Example-1: 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.
Example-2: 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 |
Example-3: 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) [x-y] |
(x-y)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 (x-y)2 = 0.0182 |
Average mean value (x) = 7.13
No. of value (n) = 10
Standard deviation = square root of: sum of (x-y)2 / (n-1)
Standard deviation = square root of 0.0181/9
Standard deviation = +0.045 %