The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to:

Given: Mean = 10, Variance = 4 for $$n$$ students. One student's marks changed from 8 to 12. New mean = 10.2.

Finding n:

Old sum = $$10n$$. New sum = $$10n + 4$$.

New mean: $$\frac{10n + 4}{n} = 10.2 \implies n = 20$$

Finding old $$\Sigma x^2$$:

Variance = $$\frac{\Sigma x^2}{n} - \bar{x}^2 = 4$$

$$\frac{\Sigma x^2}{20} = 104 \implies \Sigma x^2 = 2080$$

Computing new variance:

New $$\Sigma x^2 = 2080 - 8^2 + 12^2 = 2080 - 64 + 144 = 2160$$

New variance = $$\frac{2160}{20} - (10.2)^2 = 108 - 104.04 = 3.96$$

The answer is Option C: $$3.96$$.