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ti8623er

  • 3 years ago

Statistics Question: Normal Distibution If Z is a standard normal random variable, what is: P(z^(2)<1) ??

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  1. amistre64
    • 3 years ago
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    wouldnt that be a cumulative of z<1 ?

  2. amistre64
    • 3 years ago
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    |dw:1362004635659:dw| or am i misreading this?

  3. amistre64
    • 3 years ago
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    im prolly misreading it :/

  4. ti8623er
    • 3 years ago
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    the book (well the answers in the back) rewrite it as P(-1<Z<1), but i dont know why??? :/

  5. amistre64
    • 3 years ago
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    ahh, so i had the right idea sort of

  6. ti8623er
    • 3 years ago
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    i needs explanation please :)

  7. amistre64
    • 3 years ago
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    the reason why is because when you square a number it goes postive; and all the values of z that can be squared and be less than 1 are those values that are less than 1 to begin with.

  8. amistre64
    • 3 years ago
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    (-.5)^2 = .25 (-.9999)^2 = well something between 0 and 1 :)

  9. ti8623er
    • 3 years ago
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    i think it might be me but i sill dont understand

  10. amistre64
    • 3 years ago
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    |dw:1362005251278:dw|

  11. amistre64
    • 3 years ago
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    is 2^2 less than 1?

  12. ti8623er
    • 3 years ago
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    can we reason on it by appllying some properties of inequalitys o Z^2<1 ??

  13. amistre64
    • 3 years ago
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    sure

  14. ti8623er
    • 3 years ago
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    okay....

  15. amistre64
    • 3 years ago
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    z^2 < 1 sqrt each side z < sqrt(1) or z > -sqrt(1)

  16. ti8623er
    • 3 years ago
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    would it be or or and?

  17. slaaibak
    • 3 years ago
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    P(z^(2)<1) = P(-1< Z < 1) = \[\Phi (1) - \Phi(-1)\]

  18. amistre64
    • 3 years ago
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    .... picky picky picky, fine, the intersection

  19. slaaibak
    • 3 years ago
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    Also, Z^2 is is a chi squared distribution with 1 degree of freedom. So you can find the F(1) using the cumulative distribution function of the chi squared distribution

  20. amistre64
    • 3 years ago
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    or, another idea: \[|z|=\sqrt{z^2}\] \[z^2<1\] \[\sqrt{z^2}<\sqrt{1}\] \[|z|<1\]

  21. amistre64
    • 3 years ago
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    \[|z|<1~implies~-1<z<1\]

  22. ti8623er
    • 3 years ago
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    gracias everybody

  23. amistre64
    • 3 years ago
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    good luck :)

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