document.write( "Question 1115191: Suppose the mean heights all sixth-grade students follow normal distraction with mean 50 inches and 3 inches?
\n" ); 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?
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Algebra.Com's Answer #730101 by stanbon(75887)\"\" \"About 
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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)
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\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
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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