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

Mathematics
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wouldnt that be a cumulative of z<1 ?
|dw:1362004635659:dw| or am i misreading this?
im prolly misreading it :/

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Other answers:

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

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