document.write( "Question 1011720: If the probability of being a smoker among a group of cases with lung cancer is .6, what’s the probability that in a group of 8 cases you have less than 2 smokers? More than 5? What are the expected value and variance of the number of smokers? \n" ); document.write( "
Algebra.Com's Answer #628259 by mathmate(429) You can put this solution on YOUR website! \n" ); document.write( "Question: \n" ); document.write( " If the probability of being a smoker among a group of cases with lung cancer is .6, what’s the probability that in a group of 8 cases you have less than 2 smokers? More than 5? What are the expected value and variance of the number of smokers? \n" ); document.write( " \n" ); document.write( "Solution: \n" ); document.write( "The problem can be solved using the binomial distribution, with p=0.6. \n" ); document.write( " \n" ); document.write( "For n=8, \n" ); document.write( "P(X<2)=P(X=0)+P(X=1) \n" ); document.write( "=C(8,0)*p^0*(1-p)^8 + C(8,1)*p^1*(1-p)^(8-1) \n" ); document.write( "=0.0065536+0.0078643 \n" ); document.write( "=0.008520 \n" ); document.write( " \n" ); document.write( "P(X>5)=P(X=6)+P(X=7)+P(X=8) \n" ); document.write( "=0.2090+0.0896+0.0168 \n" ); document.write( "=0.3154 \n" ); document.write( " \n" ); document.write( "Expected value = np=8*0.6=4.8 \n" ); document.write( "Variance = npq=8*0.6*(1-0.6)=1.92 \n" ); document.write( " |