Question

Calculate the mean deviation using mean and standard deviation for the following distribution.
$\begin{array}{cc}\text{Classes}& \text{Frequencies}\\ 20-40& 3\\ 40-80& 6\\ 80-100& 20\\ 100-120& 12\\ 120-140& 9\\ \text{Total}& 50\end{array}$

Answer 1

$\begin{array}{cccccccccc}\text{Classes}& \text{Frequency}& \text{m}& X-A;A=90& |d|& f|d|& d=\frac{X-A}{t}& fd& d^2& f{d}^2\\ 20-40& 3& 30& -60& 60& 180& -6& -18& 36& 180\\ 40-80& 6& 60& -30& 30& 180& -3& -18& 9& 54\\ 80-100& 20& 90& 0& 0& 0& 0& 0& 0& 0\\ 100-120& 12& 110& 20& 20& 240& 2& 24& 4& 48\\ 120-140& 9& 130& 40& 40& 360& 4& 36& 16& 144\\ & \sum f=50& & & & \sum f|d|=960& & \sum fd=24& & \sum f{d}^2=354\end{array}$
$\overline{X}=A+\frac{\sum fd}{\sum f}\times i=90+\frac{24}{50}\times 10=9.48M{D}_{\overline{X}}=\frac{\sum f|d|}{\sum f}=\frac{960}{50}=19.2$
Mean deviation from mean :

$\sigma =\sqrt{\frac{\sum f{d}^2}{\sum f}-{\left(\frac{\sum fd}{\sum f}\right)}^2}\times i=\sqrt{\frac{354}{50}-{\left(\frac{24}{50}\right)}^2}\times 10=\sqrt{\frac{354}{50}-\frac{576}{2500}}\times 10=\sqrt{6.85}\times 10=26.17M{D}_{\sigma }=\frac{\sum f|d|}{N}=\frac{998.4}{50}=19.968$