UPSC NDA 2 2022 Mathematics Solved Paper

Concept:Mean deviation about the mean is the average of absolute deviations from the mean of a data set.Explanation:First, calculate the mean ($\overline{x}$) of the given grouped data.Use class midpoints $x_i$ and frequencies $f_i$.Total frequency $N = \sum f_i = 20$.Sum of $f_i x_i = 1 \times 5 + 2 \times 15 + 4 \times 25 + 6 \times 35 + 4 \times 45 + 3 \times 55 = 5 + 30 + 100 + 210 + 180 + 165 = 690$.Mean $\overline{x} = \frac{690}{20} = 34.5$.Next, compute the absolute deviations $|x_i - \overline{x}|$ for each class:For class 0–10: $|5 - 34.5| = 29.5$; $f_i \times |x_i - \overline{x}| = 1 \times 29.5 = 29.5$.For 10–20: $|15 - 34.5| = 19.5$; $2 \times 19.5 = 39$.For 20–30: $|25 - 34.5| = 9.5$; $4 \times 9.5 = 38$.For 30–40: $|35 - 34.5| = 0.5$; $6 \times 0.5 = 3$.For 40–50: $|45 - 34.5| = 10.5$; $4 \times 10.5 = 42$.For 50–60: $|55 - 34.5| = 20.5$; $3 \times 20.5 = 61.5$.Sum of $f_i |x_i - \overline{x}| = 29.5 + 39 + 38 + 3 + 42 + 61.5 = 213$.Mean deviation about the mean = $\frac{213}{20} = 10.65$.Answer:$10.65$, which corresponds to option B.