Question :
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 |
Solution :
Let the assumed mean a = 0.10
Concentration of \(SO_2\) (in ppm) | Frequency (\(f_i\)) | Mid – Value (\(x_i\)) | \(u_i\) = \(x_i – 0.10\over 0.04\) | \(f_iu_i\) |
0.00 – 0.04 | 4 | 0.02 | -2 | -8 |
0.04 – 0.08 | 9 | 0.06 | -1 | -9 |
0.08 – 0.12 | 9 | 0.10 | 0 | 0 |
0.12 – 0.16 | 2 | 0.14 | 1 | 2 |
0.16 – 0.20 | 4 | 0.18 | 2 | 8 |
0.20 – 0.24 | 2 | 0.22 | 3 | 6 |
\(\sum f_i\) = 30 | \(\sum f_iu_i\) = -1 |
By step deviation method :
Mean = a + \(\sum f_iu_i\over \sum f_i\) \(\times\) h = 0.10 + \(-1\over 30\) \(\times\) 0.04
= 0.10 – 0.0013 = 0.0987
= 0.099 ppm