Find the mean and standard deviation using short-cut method.
| \(x_i\) | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 |
| \(f_i\) | 2 | 1 | 12 | 29 | 25 | 12 | 10 | 4 | 5 |
The data is obtained in tabular form as follows.
| \(x_i\) | \(f_i\) | \(fx_i=\frac{x_i-64}{1}\) | \(y_1^2\) | \(f_iy_i\) | \(f_iy_1^2\) |
| 60 | 2 | -4 | 16 | -8 | 32 |
| 61 | 1 | -3 | 9 | -3 | 9 |
| 62 | 12 | -2 | 4 | -24 | 48 |
| 63 | 29 | -1 | 1 | -29 | 29 |
| 64 | 25 | 0 | 0 | 0 | 0 |
| 65 | 12 | 1 | 1 | 12 | 12 |
| 66 | 10 | 2 | 4 | 20 | 40 |
| 67 | 4 | 3 | 9 | 12 | 36 |
| 68 | 5 | 4 | 16 | 20 | 80 |
| 100 | 220 | 0 | 286 |
Mean, \(\bar{x}=A\frac{\sum_{i=1}^9f_ix_i}{n}×h=64+\frac{0}{100}×1=64+0=64\)
Variance, (σ2) = \(\frac{h^2}{N^2}[N\sum_{i=1}^9f_iy_i^2-(\sum_{i=1}^9f_iy_i)^2]\)
\(=\frac{1}{100^2}[100×286-0]\)
\(=2.86\)
\(∴\,standard\,deviation\,(σ)=√2.86=1.69\)
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: