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33
3.9 a. We can think of each one of the 60 children entering kindergarten as 60 independent Bernoulli
trials with probability of success (a twin birth) of approximately
1
90
. The probability of having
5 or more successes approximates the probability of having 5 or more sets of twins entering
kindergarten. Then
X
∼
binomial(60
,
1
90
) and
P
(
X
≥
5) = 1

4
X
x
=0
±
60
x
²±
1
90
²
x
±
1

1
90
²
60

x
=
.
0006
,
which is small and may be rare enough to be newsworthy.
b. Let
X
be the number of elementary schools in New York state that have 5 or more sets
of twins entering kindergarten. Then the probability of interest is
P
(
X
≥
1) where
X
∼
binomial(310,.0006). Therefore
P
(
X
≥
1) = 1

P
(
X
= 0) =
.
1698.
c. Let
X
be the number of States that have 5 or more sets of twins entering kindergarten
during any of the last ten years. Then the probability of interest is
P
(
X
≥
1) where
X
∼
binomial(500
,.
1698). Therefore
P
(
X
≥
1) = 1

P
(
X
= 0) = 1

3
.
90
×
10

41
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 Spring '12
 Dr.Hackney
 Statistics, Bernoulli, Probability

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