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
=
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
= 0
.
5) =
∞
∞
x
·
f
X

Y
(
x

0
.
5)
dx
=
0
.
5
0
x
·
2
x
+ 2(0
.
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
(
X >
100) (since 6:40pm is 100 minutes after 5:00pm). We condition on the value of
Y
:
P
(
X >
100)
=
∞
∞
P
(
X >
100

Y
=
y
)
f
Y
(
y
)
dy
=
100
0
P
(
X >
100

Y
=
y
)
1
120
e

y
120
dy
+
∞
100
P
(
X >
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
(
X >
100

Y
=
y
) =
y

100
y
if
y >
100
0
if
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
) =
1
y
·
1
120
e

y
120
0
< y
,0
< x < y
0
o.w.
Then, we can compute (look at a picture of the support set):
P
(
X >
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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
3. (a). Let
p
i
=
P
( she makes it to the summit

she starts
i
feet above the cli
ff
). Our goal is to find
p
1
. We
condition on the first step she takes, which takes her up by 1 foot with probability 0.3 and down by 1 foot with
probability 0.7. Thus, we get a system of 3 equations in 3 unknowns:
p
1
=
(0
.
7)
·
0 + (0
.
3)
p
2
p
2
=
(0
.
7)
p
1
+ (0
.
3)
p
3
p
3
=
(0
.
7)
p
2
+ (0
.
3)
·
1
Now, solve the system of three equations in three unknowns; the final answer is
p
1
.
(b). Let
t
i
=
E
(number of steps she takes before she is done

she starts
i
feet above the cli
ff
). Condition on the
result of the first step:
t
1
=
(0
.
7)(1 + 0) + (0
.
3)(1 +
t
2
)
t
2
=
(0
.
7)(1 +
t
1
) + (0
.
3)(1 +
t
3
)
t
3
=
(0
.
7)(1 +
t
2
) + (0
.
3)(1 + 0)
Now, solve the system of three equations in three unknowns; the final answer is
t
1
.
4. Let
X
be the number of people that show up at the party. We know that
X
≥
0 and that
E
(
X
) = 25. We want to
estimate
p
=
P
(
X
≥
61); by Markov’s inequality we get
P
(
X
≥
61)
≤
E
(
X
)
61
=
25
61
. Thus, we know that 0
≤
p
≤
25
61
.
5.
Let
X
1
, . . . , X
150
be the 150 random points from (0,1).
Let
X
=
X
1
+
· · ·
X
150
.
Each
X
i
is Uniform(0,1), so
E
(
X
i
) = 0
.
5 and
var
(
X
i
) = 1
/
12. Thus,
E
(
X
) = 150
·
(0
.
5) = 75 and
var
(
X
) = 150
·
(1
/
12) = 25
/
2 (since the
X
i
’s
are independent, by assumption, so
var
(
X
) =
var
(
X
1
) +
· · ·
+
var
(
X
150
)).
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 all
 Standard Deviation, Variance, Probability theory, Chebyshev's inequality, Joe Mitchell

Click to edit the document details