Midterm.2 solution
February, 2010
MATH 203
Q1
Solution :
(i) Roughly 50% since the 12 and 16 correspond roughly to the 25th and
75th percentiles.
(ii) Roughly 12.5
(iii) Males seems to have higher variability than the females since it appears
that there is greater variability between the 25th and 75th percentiles
and the whiskers are larger.
(iv) 3
(v) The observations seem to be skewed towards the left (lower values) and
hence the mean is not a great measure of centrality.
Q2
Solution :
(i) Clearly a fishinlake problem, with
N
= 1000
, a
=
.
10
×
1000 = 100
and n = 50.
P
(
at least
3
contaminated
) = 1

P
(
less than
3
contaminated
) =
1

[
P
(0) +
P
(1) +
P
(2)] = 1

(
100
0
)(
900
50
)
(
1000
50
)
+
(
100
1
)(
900
49
)
(
1000
50
)
+
(
100
2
)(
900
48
)
(
1000
50
)
(ii) Want
P
(
N
1
∩
N
2
∩
C
3
)
=
P
(
C
3

N
1
∩
N
2
)
P
(
N
1
∩
N
2
)
=
P
(
C
3

N
1
∩
N
2
)
P
(
N
1

N
2
)
P
(
N
1
)
=
100
998
.
899
999
.
900
1000
=
.
0812
 using the multiplication rule (twice) for conditional probability and
conditioning backwards.
Q3
Solution :
Let EBV = event of exposure to EBV.
MS = event of having MS.
Given : P(EBV) = 0.95, P(MS) = .001,
P
(
EBV
∪
MS
) =
.
9505
(i)
P
(
EBV
∪
MS
) =
P
(
EBV
) +
P
(
MS
)

P
(
EBV
∩
MS
).
(Addition
rule)
Therefore,
P
(
EBV
∩
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 Spring '10
 Wolfson,D
 Conditional Probability, Probability, Probability theory, EBV, EBV M

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