Here's the question you clicked on:
hba
Can anyone help me prove this?
yes... insyaAllah
\[\huge \sum_{i=1}^{n}({X}_{i}-X \prime)^2=\sum_{i=1}^{n}{X_i}^2-n X \prime^2 \] X prime is actually x bar.|dw:1360770364042:dw|
It's Insha-Allah * @raden_zaikaria
Um,I know we can use this, |dw:1360770482862:dw|
so difficult.... is this polynomial system....
Never studied stats in my life lol. But now i have to :/
I have a physics quiz tomorrow so i can study for that,Please give me some feedback.
is this statistics?
expand the \[\left( X _{i}-\bar{X} \right)^{2}\] then use the properties of sigma
If i new it i wouldn't have been asking :/
use the properties \[ (a+b)^2 = a^2 +2ab + b^2\] let \(a = X_i\) and \(b= \bar X\)
what do you get by expanding it?
show \[ \sum_{i=1}^{n}({X}_{i}-\bar{X} )^2=\sum_{i=1}^{n}{X_i}^2-n\bar{X}^2\] where by definition \[ \bar{X}= \frac{1}{n}\sum_{i=1}^{n} X_i \] expand the square to get \[ \sum_{i=1}^{n}({X}_{i}-\bar{X} )^2= \sum_{i=1}^{n}({X}_{i}^2 - 2\bar{X} {X}_{i}+\bar{X}^2)\] or taking the sum term by term \[ = \sum_{i=1}^{n} {X}_{i}^2 - 2\bar{X}\sum_{i=1}^{n}{X}_{i} + \sum_{i=1}^{n}\bar{X}^2\] the mean \( \bar{X}^2\) is a constant, and summing n copies is \( n\bar{X}^2\) also, by the definition of \( \bar{X} \), \[ n \bar{X}= \sum_{i=1}^{n} X_i \] using these expressions, we get \[ = \sum_{i=1}^{n} {X}_{i}^2 - 2n \bar{X}^2 + n \bar{X}^2\] or \[ = \sum_{i=1}^{n} {X}_{i}^2 - n \bar{X}^2 \]