If $$\sum_{i=1}^{n}(x_i - a) = n$$ and $$\sum_{i=1}^{n}(x_i - a)^2 = na$$, $$(n, a \gt 1)$$, then the standard deviation of $$n$$ observations $$x_1, x_2, \ldots, x_n$$ is:

We have the $$n$$ observations $$x_1,\,x_2,\,\ldots ,x_n$$ with two given relations

$$\sum_{i=1}^{n}(x_i-a)=n\quad\text{and}\quad\sum_{i=1}^{n}(x_i-a)^2=na,$$

where $$n\gt 1,\;a\gt 1.$$

First we extract the arithmetic mean of the observations. Writing the first relation explicitly,

$$\sum_{i=1}^{n}(x_i-a)=\sum_{i=1}^{n}x_i-\sum_{i=1}^{n}a =\sum_{i=1}^{n}x_i-na=n.$$

So

$$\sum_{i=1}^{n}x_i=na+n=n(a+1).$$

Dividing by the number of observations gives the mean

$$\overline x=\frac1n\sum_{i=1}^{n}x_i =\frac{n(a+1)}{n}=a+1.$$

Now we need the standard deviation. By definition, the variance is

$$\sigma^2=\frac1n\sum_{i=1}^{n}(x_i-\overline x)^2.$$

Because $$\overline x=a+1,$$ we must convert the given sum $$\sum_{i=1}^{n}(x_i-a)^2$$ to the required sum $$\sum_{i=1}^{n}(x_i-(a+1))^2.$$

For each term we note the algebraic identity

$$(x_i-(a+1))^2=\bigl[(x_i-a)-1\bigr]^2 =(x_i-a)^2-2(x_i-a)+1.$$

Summing this expression from $$i=1$$ to $$n$$ yields

$$\sum_{i=1}^{n}(x_i-(a+1))^2 =\sum_{i=1}^{n}(x_i-a)^2 -2\sum_{i=1}^{n}(x_i-a) +\sum_{i=1}^{n}1.$$

Substituting the given values step by step:

$$\sum_{i=1}^{n}(x_i-a)^2=na,$$

$$\sum_{i=1}^{n}(x_i-a)=n,$$

$$\sum_{i=1}^{n}1=n.$$

Hence

$$\sum_{i=1}^{n}(x_i-(a+1))^2 =na-2n+n =n(a-1).$$

So the variance is

$$\sigma^2=\frac1n\sum_{i=1}^{n}(x_i-\overline x)^2 =\frac1n\bigl[n(a-1)\bigr] =a-1.$$

Taking the positive square root (because standard deviation is non-negative) gives

$$\sigma=\sqrt{\sigma^2}=\sqrt{a-1}.$$

Hence, the correct answer is Option D.