Kcet
Statistics
KCET 2018
MCQ+1 / -02018
For the probability distribution given by
$$ \begin{array}{|c|c|c|c|} \hline X=x_i & 0 & 1 & 2 \\ \hline P_i & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\ \hline \end{array} $$
the standard deviation $(\sigma)$ is
$$ \begin{array}{|c|c|c|c|} \hline X=x_i & 0 & 1 & 2 \\ \hline P_i & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\ \hline \end{array} $$
the standard deviation $(\sigma)$ is
