3. Roll a die 100 times and let
X
be the sum of all of the rolls.
(a) [4 points] What is
E
(
X
) and
SD
(
X
)?
Solution:
The mean and standard deviation of a single die roll are 3
.
5 and
q
35
12
,
respectively. So
E
(
X
) = 100
·
3
.
5 = 350 and
SD
(
X
) =
√
100
q
35
12
= 5
q
35
3
.
(b) [4 points] Why can we assume that
X
is approximately normal?
Solution:
It is the sum of several (100) independent identically distributed
random variables, hence approximately normal by the
Central Limit Theorem
.
(c) [6 points] Approximate
P
(200
≤
X
≤
400).
Solution:
With the continuity correction,
P
(200
≤
X
≤
400)
≈
P
200

350

1
/
2
5
q
35
3
≤
X
*
≤
400

350 + 1
/
2
5
q
35
3
≈
Φ
400

350 + 1
/
2
5
q
35
3

Φ
200

350

1
/
2
5
q
35
3
≈
99
.
84%
.
(I did not require continuity correction, though it is applicable. Without conti
nuity correction, we would have gotten approximately 99.83%.)
(d) [8 points] Now let
Y
be the number of heads from flipping a coin 500 times. Ap
proximate
P
(
X
≤
3
Y
).
Solution:
Note that
E
(
Y
) = 250 and
V ar
(
Y
) = 500
·
1
2
·
1
2
= 125.
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Math 431: Final Exam
Let
Z
=
X

3
Y
.
Then
E
(
Z
) =
E
(
X
)

3
E
(
Y
) = 350

3
·
250 =

400
and
V ar
(
Z
) =
V ar
(
X
) + 9
V ar
(
Y
), because
X
and
Y
are independent.
So
V ar
(
Z
) = 100
·
35
12
+ 9
·
125 =
4250
3
and
SD
(
Z
) = 5
q
170
3
. Therefore, with the
continuity correction,
P
(
X
≤
3
Y
) =
P
(
Z
≤
0)
≈
P
Z
*
≤
400 + 1
/
2
5
q
170
3
≈
Φ
400 + 1
/
2
5
q
170
3
≈
100%
.
(Again, I did not require continuity correction. I meant to ask for
P
(2
X
≤
3
Y
),
which would have been approximately 85.43%.)
4. Men enter a store at a rate of 7 per hour. Women enter at a rate of 5 per hour. Assume
people arrive randomly and independently.
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 Spring '12
 Miller
 Probability, Probability theory, dy

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