Second Edition
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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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.
 Spring '12
 Dr.Hackney
 Statistics, Bernoulli, Probability

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