| \[{{x}_{i}}\] | \[5\] | \[10\] | \[15\] | \[20\] | \[25\] |
| \[{{f}_{i}}\] | \[7\] | \[4\] | \[6\] | \[3\] | \[5\] |
Solution:-
We have to make the table of the given data and append other columns after calculations.
| \[{{x}_{i}}\] | \[{{f}_{i}}\] | \[{{f}_{i}}{{x}_{i}}\] | \[\left| {{x}_{i}}-\overline{x} \right|\] | \[{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|\] |
| \[5\] | \[7\] | \[35\] | \[9\] | \[63\] |
| \[10\] | \[4\] | \[40\] | \[4\] | \[16\] |
| \[15\] | \[6\] | \[90\] | \[1\] | \[6\] |
| \[20\] | \[3\] | \[60\] | \[6\] | \[18\] |
| \[25\] | \[5\] | \[125\] | \[11\] | \[55\] |
| \[25\] | \[350\] | \[158\] |
The sum of calculated data,
N = \[\sum\limits_{i=1}^{5}{{{f}_{i}}}=25\], \[\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}}=350\]
Find mean by using the below formula
\[\overline{x}\] =\[\frac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}}=\frac{350}{25}\]= \[14\]
From the table, \[\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}=158\]
Therefore, mean deviation = \[\frac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}\]
=\[(1/25)\times 158\]
=\[6.32\]
Therefore, the mean deviation about the mean is \[6.32\]