document.write( "Question 1197411: Suppose we roll a fair die two times.
\n" ); document.write( "How many different samples are there? I figured this as 36.
\n" ); document.write( "List each of the possible samples and compute the mean. I completed an excel sheet for this and the mean is 3.5
\n" ); document.write( "Compute the mean and the standard deviation of each distribution and compare them.
\n" ); document.write( "Mean
\n" ); document.write( "Standard deviation of individual rolls
\n" ); document.write( "Standard deviation of sample means
\n" ); document.write( "I put all the figures in an excel spreadsheet and used the formula for standard deviation for a population and a sample. The calculation showed population standard deviation as 2.415 and the sample standard deviation as 1.225. This shows not correct in the assignment and I am lost. Could you please help/explain how to get the standard deviation of individual rolls and the standard deviation of the sample means?
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Algebra.Com's Answer #830692 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "a fair six-sided die rolled  {1,2,3,4,5,6} outcomes\r\n" );
document.write( " μ=  = 3.5\r\n" );
document.write( "Variance:  = 2.9167\r\n" );
document.write( "Standard deviation =   σ=1.7078 Standard deviation of individual rolls\r\n" );
document.write( "\r\n" );
document.write( "Standard deviation of sample means\r\n" );
document.write( "1	3.5	-2.5	6.25\r\n" );
document.write( "2	3.5	-1.5	2.25\r\n" );
document.write( "3	3.5	-0.5	0.25\r\n" );
document.write( "4	3.5	0.5	0.25\r\n" );
document.write( "5	3.5	1.5	2.25\r\n" );
document.write( "6	3.5	2.5	6.25\r\n" );
document.write( "		0	17.5\r\n" );
document.write( "		  	1.870828693   √(17.5/5)\r\n" );
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