AMS 311
Joe Mitchell
Practice Exam 2 – Solution Notes
Statistics:
n
= 35,
μ
= 72
.
8, median 86,
σ
= 28
.
0; score range: 8–100
1. (a).
P
(
X>
0
.
75) =
±
∞
0
.
75
f
(
x
)
dx
=
±
1
0
.
75
xdx
+
±
2
1
1
2
dx
=
23
32
(or write
P
(
0
.
75) = 1

²
0
.
75
0
xdx
)
(b). The area is given by
A
=
X
2
, since it is an
X
by
X
square. Thus, we compute:
var
(
X
2
)=
E
(
X
4
)

[
E
(
X
2
)]
2
=
±
1
0
x
4
·
xdx
+
±
2
1
x
4
·
1
2
dx

³ ±
1
0
x
2
·
xdx
+
±
2
1
x
2
·
1
2
dx
´
2
(c). (First, note that since
X
is always in (0,1) or (1,2), then
Y
=5
X
is always in (0,5) or (5,10).) We
want to compute
F
Y
(
y
P
(
Y
≤
y
P
(5
X
≤
y
P
(
X
≤
y
5
F
X
(
y
5
), which requires looking at cases:
F
Y
(
y
0i
f
y
5
≤
0; i.e., if
y
≤
0
²
y/
5
0
xdx
=
y
2
50
if 0
≤
y
5
≤
1; i.e., if 0
≤
y
≤
5
²
1
0
xdx
+
²
5
1
1
2
dx
=
y
10
if 1
≤
y
5
≤
2; i.e., if 5
≤
y
≤
10
1i
f
y
5
≥
2; i.e., if
y
≥
10
2.
X
is exponential with parameter
1
5
years

1
(since
E
(
X
) = 5 years).
(a). The probability that a radio lasts more than 15 years is
P
(
15) =
²
∞
15
1
5
e

x/
5
dx
=
e

15
/
5
=
e

3
(b). Let
Y
be the number (among the 1000) that last more than 15 years. Then,
Y
is Binomial(1000,
p
),
where
p
=
e

3
was found in part (a). We want
P
(
Y
≥
4) =
1000
µ
i
=4
¶
1000
i
·
p
i
(1

p
)
1000

i
(c). Since 1000 is “large” and
e

3
is “small”, we know that
Y
has a distribution that is approximated
by that of
Z
, a Poisson(1000
p
) random variable. Thus,
P
(
Y
≥
4)
≈
P
(
Z
≥
4) =
∞
µ
i
=4
e

1000
p
(1000
p
)
i
i
!
(Note: you could also stop the summation at
i
= 1000; it does not really matter for the approximation.)
3.
X
is Normal(75,2
2
). We want
P
(74
.
2
≤
X
≤
75
.
6) =
P
(
74
.
2

75
2
≤
X

75
2
≤
75
.
6

75
2
P
(

0
.
4
≤
Z
≤
0
.
3)
=Φ(
0
.
3)

Φ(

0
.
4) = Φ(0
.
3)

(1

Φ(0
.
4)) =
.
6179

(1

.
6554) =
.
2733
4. (20 points) We begin by plotting the support set (below, left, in red).
y=x
y=3
−
x
y=2x
y=3
−
x
y=x
1 3/2
3/2
3
3
3
3
1