University Statistics and Probability Homework: Solutions and Analysis

This document contains the solutions to a statistics and probability homework assignment. The solutions cover various probability concepts, including the calculation of probabilities using integrals, solving for unknown parameters in probability distributions, and applying integration techniques to sol...Read More

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