# The proper way to find standard error [closed]

I had a lab where calculating standard error was explained. My instructor did a cross multiplication move and ended up with range / n

How exactly does this work?

Are there other ways of representing standard error?

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 You should say that in calculating the standard error your instructor ended up with the population standard deviation divided by the square root of n. – Michael Chernick Aug 11 '12 at 20:20

## closed as not a real question by William, Qiaochu YuanAug 11 '12 at 20:37

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

I assume you are talking about the standard error of the mean which is the standard deviation of the estimate of the mean. It is not equal to the range divide by n. Let s$^2$ be the variance for each of a set of n independent identically distributed observations X$_i$ i=1,2,...,n.
Let X$_b$ = ∑X$_i$/n Then since Var(cY)=c$^2$Var(Y) where c is constant and Y is a random variable and if Y and Z are independent random variables the Var(Y+Z)=Var(Y)+Var(Z),
Var(∑X$_i$/n ) = ∑Var(X$_i$)/n$^2$ =n s$^2$ /n$^2$=s$^2$/n. Since the variance is s$^2$/n the standard error is s/√n. So contrary to your statement the standard error is the population standard deviation divided by the square root of n.