Question

In: Statistics and Probability

8% of all Americans live in poverty. If 44 Americans are randomly selected, find the probability...

8% of all Americans live in poverty. If 44 Americans are randomly selected, find the probability that

a. Exactly 1 of them live in poverty.   ____
b. At most 5 of them live in poverty. ____  
c. At least 4 of them live in poverty.   ____
d. Between 1 and 7 (including 1 and 7) of them live in poverty.  ____

Solutions

Expert Solution

X ~ B ( n = 44 , P = 0.08 )


Part a)


Part b)





P ( X <= 5 ) = 0.8634


Part c)
P ( X >= 4 ) = 1 - P ( X <= 3 )




P ( X >= 4 ) = 1 - 0.5277
P ( X >= 4 ) = 0.4723


Part d)







P ( 1 <= X <= 7 ) = 0.9524


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