Question

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

$$\begin{array}{ccccccccc}\text{wages upto (Rs.)}& 15& 30& 45& 60& 75& 90& 105& 120\\ \text{Number of workers}& 12& 30& 65& 107& 157& 202& 222& 230\end{array}$$

Answer 1

Firstly, we will prepare the frequency distribution as given below

$\begin{array}{cccccccc}\text{Class Interval}& Cumulative& \text{Mid-value}& \text{Frequency}& d_i=\frac{x_i-67.5}{15}& d_i^2& f_i{d}_i& f_i{d}_i^2\\ & Frequency& x_i& f_i& & & & \\ 0-15& 12& 7.5& 12& -4& 16& -48& 192\\ 15-30& 30& 22.5& 18& -3& 9& -54& 162\\ 30-45& 65& 37.5& 35& -2& 4& -70& 140\\ 45-60& 107& 52.5& 42& -1& 1& -42& 42\\ 60-75& 157& 67.5& 50& 0& 0& 0& 0\\ 75-90& 202& 82.5& 45& 1& 1& 45& 45\\ 90-105& 222& 97.5& 20& 2& 4& 40& 80\\ 105-120& 230& 112.5& 8& 3& 9& 24& 72\\ & & & \sum f_i=230& & \sum f_i{d}_i=-105& & \sum f_i{d}_i^2=733\end{array}$

Here, $A=67.5,h=15,\sum f_i=230$

$\sum f_i{d}_i=-105,\sum f_i{d}_i^2=733$

$\therefore Mean=A+h\left(\frac{\sum f_i{d}_i}{\sum f_i}\right)=67.5+15\left(\frac{-105}{230}\right)$

=$67.5-6.85=60.65$

and var(x) =$h^2[\frac{\sum f_i{d}_i^2}{\sum f_i}-{\left(\frac{\sum f_i{d}_i}{\sum f_i}\right)}^2]$

=$225[\frac{733}{230}-{\left(\frac{-105}{230}\right)}^2]$

=$225(3.18-02.25)$

=$669.937$

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

=$\sqrt{669.937}$

=$25.88$