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lilmissr

New member
Joined
Oct 11, 2009
Messages
3
Suppose fy (y)=4y^3 , 0<y<1
Find P(0<Y<1/2)

I started to draw it but I don't really get how to do continous random variables.
 

arthur ohlsten

Full Member
Joined
Feb 20, 2005
Messages
854
If I understand your notation then
INT(0to1) 4y^3 dy = y^4 evaluated 1,0
int (0 to 1) 4y^3= 1

Thus the area under the curve is 1

int(0 to 1/2) 4y^3 dy = y^4 evaluated (1/2,0)
int(0 to 1/2) 4y^3= [1/2]^4 = 1/16

P(0<= y <= 1/2) = [1/16]/1
P(0<= y <= 1/2)=1/16 answer

I believe this is what your instructor wants

First find the area beneath the curve over its domain, and then the area beneath the curve over the domain of interest.

Arthur

Arthur
 
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