Problem:
I'm really struggling with a really specific topic in statistics. It is working out the sample standard deviation when given summary statistics. I understand how to do the population standard deviation.
Example Question:
What is the sample standard deviation given the following information:
$$\begin{align} \sum_{k=0}^n x_k &= 80789\\ \sum_{k=0}^n x_k^2 &=113666365\\ n&=60 \end{align}$$
Could someone please take me through the process of answering the question?
Per the comments in the OP, I believe this is a freshman undergraduate-level statistics course, where less emphasis is placed on algebraic derivations and solely on the algorithms for calculations. Thus, I will present the formula. If the algebra doesn't make any sense, don't worry about the algebra; the result is what matters.
The sample variance can be written as $$\begin{align} S^2 &= \dfrac{1}{n-1}\sum_{k=1}^{n}(x_k - \bar{x})^2 \\ &= \dfrac{1}{n-1}\sum_{k=1}^{n}(x_k^2-2x_k\bar{x}+\bar{x}^2) \\ &= \dfrac{1}{n-1}\left(\sum_{k=1}^{n}x_k^2 - 2\bar{x}\sum_{k=1}^{k}x_k+ n\bar{x}^2\right) \\ &= \dfrac{1}{n-1}\left(\sum_{k=1}^{n}x_k^2 - 2n\bar{x}^2+ n\bar{x}^2\right) \\ &= \dfrac{1}{n-1}\left(\sum_{k=1}^{n}x_k^2 - n\bar{x}^2\right)\text{.} \end{align}$$ Thus, $$S^2 = \dfrac{1}{n-1}\left(\sum_{k=1}^{n}x_k^2-n\bar{x}^2 \right)$$ is the so-called "shortcut" formula for the sample variance. Note that $\bar{x} = \dfrac{\sum_{k=1}^{n}x_k}{n}$ is the sample mean. Simply put in the values you're provided to get the sample variance, and take the square root of the sample variance to get the sample standard deviation.
As a side note, you can obtain a similar formula for the population variance. Note the relationship $$\sigma^2 = \dfrac{n-1}{n}S^2\text{.}$$ Thus, $$\sigma^2 = \dfrac{n-1}{n} \cdot \dfrac{1}{n-1}\left(\sum_{k=1}^{n}x_k^2-n\bar{x}^2 \right) = \dfrac{1}{n}\left(\sum_{k=1}^{n}x_k^2-n\bar{x}^2 \right)$$ Then, take the square root of $\sigma^2$ to obtain the population standard deviation.