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

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.1wouldnt that be a cumulative of z<1 ?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dw:1362004635659:dw or am i misreading this?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1im prolly misreading it :/

ti8623er
 one year ago
Best ResponseYou've already chosen the best response.0the book (well the answers in the back) rewrite it as P(1<Z<1), but i dont know why??? :/

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1ahh, so i had the right idea sort of

ti8623er
 one year ago
Best ResponseYou've already chosen the best response.0i needs explanation please :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.1(.5)^2 = .25 (.9999)^2 = well something between 0 and 1 :)

ti8623er
 one year ago
Best ResponseYou've already chosen the best response.0i think it might be me but i sill dont understand

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dw:1362005251278:dw

ti8623er
 one year ago
Best ResponseYou've already chosen the best response.0can we reason on it by appllying some properties of inequalitys o Z^2<1 ??

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1z^2 < 1 sqrt each side z < sqrt(1) or z > sqrt(1)

ti8623er
 one year ago
Best ResponseYou've already chosen the best response.0would it be or or and?

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1.... picky picky picky, fine, the intersection

slaaibak
 one year ago
Best ResponseYou've already chosen the best response.0Also, 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
 one year ago
Best ResponseYou've already chosen the best response.1or, another idea: \[z=\sqrt{z^2}\] \[z^2<1\] \[\sqrt{z^2}<\sqrt{1}\] \[z<1\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[z<1~implies~1<z<1\]
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