JEE Main 2026 (Online) 5th April Morning ShiftQuestion 1
The mean deviation about the mean for the data
$\begin{array}{|c|c|c|c|c|c|c|}\hline x_i&5&7&9&10&12&15\\\hline f_i&8&6&2&2&2&6\\\hline\end{array}$
is equal to :
Statistics
Solution
$n=26$, $\sum fx=234$, mean $=9$. $\sum f|x-9|=32+12+0+2+6+36=88$. Mean deviation $=\frac{88}{26}=\frac{44}{13}$.