JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Statistics

Let the two unknown observations be $a$ and $b$. The mean of $8$ values being $10$ gives $5+7+10+12+14+15+a+b=80$, so $a+b=17$. For the variance, measure from $10$: the six known values become $-5,-3,0,2,4,5$ with squared sum $79$, and the condition $\frac{79+(a-10{)}^2+(b-10{)}^2}{8}=13.5$ gives $(a-10{)}^2+(b-10{)}^2=29$, which expands to $a^2+b^2=169$. From $(a+b{)}^2=a^2+b^2+2ab$, $289=169+2ab$, so $ab=60$. Then $|a-b|=\sqrt{(a+b{)}^2-4ab}=\sqrt{289-240}=\sqrt{49}=7$.