Find the mean and variance for the data
| \(x_i\) | 62 | 93 | 97 | 98 | 102 | 104 | 106 |
| \(f_i\) | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
The data is obtained in tabular form as follows.
| \(x_i\) | \(f_i\) | \(fx_i\) | \((x_i-\bar{x})\) | \((x_i-\bar{x})^2\) | \(f_i(x_i-\bar{x})^2\) | \(f_i(x_i-\bar{x})\) |
| 92 | 3 | 276 | 8 | 64 | 192 | 192 |
| 93 | 2 | 186 | 7 | 49 | 98 | 98 |
| 97 | 3 | 291 | 3 | 9 | 27 | 27 |
| 98 | 2 | 196 | 2 | 4 | 8 | 8 |
| 102 | 6 | 612 | 2 | 4 | 24 | 24 |
| 104 | 3 | 312 | 4 | 16 | 48 | 48 |
| 109 | 3 | 327 | 9 | 81 | 243 | 243 |
| 32 | 2200 | 640 | 640 |
Here, N = 22, \(\sum_{i=1}^7f_ix_i=2200\)
\(∴\bar{x}=\frac{1}{N}\frac{\sum_{i=1}^7f_ix_i}{n}=\frac{1}{22}×2200=100\)
Variance(σ2) = \(\frac{1}{n}\sum_{i=1}^7f_i(x_i-\bar{x})^2=\frac{1}{22}×640=29.09\)
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then \(a + b + ab\) is equal to: