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	499242

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.