If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then $$a + b + ab$$ is equal to :

The eight observations are $$6,\,4,\,a,\,8,\,b,\,12,\,10,\,13$$. Number of observations $$n = 8$$ and the given mean (arithmetic average) is $$\mu = 9$$.

Formula for the mean: $$\mu = \dfrac{\Sigma x}{n}$$. Hence $$\Sigma x = n\mu = 8 \times 9 = 72$$.

Adding the six known numbers: $$6 + 4 + 8 + 12 + 10 + 13 = 53$$. Therefore $$a + b = 72 - 53 = 19$$ $$-(1)$$

The variance (population form) is given to be $$\sigma^{2} = 9.25$$. For a population, $$\sigma^{2} = \dfrac{\Sigma(x - \mu)^2}{n} = \dfrac{\Sigma x^{2}}{n} - \mu^{2}$$.

Substituting the known values:
$$9.25 = \dfrac{\Sigma x^{2}}{8} - 9^{2}$$

Re-arranging:
$$\dfrac{\Sigma x^{2}}{8} = 9.25 + 81 = 90.25$$
$$\Sigma x^{2} = 90.25 \times 8 = 722$$

Squares of the six known numbers:
$$6^{2} = 36,\;4^{2} = 16,\;8^{2} = 64,\;12^{2} = 144,\;10^{2} = 100,\;13^{2} = 169$$
Sum $$= 36 + 16 + 64 + 144 + 100 + 169 = 529$$.

Hence
$$a^{2} + b^{2} = 722 - 529 = 193$$ $$-(2)$$

Square of $$a + b$$ from $$(1)$$:
$$(a + b)^{2} = 19^{2} = 361$$
But $$(a + b)^{2} = a^{2} + b^{2} + 2ab$$.
Substituting $$(2)$$:
$$361 = 193 + 2ab$$
$$2ab = 361 - 193 = 168$$
$$ab = 84$$.

Required expression:
$$a + b + ab = 19 + 84 = 103$$.

Therefore, the correct value is $$103$$   ⟹   Option B.