Question

Find the mean deviation from the mean and from median of the following distribution:

Marks	0-10	10-20	20-30	30-40	40-50
No. of students	5	8	15	16	6

Answer 1

Computation of mean distribution from the median: 

Marks	Number of Students \[f_i\]	Cumulative Frequency	Midpoints \[x_i\]	\[\left| d_i \right| = \left| x_i - 28 \right|\]	\[f_i \left| d_i \right|\]	\[f_i x_i\]	\[\left| x_i - 27 \right|\]	\[f_i \left| x_i - 27 \right|\]
0−10	5	5	5	23	115	25	22	110
10−20	8	13	15	13	104	120	12	96
20−30	15	28	25	3	45	375	2	30
30−40	16	44	35	7	112	560	8	128
40−50	6	50	45	17	102	270	18	108
	\[N = 50\]				\[\sum^5_{i = 1} f_i \left| d_i \right| = 478\]	1350		\[\sum^5_{i = 1} f_i \left| x_i - 27 \right| = 472\]

 

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

The cumulative frequency just greater than  \[\frac{N}{2} = 25\] is 28 and the corresponding class is 20−30.
Thus, the median class is 20−30.

Using formula:

\[\therefore l = 20, F = 13, f = 15, h = 10 \]
\[ \text{ Median }  = l + {\frac{\frac{N}{2} - F}{f}} \times h \]
\[\text{ Substituting the values: } \]
\[\text{ Median }  = 20 + {\frac{25 - 13}{15}} \times 10 \]
\[ = 20 + 8 \]
\[ = 28\]

\[\text{ Mean distribution from the median } = {\frac{\sum^5_{i = 1} f_i \left| d_i \right|}{N}} \]
\[ = {\frac{478}{50}}\]
\[ = 9 . 56\]

\[ \text{ Mean } (\bar  {X}) = {\frac{\sum^5_{i = 1} f_i x_i}{N}}\]
\[ = {\frac{1350}{50}}\]

\[ = 27\]
\[\text{ Mean deviation from the mean } ={\frac{1}{50}} \times \sum^5_{i = 1} f_i \left|  {x_i - 27} \right|\]
\[ ={\frac{472}{50}}\]
\[ = 9 . 44\]

 Mean deviation from the median and the mean are 9.56 and 9.44, respectively.