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1.a: [15pt]
145
±
1
.
645
×
8
√
35
= 145
±
2
.
22 = [142
.
78
,
147
.
22]
1.b: [15pt]
H
0
:
μ
= 142 vs.
H
a
:
μ >
142.
Z
=
145

142
8
/
√
35
= 2
.
22.
p
value is 0.0132. Reject
H
0
at 5%
signiﬁcance level.
1.c:[15pt]
n
=
(
1
.
96
×
8
1
.
5
)
2
= 109
.
3
≈
110
2.a: [5pt]
26
3
×
10
3
= 17
,
576
,
000
2.b: [5pt]
26
×
25
×
24
×
10
×
9
×
8
/
17
,
576
,
000 = 0
.
6390533
3.a: [15pt]
For each passenger’s baggage, the probability of being overweight is
P
(
X
≥
100) =
P
(
Z
≥
100

95
35
) =
P
(
Z
≥
0
.
14) = 0
.
4443. Hence, the probability that all three passengers have
overweight baggages is 0
.
4443
3
= 0
.
0877 (by multiplication rule for independent events).
3.b: [15pt]
Since
n
= 200
>
30, by CLT (of sample sum), we have
200
X
i
=1
x
i
˙
∼
N
(95
×
200
,
√
200
×
35).
Hence, the probability of exceed the 20,000pound limit is
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This note was uploaded on 12/28/2011 for the course EXST 7003 taught by Professor Moser,e during the Fall '08 term at LSU.
 Fall '08
 Moser,E

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