Let the six numbers $$\mathrm{a_1,a_2,a_3,a_4,a_5,a_6}$$, be in A.P. and $$\mathrm{a_1+a_3=10}$$. If the mean of these six numbers is $$\frac{19}{2}$$ and their variance is $$\sigma^2$$, then 8$$\sigma^2$$ is equal to :

If the mean deviation about median for the numbers 3, 5, 7, 2k, 12, 16, 21, 24, arranged in the ascending order, is 6 then the median is :

The number of values of a $$\in$$ N such that the variance of 3, 7, 12, a, 43 $$-$$ a is a natural number is :

Let the mean and the variance of 5 observations x₁, x₂, x₃, x₄, x₅ be $${24 \over 5}$$ and $${194 \over 25}$$ respectively. If the mean and variance of the first 4 observation are $${7 \over 2}$$ and a respectively, then (4a + x₅) is equal to: