The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is

Given: Mean $$\bar{x} = 20$$, standard deviation $$\sigma = 8$$, $$n = 10$$.

$$\sum x_i = n\bar{x} = 200$$

$$\sigma^2 = 64$$, so $$\frac{\sum x_i^2}{n} - \bar{x}^2 = 64$$

$$\frac{\sum x_i^2}{10} = 64 + 400 = 464$$

$$\sum x_i^2 = 4640$$

Correcting: Replace 50 with 40.

New $$\sum x_i = 200 - 50 + 40 = 190$$

New $$\bar{x} = \frac{190}{10} = 19$$

New $$\sum x_i^2 = 4640 - 50^2 + 40^2 = 4640 - 2500 + 1600 = 3740$$

New variance:

$$\sigma^2 = \frac{3740}{10} - 19^2 = 374 - 361 = 13$$

The correct variance is $$\mathbf{13}$$.