UPSC Combined Geo-Scientist, Mains · 2023 · Mains Paper III
Q1(a). Relative standard deviation of replicate results
Question
The following results were obtained in the replicate determinations of Fe content in a mineral ore (in mg): 10.5, 10.6, 10.3, 10.6, 10.2 and 10.4. Calculate the relative standard deviation (RSD) of this data.
Answer in brief
x̄ = 10.43 mg, s = 0.163 mg, RSD = 100 s/x̄ = 1.57 % (15.7 parts per thousand).
Key points
- Mean: x̄ = Σx/n = 62.6/6 = 10.433 mg.
- Sum of squared deviations Σ(x − x̄)² = 0.1333 mg²; s = √(0.1333/5) = 0.1633 mg (n − 1 = 5 degrees of freedom).
- RSD = (s/x̄) × 100 = (0.1633/10.433) × 100 = 1.57 % = 15.7 parts per thousand.
- The RSD in per cent is the coefficient of variation: a measure of precision, not of accuracy.
- Result to report: 10.4 ± 0.2 mg Fe (n = 6); division by n instead of n − 1 would give 1.43 %.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.

Coversrelative standard deviationstandard deviationcoefficient of variationprecisionreplicate analysis
All parts of Question 1
- 1(a) Relative standard deviation of replicate results5 marks
- 1(b) HPLC against GLC; particle size of packed GLC columns5 marks
- 1(c) Sources of radiation for infrared spectrometry5 marks
- 1(d) Graphite-furnace AAS compared with flame AAS5 marks
- 1(e) Calibration of the temperature axis in thermogravimetry5 marks
- 1(f) K and L lines in XRF; wavelength-dispersive spectrum of Zn5 marks
- 1(g) Interference of calcium in the ICP–MS determination of iron5 marks
- 1(h) Type metal: constituents and non-destructive analysis5 marks