JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Statistics

Let the two unknown observations be $p$ and $q$. The mean gives $\frac{80+p+q}{10}=9$, so $p+q=10$. The variance gives $\frac{\sum x_i^2}{10}-9^2=34.2$, so $\sum x_i^2=1152$; subtracting the known $1094$ leaves $p^2+q^2=58$, whence $pq=21$ and $\{p,q\}=\{3,7\}$. Ordering all ten values $2,3,3,5,7,10,11,13,15,21$ gives the median $\frac{7+10}{2}=8.5$. The absolute deviations from $8.5$ are $6.5,5.5,5.5,3.5,1.5,1.5,2.5,4.5,6.5,12.5$, summing to $50$, so the mean deviation about the median is $\frac{50}{10}=5$.