Question

Calculate the mean and the standard deviation for the following distribution.

$\begin{array}{cccccccc}\text{Marks}& 20-30& 30-40& 40-50& 50-60& 60-70& 70-80& 80-90\\ \text{Number of students}& 3& 6& 13& 15& 14& 5& 4\end{array}$

Answer 1

Table for mean and standard deviation is given below

$\begin{array}{ccccccc}\text{Class}& & & u_i& u_i^2& f_i{u}_i& f_i{u}_i^2\\ \text{interval}& \left(f_i\right)& \left(x_i\right)& & & & \\ 20-30& 3& 25& -3& 9& -9& 27\\ 30-40& 6& 35& -2& 4& -12& 24\\ 40-50& 13& 45& -1& 1& -13& 13\\ 50-60& 15& 55=A& 0& 0& 0& 0\\ 60-70& 14& 65& 1& 1& 14& 14\\ 70-80& 5& 75& 2& 4& 10& 20\\ 80-90& 4& 85& 3& 9& 12& 36\end{array}$

Here, $A=55,N=\sum f_i=60,\sum f_i{u}_i=2,\sum f_i{u}_i^2=134$ and $h=10$

$\therefore$ Mean $\left(\overline{x}\right)=A+h(\frac{1}{N}\sum f_i{u}_i)=55+10\left(\frac{2}{60}\right)=55+0.333=55333$

$\Rightarrow$ $Var\left(x\right)=h^2[\frac{1}{N}\sum f_i{u}_i^2-{(\frac{1}{N}\sum f_i{u}_i)}^2]={10}^2[\frac{134}{60}-{\left(\frac{2}{60}\right)}^2]$

= $100(\frac{134}{60}-\frac{4}{3600})$

= $100\left(\frac{8040-4}{3600}\right)=\frac{8036}{36}=223.2$

SD = $\sqrt{Var\left(x\right)}=\sqrt{223.2}=14.94$