Computing the Standard Deviation

Obvious how to do it in multiple passes through the list.

But...

• $\sum_{i = 0}^{n-1} (L[i]-\mu)^2$
• $= \sum_{i = 0}^{n-1} (L[i]^2 -2 \mu L[i] + \mu^2)$
• $= \sum_{i = 0}^{n-1} L[i]^2 -2\mu \sum_{i = 0}^{n-1} L[i] + \mu^2 \sum_{i = 0}^{n-1} 1$
• $= \sum_{i = 0}^{n-1} L[i]^2 - 2 n \mu^2 + \mu^2 n$
• $= \sum_{i = 0}^{n-1} L[i]^2 - n \mu^2$

Standard deviation is easily computed from this.

So, we can compute the sum of the list elements and the sum of the squares of the list elements in a single pass.