## ti8623er Group Title Statistics Question: Normal Distibution If Z is a standard normal random variable, what is: P(z^(2)<1) ?? one year ago one year ago

• This Question is Open
1. amistre64 Group Title

wouldnt that be a cumulative of z<1 ?

2. amistre64 Group Title

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

3. amistre64 Group Title

4. ti8623er Group Title

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

5. amistre64 Group Title

ahh, so i had the right idea sort of

6. ti8623er Group Title

7. amistre64 Group Title

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 Group Title

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

9. ti8623er Group Title

i think it might be me but i sill dont understand

10. amistre64 Group Title

|dw:1362005251278:dw|

11. amistre64 Group Title

is 2^2 less than 1?

12. ti8623er Group Title

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

13. amistre64 Group Title

sure

14. ti8623er Group Title

okay....

15. amistre64 Group Title

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

16. ti8623er Group Title

would it be or or and?

17. slaaibak Group Title

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

18. amistre64 Group Title

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

19. slaaibak Group Title

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 Group Title

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

21. amistre64 Group Title

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

22. ti8623er Group Title

gracias everybody

23. amistre64 Group Title

good luck :)