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Solution to Homework 7
Section 4.1
1.
Treat the density function as constant over those small intervals.
a)
(0.001)(
1
2
π
)=0.000399
b)
(0.001)(
1
2
)(
0.5
e

)=0.00025
2.
a)
4
1
1
cx dx
c

=
c = 3
b) mean = 3/2 c) variance = ¾
3.
a)
1
0
(1
)
1
cx
x dx

=
c = 6
b) By symmetry,
(
1/ 2)
(
1/ 2)
1/ 2
P X
P X
d
=
=
c)
1/3
0
(
1/ 3)
6 (1
)
P X
x
x dx
c
=

c
=7/27
d) 1/27/27=13/54
e) Mean = ½
Variance =1/20
5.
a)
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2
2
1
1
1
( 1
2)
7 /12
2 (1  )
P
X
dx
x

 <
<
=
=
+
c)
2
1
1
(
 1)
2 (
1)
1/ 2
(1
)
P X
P X
dx
x
c
=
=
=
+
d) Expectation doesn’t exist because
 
( )
x f x dx
c

=
.
6.
a)
(
1)
2/3
(
1)
1/ 3
P X
P X
== =
, also
(
0)
1/3
P X c
=
, by symmetry we have
0.5
μ
=
.
Since
0 0.5
1
(
0)
(
)
(
)
1/ 3
2
P X
P Z
P Z
σ

=
=

=
, we have:
From table
1
0.4303
2

= 
, so
1.162
=
b)
1
1
(
1)
3/ 4
(
)
0.6742
P X
P Z


=
=
=
,
(
0)
1/3
(
)
0.4303
P X
P Z


=
=
= 
so we have
0.3896,
0.904
=
=
8.
a)
11.8 12
12.2 12
2
2
(11.8
12.2)
(
)
(
)
1.1
1.1
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This note was uploaded on 09/14/2009 for the course STAT 134 taught by Professor Aldous during the Fall '03 term at University of California, Berkeley.
 Fall '03
 aldous
 Variance

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