Let M denote the median of the following frequency distribution

Class	0 - 4	4 - 8	8 - 12	12 - 16	16 - 20
Frequency	3	9	10	8	6

Then 20M is equal to :

Let $\mathrm{a}_1, \mathrm{a}_2, \ldots \mathrm{a}_{10}$ be 10 observations such that $\sum\limits_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50$ and $\sum\limits_{\forall \mathrm{k} < \mathrm{j}} \mathrm{a}_{\mathrm{k}} \cdot \mathrm{a}_{\mathrm{j}}=1100$. Then the standard deviation of $\mathrm{a}_1, \mathrm{a}_2, \ldots, \mathrm{a}_{10}$ is equal to :

The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is :