Let the Mean and Variance of five observations $$x_1 = 1, x_2 = 3, x_3 = a, x_4 = 7$$ and $$x_5 = b$$, $$a \gt b$$, be 5 and 10 respectively. Then the Variance of the observations $$n + x_n$$, $$n = 1, 2, \ldots, 5$$ is

The five given observations are $$x_1 = 1,\; x_2 = 3,\; x_3 = a,\; x_4 = 7,\; x_5 = b$$ with $$a \gt b$$.

Their mean is given to be $$5$$, so

$$\frac{1 + 3 + a + 7 + b}{5} = 5 \; \Longrightarrow \; 11 + a + b = 25 \; \Longrightarrow \; a + b = 14 \; -(1)$$

The variance is given to be $$10$$. For a population of five values,

$$\text{Variance} = \frac{1}{5}\sum_{i=1}^{5}(x_i - \mu)^2 = 10,\quad \mu = 5.$$

Compute the three known squared deviations:

$$(1-5)^2 = 16,\; (3-5)^2 = 4,\; (7-5)^2 = 4.$$

Let $$S = (a-5)^2 + (b-5)^2$$. Then

$$\frac{16 + 4 + 4 + S}{5} = 10 \; \Longrightarrow \; 24 + S = 50 \; \Longrightarrow \; S = 26 \; -(2)$$

Expand $$S$$ using $$a + b = 14$$:

$$S = (a-5)^2 + (b-5)^2 = a^2 - 10a + 25 + b^2 - 10b + 25 \\ = (a^2 + b^2) - 10(a + b) + 50.$$

Since $$a^2 + b^2 = (a + b)^2 - 2ab = 196 - 2ab$$,

$$S = 196 - 2ab - 140 + 50 = 106 - 2ab.$$

Using $$S = 26$$ from $$(2)$$,

$$106 - 2ab = 26 \; \Longrightarrow \; 2ab = 80 \; \Longrightarrow \; ab = 40 \; -(3)$$

Equations $$(1)$$ and $$(3)$$ give the quadratic

$$t^2 - 14t + 40 = 0 \quad (\text{with roots } a,\,b).$$

Discriminant $$D = 196 - 160 = 36 \;\;\Longrightarrow\;\; t = \frac{14 \pm 6}{2} = 10 \text{ or } 4.$$

Since $$a \gt b$$, we have $$a = 10,\; b = 4.$$

The original data are therefore $$1,\,3,\,10,\,7,\,4.$$

Now form the new observations $$n + x_n$$ for $$n = 1,2,\ldots,5$$:

$$\begin{aligned} n=1 &: 1 + 1 = 2\\ n=2 &: 2 + 3 = 5\\ n=3 &: 3 + 10 = 13\\ n=4 &: 4 + 7 = 11\\ n=5 &: 5 + 4 = 9 \end{aligned}$$

Thus the new set is $$2,\,5,\,13,\,11,\,9.$$

Mean of the new set:

$$\bar{y} = \frac{2 + 5 + 13 + 11 + 9}{5} = \frac{40}{5} = 8.$$

Compute squared deviations from $$8$$:

$$ (2-8)^2 = 36,\; (5-8)^2 = 9,\; (13-8)^2 = 25,\; (11-8)^2 = 9,\; (9-8)^2 = 1.$$

Sum of squared deviations $$= 36 + 9 + 25 + 9 + 1 = 80.$$

Variance (population) $$= \frac{80}{5} = 16.$$

Therefore, the variance of the observations $$n + x_n$$ is $$16$$.

Option D is correct.