1. Let the mean of the observations $x_1,x_2,\dots,x_n$ be $\bar{x}$.

Given: $\sum_{i=1}^n (x_i-a)=n$ So, $\sum_{i=1}^n x_i-na=n$ $\sum_{i=1}^n x_i=n(a+1)$ Hence, $\bar{x}=\frac{1}{n}\sum_{i=1}^n x_i=a+1$

2. We are also given: $\sum_{i=1}^n (x_i-a)^2=na$

Now write $x_i-a=(x_i-\bar{x})+(\bar{x}-a)$ Since $\bar{x}=a+1$, we get $x_i-a=(x_i-\bar{x})+1$

Therefore, $\sum_{i=1}^n (x_i-a)^2=\sum_{i=1}^n \big((x_i-\bar{x})+1\big)^2$ Expand: $=\sum_{i=1}^n (x_i-\bar{x})^2+2\sum_{i=1}^n (x_i-\bar{x})+\sum_{i=1}^n 1$ But, $\sum_{i=1}^n (x_i-\bar{x})=0$ So, $\sum_{i=1}^n (x_i-a)^2=\sum_{i=1}^n (x_i-\bar{x})^2+n$ Given left side is $na$, hence $na=\sum_{i=1}^n (x_i-\bar{x})^2+n$ $\sum_{i=1}^n (x_i-\bar{x})^2=n(a-1)$

3. Variance is $\sigma^2=\frac{1}{n}\sum_{i=1}^n (x_i-\bar{x})^2$ So, $\sigma^2=\frac{1}{n}\cdot n(a-1)=a-1$ Thus standard deviation is $\sigma=\sqrt{a-1}$

4. Checking options:

• A: $a-1$ is the variance, not standard deviation.
• B: $n\sqrt{a-1}$ is incorrect.
• C: $\sqrt{n(a-1)}$ is incorrect.
• D: $\sqrt{a-1}$ is correct.

Therefore, the correct answer is D.