Statistics | Exercise 13.1

Question 7

To find out the concentration of $\text{SO}_2$ 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 $\text{SO}_2$ in the air.

Solution

Understand the Question

We are given grouped frequency distribution data for the concentration of $\text{SO}_2$ across $30$ localities.
To find the mean concentration using the direct method:
1. Calculate the class mark ( $x_i$ ) for each interval: $x_i = \dfrac{\text{Upper limit} + \text{Lower limit}}{2}$ .
2. Compute the product $f_i x_i$ for each class interval.
3. Sum the frequencies ( $\sum f_i$ ) and products ( $\sum f_i x_i$ ).
4. Find the mean using $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$ .

Step 1 · Find Class Marks

The class mark ( $x_i$ ) is the midpoint of each class interval: $x_i = \dfrac{\text{Upper limit} + \text{Lower limit}}{2}$ . Diagram 1

\begin{aligned} x_1 &= \dfrac{0.00 + 0.04}{2} = 0.02 \\[0.6em] x_2 &= \dfrac{0.04 + 0.08}{2} = 0.06 \\[0.6em] x_3 &= \dfrac{0.08 + 0.12}{2} = 0.10 \\[0.6em] x_4 &= \dfrac{0.12 + 0.16}{2} = 0.14 \\[0.6em] x_5 &= \dfrac{0.16 + 0.20}{2} = 0.18 \\[0.6em] x_6 &= \dfrac{0.20 + 0.24}{2} = 0.22 \end{aligned}

Step 2 · Calculate $f_i x_i$

Multiply each frequency ( $f_i$ ) by its corresponding class mark ( $x_i$ ). Diagram 2

\begin{aligned} f_1 x_1 &= 4 \times 0.02 = 0.08 \\ f_2 x_2 &= 9 \times 0.06 = 0.54 \\ f_3 x_3 &= 9 \times 0.10 = 0.90 \\ f_4 x_4 &= 2 \times 0.14 = 0.28 \\ f_5 x_5 &= 4 \times 0.18 = 0.72 \\ f_6 x_6 &= 2 \times 0.22 = 0.44 \end{aligned}

Step 3 · Find Sum of Frequencies and Products

Sum all the frequencies ( $\sum f_i$ ) and products ( $\sum f_i x_i$ ). Diagram 3

\begin{aligned} \sum f_i &= 4 + 9 + 9 + 2 + 4 + 2 \\ &= 30 \end{aligned}

\begin{aligned} \sum f_i x_i &= 0.08 + 0.54 + 0.90 + 0.28 + 0.72 + 0.44 \\ &= 2.96 \end{aligned}

Step 4 · Calculate the Mean

Using the direct method formula $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$

\begin{aligned} \text{Mean} &= \dfrac{2.96}{30} \\[0.6em] &= 0.09866\dots \\ &\approx 0.099 \text{ ppm} \end{aligned}

Answer

$0.099 \text{ ppm}$

Common Mistakes

Decimal Arithmetic Errors: Misplacing decimal points during multiplication (e.g., writing $4 \times 0.02 = 0.8$ instead of $0.08$ ).
Class Mark Confusion: Using interval boundaries (lower or upper limits) instead of the midpoint $x_i = \dfrac{\text{Lower limit} + \text{Upper limit}}{2}$ .
Rounding Off: Forgetting to round off the non-terminating decimal $0.09866\dots$ to $3$ decimal places as $0.099$ .

Question 7

More questions in Exercise 13.1