To find out the concentration of SO_{2} in th | vedclass

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(B) To find the class marks $(x_{i})$ for each interval,we use the formula:$x_{i} = \frac{	ext{Upper class limit} + 	ext{Lower class limit}}{2}$The

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$0.099$

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(B) To find the class marks $(x_{i})$ for each interval,we use the formula:$x_{i} = \frac{	ext{Upper class limit} + 	ext{Lower class limit}}{2}$The class size $(h)$ is $0.04 - 0.00 = 0.04$.Let the assumed mean $(a)$ be $0.14$.We calculate $d_{i} = x_{i} - a$ and $u_{i} = \frac{d_{i}}{h}$.Concentration of $SO_{2}$ $(ppm)$Frequency $(f_{i})$Class mark $(x_{i})$$d_{i} = x_{i} - 0.14$$u_{i} = \frac{d_{i}}{0.04}$$f_{i}u_{i}$$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$$4$Total$\Sigma f_{i} = 30$$\Sigma f_{i}u_{i} = -31$Using the step-deviation method:Mean $\bar{x} = a + \left( \frac{\Sigma f_{i}u_{i}}{\Sigma f_{i}} ight) 	imes h$$\bar{x} = 0.14 + \left( \frac{-31}{30} ight) 	imes 0.04$$\bar{x} = 0.14 - \frac{1.24}{30}$$\bar{x} = 0.14 - 0.04133...$$\bar{x} \approx 0.09867 \approx 0.099\, ppm$.Thus,the mean concentration of $SO_{2}$ in the air is $0.099\, ppm$.

Class	$200-299$	$300-399$	$400-499$	$500-599$	$600-699$	$700-799$	$800-899$
Frequency	$3$	$61$	$118$	$139$	$126$	$151$	$2$

Class	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$5$	$3$	$4$	$f$	$2$	$6$	$13$

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Concentration of $SO_{2}$ (in $ppm$)	Frequency
$0.00-0.04$	$4$
$0.04-0.08$	$9$
$0.08-0.12$	$9$
$0.12-0.16$	$2$
$0.16-0.20$	$4$
$0.20-0.24$	$2$