AMS 311
Joe Mitchell
Practice Exam 3 – Solution Notes
Statistics:
n
=30,
μ
=64
.
1, median 68.5,
σ
=22
.
9; score range: 23–99
1. (a).
f
Y
(
y
)=
±
∞
∞
f
(
x, y
)
dx
=
²
³
y
0
(2
x
+2
y
)
dx
=3
y
2
if 0
<y<
1
0
otherwise
(b).
P
(
X>
0
.
1

Y
=0
.
5) =
±
∞
0
.
1
f
X

Y
(
x

0
.
5)
dx
=
±
∞
0
.
1
f
(
x,
0
.
5)
f
Y
(0
.
5)
dx
=
±
0
.
5
0
.
1
2
x
+2(0
.
5)
3(0
.
5)
2
dx
=
64
75
,
where we have used the answer to part (a) to obtain
f
Y
(0
.
5) = 3(0
.
5)
2
.
(c).
E
(
X

Y
.
5) =
±
∞
∞
x
·
f
X

Y
(
x

0
.
5)
dx
=
±
0
.
5
0
x
·
2
x
.
5)
3(0
.
5)
2
dx
=
5
18
2. [This is similar to Problem 5, HW7 and example 8.17 of the suggested reading in the text.] We are told that
Y
is exponential with mean 120 minutes. (Note, since
Y
is the lifetime in MINUTES, its mean is 120, NOT 2.) Thus,
Y
is exponential(
1
120
).
Let
X
be the number of minutes after 5:00pm when Joe enters the room. Then, given the
Y
=
y
,
X
is
Uniform(0
,y
); thus,
E
(
X

Y
=
y
y/
2, for any
y>
0.
(a). We can compute
E
(
X
) by conditioning on
Y
:
E
(
X
±
∞
∞
E
(
X

Y
=
y
)
f
Y
(
y
)
dy
=
±
∞
0
y
2
·
1
120
e

y
120
dy
=60
Thus, I expect Joe to enter at 60 minutes after 5:00pm, i.e., at time 6:00pm.
(b). We want to compute
P
(
100) (since 6:40pm is 100 minutes after 5:00pm). We condition on the value of
Y
:
P
(
100)
=
±
∞
∞
P
(
100

Y
=
y
)
f
Y
(
y
)
dy
=
±
100
0
P
(
100

Y
=
y
)
1
120
e

y
120
dy
+
±
∞
100
P
(
100

Y
=
y
)
1
120
e

y
120
dy
=
±
100
0
0
·
1
120
e

y
120
dy
+
±
∞
100
y

100
y
·
1
120
e

y
120
dy
where we have used the fact that
P
(
100

Y
=
y
²
y

100
y
if
100
0i
f
y<
100
Alternatively, both parts can be solved directly, using the joint density (see Example 8.17):
f
(
x, y
f
X

Y
(
x

y
)
·
f
Y
(
y
²
1
y
·
1
120
e

y
120
0
<y
,0
<x<y
0o
.
w
.
Then, we can compute (look at a picture of the support set):
P
(
100) =
±
∞
100
±
∞
x
1
y
·
1
120
e

y
120
dydx
and
E
(
X
±
∞
0
±
∞
x
x
·
1
y
·
1
120
e

y
120
dydx
1