## hba Group Title Can anyone help me prove this? one year ago one year ago

yes... insyaAllah

2. hba

$\huge \sum_{i=1}^{n}({X}_{i}-X \prime)^2=\sum_{i=1}^{n}{X_i}^2-n X \prime^2$ X prime is actually x bar.|dw:1360770364042:dw|

3. hba

It's Insha-Allah * @raden_zaikaria

4. hba

Um,I know we can use this, |dw:1360770482862:dw|

so difficult.... is this polynomial system....

6. hba

Never studied stats in my life lol. But now i have to :/

7. hba

I have a physics quiz tomorrow so i can study for that,Please give me some feedback.

8. chihiroasleaf

is this statistics?

9. hba

Yes it is.

10. chihiroasleaf

expand the $\left( X _{i}-\bar{X} \right)^{2}$ then use the properties of sigma

11. hba

If i new it i wouldn't have been asking :/

12. hba

knew*

13. chihiroasleaf

use the properties $(a+b)^2 = a^2 +2ab + b^2$ let $$a = X_i$$ and $$b= \bar X$$

14. chihiroasleaf

what do you get by expanding it?

15. phi

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 $= \sum_{i=1}^{n} {X}_{i}^2 - n \bar{X}^2$