Calculate the mean deviation from the median of the following data:
Class interval $0-6$ $6-12$ $12-18$ $18-24$ $24-30$ Frequency $4$ $5$ $3$ $6$ $2$
| Class interval | $0-6$ | $6-12$ | $12-18$ | $18-24$ | $24-30$ |
| Frequency | $4$ | $5$ | $3$ | $6$ | $2$ |
(7) Step $1$: Find the cumulative frequency $(cf)$ and total frequency $(N)$.
Step $2$: Find the median class.
Since $N = 20$,$\frac{N}{2} = 10$. The cumulative frequency just greater than $10$ is $12$,so the median class is $12-18$.
Step $3$: Calculate the median $(M_d)$.
$M_d = l + \left( \frac{\frac{N}{2} - C}{f} \right) \times h = 12 + \left( \frac{10 - 9}{3} \right) \times 6 = 12 + 2 = 14$.
Step $4$: Calculate mean deviation $(MD)$.
$MD = \frac{\sum f_i |x_i - M_d|}{N} = \frac{4|3-14| + 5|9-14| + 3|15-14| + 6|21-14| + 2|27-14|}{20} = \frac{4(11) + 5(5) + 3(1) + 6(7) + 2(13)}{20} = \frac{44 + 25 + 3 + 42 + 26}{20} = \frac{140}{20} = 7$.
| Class interval | $f_i$ | $x_i$ | $cf$ |
| $0-6$ | $4$ | $3$ | $4$ |
| $6-12$ | $5$ | $9$ | $9$ |
| $12-18$ | $3$ | $15$ | $12$ |
| $18-24$ | $6$ | $21$ | $18$ |
| $24-30$ | $2$ | $27$ | $20$ |
Step $2$: Find the median class.
Since $N = 20$,$\frac{N}{2} = 10$. The cumulative frequency just greater than $10$ is $12$,so the median class is $12-18$.
Step $3$: Calculate the median $(M_d)$.
$M_d = l + \left( \frac{\frac{N}{2} - C}{f} \right) \times h = 12 + \left( \frac{10 - 9}{3} \right) \times 6 = 12 + 2 = 14$.
Step $4$: Calculate mean deviation $(MD)$.
$MD = \frac{\sum f_i |x_i - M_d|}{N} = \frac{4|3-14| + 5|9-14| + 3|15-14| + 6|21-14| + 2|27-14|}{20} = \frac{4(11) + 5(5) + 3(1) + 6(7) + 2(13)}{20} = \frac{44 + 25 + 3 + 42 + 26}{20} = \frac{140}{20} = 7$.
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