Case Study on Air Pollution Data

Question 1

What is the lower limit of median class ?

Accepted Answer

0.08. $N = 30 \implies \frac{N}{2} = 15$. Cumulative frequencies: 4, 12, 22, 24, 28, 30. Median class is 0.08 – 0.12. Lower limit = 0.08.

Question 2

What is the modal class of the above data ?

Accepted Answer

Maximum frequency is 10, which corresponds to class interval 0.08 – 0.12. Modal class = 0.08 – 0.12.

Question 3

(a) What is the mean concentration of $SO_2$ in air ?

Accepted Answer

0.099. Midpoints $x_i$: 0.02, 0.06, 0.10, 0.14, 0.18, 0.22. $f_i x_i$: 0.08, 0.48, 1.00, 0.28, 0.72, 0.44. $\sum f_i x_i = 3.00$. Mean = $\frac{3.00}{30} = 0.10	ext{ ppm}$ (or $0.099	ext{ ppm}$).

Question 4

**OR** (b) Find the median of the above data.

Accepted Answer

$	ext{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}ight) 	imes h = 0.08 + \left(\frac{15 - 12}{10}ight) 	imes 0.04 = 0.08 + 0.012 = 0.092	ext{ ppm}$.

Concentration of $SO_2$ (in ppm)	Frequency
0.00 – 0.04	4
0.04 – 0.08	8
0.08 – 0.12	10
0.12 – 0.16	2
0.16 – 0.20	4
0.20 – 0.24	2

Case Study on Air Pollution Data

Passage / data set

Question 1 of 4

Explanation

Question 2 of 4

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Question 3 of 4

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Question 4 of 4

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