Find the mean deviation about the mean for the data.
| Income per day | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 |
| Number of persons | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |
The following table is formed.
| Income per day | Number of persons \(f_i\) | Mid-point \(x_i\) | \(f_ix_i\) | \(|x_i-\bar{x}\) | \(f_i|x_i-\bar{x}\)| |
| 0-100 | 4 | 50 | 200 | 308 | 1232 |
| 100-200 | 8 | 150 | 1200 | 208 | 1664 |
| 200-300 | 9 | 250 | 2250 | 108 | 972 |
| 300-400 | 10 | 350 | 3500 | 8 | 80 |
| 400-500 | 7 | 450 | 3150 | 92 | 644 |
| 500-600 | 5 | 550 | 2750 | 192 | 960 |
| 600-700 | 4 | 650 | 2600 | 292 | 1168 |
| 700-800 | 3 | 750 | 2250 | 392 | 1176 |
| 50 | 17900 | 7896 |
Here, \(\sum_{I=1}^{8}f_i=50\sum_{1=i}^8f_ix_i=17900\)
∴ \(\bar{x}=\frac{1}{N}\sum_{i=1}^{8}f_ix_i=\frac{1}{50}×17900=358\)
\(M.D(\bar{x})=\frac{1}{N}\sum_{i=1}^{8}f_i|x_i-\bar{x_i}|=\frac{1}{50}×7896=157.92\)