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

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amistre64Best ResponseYou've already chosen the best response.1
wouldnt that be a cumulative of z<1 ?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
dw:1362004635659:dw or am i misreading this?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
im prolly misreading it :/
 one year ago

ti8623erBest ResponseYou've already chosen the best response.0
the book (well the answers in the back) rewrite it as P(1<Z<1), but i dont know why??? :/
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
ahh, so i had the right idea sort of
 one year ago

ti8623erBest ResponseYou've already chosen the best response.0
i needs explanation please :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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.
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
(.5)^2 = .25 (.9999)^2 = well something between 0 and 1 :)
 one year ago

ti8623erBest ResponseYou've already chosen the best response.0
i think it might be me but i sill dont understand
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
dw:1362005251278:dw
 one year ago

ti8623erBest ResponseYou've already chosen the best response.0
can we reason on it by appllying some properties of inequalitys o Z^2<1 ??
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
z^2 < 1 sqrt each side z < sqrt(1) or z > sqrt(1)
 one year ago

ti8623erBest ResponseYou've already chosen the best response.0
would it be or or and?
 one year ago

slaaibakBest ResponseYou've already chosen the best response.0
P(z^(2)<1) = P(1< Z < 1) = \[\Phi (1)  \Phi(1)\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
.... picky picky picky, fine, the intersection
 one year ago

slaaibakBest ResponseYou've already chosen the best response.0
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
or, another idea: \[z=\sqrt{z^2}\] \[z^2<1\] \[\sqrt{z^2}<\sqrt{1}\] \[z<1\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[z<1~implies~1<z<1\]
 one year ago
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