Question

Calculate mean, variance and standard deviation for the following frequency distribution:

$\begin{array}{cccccccc}\text{Class}& 0-30& 30-60& 60-90& 90-120& 120-150& 150-180& 180-210\\ \text{Frequency}& 2& 3& 5& 10& 3& 5& 2\end{array}$

Answer 1

Here h = 30. Let the assumed mean be A = 105.

Now, we prepare the table given below.

$\begin{array}{ccccccc}\text{Class}& \text{Frequency}& \text{Midpoint}& y_i=\frac{(x_i-105)}{30}& y_i^2& f_i{y}_i& f_i{y}_i^2\\ & f_i& x_i& & & & \\ 0-30& 2& 15& -3& 9& -6& 18\\ 30-60& 3& 45& -2& 4& -6& 12\\ 60-90& 5& 75& -1& 1& -5& 5\\ 90-120& 10& 105=A& 0& 0& 0& 0\\ 120-150& 3& 135& 1& 1& 3& 3\\ 150-180& 5& 165& 2& 4& 10& 20\\ 180-210& 2& 195& 3& 9& 6& 18\\ \text{Total}& N=30& & & & 2& 76\end{array}$

$\therefore A=105,h=30,N=\sum f_i=30,\sum f_i{y}_i=2$ and $\sum f_i{y}_i^2=76$

$\therefore \overline{x}=(A+\frac{\sum f_i{y}_i}{N}\times h)\Rightarrow \overline{x}=(105+\frac{2}{30}\times 30)=(105+2)=107$

Thus, mean = 107.

Vriance, ${\sigma }^2=\frac{h^2}{N^2}.\{N.\sum f_i{y}_i^2-(\sum f_i{y}_i{)}^2\}$

$=\frac{(30{)}^2}{(30{)}^2}.\{3\times 76-(2{)}^2\}=(2280-4)=2276$

Standard deviation, $\sigma =\sqrt{2276}=47.71$