Solutions
Stat652
Homework 6
1. Navidi 7.1 #1
Sln:
n
n
n
x = y = 4, ( xi x ) = 28, ( yi y ) = 28, ( xi x ) ( yi y ) = 23.
2
i =1
2
i =1
i =1
n
r=
( x x )( y y )
i
i =1
n
i
n
( x x ) ( y y )
i =1
2
i
i =1
= 0.8214.
2
i
2. Navidi 7.1 #2
Sln:
a) g ( x ) = X
Solutions
Stat652
Homework 1
Bernoulli distribution: Bernoulli(p). f ( x | p ) = p x (1 p )
1 x
, x = 0,1; p [ 0,1]
E ( X ) = 1 p + 0 (1 p ) = p;
2
2
2
var ( X ) = E ( X ) = p (1 p ) + (1 p )( 0 p ) = p (1 p ) .
Binomial distribution: Bin(p). f ( x | p )
STATS 452/652, Spring 2013
Midterm 1 (February 26, 2013)
Name:
Level:
Read assignments carefully
If you do not see a solution go to the
next problem, good ideas may come later
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1
2
3
4
5
6
7
8
Total
(6 pt)
(6 pt)
(8 pt)
(6 pt)
(6 pt)
(
Solutions
Stat652
Sln 1: Fn =
Fn ( 1) =
Fn ( 5 ) =
n
1cfw_xi < x
i =1
n
n
1cfw_xi <1
i =1
n
n
1cfw_xi <0
i =1
n
Homework 1
.
= 0; Fn ( 0 ) =
n
1cfw_xi <0
i =1
n
1
= ; Fn ( 2 ) =
4
n
1cfw_xi <0
i =1
n
=
21
=;
42
3
=;
4
if x < 1,
0
1
4
( x) = 1
Therefore,
Solutions
Stat652
Homework 7
1. Sln:
A
B
C
D
E
F
Pearson coefficient
sensitive to outlier
0.6
sensitive to outlier
0.6
sensitive to outlier
0.4
strong positive assoc.
0.9
nonlinear association
0
pos cluster association
0.7
Kendall coefficient
insens
Solutions
STAT452/652
HW1
Problem 3
Let X1 , ., X100
be the heights of the 100 men. Then X is approximately normally
distributed with mean X = 70 and standard deviation X = 2.5 / 100 = 0.25 . The z-score
of 69 is (69 70)/0.25 = 4.00. The area under the st