UPSC NDA 2 2022 Mathematics Solved Paper

Concept:Mean deviation about the median for grouped data is $\frac{\sum f_i |x_i - M|}{N}$, where $M$ is the median, $x_i$ are class midpoints, $f_i$ are frequencies, and $N$ is total frequency.Explanation:First, find the median class. The cumulative frequencies are $1, 3, 7, 13, 17, 20$. Total $N = 20$, so $\frac{N}{2} = 10$. The $10$th value lies in class $30-40$.Median $M = L + \frac{\frac{N}{2} - CF}{f} \times h = 30 + \frac{10-7}{6} \times 10 = 35$.Now compute absolute deviations $|x_i - M|$ and the product $f_i |x_i - M|$ for each class:Class $0-10$: midpoint $5$, $|5-35|=30$, product $30$.Class $10-20$: midpoint $15$, $|15-35|=20$, product $40$.Class $20-30$: midpoint $25$, $|25-35|=10$, product $40$.Class $30-40$: midpoint $35$, $|35-35|=0$, product $0$.Class $40-50$: midpoint $45$, $|45-35|=10$, product $40$.Class $50-60$: midpoint $55$, $|55-35|=20$, product $60$.Sum of products $\sum f_i |x_i - M| = 30+40+40+0+40+60 = 210$.Mean deviation about median $= \frac{210}{20} = 10.5$.Answer:10.5 (Option D).