If $\sum_{i=1}^{9}{(x_{i}-99)}=27$ and $\sum_{i=1}^{9}{(x_{i}-99)^{2}}=306$ then the mean of the squares of the deviation of $x_{1}, x_{2},..........., x_9$ from their $AM$ is
Reveal answer
Fill a bubble to check yourself
If $\sum_{i=1}^{9}{(x_{i}-99)}=27$ and $\sum_{i=1}^{9}{(x_{i}-99)^{2}}=306$ then the mean of the squares of the deviation of $x_{1}, x_{2},..........., x_9$ from their $AM$ is