Dr. Hackney STA Solutions pg 30

Dr. Hackney STA Solutions pg 30 - Second Edition 3-3 3.9 a....

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Second Edition 3-3 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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