Question

Question 7
To find out the concentration of $S{O}_2$ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

$\begin{array}{cc}\text{Concentration of}S{O}_2\text{(in pmm)}& \text{Frequency}\\ 0.00-0.04& 4\\ 0.04-0.08& 9\\ 0.08-0.12& 9\\ 0.12-0.16& 2\\ 0.16-0.20& 4\\ 0.20-0.24& 2\end{array}$

Find the mean concentration of $S{O}_2$ in the air.

Answer 1

We may find the class marks for each interval by using the relation:

$x_i=\frac{\text{Upper class limit + lower class limit}}{2}$

Class size of this data = 0.04

Now, taking 0.14 as assumed mean $a$, we may calculate $d_i,u_i,f_i,u_i$ as following:

$\begin{array}{cccccc}\text{Concentration of}S{O}_2\text{(in pmm)}& \text{Frequency}& \text{Class mark}{x}_i& d_i=x_i-0.14& x_i=\frac{x_i-0.14}{h}& f_i{u}_i\\ 0.00-0.04& 4& 0.02& -0.12& -3& -12\\ 0.04-0.08& 9& 0.06& -0.08& -2& -18\\ 0.08-0.12& 9& 0.10& -0.04& -1& -9\\ 0.12-0.16& 2& 0.14& 0& 0& 0\\ 0.16-0.20& 4& 0.18& 0.04& 1& 4\\ 0.20-0.24& 2& 0.22& 0.08& 2& 4\\ Total& 30& & & & -31\end{array}$

From the table, we may observe that:

$\sum f_i=30\sum f_i{u}_i=-31Mean\overline{x}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)\times h=0.14+\left(\frac{-31}{30}\right)\left(0.04\right)=0.14-0.04133=0.099ppm$

So, mean concentration of $S{O}_2$ in the air is 0.099 ppm.