document.write( "Question 495768: A producer of a juice drink advertises that it contains 10% real fruit juices. A sample of 75 bottles of the drink is analyzed and the percent of real fruit juices is found to be 6.9%. If the true proportion is actually 0.10, what is the probability that the sample percent will be 6.9% or less? \r
\n" );
document.write( "\n" );
document.write( " \n" );
document.write( "
Algebra.Com's Answer #336192 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A producer of a juice drink advertises that it contains 10% real fruit juices. A sample of 75 bottles of the drink is analyzed and the percent of real fruit juices is found to be 6.9%. If the true proportion is actually 0.10, what is the probability that the sample percent will be 6.9% or less? \n" ); document.write( "------------------------------------------ \n" ); document.write( "z(0.069) = (0.069-0.10)/sqrt[0.1*0.9/75] = -0.8949 \n" ); document.write( "--- \n" ); document.write( "P(p-hat < 0.069) = P(z < -0.8949) = 0.1854 \n" ); document.write( "================================================ \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |