Question

Why would the standard deviation for N trials be \[\sqrt{NPQ}\] http://prntscr.com/60h4lt

Answer 1

@perl

Answer 2

nbar= 1/6*1+1/6*2....
=1/6(1+2+3..+6) = 3.5

Answer 3

Well I know you can look at the standard deviation as being this and I can show how to derive it and it appears to be an almost identical statement to what you have written there but the context in which I learned it was quantum mechanics so I'm not sure what exactly the notation there says. \[\Large \sigma^2 = \langle x^2 \rangle - \langle x \rangle ^2\]

Answer 4

ya its a bit weird im pretty sure them imply the sum statement there