An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2. The variance of marks obtained by 30 girls is also 2. The average marks of all 50 candidates is 15. If $$\mu$$ is the average marks of girls and $$\sigma^2$$ is the variance of marks of 50 candidates, then $$\mu + \sigma^2$$ is equal to _________.

Let us first interpret the data. There are in all 50 candidates. Out of these, the number of boys is 20 and the number of girls is 30.

We are told that the average (mean) marks of the boys is $$\mu_1 = 12$$ and the variance of the boys’ marks is $$\sigma_1^{2}=2.$$ Likewise, the variance of the girls’ marks is given to be $$\sigma_2^{2}=2,$$ while their mean is presently unknown; let us denote it by $$\mu_2 = \mu.$$ The overall mean (average of all 50 candidates taken together) is given as $$\bar{x}=15.$$

We begin by finding the girls’ mean $$\mu.$$ Using the definition of mean, the total marks obtained by all 50 candidates equals the sum of total marks of boys and girls. Hence

$$\underbrace{50}_{\text{total students}}\times\underbrace{15}_{\text{overall mean}} \;=\; \underbrace{20}_{\text{boys}}\times\underbrace{12}_{\text{boys’ mean}} \;+\; \underbrace{30}_{\text{girls}}\times\mu.$$

Carrying out the multiplication, we have

$$750 \;=\; 240 \;+\; 30\mu.$$

Now subtract 240 from both sides:

$$750 - 240 \;=\; 30\mu.$$ $$510 \;=\; 30\mu.$$

Divide both sides by 30:

$$\mu \;=\; \frac{510}{30} \;=\; 17.$$

Thus the average marks of the girls is $$\mu_2 = 17.$$

Next we compute the variance of marks for all 50 candidates together. For two groups, the formula for the combined variance $$\sigma^{2}$$ is

$$\sigma^{2} \;=\; \frac{n_1\Big(\sigma_1^{2} + (\mu_1 - \bar{x})^{2}\Big) \;+\; n_2\Big(\sigma_2^{2} + (\mu_2 - \bar{x})^{2}\Big)}{n_1 + n_2},$$

where $$n_1,\,\mu_1,\,\sigma_1^{2}$$ refer to the first group (boys) and $$n_2,\,\mu_2,\,\sigma_2^{2}$$ refer to the second group (girls), while $$\bar{x}$$ is the overall mean.

Substituting the known numbers:

$$\sigma^{2} \;=\; \frac{20\Big(2 + (12 - 15)^{2}\Big) + 30\Big(2 + (17 - 15)^{2}\Big)}{20 + 30}.$$

Let us evaluate each bracket step by step.

For boys: $$(12 - 15)^{2} = (-3)^{2} = 9,$$ so $$2 + 9 = 11.$$ Multiplying by the number of boys: $$20 \times 11 = 220.$$

For girls: $$(17 - 15)^{2} = (2)^{2} = 4,$$ so $$2 + 4 = 6.$$ Multiplying by the number of girls: $$30 \times 6 = 180.$$

Add these two totals:

$$220 + 180 = 400.$$

Finally, divide by the total number of candidates (50):

$$\sigma^{2} = \frac{400}{50} = 8.$$

Thus the variance of marks of the entire set of 50 candidates is $$\sigma^{2}=8.$$

We are asked to find $$\mu + \sigma^{2}.$$ We already have $$\mu = 17$$ and $$\sigma^{2}=8,$$ therefore

$$\mu + \sigma^{2} = 17 + 8 = 25.$$

So, the answer is $$25.$$