If $\bar{x}$ is the mean of $n$ observations $x_1, x_2, \ldots ., x_n$ then the mean of the absolute deviations of these observations from $\bar{x}$ is

In a discrete data $\frac{1 \text { th }}{4}$ of the observations are equal to $a$, another $\frac{1 \text { th }}{4}$ of the observations are equal to $-a$. Out of the remaining, half of them are equal to $b$ and the rest are equal to $-b$. If the variance of all the observations is $(a b)$, then

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 $\sum_{i=1}^{100} x_i^2=$