Example1: Determine the accuracy and precision of the mass of a piece of a metal performed by three different students, where mass of a piece of metal is 0.520gm. Data obtained by each student are recorded as follow.
Student A: 0.521, 0.521, 0.509 Average: 0.515
Student B: 0.516, 0.515, 0.514 Average: 0.515
Student C: 0.521, 0.520, 0.520 Average: 0.520
Result:
Student A: Data are not accurate and not precise because the individual values differ widely from each other and average value is not matching with true value.
Student B: Data are not accurate but precise because the individual values deviate but little from one another, however, the average value is still not matching with true value.
Student C: Data are accurate and precise as the individual value is almost same and average value is matching with true value.
Example2: Determine the significant figure of 0.230, 0.002, 005.2, and 10.0.
Result:
Values 
Significant figure 
0.230 
3 
0.002 
3 
005.2 
2 
10.0 
2 
Example3: Iron samples are analyzed and the values obtained are 7.08, 7.21, 7.12, 7.09, 7.16, 7.14, 7.07, 7.14, 7.18, and 7.11. Calculate the mean and standard deviation.
Value (x) 
Value – mean (y) [xy] 
(xy)^{2} 
7.08 
0.05 
0.0025 
7.21 
0.08 
0.0064 
7.12 
0.01 
0.0001 
7.09 
0.04 
0.0016 
7.16 
0.03 
0.0009 
7.14 
0.01 
0.0001 
7.07 
0.06 
0.0036 
7.14 
0.01 
0.0001 
7.18 
0.05 
0.0025 
7.11 
0.02 
0.0004 
Mean value = 7.13 

Sum of (xy)2 = 0.0182 
Average mean value (x) = 7.13
No. of value (n) = 10
Standard deviation = square root of: sum of (xy)^{2} / (n1)
Standard deviation = square root of 0.0181/9
Standard deviation = +0.045 %