document.write( "Question 775571: The algebraic sum of debiations of 20 observations measured from 30 is 2.then the mean of these observatios is \n" ); document.write( "

Algebra.Com's Answer #473023 by KMST(5328) $\"\"$ $\"About$

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$\"30%2B2%2F20=30%2B0.1=30.1\"$
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\n" ); document.write( "Lets say those 20 deviations are $\"a\"$ , $\"b\"$ , $\"c\"$ , $\"%22...%22\"$ $\"r\"$ , $\"s\"$ , and $\"t\"$ .
\n" ); document.write( "The 20 observations must be $\"30%2Ba\"$ , $\"30%2Bb\"$ , $\"30%2Bc\"$ , $\"%22...%22\"$ $\"30%2Br\"$ , $\"30%2Bs\"$ , and $\"30%2Bt\"$ .
\n" ); document.write( "The sum of those 20 observations is
\n" ); document.write( " $\"%2830%2Ba%29%2B%2830%2Bb%29%2B%2830%2Bc%29%2B%22...%2B%22\"$

\n" ); document.write( "The average of those 20 observations is the sum divided by 20, so it is
\n" ); document.write( " $\"%2820%2A30%2B2%29%2F20=20%2A30%2F20%2B2%2F20=30%2B2%2F20=30%2B0.1=30.1\"$
\n" ); document.write( "When averaging large numbers that are close together, it is often easier to do the calculations like that.
\n" ); document.write( "For example, if most of the grades of the students in a class are 80 or above, to calculate the average it is easier to add the deviations from 80, and then add their average to 80.
\n" ); document.write( "You would use the deviation 15 for that top-scorer who got a 95.
\n" ); document.write( "You probably would have single digit numbers for most of the other students, and you would have a few negative numbers, like the -6 for the guy who scored 74, and the -20 for that low-scorer who got a 60. \n" ); document.write( "

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