document.write( "Question 1200736: 16% of all Americans live in poverty. If 47 Americans are randomly selected, find the probability that\r
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document.write( "a. Exactly 7 of them live in poverty. _______\r
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document.write( "b. At most 6 of them live in poverty. ________\r
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document.write( "c. At least 8 of them live in poverty. _________\r
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document.write( "d. Between 5 and 10 (including 5 and 10) of them live in poverty. ________
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Algebra.Com's Answer #835005 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! a. This is 47C7*0.16^7*0.84^40=0.1580 \n" ); document.write( "b. this is 0.3572 using binomial cdf (47,0.16,6) \n" ); document.write( "check for 0 it is 0.84^47=0.0003 \n" ); document.write( "for 1 it is 47*0.16*0.84^46=0.0025 \n" ); document.write( "for 2 it is 47C2*0.16^2*0.84^45=0.0108 \n" ); document.write( "for 3 it is 0.0309 \n" ); document.write( "for 4 it is 0.0648 \n" ); document.write( "for 5 it is 0.1062 \n" ); document.write( "for 6 it is 0.1416. They add to 0.3571 (rounding is the 0.0001 difference; use 0.3572 as the answer. \n" ); document.write( "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. \n" ); document.write( "d. for at most 4, the answer is 0.1094. \n" ); document.write( "for at most 10, the answer is 0.8797 \n" ); document.write( "Subtracting the second from the first and the answer is 0.7703, which its between 5 and 10. \n" ); document.write( " |