The mean and standard deviation of 15 observations are found to be $$8$$ and $$3$$ respectively. On rechecking it was found that, in the observations, $$20$$ was misread as $$5$$. Then, the correct variance is equal to ______

Given: $$n = 15$$, mean $$\bar{x} = 8$$, standard deviation $$\sigma = 3$$.

$$\sum x_i = n \cdot \bar{x} = 15 \times 8 = 120$$

$$\sigma^2 = 9 = \frac{\sum x_i^2}{n} - \bar{x}^2 \implies \frac{\sum x_i^2}{15} = 9 + 64 = 73$$

$$\sum x_i^2 = 1095$$

The observation 20 was misread as 5. Correcting:

$$\sum x_i^{\text{correct}} = 120 - 5 + 20 = 135$$

$$\sum (x_i^{\text{correct}})^2 = 1095 - 25 + 400 = 1470$$

Correct mean: $$\bar{x}_c = \frac{135}{15} = 9$$

Correct variance:

$$\sigma_c^2 = \frac{\sum (x_i^{\text{correct}})^2}{n} - \bar{x}_c^2 = \frac{1470}{15} - 81 = 98 - 81 = 17$$

Hence the correct variance is $$\boxed{17}$$.