Question

To find out the concentration of SO\(_2\) 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:

Concentration of SO\(\bf{_2}\) (in ppm)	Frequency
0.00 - 0.04  0.04 - 0.08  0.08 - 0.12  0.12 - 0.16  0.16 - 0.20  0.20 - 0.24	4 9 9 2 4 2

Answer 1

To find the class mark (\(x_i\)) for each interval, the following relation is used.  

Class mark  \((x_i)\) = \(\frac {\text{Upper \,limit + Lower \,limit}}{2}\)

class size (h) of the data = 0.04

Taking 0.14 as assumed mean (a), \(d_i\), \(u_i\), and \(f_iu_i\) can be calculated as follows.   

Concentration of SO\(\bf{_2}\) (in  ppm)	\(\bf{f_i}\)	\(\bf{x_i}\)	\(\bf{d_i = x_i -0.14}\)	\(\bf{u_i = \frac{d_i}{0.04}}\)	\(\bf{f_iu_i}\)
0.00 - 0.04	4	0.02	-0.12	-4	-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

From the table, it can be observed that  

\(\sum f_i = 30\)
\(\sum f_iu_i = -31\)

Mean, \(\overset{-}{x} = a + (\frac{\sum f_iu_i}{\sum f_i}) \times h\)
x = \(0.14  + (\frac{-31 }{30})\times (0.04)\)
x = 0.14 - 0.04133
x = 0.09867
x = 0.099 ppm

Therefore, the mean concentration of SO\(_2\) in the air is 0.099 ppm.