ti8623er
Statistics Question: Normal Distibution
If Z is a standard normal random variable, what is:
P(z^(2)<1) ??
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amistre64
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wouldnt that be a cumulative of z<1 ?
amistre64
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|dw:1362004635659:dw|
or am i misreading this?
amistre64
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im prolly misreading it :/
ti8623er
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the book (well the answers in the back) rewrite it as P(-1<Z<1), but i dont know why??? :/
amistre64
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ahh, so i had the right idea sort of
ti8623er
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i needs explanation please :)
amistre64
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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.
amistre64
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(-.5)^2 = .25
(-.9999)^2 = well something between 0 and 1 :)
ti8623er
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i think it might be me but i sill dont understand
amistre64
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|dw:1362005251278:dw|
amistre64
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is 2^2 less than 1?
ti8623er
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can we reason on it by appllying some properties of inequalitys o Z^2<1 ??
amistre64
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sure
ti8623er
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okay....
amistre64
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z^2 < 1
sqrt each side
z < sqrt(1) or z > -sqrt(1)
ti8623er
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would it be or or and?
slaaibak
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P(z^(2)<1) = P(-1< Z < 1) = \[\Phi (1) - \Phi(-1)\]
amistre64
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.... picky picky picky, fine, the intersection
slaaibak
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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
amistre64
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or, another idea:
\[|z|=\sqrt{z^2}\]
\[z^2<1\]
\[\sqrt{z^2}<\sqrt{1}\]
\[|z|<1\]
amistre64
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\[|z|<1~implies~-1<z<1\]
ti8623er
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gracias everybody
amistre64
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good luck :)