For a statistical data $$x_1,x_2,\ldots,x_{10}$$ 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:

Mean = 5.5 for 10 values, $$\sum x_i^2 = 371$$. Two values 4,5 should be 6,8. Find corrected variance.

The original sum of the observations is $$\sum x_i = 10 \times 5.5 = 55$$, and after replacing 4 and 5 with 6 and 8 the corrected sum becomes $$55 - 4 - 5 + 6 + 8 = 60$$ so that the corrected mean is $$60/10 = 6$$.

The original sum of squares is 371, and after removing 16 and 25 and adding 36 and 64 the corrected sum of squares becomes $$371 - 16 - 25 + 36 + 64 = 430$$.

The corrected variance is then given by $$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{430}{10} - 36 = 43 - 36 = 7$$.

The correct answer is Option 3: 7.