hba
  • hba
Can anyone help me prove this?
Statistics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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hba
  • hba
Can anyone help me prove this?
Statistics
schrodinger
  • schrodinger
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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raden_zaikaria
  • raden_zaikaria
yes... insyaAllah
hba
  • 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|
hba
  • hba
It's Insha-Allah * @raden_zaikaria

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hba
  • hba
Um,I know we can use this, |dw:1360770482862:dw|
raden_zaikaria
  • raden_zaikaria
so difficult.... is this polynomial system....
hba
  • hba
Never studied stats in my life lol. But now i have to :/
hba
  • hba
I have a physics quiz tomorrow so i can study for that,Please give me some feedback.
chihiroasleaf
  • chihiroasleaf
is this statistics?
hba
  • hba
Yes it is.
chihiroasleaf
  • chihiroasleaf
expand the \[\left( X _{i}-\bar{X} \right)^{2}\] then use the properties of sigma
hba
  • hba
If i new it i wouldn't have been asking :/
hba
  • hba
knew*
chihiroasleaf
  • chihiroasleaf
use the properties \[ (a+b)^2 = a^2 +2ab + b^2\] let \(a = X_i\) and \(b= \bar X\)
chihiroasleaf
  • chihiroasleaf
what do you get by expanding it?
phi
  • 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 \[ = \sum_{i=1}^{n} {X}_{i}^2 - n \bar{X}^2 \]

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