Question

The lengths (in cm) of 10 rods in a shop are given below :

40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2

(i) Find mean deviation from median

(ii) Find mean deviation from the median also.

Answer 1

(i) Formula for the mean deviation from the median is as follows.

$M.D.=M.D.=\frac{1}{n}{\sum }_{i=1}^n|d_i|$, where $|d_i|=|x_i-M|$

Arranging the data in ascending order for finding the mean

15.2, 27.9, 30.2, 32.5, 40, 52.3, 52.8, 55.5, 72.9, 79

Here, n = 10

Therefore, median is the average of the fifth and the sixth observations.

$M=\frac{40+52.3}{2}=46.15$

$\begin{array}{cc}{x}_i& |d_i|=|x_i-46.15|\\ 40& 6.15\\ 52.3& 6.15\\ 55.2& 9.05\\ 72.9& 26.75\\ 52.8& 6.65\\ 79& 32.85\\ 32.5& 13.65\\ 15.2& 30.95\\ 27.9& 18.25\\ 32& 14.15\\ \text{Total}& 164.6\end{array}$

$M.D.=\frac{1}{10}\times 164.6=16.46$

Mean deviation from median in 16.4 cm.

(ii) Let $\overline{x}$ be the mean of the given data set.

$\overline{=}\frac{40+52.3+55.2+72.9+52.8+52.8+79+32.5+15.2+27.9+30.2}{10}=45.98$

$\begin{array}{cc}{x}_i& |d_i|=|x_i-45.98|\\ 40& 5.98\\ 52.3& 6.32\\ 55.2& 9.22\\ 72.9& 26.92\\ 52.8& 6.82\\ 79& 33.02\\ 32.5& 13.48\\ 15.2& 30.78\\ 27.9& 18.08\\ 32& 13.98\\ \text{Total}& 164.6\end{array}$

$M.D.=\frac{1}{10}\times 164.6=16.46$

Mean deviation from the mean is 16.4 cm.