Question

Compute the mean deviation from the median of the following distribution:

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	10	20	5	10

Answer 1

Class	Frequency   \[f_i\]	Cumulative frequency	\[x_i\]	\[\left| d_i \right| = \left| x_i - 25 \right|\]	\[f_i \left| d_i \right|\]
0−10	5	5	5	20	100
10−20	10	15	15	10	100
20−30	20	35	25	0	0
30−40	5	40	35	10	50
40−50	10	50	45	20	200
	\[N = \sum ^5_{i = 1} f_i= 50\]				\[N = \sum f_i5_{i = 1} = 50\ \[\sum f_i \left| d_i \right|^5_{i = 1} = 450\]

\[Here, N = 50 \]
\[ \Rightarrow \frac{N}{2} = 25\]

The cumulative frequency greater than \[\frac{N}{2} = 25\] is 35 and the corresponding class is 20−30.
Therefore, the median class is  20−30.

\[ \therefore l = 20, f = 20, F = 15, N = 50, h = 10\]
\[ \therefore \text{ Median } = l + \frac{\left( \frac{N}{2} - F \right)}{f} \times h \]
\[ = 20 + \frac{\left( \frac{50}{2} - 15 \right)}{20} \times 10 \]
\[ = 20 + \frac{\left( 25 - 15 \right)}{20} \times 10 \]
\[ = 25\]

\[\text{ Mean deviation from the median } = \frac{\sum^5_{i = 1} f_i \left| d_i \right|}{N} = \frac{450}{50} = 9\]