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hba

  • 3 years ago

Can anyone help me prove this?

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  1. raden_zaikaria
    • 3 years ago
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    yes... insyaAllah

  2. hba
    • 3 years ago
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    \[\huge \sum_{i=1}^{n}({X}_{i}-X \prime)^2=\sum_{i=1}^{n}{X_i}^2-n X \prime^2 \] X prime is actually x bar.|dw:1360770364042:dw|

  3. hba
    • 3 years ago
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    It's Insha-Allah * @raden_zaikaria

  4. hba
    • 3 years ago
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    Um,I know we can use this, |dw:1360770482862:dw|

  5. raden_zaikaria
    • 3 years ago
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    so difficult.... is this polynomial system....

  6. hba
    • 3 years ago
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    Never studied stats in my life lol. But now i have to :/

  7. hba
    • 3 years ago
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    I have a physics quiz tomorrow so i can study for that,Please give me some feedback.

  8. chihiroasleaf
    • 3 years ago
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    is this statistics?

  9. hba
    • 3 years ago
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    Yes it is.

  10. chihiroasleaf
    • 3 years ago
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    expand the \[\left( X _{i}-\bar{X} \right)^{2}\] then use the properties of sigma

  11. hba
    • 3 years ago
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    If i new it i wouldn't have been asking :/

  12. hba
    • 3 years ago
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    knew*

  13. chihiroasleaf
    • 3 years ago
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    use the properties \[ (a+b)^2 = a^2 +2ab + b^2\] let \(a = X_i\) and \(b= \bar X\)

  14. chihiroasleaf
    • 3 years ago
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    what do you get by expanding it?

  15. phi
    • 3 years ago
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    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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