19 SPRING STAT 630 (Texas A&M) Exam 1 Solutions: Math Stat

This document provides a complete solution set for the STAT 630 Exam 1, focusing on the overview of mathematical statistics. The solutions cover a range of probability and statistics topics, including binomial distributions, mutual exclusivity, conditional probability, and the analysis of random variables. Specific problems addressed include calculating probabilities using binomial distributions, determining independence and mutual exclusivity, and solving problems related to probability distributions. The solutions also include calculations for conditional probabilities, and the analysis of continuous random variables. The document is designed to help students understand and solve complex statistical problems, providing a detailed breakdown of each solution step. This assignment is particularly relevant for students at Texas A&M University enrolled in the 19 SPRING STAT 630 course.

PROBLEM I
1) θ=0.2 and n=7
P ( X=3 ) =7C3
( θ ) 3 (1−θ)4
P ( X=3 ) Where X ~ Binomial (n=7, θ=0.2 ¿
Answer) a)
2) P ( A )=0.6∧P ( B )=0.1
P ( A ∩B )=0.08
P ( A ∪B ) =P ( A ) + P ( B ) −P ( A ∩ B )
P ( A ∪B ) =0.62
Since,
P ( A ∩B ) ≠ P ( A ) × P ( B )
And,
P ( A ∪ B ) ≠ P ( A )+ P ( B )
So it is neither independent nor mutually exclusive.
Answer) e)
3) P ( X ≥2 ) =1−P ( X=0 )−P ( X =1 )
P ( X ≥2 ) =1−0.1−0.3=0.6
Answer) d)
4) f X ( x )=
{ d
dx
x
4 0 ≤ x <1
d
dx
x4 + 4
20 1 ≤ x<2
d
dx 1 x ≥2
f X ( x ) =
{ 1
4 0 ≤ x <1
x3
5 1 ≤ x <2
0 x ≥2
Answer) b)

5) 6
20 × 5
19 × 4
18
(6
3 )
(20
3 )
Answer) d)
6) P ( Good Risk ) =0.4
P ( Bad Risk ) =0.6
P ( NoCitation∨Good Risk ) =1−0.1=0.9
P ( NoCitation∨Bad Risk )=1−0.5=0.5
P ( NoCitation ) =P ( No Citation∨Good Risk ) ×
P ( Good Risk ) +P ( No Citation∨Bad Risk ) × P ( Bad Risk ) = ( 0.4 ×0.9 ) + ( 0.6 × 0.5 ) =0.66
Answer) b)
7) f X ( x )= { x
2 0 ≤ x< 2
0 x ≥ 2
∫
0
x
x
2 dx=0.75
x2
4 =0.75
x= √4 ×0.75
Answer) c)
8) P ( X >Y )= 4
24 + 2
24 + 1
24 = 7
24
Answer) a)
PROBLEM II

1) P ( Goalkeeper∨Sam )=0.5
P ( Goalkeeper∨ Alex )=0.3
P ( NoGoalkeeper ∨Sam )=0.5
P ( NoGoalkeeper ∨Alex )=0.7
P ( S am ) =0. 6
P ( Alex ) =0. 4
P ( GoalKeeper ) =P ( Goalkeeper ∨Sam ) × P ( Sam ) + P (Goalkeeper ∨ Alex ) ×
P ( Alex ) = ( 0.5 ×0. 6 ) + ( 0.3 × 0. 4 ) =0.42
P ( NoGoalKeeper ) =P ( NoGoalkeeper ∨Sam ) × P ( Sam ) + P ( No Goalkeeper ∨Alex ) ×
P ( Alex ) = ( 0.5 ×0.6 )+ ( 0.7 ×0.4 )=0.58
P ( Alex∨NoGoalkeeper )= P ( No Goalkeeper∩ Alex )
P ( NoGoalKeeper )
P ( Alex∨NoGoalkeeper )= P ( No Goalkeeper∨ Alex ) × P ( Alex )
P ( NoGoalKeeper )
P ( Alex∨NoGoalkeeper )= 0.7 ×0.4
0.58 =0.28
0.58 = 14
29
2) f X Y (x , y )= { x + y
8 0≤ x <2 ,0 ≤ y <2
0 otherwise
f X ( x )=∫
0
2
x+ y
8 dy=¿∫
0
2
x
8 dy +∫
0
2
y
8 dy= x
4 + 1
4 ¿
f Y ( y)=∫
0
2
x + y
8 dx=¿∫
0
2
x
8 dx +∫
0
2
y
8 dx= y
4 + 1
4 ¿
Since,
f XY ( x , y ) ≠ f X ( x ) × f Y ( y )

We can say that X and Y are not independent.
3) a) If mutually exclusive then,
P ( A ∪ B ) =P ( A )+ P ( B )=0.5+0.3=0.8
b) If mutually independent then,
P ( A ∩B )=P ( A ) × P ( B ) =0.5 ×0.3=0.15
P ( A ∪ B ) =P ( A )+ P ( B )−P ( A ∩ B )=0.5+ 0.3−0.15=0.65
c) B⊂ A
Then we can find the value as,
P ( A ∪ B ) =P ( A ) =0.5
d) P ( A ∩B )=0.2
P ( A ∪ B ) =P ( A )+P ( B )−P ( A ∩ B )=0.5+0.3−0.2=0.6
4)
f X ( x ) =e− x
P ( Y < y )=P ( e5 x < y ) =P ¿
FY ( y ) = ∫
0
ln y
5
e−x dx=−e− x∨¿0
ln y
5 =1−e
−ln y
5 =1− y
−1
5 ¿
f Y ( y ) = d
dy FY ( y ) = d
dy ( 1− y
−1
5 ) =1
5 y
−6
5
Therefore the distribution of Y is
f Y ( y )= 1
5 y
−6
5