Mean and Standard Deviation of Y.?

both standard deviation and mean = 4 

mean =∫ integral y*f(y) dx  

mean = from 0 to infinity ∫0.25y *e^(-y/4) dy   

mean = from 0 to infinity  0.25 ∫y *e^(-y/4) dy   

integration by parts

let u = y 

du = 1 

let v' = e^(-y/4) 

v = 4e^(-y/4)   

∫udv = uv - ∫vdu 

∫y *e^(-y/4) dy    =from 0 to inf. | (y*4e^(-y/4)   - ∫e^(-y/4) dy    

∫y *e^(-y/4) dy =  from 0 to inf. | y*4e^(-y/4)  - 16e^(-y/4)   

∫y *e^(-y/4) dy =  infintiy*e^(-infinity/4) - 16e^(-infinity/4)- (-16e^(0)) 

- 16e^(-infinity/4) =0 

 infinty*e^(-infinity/4)= infinity * zero 

so you must use limit to show , it is really zero

y/ e^(y/4)  = l'hopital rule 1/e^(y/4) = e^(-y/4) = 0 

∫y *e^(-y/4) dy = 16e^0 = 16  

mean = from 0 to infinity 0.25*16 = 4  

mean = 4 

variance  = E(y^2) -(mean)^2 =   E(y^2) -16 

E(y^2) =0.25∫y ^2*e^(-y/4) dy 

if you 

let u = y^2 

du = 2y 

v' = e^(-y/4)

v = -4e^(-y/4) 

E(x^2)   =  0.25*[ y^2*-4e^(-y/4) -  ∫2y*-4e^(-y/4) dy ]

E(x^2) = 0.25*[ y^2*-4e^(-y/4) - 8∫y*e^(-y/4) dy ]

we already  did ∫y *e^(-y/4) dy  and know it equals -16  

and again the first term contributes 0 if you work it out again 

E(x^2)  =    0.25 *[0  -(-8*16) ]  =  32  

Variance  = 32 -16 = 16 

standard deviation = sqrt(variance)  =  sqrt(16)  = 4