Kata Bognar
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Economics 41
Statistics for Economists
UCLA
Winter 2010
Final Exam
 suggested solutions 
Part I  Multiple Choice Questions (3 points each)
1. Which of the following is an observational study?
B
(a) You investigate the effect of aspirin on preventing heart attacks by giving one group of people
a daily dose of aspirin and another group a placebo. Then you compare the rates of heart
attacks between the two groups.
(b)
You investigate the effect of college GPA on starting salary by questioning ran
dom recent graduates what their starting salary is and you find their GPA from
school records.
(c) You are the dictator of a country and you decide to investigate the effect of smoking on lung
cancer.
You force half of your population to smoke and the other half not to.
Then you
compare the rates of lung cancer.
(d) Both (a) and (c).
2. Which of the following is an example of a discrete variable?
C
3. Which of the following is true?
C
)
= 5
4. Assume there are two urns. Urn A has one white and three black balls and Urn B has two black
and two white balls. You randomly take out two balls (without replacement) from Urn A and
(independently so) one ball from Urn B. What is the probability of the event among the three
balls taken out of the urns there are exactly two black ones?
C
Explanation
: One can select 2 balls form Urn A and one ball from Urn B in
(
4
2
)(
4
1
)
different
ways. One can select two balls form Urn A and one ball from Urn B such that there are exactly
two black balls in
(
3
1
)(
2
1
)
+
(
3
2
)(
2
0
)
.
Notice that one can either pick exactly one black ball from each
urn or exactly two form Urn A and none form Urn B.
5. A random variable X that takes on four different values has the following distribution:
P
(
X
=

1) = 2
c, P
(
X
= 0) = 3
c, P
(
X
= 1) = 0
.
4 and
P
(
X
= 2) = 0
.
1
.
The value of the constant c is:
B
(a) 1
1
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(b)
0.1
(c) 0.3
(d) 2.5
Explanation
: Since
X
can only take the for values above, the total probability of those must
sum up to 1. Thus, 2
c
+ 3
c
+ 0
.
4 + 0
.
1 = 1 and
c
= 0
.
1
.
6. Suppose that
A
and
B
are events such that
P
(
A
) = 1
/
2
, P
(
B
) = 1
/
4 and
P
(
A

B
)+
P
(
B

A
) = 2
/
3
,
then
P
(
A
∪
B
) is
D
Explanation
: Recall the addition rule:
P
(
A
∪
B
) =
P
(
A
)+
P
(
B
)

P
(
A
∩
B
)
. P
(
A
) and
P
(
B
) are
given. How can you get
P
(
A
∩
B
)? According to the conditional probability rule
P
(
A

B
) =
P
(
A
∩
B
)
P
(
B
)
and
P
(
B

A
) =
P
(
A
∩
B
)
P
(
A
)
.
Therefore,
P
(
A

B
) +
P
(
B

A
) =
P
(
A
∩
B
)
P
(
A
)
+
P
(
A
∩
B
)
P
(
B
)
.
Denote by
x
the
probability of
A
∩
B.
Then,
x
1
/
2
+
x
1
/
4
=
2
3
and
x
=
1
9
.
Finally,
P
(
A
∪
B
) =
1
2
+
1
4

1
9
.
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 Fall '07
 Guggenberger
 Economics, Normal Distribution, Standard Deviation, Probability theory, Urn A

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