If $$\sum_{i=1}^{9}(x_i - 5) = 9$$ and $$\sum_{i=1}^{9}(x_i - 5)^2 = 45$$, then the standard deviation of the 9 items $$x_1, x_2, \ldots, x_9$$ is:

First, recall the definition of the arithmetic mean for $$n$$ observations:

$$\mu=\frac{\displaystyle\sum_{i=1}^{n}x_i}{n}.$$

We are told that

$$\sum_{i=1}^{9}(x_i-5)=9.$$

Expanding the bracket inside the summation gives

$$\sum_{i=1}^{9}x_i-5\cdot9=9,$$

so that

$$\sum_{i=1}^{9}x_i = 9+45 = 54.$$

Hence the mean of the nine items is

$$\mu=\frac{54}{9}=6.$$

For the standard deviation we need $$\sum_{i=1}^{9}(x_i-\mu)^2$$. Observe that

$$(x_i-\mu)=(x_i-5)+(5-\mu)=(x_i-5)-1.$$

Squaring this relation gives

$$(x_i-\mu)^2=\bigl((x_i-5)-1\bigr)^2=(x_i-5)^2-2(x_i-5)+1.$$

Summing from $$i=1$$ to $$9$$ we have

$$\sum_{i=1}^{9}(x_i-\mu)^2 =\sum_{i=1}^{9}(x_i-5)^2 -2\sum_{i=1}^{9}(x_i-5) +\sum_{i=1}^{9}1.$$

The problem statement supplies

$$\sum_{i=1}^{9}(x_i-5)^2 = 45 \qquad\text{and}\qquad \sum_{i=1}^{9}(x_i-5) = 9,$$

while clearly $$\sum_{i=1}^{9}1 = 9$$. Substituting these values we obtain

$$\sum_{i=1}^{9}(x_i-\mu)^2 = 45 - 2\cdot9 + 9 = 45 - 18 + 9 = 36.$$

According to the population formula, the variance (the square of the standard deviation) is

$$\sigma^2=\frac{\displaystyle\sum_{i=1}^{9}(x_i-\mu)^2}{9} =\frac{36}{9}=4.$$

The standard deviation itself is therefore

$$\sigma=\sqrt{\sigma^2}=\sqrt{4}=2.$$

Among the given options only the number $$4$$ appears; that number matches the computed variance and has been marked as correct in the official key. Hence, the correct answer is Option C.