Find the mean, variance and standard deviation using short-cut method.
| Height in cms | 70-75 | 75-80 | 80-85 | 85-90 | 90-95 | 95-100 | 100-105 | 105-110 | 110-115 |
| No. of children | 3 | 4 | 7 | 7 | 15 | 9 | 6 | 6 | 3 |
| Class Interval | Frequency \(f_i\) | \(mid-point\,x_i\) | \(y_i=\frac{x_i-92.5}{5}\) | \(f_iy_i\) | \(f_i^2\) | \(f_iy_1^2\) |
| 70-75 | 3 | 72.5 | -4 | -12 | 16 | 48 |
| 75-80 | 4 | 77.5 | -3 | -12 | 9 | 36 |
| 80-85 | 7 | 82.5 | -2 | -14 | 4 | 28 |
| 85-90 | 7 | 87.5 | 1 | -7 | 1 | 7 |
| 90-95 | 15 | 92.5 | 0 | 0 | 0 | 0 |
| 95-100 | 9 | 97.5 | 1 | 9 | 9 | 9 |
| 100-105 | 6 | 102.5 | 2 | 12 | 12 | 24 |
| 105-110 | 6 | 107.5 | 3 | 18 | 18 | 54 |
| 110-115 | 3 | 112.5 | 4 | 12 | 12 | 48 |
| 60 | 6 | 6 | 254 |
Mean, \(\bar{x}=A\frac{\sum_{i=1}^9f_ix_i}{n}×h=92.5+\frac{6}{60}×5=92.5+0.5=93\)
Variance (σ2) = \(\frac{h^2}{N^2}[N\sum_{i=1}^9f_iy_i^2-(\sum_{i=1}^9f_iy_i)^2]\)
\(=\frac{(5)^2}{(60)^2}[60×254-(6)^2]\)
\(=\frac{25}{3600}(15204)=105.58\)
\(Standard\,\,deviation\.(σ) =√105.58=10.27\)
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: