Standard deviation is a way of measuring how spread out a set of data happens to be. More specifically, its a measure of much the data deviates from the mean. Data that is tightly clumped will have a smaller standard deviation, compared to a larger one for data that is more spread out.
There are a number of steps involved in calculating standard deviation, so you won’t be asked to do it by hand too often. And also, these steps are for finding standard deviation of a sample, not a population. There’s only one minor difference, but
With math symbols, the above looks like this
\[s = \sqrt{\frac{\Sigma (x-\bar{x})^2}{n-1}}\]The square root of the sum of all the squares of the deviations divided by one less than the number of data points.
Find the mean first (often given to you in the problems).
\[\bar{x}=6\]I’m going to combine steps 2 through 4 here, mostly since this is how it will be presented to you in the problems. We’ll be finding all the differences, squaring them, and adding them up.
\[\begin{align*} \Sigma(x-\bar{x})^2 &= (4-6)^2 + (3-6)^2 + (5-6)^2 + (7-6)^2 + (2-6)^2 + (9-6)^2 + (11-6)^2 + (7-6)^2\\ &= (-2)^2 + (-3)^2 + (-1)^2 + 1^2 + (-4)^2 + 3^2 + 5^2 + 1^2 \\ &= 4 + 9 + 1 + 1 + 16 + 9 + 25 + 1 \\ &= 66 \end{align*}\]Variance, or step 5, is next. This is where we take our sum and divide is by one less than the number of entries.
\[s^2 =7 frac{\Sigma(x-\bar{x})^2}{n-1} = \frac{66}{7} \approx 9.43\]Lastly, we square root that result to get our standard deviation.
\[s = \sqrt{\frac{\Sigma (x-\bar{x})^2}{n-1}} = \sqrt{9.43} \approx 3.07\]