Calculate the mean and standard deviation for the following table, if the age distribution of a group of people are given below.

$\begin{array}{cccccccc}\text{Age}& 20-30& 30-40& 40-50& 50-60& 60-70& 70-80& 80-90\\ \text{Number of persons}& 3& 51& 122& 141& 130& 51& 2\end{array}$

We make the table from the given data

$\begin{array}{ccccccc}\text{Age}& \text{Mid-value}{X}_1& \text{Number of persons}{f}_1& d_1=\frac{X_1-55}{10}& d_1^2& f_1,d_1& f_1{d}_1^2\\ 20-30& 25& 3& -3& 9& -9& 27\\ 30-40& 35& 51& -2& 4& -102& 204\\ 40-50& 45& 122& -1& 1& -122& 122\\ 50-60& 55& 141& 0& 0& 0& 0\\ 60-70& 65& 130& 1& 1& 130& 130\\ 70-80& 75& 51& 2& 4& 102& 204\\ 80-90& 85& 2& 3& 9& 6& 18\\ \text{Total}& & N=\sum f_i=500& & & \sum f_i,d_i=5& \sum f_i,d_i^2=705\end{array}$

Here $A=55,h=10$

Therefore, Mean,$\overline{X}=A+\left(\frac{\sum f_i{d}_i}{\sum f_i}\right)\times h=55+\frac{5}{500}\times 10=55+0.1=55.1$

Variance, ${\sigma }^2=h^2[\frac{\sum f_i{d}_i^2}{\sum f_i}-{\left(\frac{\sum f_i{d}_i}{\sum f_i}\right)}^2]=100[\frac{705}{500}-{\left(\frac{5}{100}\right)}^2]$

$=100\left(\frac{352500-25}{250000}\right)=\frac{35247500}{250000}=\frac{14099}{100}$

and standard deviation , $\sigma =\sqrt{\frac{14099}{100}}=\frac{118.739}{10}=11.8739$

Hence, mean$\left(\overline{X}\right)=55.1$and standard deviation $\left(\sigma \right)=11.8739$