The diameters of circles (in mm) drawn in a design are given below
| Diameters | No. of children |
| 33-36 | 15 |
| 37-40 | 17 |
| 41-44 | 21 |
| 45-48 | 22 |
| 49-52 | 25 |
| Class Interval | Frequency \(f_i\) | \(mid-point\,x_i\) | \(y_i=\frac{x_i-42.5}{4}\) | \(f_i^2\) | \(f_iy_i\) | \(f_iy_1^2\) |
| 32.5-36.5 | 15 | 34.5 | -2 | 4 | -30 | 60 |
| 36.5-40.5 | 17 | 38.5 | -1 | 1 | -17 | 17 |
| 40.5-44.5 | 21 | 42.5 | 0 | 0 | 0 | 0 |
| 44.5-48.5 | 22 | 46.5 | 1 | 1 | 22 | 22 |
| 48.5-52.5 | 25 | 50.5 | 2 | 4 | 50 | 100 |
| 100 | 25 | 199 |
Here, N = 100, h = 4
Let the assumed mean, A, be 42.5
Mean, \(\bar{x}=A\frac{\sum_{i=1}^5f_ix_i}{n}×h=42.5+\frac{25}{100}×4=43.5\)
Variance (σ2) = \(\frac{h^2}{N^2}[N\sum_{i=1}^5f_iy_i^2-(\sum_{i=1}^5f_iy_i)^2]\)
\(=\frac{16}{10000}[100×199-(25)^2]\)
\(=\frac{16}{10000}[19900-625]\)
\(=\frac{16}{1000}×19275\)
\(=30.84\)
\(Standard\,\,deviation\.(σ) =5.55\)
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: