anonymous
  • anonymous
Statistics Question: Normal Distibution If Z is a standard normal random variable, what is: P(z^(2)<1) ??
Mathematics
schrodinger
  • schrodinger
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amistre64
  • amistre64
wouldnt that be a cumulative of z<1 ?
amistre64
  • amistre64
|dw:1362004635659:dw| or am i misreading this?
amistre64
  • amistre64
im prolly misreading it :/

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anonymous
  • anonymous
the book (well the answers in the back) rewrite it as P(-1
amistre64
  • amistre64
ahh, so i had the right idea sort of
anonymous
  • anonymous
i needs explanation please :)
amistre64
  • amistre64
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.
amistre64
  • amistre64
(-.5)^2 = .25 (-.9999)^2 = well something between 0 and 1 :)
anonymous
  • anonymous
i think it might be me but i sill dont understand
amistre64
  • amistre64
|dw:1362005251278:dw|
amistre64
  • amistre64
is 2^2 less than 1?
anonymous
  • anonymous
can we reason on it by appllying some properties of inequalitys o Z^2<1 ??
amistre64
  • amistre64
sure
anonymous
  • anonymous
okay....
amistre64
  • amistre64
z^2 < 1 sqrt each side z < sqrt(1) or z > -sqrt(1)
anonymous
  • anonymous
would it be or or and?
slaaibak
  • slaaibak
P(z^(2)<1) = P(-1< Z < 1) = \[\Phi (1) - \Phi(-1)\]
amistre64
  • amistre64
.... picky picky picky, fine, the intersection
slaaibak
  • slaaibak
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
amistre64
  • amistre64
or, another idea: \[|z|=\sqrt{z^2}\] \[z^2<1\] \[\sqrt{z^2}<\sqrt{1}\] \[|z|<1\]
amistre64
  • amistre64
\[|z|<1~implies~-1
anonymous
  • anonymous
gracias everybody
amistre64
  • amistre64
good luck :)

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