# Ex.14.1 Q7 Statistics Solution - NCERT Maths Class 10

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## 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 $$SO_2$$ (in ppm) Frequency $$0.00 - 0.04$$ $$4$$ $$0.04 - 0.08$$ $$9$$ $$0.08 - 0.12$$ $$9$$ $$0.12 - 0.16$$ $$2$$ $$0.16 - 0.20$$ $$4$$ $$0.20 - 0.24$$ $$2$$

Find the mean concentration of $$SO_2$$ in the air.

## Text Solution

What is known?

The Concentration of $$SO_2$$ in the air (in parts per million, i.e., ppm) for $$30$$ localities in a certain city:

What is unknown?

The mean concentration of $$SO_2$$in the air

Reasoning:

We solve this question by step deviation method.

Mean, $$\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h$$

Steps:

We know that,

Class mark,$${x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}$$

Class size$$, h = 0.04$$

Taking assumed mean$$,a = 0.14$$

 Content of SO2 Frequency $$(f_i)$$ $$x_i$$ $$d_i = x_i -a$$ \begin{align}{u_i}=\frac{ d_i}{h}\end{align} $$f_iu_i$$ $$0.00-0.04$$ $$4$$ $$0.02$$ $$-0.12$$ $$-3$$ $$-12$$ $$0.04-0.08$$ $$9$$ $$0.06$$ $$-0.08$$ $$-2$$ $$-18$$ $$0.08-0.12$$ $$9$$ $$0.10$$ $$-0.04$$ $$-1$$ $$-9$$ $$0.12-0.16$$ $$2$$ $$0.14(a)$$ $$0$$ $$0$$ $$0$$ $$0.16-0.20$$ $$4$$ $$0.18$$ $$0.04$$ $$1$$ $$4$$ $$0.20-0.24$$ $$2$$ $$0.22$$ $$0.08$$ $$2$$ $$4$$ $$\Sigma f_i=30$$ $$\Sigma f_iu_i=-31$$

From the table, we obtain

$\begin{array}{l} \sum {{f_i} = 30} \\ \sum {{f_i}{u_i}} = - 31 \end{array}$

\begin{align} \text { Mean }(\overline{{x}}) &={a}+\left(\frac{\sum {f}_{{i}} {u}_{{i}}}{{f}_{{i}}}\right) {h} \\ \overline{{x}} &=0.14+\left(\frac{-31}{30}\right) 0.04 \\ \overline{{x}} &=0.14-| 0.04133 \\ \overline{{x}} &=0.099 \end{align}

The mean concentration of $$SO_2$$ in the air is $$0.099$$.

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