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Q7:

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: 

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

Solution :

Given: The frequency distribution of the concentration of $SO_2$ (in ppm) for 30 localities.

Concentration of $SO_2$ (ppm) Number of Localities ($f_i$)
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

To Find: The mean concentration of $SO_2$ in the air.

Step 1: Determine the Class Marks ($x_i$)
The class mark ($x_i$) for each interval is calculated using the formula: $x_i = \frac{\text{Upper Class Limit} + \text{Lower Class Limit}}{2}$.

Concentration ($SO_2$) Frequency ($f_i$) Class Mark ($x_i$) $f_i \cdot x_i$
0.00 - 0.04 4 0.02 0.08
0.04 - 0.08 9 0.06 0.54
0.08 - 0.12 9 0.10 0.90
0.12 - 0.16 2 0.14 0.28
0.16 - 0.20 4 0.18 0.72
0.20 - 0.24 2 0.22 0.44
Total $\sum f_i = 30$ - $\sum f_i x_i = 2.96$

Step 2: Apply the Direct Method Formula for Mean
The mean ($\bar{x}$) is given by the formula: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

Step 3: Perform the Calculation
Substitute the values obtained from the table into the formula:
$\sum f_i = 4 + 9 + 9 + 2 + 4 + 2 = 30$
$\sum f_i x_i = 0.08 + 0.54 + 0.90 + 0.28 + 0.72 + 0.44 = 2.96$
$\bar{x} = \frac{2.96}{30}$

To simplify the division:
$\bar{x} = \frac{2.96}{30} = \frac{296}{3000}$
$\bar{x} = \frac{148}{1500} = \frac{74}{750} = \frac{37}{375}$
Performing the division $37 \div 375 \approx 0.09866...$

Rounding to four decimal places, we get $\bar{x} \approx 0.0987$.

Final Answer: The mean concentration of $SO_2$ in the air is 0.0987 ppm.


More Questions from Class 10 Mathematics Statistics EXERCISE 13.1


CBSE Solutions for Class 10 Mathematics Statistics


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