Given that the mean \(\bar{x} = 56\).
Given that the variance \(\sigma^2 = 66.2\).
Using the formula for variance with mean and variance values:
\[ \frac{\alpha^2 + \beta^2 + 25678}{10} - (56)^2 = 66.2 \]
Rearranging, we find:
\[ \alpha^2 + \beta^2 = 6344 \]
So, the correct answer is: 6344
For a statistical data \( x_1, x_2, \dots, x_{10} \) of 10 values, a student obtained the mean as 5.5 and \[ \sum_{i=1}^{10} x_i^2 = 371. \] He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively.
The variance of the corrected data is: