document.write( "Question 332582: for the following data set, approximate the sample standard deviation.\r
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document.write( "miles (per day)---Frequency\r
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document.write( "1-2--------------- 9
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document.write( "3-4--------------- 22
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document.write( "5-6--------------- 28
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document.write( "7-8--------------- 15
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document.write( "9-10-------------- 14 \n" );
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Algebra.Com's Answer #238367 by jrfrunner(365)![]() ![]() You can put this solution on YOUR website! Since you dont have the exact value, but instead have a range of values \n" ); document.write( "the first thing to do is to get a surrogate (proxy) value. This is usually set a the mid point of the interval\r \n" ); document.write( "\n" ); document.write( "1.5 \n" ); document.write( "3.5 \n" ); document.write( "etc... \n" ); document.write( "--- \n" ); document.write( "next set up a data table as follows\r \n" ); document.write( "\n" ); document.write( "Let X=midpoint, F=frequency \n" ); document.write( "-- \n" ); document.write( " X ------ F ---- X*F \n" ); document.write( "1.5 ----- 9 ---- 13.5 \n" ); document.write( "3.5 ----- 22 ---- 77.0 \n" ); document.write( "5.5 ----- 28 ---- 154.0 \n" ); document.write( "7.5 ----- 15 ---- 112.5 \n" ); document.write( "9.5 ----- 14 ---- 133.0 \n" ); document.write( "--- \n" ); document.write( "Compute the average = Xbar = \n" ); document.write( "-- \n" ); document.write( "for each interval compute a column (Xi-Xbar)^2 \n" ); document.write( "and then sum this column \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "this sum represents the sum of squared deviations. \n" ); document.write( "--- \n" ); document.write( "Find the Variance \n" ); document.write( "Variance = \n" ); document.write( "--- \n" ); document.write( "Standard dev = Sqrt(Variance) = Sqrt(5.926)=2.434\r \n" ); document.write( "\n" ); document.write( "=== \n" ); document.write( "There is another form of the variance formula that you could use\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Variance = |