Page 50 - Quality control of pharmaceuticals (07-PA 704)
P. 50
Calculation of standard deviation of the data from the previous
Example (1):
No Fe concentration, ppm (Xi- ̅ ) (Xi- ̅ )2
Xi 0.38
0.28 0.1444
X1 19.4 0.18 0.0784
X2 19.5 0.02 0.0324
X3 19.6 0.32 0.0004
X4 19.8 0.52 0.1024
X5 20.1 Ʃ (Xi- ̅ )2 0.2704
X6 20.3 0.6284
Ʃ Xi 118.7
Standard deviation, S = √∑ = 1 ( − ̅ )2 = √0.6284
−1 6−1
= √0.1257 = 0.354 = 0.35ppm Fe
Variance, S2 =0.1257 = 0.13 (ppm Fe)2
Relative standard deviation = × 1000 = 18 ppt
̅
Coefficient of variation, CV % = × 100 = 0.354 × 100 = 1.79 ≃1.8%
̅ 19.8
