For this example, use the assigned data set of 6, 7, 10, 12, 13, 4, 8 and 12. This set is small enough to count by hand to find that there are eight numbers in the set. In statistical work, the variable N{\displaystyle N} or n{\displaystyle n} is commonly used to represent the number of data values.
Σx=6+7+10+12+13+4+8+12=72{\displaystyle \Sigma x=6+7+10+12+13+4+8+12=72}
μ=ΣxN=728=9{\displaystyle \mu ={\frac {\Sigma x}{N}}={\frac {72}{8}}=9}
Fill the first column with the data points for your calculation.
For the sample data set, these deviations will be: 6−9=−3{\displaystyle 6-9=-3} 7−9=−2{\displaystyle 7-9=-2} 10−9=1{\displaystyle 10-9=1} 12−9=3{\displaystyle 12-9=3} 13−9=4{\displaystyle 13-9=4} 4−9=−5{\displaystyle 4-9=-5} 8−9=−1{\displaystyle 8-9=-1} 12−9=3{\displaystyle 12-9=3} To check the validity of your calculations, the sum of the values in this deviation column should be 0. If you add them up and get something other than 0, then either your mean is incorrect or you made an error in calculating one or more of the deviations. Go back and check your work.
Absolute value is a mathematical tool used to measure distance or size, regardless of direction. To find absolute value, just drop the negative sign from each number in the second column. Thus, fill the third column with the absolute values as follows: x. . . . . (x−μ). . . . . |(x−μ)|{\displaystyle x. . . . . (x-\mu ). . . . . |(x-\mu )|} 6. . . . . . . . . . −3. . . . . . . . . . 3{\displaystyle 6. . . . . . . . . . -3. . . . . . . . . . 3} 7. . . . . . . . . . −2. . . . . . . . . . 2{\displaystyle 7. . . . . . . . . . -2. . . . . . . . . . 2} 10. . . . . . . . . . 1. . . . . . . . . . 1{\displaystyle 10. . . . . . . . . . 1. . . . . . . . . . 1} 12. . . . . . . . . . 3. . . . . . . . . . 3{\displaystyle 12. . . . . . . . . . 3. . . . . . . . . . 3} 13. . . . . . . . . . 4. . . . . . . . . . 4{\displaystyle 13. . . . . . . . . . 4. . . . . . . . . . 4} 4. . . . . . . . . . −5. . . . . . . . . . 5{\displaystyle 4. . . . . . . . . . -5. . . . . . . . . . 5} 8. . . . . . . . . . −1. . . . . . . . . . 1{\displaystyle 8. . . . . . . . . . -1. . . . . . . . . . 1} 12. . . . . . . . . . 3. . . . . . . . . . 3{\displaystyle 12. . . . . . . . . . 3. . . . . . . . . . 3}
For this data set, this final calculation will be: 3+2+1+3+4+5+1+38=228=2. 75{\displaystyle {\frac {3+2+1+3+4+5+1+3}{8}}={\frac {22}{8}}=2. 75}
For example, with this data set, you can say that the mean is 9 and the average distance from that mean is 2. 75. Note that some numbers are closer than 2. 75 and some are farther. But that is the average distance.
