Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

hba

  • one year ago

Can anyone help me prove this?

  • This Question is Closed
  1. raden_zaikaria
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes... insyaAllah

  2. hba
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. hba
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It's Insha-Allah * @raden_zaikaria

  4. hba
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. raden_zaikaria
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  6. hba
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  7. hba
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  8. chihiroasleaf
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is this statistics?

  9. hba
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes it is.

  10. chihiroasleaf
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  11. hba
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  12. hba
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    knew*

  13. chihiroasleaf
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. chihiroasleaf
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what do you get by expanding it?

  15. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  16. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.