Question

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

$\begin{array}{ccccccc}\text{Marks obtained}& 10-20& 20-30& 30-40& 40-50& 50-60& 60-70\\ \text{Number of students}& 8& 6& 12& 5& 2& 7\end{array}$

Answer 1

Here h = 10. Let the assumed mean be A = 35.

Then, we prepared the table given below.

$\begin{array}{ccccccc}\text{Marks obtained}& \text{Number of students}\left(f_i\right)& \text{Midpoint}{x}_i& d_i=\frac{x_i-35}{10}& f_i{d}_i& |x_i-\overline{x}|& f_i\times |x_i-\overline{x}|\\ 10-20& 8& 15& -2& -16& 22& 176\\ 20-30& 6& 25& -1& -6& 12& 72\\ 30-40& 10& 35=A& 0& 0& 2& 24\\ 40-50& 5& 45& 1& 5& 8& 40\\ 50-60& 2& 55& 2& 4& 18& 36\\ 60-70& 7& 65& 3& 21& 28& 196\\ & N=\sum f_i=40& & & \sum f_i{d}_i=8& & \sum f_i\times |x_i-\overline{x}|=544\end{array}$

$N=\sum f_i=40,\overline{x}=A+\{\frac{\sum f_i{d}_i}{N}\times h\}=35+\{\frac{8}{40}\times 10\}=37$

$\therefore MD\left(\overline{x}\right)=\frac{\sum f_i\times |x_i-\overline{x}|}{N}=\frac{544}{40}=\frac{136}{10}=13.6$