Solution Q1a). { ( ) 2. Fy(y) =. −y , y >0 2 0 , otherw

Solution
Q1a)
Fy(y) = { yexp ( − y2
2 ) , y >0
0 , otherwise
P(Y > y) = ∫
y
∞
f y(y)
= ∫
y
∞
yexp ( − y2
2 ) dy
Substitute
u = − y2
2 → dx=−1
y du
= −∫eu du
∫ au du= au
ln ( a ) , with a = e
= eu
Solve integrals
= ∫ eu du=−eu
u = − y2
2 , then −e
y2
2 +c
[
−e
y2
2 +c ]y∞
= −e
∞2
2 −(−e
y2
2 )
= 0 + e
y2
2
= e
y2
2

b)
P (0 < y< 1) = ∫
0
1
f y(y)
= ∫
0
1
yexp ( − y2
2 ) dy
[
−e
y2
2 +c ]01
= −e
12
2 −(−e
02
2 )
-1.64872 + 1
= -0.64872
Q2a)
P(x ≤ m¿=0.5=¿∫
0
m
λ
2 exp (−λx)dx
0.5 => [e− λx]0m
m = ln ( 0.5 )
−λ
b)
P(0.25 < x< 0.75) = ∫
0.25
0.75
f x(x)dx
= ∫
0.25
0.75
8
( x+2 ) 3 dx
Ley u = x + 2 → dx=du

= 8 ∫
0.25
0.75
1
u3 du
Solving
∫
0.25
0.75
1
u3 du
∫
0.25
0.75
u2 du = un +1
n+1 with n=−3
= 1
2u2
= 8∫ 1
u3 du=−4
u2
Substitute u = x + 2
= −4
( x+2 )2
= [ −4
( x+2 )2 +C ]0.250.75
= −4
( 0.75+2 ) 2 − −4
( 0.75+ 2 ) 2
= -0.5289 + 0.79012
= 0.26122
Q3a)
P(Z ≤ z ¿=∫
0
∞
−l og(1− x)dx
Let, u = 1-x → dx=−du
∫ log (u)du
Integrate by parts, ∫ fg ' =fg−∫f ' g

f = log(u), g’ = 1
f’ = 1/u, g= u
= ulog(u) - ∫1 du
Applying constant rule
= −∫1 du=u
Solve integration
ulog(u) - ∫1 du
= ulog(u) = u
Substituting
u = 1 – x
x + log(1-x)(1-x)-1
[x + log(1-x)(1-x)-1]0∞
= 0 – 0 = 0
b)
pdf of ∫
−∞
∞
−l og (1−x)dx = 1
distribution of Z
fz(Z) = {−log ( 1−x ) , z >0
0 ,otherwise