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3.2.16
problem 36
(a)
P
(
X
=
n
2
)
=
(
n
n
2
)
(
1
2
)
n
(b)
Using Sterling’s formula
(
n
n
2
)
=
√
2
πn
(
n
e
)
n
√
2
π
n
2
(
n
2
e
)
n
2
√
2
π
n
2
(
n
2
e
)
n
2
=
√
2
2
n
√
πn
Thus,
P
(
X
=
n
2
)
=
√
2
πn
2
n
1
2
n
=
1
√
πn
2
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