Let $$X= \left\{x\in N:1\leq x\leq19 \right\}$$ and for some $$a,b \in \mathbb R, Y = \left\{ax+b:x\in X\right\}.$$ If the mean and variance of the elements of Y are 30 and 750, respectively, then the sum of all possible values of b is

$$X = \{1,2,...,19\}$$, $$Y = \{ax+b : x \in X\}$$. Mean of Y = 30, Variance of Y = 750.

Mean of X: $$\bar{X} = 10$$. Variance of X: $$\sigma_X^2 = \frac{n^2-1}{12} = \frac{361-1}{12} = 30$$ (for uniform $$\{1,...,n\}$$).

Mean of Y: $$a\bar{X}+b = 10a+b = 30$$.

Variance of Y: $$a^2 \sigma_X^2 = a^2 \times 30 = 750$$, so $$a^2 = 25$$, giving $$a = \pm 5$$.

If $$a = 5$$: $$50+b = 30$$, $$b = -20$$.

If $$a = -5$$: $$-50+b = 30$$, $$b = 80$$.

Sum of all possible values of b: $$-20 + 80 = 60$$.

The correct answer is Option 3: 60.