ | $6$ | $12$ | $14$ | $26$ | $12$ | $16$ | $9$ |
Difficult
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