SOLUTION: Answer the following questions and round your answers to 2 decimal places. 13% of all Americans live in poverty. If 45 Americans are randomly selected, find the probability that

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Question 1153245: Answer the following questions and round your answers to 2 decimal places. 13% of all Americans live in poverty. If 45 Americans are randomly selected, find the probability that
A.Exactly 5 of them live in poverty.
B.At most 5 of them live in poverty .
C.At least 5 of them live in poverty.
D. Between 3 and 6 (including 3 and 6) of them live in poverty.
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
Look at 0 to 6 live in poverty
for 0: 45C0*0.13^0*0.87^45, per the binomial formula. Prob=0.0019
for 1: 45*0.13*0.87^44=0.0128
for 2: 45C2*0.13^2*0.87^43=0.0420
for 3: 45C3*0.13^3*0.87^42=0.0899
for 4: 0.1410
for 5: 0.1728
for 6: 0.1721
From here, can get exactly 5 0.1728
At most 5, the sum of the 0-6, or 0.4604
At least 5 would be 1-the probabilities from 0 to 4 or 1-0.2876=0.7124
Between 3 and 6 would be probability 0.5758.