Question

Calculate the mean, median and standard deviation of the following distribution :

$\begin{array}{ccccccccc}textClass-interval:& 31-35& 36-40& 41-45& 46-50& 51-55& 56-60& 61-65& 66-70\\ \text{Frequency}:& 2& 3& 8& 12& 16& 5& 2& 3\end{array}$

Answer 1

$\begin{array}{ccccccc}CI& f_i& \text{Mid point}{x}_i& u_i=\frac{x_i-53}{4}& u_i^2& f_i{u}_i& f_i{u}_i^2\\ 31-35& 2& 33& -5& 25& -10& 50\\ 36-40& 3& 38& -3.75& 14.06& -11.25& 42.18\\ 41-45& 8& 43& -2.5& 6.25& -20& 50\\ 46-50& 12& 48& -1.25& 1.56& -15& 18.72\\ 51-55& 16& 53& 0& 0& 0& 0\\ 56-60& 5& 58& 1.25& 1.56& 6.25& 7.8\\ 61-65& 2& 63& 2.5& 6.25& 5& 12.5\\ 66-70& 3& 68& 3.75& 14.06& 11.25& 42.18\\ & N=51& & & & {\sum }_{i=1}^n=f_i{u}_i& {\sum }_{i=1}^n=f_i{u}_i^2\\ & & & & & =-33.75& =223.38\end{array}$

$\overline{X}=a+h\left(\frac{{\sum }_{i=1}^n{f}_i{u}_i}{N}\right)=53+4\left(\frac{-33.75}{51}\right)=50.36$

${\sigma }^2=h^2\left(\frac{{\sum }_{i=1}^n{f}_i{u}_i}{N}\right)=53+4\left(\frac{-33.75}{51}\right)=50.36$

${\sigma }^2=h^2(\frac{{\sum }_{i=1}^n{f}_i{u}_i^2}{N}-{\left(\frac{{\sum }_{i=1}^n{f}_i{u}_i}{N}\right)}^2)=16(\frac{223.38}{51}-\frac{1139.06}{2601})=63.07$

$\sigma =\sqrt{63.07}=7.94$

$\begin{array}{cc}{f}_i& C.F\\ 2& 2\\ 3& 5\\ 8& 15\\ 12& 25\\ 16& 41\\ 5& 46\\ 2& 48\\ 3& 51\\ \sum f_i=51=N& \end{array}$

$\frac{N}{2}=25.5$

Median class interval is 51-55

L = 51, F = 25, f = 16

h =4

Median $L+\frac{\frac{N}{2}-F}{f}\times h=\frac{51+25.5-25}{16}\times 4=51+\frac{0.5}{4}=51.125$