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
π
)(
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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b)
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
σ

=
=

=
, we have:
From table
1
0.4303
2
σ

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


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


=
=
= 
�
�
�
so we have
0.3896,
0.904
μ
σ
=
=
8.
a)
11.8
12
12.2
12
2
2
(11.8
12.2)
(
)
(
)
1.1
1.1
11
11
P
X
P
Z
P
Z


<
<
=
<
<
=

<
<
=0.1443
b)
11.8
12
12.2
12
0.2
0.2
(11.8
12.2)
(
)
(
)
11
11
1.1/
100
1.1/
100
P
X
P
Z
P
Z


<
<
=
<
<
=

<
<
=0.9307. The
measurements need not follow normal distribution individually; as the CLT says their
sum/average would be approximately normal as long as they are ‘iid’ [independent and
identically distributed] with finite variance.
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 Fall '03
 aldous
 Normal Distribution, Variance, Probability theory, probability density function, dx, small neighborhood dx

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