document.write( "Question 334154: The ages (in years) of the employees at a particular computer store are \r
\n" ); document.write( "\n" ); document.write( " 36,40,24,24,31\r
\n" ); document.write( "\n" ); document.write( "Assuming that these ages constitute an entire population, find the standard deviation of the population. Round your answer to at least two decimal places.\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #239445 by jrfrunner(365)\"\" \"About 
You can put this solution on YOUR website!
1. compute the average: add the values and divide by 5
\n" ); document.write( "Average =(36+ 40+24+24+31)/5=31.0
\n" ); document.write( "--
\n" ); document.write( "2. Compute the deviations from the average
\n" ); document.write( "dev: (36-31)=5, (40-31)=9,etc
\n" ); document.write( "--
\n" ); document.write( "3. Square the deviations and add
\n" ); document.write( "sum (dev^2): 5^2+9^+...+0^2=204
\n" ); document.write( "--
\n" ); document.write( "4. Divide step #3 by the sample size=5 (typically you divide by sample size-1 to get the sample standard deviation, but you are assuming the 5 values are the population, so no need to subtract 1, from the sample size. This result is the variance
\n" ); document.write( "Variance =204/5=40.8
\n" ); document.write( "--
\n" ); document.write( "5. Standard deviation = sqrt(variance)
\n" ); document.write( "standard deviation= sqrt(40.8)
\n" ); document.write( "
\n" );