It is possible to use Mean Absolute Deviation instead! And maybe in some alternate universe, that would be the value chosen to be the "standard deviation". But squaring instead of absolute-value-ing gives us a bunch of nice properties. Absolute value is hard to work with due to its "pointiness". Squaring is easy to work with. For instance, we can think about least squares regression - this is where we have a bunch of data points, and we want to fit a line to them. The line predicts a certain y-value for every x-value, but it might not be exactly the same. We can look at the 'error' in each of our predictions - this gives us a data set. We want this data set to have a mean of 0, and a small deviation from 0, to get the best possible fit. It turns out that it's very easy to do this if we choose standard deviation as our measurement of deviation: there's a nice formula involving a few matrix multiplications. It's easy to do on a computer. But if we chose MAD, there's no nice and easy formula. A bunch of other similar things go the same way: with squares, they're [relatively] easy, and with absolute value, they may not even have a 'nice' solution. More on reddit.com

The second formula is the most general. It says: if you have any random variable X with expected value E(X), its standard deviation is given by √(E(X-μ)²). In the special case where X~B(n;p), i.e. X follows a binomial distribution , it can be readily shown that Var(X) = np(1-p), hence you get your third formula. (This is because a binomial is a sum of n independent Bernouillis, each with variance p(1-p), as can be computed easily.) When n is large, the binomial distribution is well approximated by a normal distribution, so you may see this formula pop up when doing statistical inference on a proportion. Your first formula is a statistical biased estimator of the (generally unknown) standard deviation σ in a population. You shouldn't use σ to denote it but rather s or some variant (generally, you use the square root of the unbiased variance estimator: s = √((xᵢ-xbar)²/(n-1))). See also the Bessel correction . In the special case where X follows a finite uniform distribution, your first formula holds, but there you should write μ and not xbar for the actual mean and not the sample mean. More on reddit.com

At least for OCR that's a further stats topic, so I wouldn't be surprised if it's the same here. Look through your spec to see if it includes it, else you don't need to know it More on reddit.com

if i remember correctly just using the calculator will do. maybe your teacher wants you to fully understand how to do standard devi thats why they asked you to manually work it out? More on reddit.com