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Math 302.102 Fall 2010
Solutions to Assignment #8
1.
Since
X
has a binomial distribution with parameters
n
= 100 and
p
= 4
/
7, we can use the
central limit theorem to approximate
P
{
X
≥
50
}
. That is, we know that if
X
∼
Bin(
n,p
), then
X

np
p
np
(1

p
)
has an approximately normal
N
(0
,
1) distribution as long as both
np
and
n
(1

p
) are not too small.
Solution 1
. Using the continuity correction, we ﬁnd
P
{
X
≥
50
}
=
P
{
X >
49
.
5
}
=
P
X

100
·
4
7
q
100
·
4
7
·
3
7
>
49
.
5

100
·
4
7
q
100
·
4
7
·
3
7
≈
P
{
Z >

1
.
544
}
where
Z
∼ N
(0
,
1). Using a table of normal probabilities, we ﬁnd
P
{
Z >

1
.
544
}
.
= 0
.
9382. Thus,
P
{
X
≥
50
} ≈
0
.
9382
.
Solution 2
. Observe that the largest
X
could be is 100. Thus,
P
{
X
≥
50
}
is really equivalent to
P
{
50
≤
X
≤
100
}
, and so using the continuity correction, we ﬁnd
P
{
X
≥
50
}
=
P
{
50
≤
X
≤
100
}
=
P
{
49
.
5
< X <
100
.
5
}
=
P
49
.
5

100
·
4
7
q
100
·
4
7
·
3
7
<
X

100
·
4
7
q
100
·
4
7
·
3
7
<
100
.
5

100
·
4
7
q
100
·
4
7
·
3
7
≈
P
{
1
.
544
< Z <
8
.
761
}
=
P
{
Z <
8
.
761
} 
P
{
Z <

1
.
544
}
where
Z
∼ N
(0
,
1).
Using a table of normal probabilities, we ﬁnd
P
{
Z <
8
.
761
}
= 1 and
P
{
Z <

1
.
544
}
.
= 0
.
0618. Thus,
P
{
X
≥
50
} ≈
1

0
.
0618 = 0
.
9382
.
Solution 3
. Without using the continuity correction, we ﬁnd
P
{
X
≥
50
}
=
P
X

100
·
4
7
q
100
·
4
7
·
3
7
≥
50

100
·
4
7
q
100
·
4
7
·
3
7
≈
P
{
Z >

1
.
443
}
where
Z
∼ N
(0
,
1). Using a table of normal probabilities, we ﬁnd
P
{
Z >

1
.
443
}
.
= 0
.
9251. Thus,
P
{
X
≥
50
} ≈
0
.
9251
.
Remark.
The exact probability can be determined using a computer. It is 0
.
9381732. Note that
the approximations provided by Solutions 1 and 2 are remarkably accurate.
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View Full Document2.
Let
X
j
denote the outcome of the
j
th game so that
Y
j
=
X
j

4 denotes the net winnings on
the
j
th game. The total net winnings are therefore
Y
1
+
···
+
Y
100
=
X
1
+
···
+
X
100

400
.
Let
Y
=
Y
1
+
···
+
Y
100
100
and
X
=
X
1
+
···
+
X
100
100
so that
Y
=
Y
1
+
···
+
Y
100
100
=
X
1
+
···
+
X
100

400
100
=
X
1
+
···
+
X
100
100

400
100
=
X

4
.
It is equivalent to work with either
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 Spring '08
 ISRAEL
 Binomial, Central Limit Theorem, Probability

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