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

The class mark (\(x_i\)) for each interval is calculated using the formula: Class mark \((x_i)\) = \(\frac {\text{Upper \,limit + Lower \,limit}}{2}\). The class size (h) of the data is 0.04.

With an assumed mean (a) of 0.14, the values for \(d_i\), \(u_i\), and \(f_iu_i\) are computed as shown in the table below.

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

The table shows that \(\sum f_i = 30\) and \(\sum f_iu_i = -31\).

The mean (\(\overset{-}{x}\)) is calculated as: \(\overset{-}{x} = a + (\frac{\sum f_iu_i}{\sum f_i}) \times h\).

Substituting the values: \(\overset{-}{x} = 0.14 + (\frac{-31}{30}) \times (0.04)\).

\(\overset{-}{x} = 0.14 - 0.04133\)

\(\overset{-}{x} = 0.09867\), which rounds to 0.099 ppm.

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