document.write( "Question 1115191: Suppose the mean heights all sixth-grade students follow normal distraction with mean 50 inches and 3 inches?
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document.write( "Suppose you take a sample of 250 students. Give the shape, center and standard-deviation of the sampling distribution of samples means. Find the probability that a sample mean is below 52.5 inches? \n" );
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Algebra.Com's Answer #730101 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Suppose the mean heights all sixth-grade students follow normal distraction with mean 50 inches and standard distribution = 3 inches? \n" ); document.write( "Suppose you take a sample of 25 students. \n" ); document.write( "Give the shape:: normal distribution \n" ); document.write( "center:: mean of the sample means = mean of the population = 50 \n" ); document.write( "standard-deviation of the sample means = 3/sqrt(25) \n" ); document.write( "-------- \n" ); document.write( " Find the probability that a sample mean is below 52.5 inches? \n" ); document.write( "z(52.5) = (52.5-50)/(3/sqrt(25)) = 2.5*sqrt(25)/3 = 4.167 \n" ); document.write( "P(x-bar < 52.5) = P(z < 4.167) = normalcdf(-100,4.167) = 0.99998 \n" ); document.write( "------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "------------ \n" ); document.write( " |