UPSC NDA I 2024 Mathematics Paper

Concept:Variance measures the spread of data points around their mean by averaging the squared differences from the mean.Explanation:First, sort the given marks in ascending order:15, 18, 19, 21, 24, 32, 35, 41, 47, 50.The largest five observations are: 32, 35, 41, 47, 50.Calculate the mean of these five numbers:$\mu = \frac{32+35+41+47+50}{5} = \frac{205}{5} = 41$.Now, compute each squared deviation from the mean:$(32-41)^2 = (-9)^2 = 81$,$(35-41)^2 = (-6)^2 = 36$,$(41-41)^2 = 0^2 = 0$,$(47-41)^2 = 6^2 = 36$,$(50-41)^2 = 9^2 = 81$.Sum of squared deviations = $81 + 36 + 0 + 36 + 81 = 234$.Variance = $\frac{234}{5} = 46.8$.Answer:46.8 (Option D)