## ti8623er 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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1. amistre64

wouldnt that be a cumulative of z<1 ?

2. amistre64

|dw:1362004635659:dw| or am i misreading this?

3. amistre64

4. ti8623er

the book (well the answers in the back) rewrite it as P(-1<Z<1), but i dont know why??? :/

5. amistre64

ahh, so i had the right idea sort of

6. ti8623er

7. amistre64

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.

8. amistre64

(-.5)^2 = .25 (-.9999)^2 = well something between 0 and 1 :)

9. ti8623er

i think it might be me but i sill dont understand

10. amistre64

|dw:1362005251278:dw|

11. amistre64

is 2^2 less than 1?

12. ti8623er

can we reason on it by appllying some properties of inequalitys o Z^2<1 ??

13. amistre64

sure

14. ti8623er

okay....

15. amistre64

z^2 < 1 sqrt each side z < sqrt(1) or z > -sqrt(1)

16. ti8623er

would it be or or and?

17. slaaibak

P(z^(2)<1) = P(-1< Z < 1) = $\Phi (1) - \Phi(-1)$

18. amistre64

.... picky picky picky, fine, the intersection

19. slaaibak

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

20. amistre64

or, another idea: $|z|=\sqrt{z^2}$ $z^2<1$ $\sqrt{z^2}<\sqrt{1}$ $|z|<1$

21. amistre64

$|z|<1~implies~-1<z<1$

22. ti8623er

gracias everybody

23. amistre64

good luck :)