document.write( "Question 178753: Below are the times (in days) it takes for a sample of 5 customers from Jack's computer store to pay their invoices. \r
\n" ); document.write( "\n" ); document.write( "44,41,20,24,41\r
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\n" ); document.write( "\n" ); document.write( " Find the standard deviation of this sample of times. Round your answer to at least two decimal places.\r
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Algebra.Com's Answer #133750 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
First find the average.
\n" ); document.write( "\"x%5Bave%5D=%2844%2B41%2B20%2B24%2B41%29%2F5=34\"
\n" ); document.write( "Now for each sample, find the difference between the value and the average.
\n" ); document.write( "\"x%5B1%5D-x%5Bave%5D=+44-34=10\" then square it \"100\"
\n" ); document.write( "\"x%5B2%5D-x%5Bave%5D=+41-34=7\" then square it \"49\"
\n" ); document.write( "\"x%5B3%5D-x%5Bave%5D=+20-34=-14\" then square it \"256\"
\n" ); document.write( "\"x%5B4%5D-x%5Bave%5D=+24-34=-10\" then square it \"100\"
\n" ); document.write( "\"x%5B5%5D-x%5Bave%5D=+41-34=7\" then square it \"49\"
\n" ); document.write( "Sum those values,
\n" ); document.write( "\"100%2B49%2B256%2B100%2B49=494\"
\n" ); document.write( "The standard deviation is then,
\n" ); document.write( "\"SD=sqrt%28%28sum%28%28x-x%5Bave%5D%29%5E2%29%29%2F%28n%29%29=sqrt%28494%2F%285%29%29=9.94\"
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