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Solutions to Assigned Problems
3.82
[6 pts (2,2,2)]
(a)
We are assuming that
X
∼
Pois(
λ
= 0
.
2). The probability that the disk has exactly one
missing pulse is
P
(
X
= 1) =
(0
.
2)
1
e

0
.
2
1!
= 0
.
2
e

0
.
2
= 0
.
1637
.
(b)
It is easier to work with the complement:
P
(
X
≥
2) = 1

P
(
X <
2)
= 1

P
(
X
= 0)

P
(
X
= 1)
= 1

e

0
.
2

(0
.
2)
e

0
.
2
= 1

0
.
982477
= 0
.
0175
.
(c)
Now we have
X
i
∼
Pois(0
.
2) for
i
= 1
,
2 and we know that
X
1
and
X
2
are independent.
“Neither” means that both do not contain a missing pulse. Because they are independent, we can
compute
P
(
X
1
= 0 and
X
2
= 0) =
P
(
X
1
= 0)
×
P
(
X
2
= 0)
=
±
e

0
.
2
²
×
[
e

0
.
2
]
=
e

0
.
4
= 0
.
67032
.
3.83
[6 pts (3,3)]
(a)
Here
n
= 1000
>
50 and
np
= 1000
/
200 = 5, which is not strictly less than ﬁve, but we
will use the Poisson approximation to the binomial distribution anyway:
X
∼
Pois(5). We are
asked to compute
P
(5
≤
X
≤
8) =
8
X
x
=5
5
x
e

5
x
!
=
F
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This note was uploaded on 04/07/2010 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.
 Spring '08
 Staff
 Statistics, Probability

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