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

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

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

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

ti8623er
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1ahh, so i had the right idea sort of

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

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

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

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

ti8623er
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1z^2 < 1 sqrt each side z < sqrt(1) or z > sqrt(1)

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

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

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

slaaibak
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1\[z<1~implies~1<z<1\]
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