Solution:
Note that
P
(
X
= 0) =
2
4
×
1
3
=
1
6
,
P
(
X
= 1) = 2
2
4
×
2
3
=
2
3
, and
P
(
X
= 2) =
2
4
×
1
3
=
1
6
.
Therefore,
E
(
X
) = 0
×
1
6
+ 1
×
2
3
+ 2
×
1
6
= 1
,
E
(
X
2
) = 0
×
1
6
+ 1
×
2
3
+ 4
×
1
6
=
4
3
,
V ar
(
X
) =
E
(
X
2
)

(
E
(
X
))
2
=
4
3

1
2
=
1
3
, and finally
SD
(
X
) =
p
V ar
(
X
) =
1
√
3
.
Page 3
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Math 431: Exam 1
5. [10 points] Ten people sit randomly around a circular table. Six are men and four are
women. How many men do you expect are not sitting next to a woman?
Solution:
Let
I
n
be the indicator that the person in the
i
th chair is a man and he
has men sitting on both sides. Then, for each
n
∈ {
1
, . . . ,
10
}
E
(
I
n
) =
P
(
I
n
= 1) =
6
10
×
5
9
×
4
8
=
1
6
.
The number of men not sitting next to a woman is
X
=
I
1
· · ·
I
10
and
E
(
X
) =
10
E
(
I
1
) =
10
6
=
5
3
.
6. Let
X
and
Y
be independent random variables with
E
(
X
) = 4,
SD
(
X
) = 3,
E
(
Y
) = 6,
and
SD
(
Y
) = 4.
(a) [4 points] Find
E
(
X
+
Y
) and
SD
(
X
+
Y
).
Solution:
E
(
X
+
Y
) =
E
(
X
) +
E
(
Y
) = 10.
Because
X, Y
are independent
V ar
(
X
+
Y
) =
V ar
(
X
) +
V ar
(
Y
) = 3
2
+ 4
2
= 25, so
SD
(
X
+
Y
) =
√
25 = 5.
(b) [8 points] Find an upper bound for
P
(
X
+
Y
≥
25).
Solution:
Let
μ
=
E
(
X
+
Y
) and
σ
=
SD
(
X
+
Y
).
Using Chebychev’s in
equality,
P
(
X
+
Y
≥
25) =
P
(
X
+
Y

μ
≥
15)
≤
P
(

X
+
Y

μ
 ≥
15)
=
P
(

X
+
Y

μ
 ≥
3
σ
)
≤
1
3
2
=
1
9
.
(c) [2 points] Assuming that
X
and
Y
are symmetric (hence also
X
+
Y
), improve the
answer to part (b).
Solution:
By symmetry (meaning symmetry around
μ
),
P
(
X
+
Y

μ
≥
3
σ
) =
P
(
X
+
Y

μ
≤ 
3
σ
). Since
P
(
X
+
Y

μ
≥
3
σ
) +
P
(
X
+
Y

μ
≤ 
3
σ
) =
P
(

X
+
Y

μ
 ≥
3
σ
)
≤
1
9
,
each of these probabilities is at most 1
/
18.
7. [10 points] Assume that each job you have lasts an average of 5 years with a standard
deviation of 2. Approximate the probability that you will have more than 10 jobs in 40
years. (
Hint
: find the probability that your first 10 jobs last less than 40 years.)
Page 4
Math 431: Exam 1
Solution:
Let
J
i
be the length of the your
i
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 Spring '12
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 Binomial, Probability, Probability theory, Binomial distribution, 1%, 5.7%, 14.38%

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