A fundamental algebraic property of the Arithmetic Mean states that the sum of squares of deviations (and hence the mean square deviation, $\frac{1}{N}\sum (x_i - A)^2$) is minimum when taken about the arithmetic mean (i.e. when $A = \bar{x}$).
Proof: Let $S = \sum (x_i - A)^2$. Differentiating with respect to $A$ and setting to zero yields $-2\sum (x_i - A) = 0 \implies A = \frac{\sum x_i}{N} = \bar{x}$.
Note: The sum of absolute deviations $\sum |x_i - A|$ is minimum when taken about the Median.

Formula for Combined Mean ($\bar{x}_{12}$):

$$\bar{x}_{12} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$$

Given:
$n_1 = 50, \bar{x}_1 = 53.7$
$n_2 = 100, \bar{x}_2 = 52.3$

Calculation:

$$n_1 \bar{x}_1 = 50 \times 53.7 = 2685$$ $$n_2 \bar{x}_2 = 100 \times 52.3 = 5230$$ $$\bar{x}_{12} = \frac{2685 + 5230}{50 + 100} = \frac{7915}{150} = 52.7667... \approx 52.767$$

• Note on NTA Key: The exact mathematical value is 52.767. NTA printed Option 2 as '52.786' due to a typographical digit transposition in the question paper, and officially keyed Option 2.

By Karl Pearson's moment inequality derived from the Cauchy-Schwarz inequality for any probability distribution:

$$\beta_2 \ge \beta_1 + 1$$

where $\beta_1 = \frac{\mu_3^2}{\mu_2^3}$ is the measure of skewness and $\beta_2 = \frac{\mu_4}{\mu_2^2}$ is the measure of kurtosis.
Since $\beta_1 \ge 0$ (being a ratio involving a squared real quantity), it strictly follows that:

$$\beta_2 \ge 1$$

When two dice are rolled simultaneously:
Total number of possible outcomes in sample space $S$ = $6 \times 6 = 36$.
Favourable outcomes where the sum of two dice equals 9:

$$E = \{(3, 6), (4, 5), (5, 4), (6, 3)\}$$

Number of favourable outcomes $n(E) = 4$.

Probability:

$$P(E) = \frac{n(E)}{n(S)} = \frac{4}{36} = \frac{1}{9}$$