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Second Edition
515
is in the form
√
n
(
(
¯
U

E
U
)
/σ
)
with
n
= 12, so
X
is approximately n(0
,
1) by the Central
Limit Theorem.
b. The approximation does not have the same range as
Z
∼
n(0
,
1) where
∞
< Z <
+
∞
,
since

6
< X <
6.
c.
E
X
= E
±
12
X
i
=1
U
i

6
!
=
12
X
i
=1
E
U
i

6 =
±
12
X
i
=1
1
2
!

6 = 6

6 = 0
.
Var
X
= Var
±
12
X
i
=1
U
i

6
!
= Var
12
X
i
=1
U
i
= 12Var
U
1
= 1
E
X
3
= 0 since
X
is symmetric about 0. (In fact, all odd moments of
X
are 0.) Thus, the ﬁrst
three moments of
X
all agree with the ﬁrst three moments of a n(0
,
1). The fourth moment
is not easy to get, one way to do it is to get the mgf of
X
. Since E
e
tU
= (
e
t

1)
/t
,
E
²
e
t
(∑
12
i
=1
U
i

6
)
³
=
e

6
t
´
e
t

1
t
µ
12
=
´
e
t/
2

e

t/
2
t
µ
12
.
Computing the fourth derivative and evaluating it at
t
= 0 gives us E
X
4
. This is a lengthy
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.
 Spring '12
 Dr.Hackney
 Statistics, Central Limit Theorem

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