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Errors in Analysis (Examples)

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 %



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