For the following distribution, the mean deviation about the median is

$$ \begin{array}{cccccccc} \hline x_i & 6 & 12 & 18 & 24 & 30 & 36 & 42 \\ \hline f_i & 4 & 7 & 9 & 18 & 15 & 10 & 5 \\ \hline \end{array} $$

The mean and standard deviation of 100 observations $x_1, x_2, \ldots, x_{100}$ were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. Then the correct value of $\sum_{i=1}^{100} x_i^2=$

The coefficient of variation of the first 5 prime numbers is

If $S_1$ and $S_2$ are the variances of the first $2 k$ and $k(k>1)$ natural numbers respectively, then ( $S_1 / S_2$ ) lies in the interval