If 10 is the mean deviation of ' $n$ ' observations $x_1, x_2, x_3, \ldots, x_n$, then the mean deviation of the observations $\frac{2 x_1+5}{3}, \frac{2 x_2+5}{3}, \frac{2 x_3+5}{3}, \ldots . \frac{2 x_n+5}{3}$ is

There are $n$ observations and all of them are negative numbers. The ascending order of these observations is $x_1, x_2, \ldots . x_n$. If the signs of the first term and last term in that order are changed, then the range of the data is

The mean deviation from the mean for the observations $1,3,5,7,11,13,17,19,23$ is

The mean deviation from the mean of the discrete data $1,3,4,7,11,18,29,47,78$ is