Question

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

(i) $\begin{array}{ccccccc}{x}_i& 5& 7& 9& 10& 12& 15\\ f_i& 8& 6& 2& 2& 2& 6\end{array}$

(ii) $\begin{array}{cccccc}{x}_i& 5& 10& 15& 20& 25\\ f_i& 7& 4& 6& 3& 5\end{array}$

(iii) $\begin{array}{cccccc}{x}_i& 10& 30& 50& 70& 90\\ f_i& 4& 24& 28& 16& 8\end{array}$

(iv) $\begin{array}{cccccc}\text{Size}& 20& 21& 22& 23& 24\\ \text{Frequency}& 6& 4& 5& 1& 4\end{array}$

(v) $\begin{array}{ccccccccc}\text{Size}& 1& 3& 5& 7& 9& 11& 13& 15\\ \text{Frequency}& 3& 3& 4& 14& 7& 4& 3& 4\end{array}$

Answer 1

$\begin{array}{ccccc}{x}_i& f_i& \text{CF}& |x_i-M|& f_i|x_i-M|\\ 5& 8& 8& |5-7|=2& 16\\ 7& 6& 14& |7-7|=0& 00\\ 9& 2& 16& |9-7|=2& 04\\ 10& 2& 18& |10-7|=3& 06\\ 12& 2& 20& |12-7|=5& 10\\ 15& 6& 26& |15-7|=8& 48\\ \text{Total}& \sum f_i=26& & & \sum f_i|x_i-M|=84\end{array}$

Here, $N=\sum f_i=26$ even

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

$\frac{\frac{26}{2}\text{th observation}+(\frac{26}{2}+1)\text{th observation}}{2}$

$=\frac{\text{13th observation + 14th observation}}{2}$

$\therefore$ Mean deviation about median $=\frac{\sum f_i|x_i-M|}{\sum f_i}=\frac{84}{26}=3.23$

$\begin{array}{ccccc}{x}_i& f_i& f_i{x}_i& |x_i-\overline{x}|& f_i|x_i-\overline{x}|\\ 5& 7& 35& |5-14|=9& 63\\ 10& 4& 40& |10-14|=4& 16\\ 15& 6& 90& |15-14|=1& 06\\ 20& 3& 60& |20-14|=6& 18\\ 25& 5& 125& |25-14|=11& 55\\ \text{Total}& \sum f_i=25& 350& & 158\end{array}$

Mean $\overline{=}\frac{\sum f_i{x}_i}{\sum f_i}=\frac{350}{25}=14$

$\therefore$ Mean deviation about mean $=\frac{\sum f_i|x_i-\overline{x}|}{\sum f_i}=\frac{158}{25}=6.32$

$\begin{array}{ccccc}{x}_i& f_i& f_i{x}_i& |x_i-\overline{x}|& f_i|x_i-\overline{x}|\\ 10& 4& 40& |10-50|=40& 160\\ 30& 24& 720& |30-50|=20& 480\\ 50& 28& 1400& |50-50|=00& 000\\ 70& 16& 1120& |70-50|=20& 320\\ 90& 8& 720& |90-50|=40& 320\\ \text{Total}& \sum f_i=80& \sum f_i{x}_i=4000& & 1280\end{array}$

Mean $\overline{=}\frac{\sum f_i{x}_i}{\sum f_i}=\frac{4000}{80}=50$

$\therefore$ Mean deviation about mean $=\frac{\sum f_i|x_i-\overline{x}|}{\sum f_i}=\frac{1280}{80}=16$

$\begin{array}{ccccc}\text{Size}& \text{Frequency}& f_i{x}_i& d_i=|x_i-x|& f_i{d}_i\\ 20& 6& 120& 1.65& 9.90\\ 21& 4& 84& 0.65& 2.60\\ 22& 5& 110& 0.35& 1.75\\ 23& 1& 23& 1.35& 1.35\\ 24& 4& 96& 2.35& 9.40\\ \text{Total}& 20& 433& & 25\end{array}$
Now, $\overline{x}=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{433}{20}=21.65$
$M.D.=\frac{\sum f_i|x_i-\overline{x}|}{\sum f_i}=\frac{25}{20}=1.25$

$\begin{array}{ccccc}{x}_i& f_i& f_i{x}_i& |x_i-\overline{x}|-|x_i-8|& f_i|x_i-\overline{x}|\\ 1& 3& 3& |1-8|=7& 21\\ 3& 3& 9& |3-8|=5& 15\\ 5& 4& 20& |5-8|=3& 12\\ 7& 14& 98& |7-8|=1& 14\\ 9& 7& 63& |9-8|=1& 7\\ 11& 4& 44& |11-8|=3& 12\\ 13& 3& 39& |13-8|=5& 15\\ 15& 4& 60& |15-8|=7& 28\\ \text{Total}& \sum f_i=42& \sum f_i{x}_i=336& & 124\end{array}$

Now, $\overline{x}=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{336}{42}=8$

$M.D.=\frac{\sum f_i|x_i-\overline{x}|}{\sum f_i}=\frac{124}{42}=2.95$