(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.)
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View Full DocumentMath 431: Exam 1
Solution:
Let
J
i
be the length of the your
i
th job, for
i
∈ {
1
,...,
10
}
. Then we
are trying to ﬁnd the probability that
X
=
J
1
+
···
+
J
10
is less than 40. We
assume that the sum is approximately normal. Note that
E
(
X
) = 10
E
(
J
1
) = 50 and
SD
(
X
) =
√
10
SD
(
J
1
) = 2
√
10. So
P
(
X <
40) =
P
±
X

50
2
√
10
<
40

50
2
√
10
²
≈
Φ
±
40

50
2
√
10
²
= 1

Φ
√
10
2
!
≈
5
.
7%
.
8. Assume that in a class of 80 students, each student independently has a 1% chance of
forgetting to take the exam.
(a) [3 points] What is the probability that two students forget?
Solution:
Using the binomial (80
,
1
/
100) distribution,
±
80
2
²±
1
100
²
2
±
99
100
²
78
≈
14
.
43%
.
(b) [7 points] Use the Poisson approximation to approximate the probability that two
students forget.
Solution:
The expected number of students who forget the exam is
μ
=
np
=
0
.
8. So the probability that two students forget is approximately
e

0
.
8
(0
.
8)
2
2!
≈
14
.
38%
.
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 Spring '12
 Miller
 Binomial, Probability, Probability theory, Binomial distribution, 1%, 5.7%, 14.38%

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