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Solution for assignment # 2
1.
a)
λ
= 2
, P
(
X
>
2) = 1

P
(
X
6
1) = 0
.
594
b)
λ
= 4
, P
(
X
= 0) =
e

4
= 0
.
018
c)
P
(
X
= 0

10 calls before) =
P
(
X
= 0) = 0
.
018, because the # of events in any
interval are independent from that of any other mutually exclusive interval
d)
If we wait khours, then
W
=# of calls in the next khours satisﬁes
W
∼
P
(4
k
)
so
P
(
W
>
1) = 1

P
(
W <
1) = 1

P
(
W
= 0) = 1

e

4
k
>
0
.
90
⇒
k
>

ln(0
.
10)
/
4 = 0
.
58, which is almost 35 minutes.
2.
a)
P
(
X
6
1
2
) =
R
1
2
0
2
e

2
x
dx
=

e

2
x
±
1
2
0
= 0
.
632
b)
P
(
X
>
1) =
R
∞
1
2
e

2
x
dx
=

e

2
x
±
∞
1
= 0
.
135
c)
P
(
1
2
6
X
6
1) =

e

2
+
e

1
= 0
.
233
d)
F
(
x
) =
R
x
0
2
e

2
y
dy
= 1

e

2
x
3.
a)
R
4
0
cxdx
=
cx
2
2
±
4
0
=
c
(8) = 1
⇒
c
=
1
8
b)
R
a
0
x
8
dx
=
x
2
16
±
a
0
=
a
2
16
= 0
.
5
⇒
a
=
√
8 = 2
.
828
c)
R
b
0
x
8
dx
=
b
2
16
= 0
.
95
⇒
b
= 3
.
899
4.
X
1
∼
N
(3
,
(0
.
1)
2
)
, X
2
∼
N
(3
.
04
,
(0
.
02)
2
)
a)
P
(2
.
9
< X
1
<
3
.
1) =
P
(
2
.
9

3
0
.
1
< Z <
3
.
1

3
0
.
1
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This note was uploaded on 09/27/2010 for the course STAT 104157 taught by Professor Jhonbran during the Spring '09 term at California Coast University.
 Spring '09
 JHONBRAN

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