Question

Find the mean deviation about the mean for the following data:

$\begin{array}{ccccccc}{x}_i& 3& 5& 7& 9& 11& 13\\ f_i& 6& 8& 15& 25& 8& 4\end{array}$

Answer 1

We have

$N={\sum }_{i=1}^6{f}_i=(6+8+15+25+8+4)=66$

$\overline{x}=\frac{{\sum }_{i=1}^6{f}_i{x}_i}{N}=\frac{(6\times 3)+(8\times 5)+(15\times 7)+(25\times 9)+(8\times 11)+(4\times 13)}{66}$

$=\frac{528}{66}=8$

Now, we prepare the table given below:

$\begin{array}{cccc}{x}_i& f_i& |x_i-\overline{x}|& {\sum }_{i=1}^6{f}_i|x_i-\overline{x}|\\ 3& 6& 5& 30\\ 5& 8& 3& 24\\ 7& 15& 1& 15\\ 9& 25& 1& 25\\ 11& 8& 3& 24\\ 13& 4& 5& 20\\ & N=66& & \text{sum = 138}\end{array}$

$\therefore MD\left(\overline{x}\right)=\frac{{\sum }_{i=1}^6{f}_i|x_i-\overline{x}|}{N}=\frac{138}{66}=\frac{23}{11}=2.09$

Hence, mean deviation about the mean = 2.09