Question 1200736: 16% of all Americans live in poverty. If 47 Americans are randomly selected, find the probability that
a. Exactly 7 of them live in poverty. _______
b. At most 6 of them live in poverty. ________
c. At least 8 of them live in poverty. _________
d. Between 5 and 10 (including 5 and 10) of them live in poverty. ________
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! a. This is 47C7*0.16^7*0.84^40=0.1580
b. this is 0.3572 using binomial cdf (47,0.16,6)
check for 0 it is 0.84^47=0.0003
for 1 it is 47*0.16*0.84^46=0.0025
for 2 it is 47C2*0.16^2*0.84^45=0.0108
for 3 it is 0.0309
for 4 it is 0.0648
for 5 it is 0.1062
for 6 it is 0.1416. They add to 0.3571 (rounding is the 0.0001 difference; use 0.3572 as the answer.
c. for 7 it is 0.1580, so adding that to 0.3572 is 0.5152. That is at most 7, so at least 8 is the complement or 0.4848.
d. for at most 4, the answer is 0.1094.
for at most 10, the answer is 0.8797
Subtracting the second from the first and the answer is 0.7703, which its between 5 and 10.
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