document.write( "Question 1189929: If a population proportion is 0.32 and if the sample size is 100, 45% of the time the sample proportion will be more than what value if you are taking random samples? \n" ); document.write( "
Algebra.Com's Answer #822112 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! The standard error is sqrt(p*(1-p)/n)=sqrt(0.32*0.68/100)=0.0466 \n" ); document.write( "45% of the time greater occurs with a z=+0.1257 \n" ); document.write( "z*SE is margin of error or in this instance the value where 45% are more than the mean of 32 with a sample size of 100. It is 0.0059 \n" ); document.write( "so the proportion is 0.326 and for 100, that would be a mean of 32.6 \n" ); document.write( "- \n" ); document.write( "can check by approximation \n" ); document.write( "mean is np=32 \n" ); document.write( "variance is np(1-p)=32*0.68=21.76 \n" ); document.write( "sd is sqrt (V)=4.66 \n" ); document.write( "0.1257 of that sd is 0.59, and add that to the mean and get 32.6 \n" ); document.write( " |