Question

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

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

Answer 1

We have

$N-{\sum }_{i-1}^6{f}_i=(6+8+15+3+8+4)=44$

$\overline{x}=\frac{{\sum }_{i=1}^6{f}_i{x}_i}{N}=\frac{(6\times 3)+(8\times 5)+(15\times 7)+(3\times 9)+(8\times 11)+(4\times 13)}{44}$

$=\frac{(18+40+105+27+88+52)}{44}=\frac{330}{44}=\frac{15}{2}=7.5$

$\begin{array}{ccccccc}{x}_i& 3& 5& 7& 9& 11& 13\\ f_i& 6& 8& 15& 3& 8& 4\\ cf& 6& 14& 29& 32& 40& 44\end{array}$

$\therefore$ N = 44, which is even.

$\therefore \text{Median}=\frac{1}{2}.\{\frac{N}{2}th\text{observation}+(\frac{N}{2}+1)th\text{observation}\}$

$=\frac{1}{2}$ $\text{{22nd observation + 23rd observation}}$

$=\frac{1}{2}(7+7)=7$

Thus, M = 7

Now, we have:

$\begin{array}{ccccccc}|x_i-M|& 4& 2& 0& 2& 4& 6\\ f_i& 6& 8& 15& 3& 8& 4\\ f_i|x_i-M|& 24& 16& 0& 6& 32& 24\end{array}$

$\therefore {\sum }_{i=1}^6{f}_i=44$ and ${\sum }_{i=1}^6{f}_i|x_i-M|=102$

$\therefore MD\left(\overline{x}\right)=\frac{{\sum }_{i=1}^6{f}_i|x_i-M|}{N}=\frac{102}{44}=2.32$

Hence, the mean deviation about the median is 2.32