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71% of all Americans live in cities with population greater than 100,000 people. If 48 Americans...

71% of all Americans live in cities with population greater than 100,000 people. If 48 Americans are randomly selected, find the probability that

a. Exactly 36 of them live in cities with population greater than 100,000 people.
b. At most 33 of them live in cities with population greater than 100,000 people.
c. At least 31 of them live in cities with population greater than 100,000 people.
d. Between 27 and 32 (including 27 and 32) of them live in cities with population greater than 100,000 people

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Answer #1

Let "p" be the proportion of all Americans live in cities with population greater than 100,000 people.

Let X be the number of Americans live in cities with population greater than 100,000 people in a random sample of n=48

Therefore,

a) The probability that exactly 36 of them live in cities with population greater than 100,000 people is given by:

    ### By using the excel function:

=BINOMDIST(36,48,0.71,0)

b) The probability that at most 33 of them live in cities with population greater than 100,000 people is given by:

    ### By using the excel function:

=BINOMDIST(33,48,0.71,1)

c) The probability that at least 31 of them live in cities with population greater than 100,000 people is given by:

    ### By using the excel function:

=1-BINOMDIST(30,48,0.71,1)

d) The probability that Between 27 and 32 (including 27 and 32) of them live in cities with population greater than 100,000 people is given by:

         ### By using the excel function:

=BINOMDIST(32,48,0.71,1)-BINOMDIST(27,48,0.71,1)

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