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)\"\" \"About 
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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
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\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
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