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Daily wages (in Rs.) | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |
Number of workers | 12 | 14 | 8 | 6 | 10 |
Answer :
The midpoint of the given interval is found by formula:
\(Midpoint \ (x_i) \ = \ \frac{upper \ limit \ + \ lower \ limit}{2} \)
In this case, the value of mid-point ( \(x_i\)) is very large,so let us assume the mean value, A = 150 and class interval is h = 20.
So, \(u_i \ = \ \frac{x_i \ - \ A}{h} \ \)
\( \Rightarrow\ u_i \ = \ \frac{x_i \ - \ 150}{20} \)
Daily wages (Class interval) | Number of workers (frequency (\(f_i\)) ) | Mid-point ( \(x_i \) ) | \( u_i \ = \ \frac{x_i \ - \ 150}{20} \) | \(f_i u_i \) |
100-120 | 12 | 110 | -2 | -24 |
120-140 | 14 | 130 | -1 | -14 |
140-160 | 8 | 150 | 0 | 0 |
160-180 | 6 | 170 | 1 | 6 |
180-200 | 10 | 190 | 2 | 20 |
Total | \( \sum f_i \ = \ 50 \) | \( \sum f_i u_i \ = \ -12 \) |
So, the formula to find out the mean is : \( \ \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)
\( = \ 150 \ + \ 20 \ × \ \frac{-12}{50} \)
\(= \ 150 \ - \ 4.8 \)
\(= \ 145.20 \)
\(\therefore \) Mean daily wage of the workers = Rs. 145.20