Solutions to Stat 134 Old Finals
Final#1
1) a)
13
4
by using 13 indicators, 1 for each card in the hand.
b) The number of hearts in the hand is hypergeometric, so
(
13
3
)(
39
10
)
(
52
13
)
2) Using indicators
I
i
, one for each card,
E
(number of records) is
E
(
10
X
i
=1
I
i
) =
10
X
i
=1
E
(
I
i
) =
10
X
i
=1
1
i
where
1
i
is the probability that the
i
th card is the largest of the cards so far,
in other words, the probability that the
i
th card is a record.
3) a) binomial(100,.4)
b)
E
(
X
) =
μ
= 40,
SD
(
X
) =
σ
=
√
npq
= 4
.
9 Since
σ >
3, normal
approximation works well, and
P
(
X
= 45)
=
P
(44
.
5
≤
X
≤
45
.
5)
=
P
(
44
.
5

40
4
.
9
≤
X

40
4
.
9
≤
45
.
5

40
4
.
9
)
=
Φ(1
.
12)

Φ(
.
92)
c) negative binomial(4,.4) This is waiting until the 4th head.
d) The negative binomial(4,.4) can be thought of as a sum of 4 geometric(.4)
random variables, each of which has expected value
1
.
4
= 2
.
5, so
E
(
Z
) = 10.
4) a)
Z
1
0
cx
2
dx
=
c
1
3
x
3

1
0
=
c
3
= 1
since the integral of the density over entire range must equal 1. Thus
c
= 3.
b) Using the density function,
E
(
X
) =
Z
1
0
x
(3
x
2
)
dx
=
3
4
x
4

1
0
=
3
4
c)
SD
(
X
) =
q
var
(
X
) =
q
E
(
X
2
)

[
E
(
X
)]
2
so we need only find
E
(
X
2
).
E
(
X
2
) =
Z
1
0
x
2
(3
x
2
)
dx
=
3
5
x
5

1
0
=
3
5
and so
SD
(
X
) =
q
3
5

9
16
=
q
3
80
=
.
194.
d) The cdf of
X
is
F
(
x
) =
P
(
X
≤
x
) =
Z
x
0
f
(
x
)
dx
=
Z
x
0
3
x
2
dx
=
x
3

x
0
=
x
3
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
e) Since
g
(
x
) is strictly increasing,
f
Y
(
y
) =
3
x
2
.
5
x

.
5
= 6
y
5
for 0
≤
y
≤
1
,
0 else.
5) a)
Z
∞
0
Z
x
0
λ
1
λ
2
e

λ
1
x
e

λ
2
y
dydx
=
Z
∞
0
λ
1
e

λ
1
x
(1

e

λ
2
x
)
dx
= 1

λ
1
λ
1
+
λ
2
=
λ
2
λ
1
+
λ
2
b)
P
(
1
2
<
X
Y
<
2) =
P
(
.
5
Y < X <
2
Y
). Figuring out what region this is in
the plane, integrate the jt density over this region.
Z
∞
0
Z
2
y
.
5
y
λ
2
e

λx

λy
dxdy
=
Z
∞
0
λe

λy
(
e

λ.
5
y

e

λ
2
y
)
dy
=

2
3
e

λ
1
.
5
y
+
1
3
e

λ
3
y

∞
0
=
1
3
c)
P
(
T <
3) =
P
(2 or more hits by time 3).
Using the Poisson, this is
1

e

6

6
e

6
= 1

7
e

6
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 MCCULLOUGH
 Normal Distribution, Poisson Distribution, Probability theory, Exponential distribution, Poisson process, WI

Click to edit the document details