Average Example 1
Problem Find the average, and average
deviation
for the following data on the length of a pen, L. We have 5
measurements,
(12.2, 12.5, 11.9,12.3, 12.2) cm.
Length (cm) 
 

12.2 
0.02 
0.0004 
12.5 
0.28 
0.0784 
11.9 
0.32 
0.1024 
12.3 
0.08 
0.0064 
12.2 
0.02 
0.0004 
Sum 61.1 
Sum 0.72 
Sum 0.1880 
Average 61.1/5 = 12.22 
Average 0.14 

To get the average sum the values and divide by the number of measurements.
To get the average deviation,
 Find the deviations, the absolute values of the quantity (value minus the average), L  L_{ave}
 Sum the absolute deviations,
 Get the average absolute deviation by dividing by the number of measurements
To get the standard deviation
 Find the deviations and square them
 Sum the squares
 Divide by (N1), the number of measurements minus 1 (here it is 4)
 Take the square root.
The pen has a length of (12.22 +/ 0.14) cm or (12.2 +/ 0.1) cm [using average deviations] or
(12.22 +/ 0.22) cm or (12.2 +/ 0.2) cm [using standard deviations].
Return (or use Browser "back"
function)
Average Example 2
Problem: Find the average and the average deviation of the following measurements of a mass.
(4.32, 4.35, 4.31, 4.36, 4.37, 4.34) grams.
Mass (grams) 


4.32 
0.0217 
0.000471 
4.35 
0.0083 
0.000069 
4.31 
0.0317 
0.001005 
4.36 
0.0183 
0.000335 
4.37 
0.0283 
0.000801 
4.34 
0.0017 
0.000003 
Sum 26.05 
0.1100 
0.002684 
Average 4.3417 
Average 0.022 

The same rules as Example 1 are applied. This time there are N = 6 measurements, so for the standard deviation we divide by (N1) = 5.
The mass is (4.342 +/ 0.022) g or (4.34 +/ 0.02) g [using average deviations] or
(4.342 +/ 0.023) g or (4.34 +/ 0.02) g [using standard deviations].
Return (or use Browser "back" function)