JEE Main 2023 (Online) 24th January Evening ShiftQuestion 1
Let the six numbers $a_1,a_2,\ldots,a_6$ be in A.P. and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^{2}$, then $8\sigma^{2}$ is equal to
Statistics
Solution
$2a+2d=10$ and $a+\frac52d=\frac{19}{2}$ give $d=3$, $a=2$. Variance $=\frac{d^2(36-1)}{12}=\frac{105}{4}$, so $8\sigma^2=210$.