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 solve probability problems. The assignment includes detailed step-by-step solutions for each question, demonstrating the application of different statistical methods and formulas. The problems involve both continuous and discrete probability distributions. The solution provides an in-depth analysis of the problems and offers insights into the underlying statistical principles. This resource is designed to help students understand and master probability concepts and improve their problem-solving skills.

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