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NTHU MATH 2810
Final Examination Solution
Jan 8, 2008
1. (
16pts, 2pts for each
)
(a) True.
(b) False. The values of a pdf can be larger than 1. However, the integration of a pdf
over any region must have values between 0 and 1.
(c) False. It must be a onetoone transformation.
(d) True.
(e) True.
(f) True.
(g) False. If
X
and
Y
are
independent
, then
E
(
X

Y
) =
E
(
X
). Zero correlation (i.e.,
uncorrelated) is a weaker condition than independence. It cannot guarantee this
property.
(h) False. When
X
and
Y
are independent,
E
(
X/Y
) =
E
(
X
)
E
(1
/Y
)
6
=
E
(
X
)
/E
(
Y
)
in general.
2. (
15pts, 3pts for each
)
(a) Normal(
μ,σ
2
) with
μ
= 68 and
σ
2
= 2.
(b) Gamma(
α,λ
) with
α
= 1000 and
λ
= 5. (An alternative answer that is acceptable
is Exponential(
λ
) with
λ
=
1
1000
/
5
=
1
200
.
)
(c) Poisson(
λ
) with
λ
= 2
×
2 = 4.
(d) Binomial(
n,p
) with
n
= 20 and
p
= 1
/
8.
(e) Uniform(
a,b
) with
a
= 0 and
b
= 360.
3. (
6pts
) Let
X
be the number of “5” that occurs in the 500 rolls, then
X
∼
Binomial(500
,
1
/
6)
since the die is fair. Therefore,
E
(
X
) = 500
×
(1
/
6) = 500
/
6
,
and
V ar
(
X
) = 500
×
(5
/
36) = 2500
/
36
.
Because
n
= 500 is large, we can use Normal approximation to evaluate
P
(
X
≥
100) as
follows:
P
(
X
≥
100) =
P
(
X
≥
99
.
5) =
P
X

500
6
q
2500
36
≥
99
.
5

500
6
q
2500
36
≈
P
Z
≥
99
.
5

500
6
q
2500
36
= 1

Φ
99
.
5

500
6
q
2500
36
= 1

Φ(1
.
94)
,
where
Z
∼
Normal(0, 1).
1
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View Full Document 4. (a) (
2pts
) Because
X
∼
Uniform(0
,L/
2),
Y
∼
Uniform(
L/
2
,L
), and they are indepen
dent, their joint pdf is
f
X,Y
(
x,y
) =
f
X
(
x
)
f
Y
(
y
) =
(
1
L/
2
1
L/
2
=
4
L
2
,
for 0
< x < L/
2 and
L/
2
< y < L,
0
,
otherwise.
(b) (
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This note was uploaded on 02/12/2012 for the course MATH 2810 taught by Professor Shaoweicheng during the Fall '11 term at National Tsing Hua University, China.
 Fall '11
 ShaoWeiCheng
 Math, Probability

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