The mean deviation about the mean of the following data is nearly

$$ \begin{array}{ccccccccc} \hline \text { Size }(x) & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\ \hline \text { Frequency }(f) & 3 & 3 & 4 & 14 & 7 & 4 & 3 & 4 \\ \hline \end{array} $$

Assertion (A) Variance of $4 x_1, 4 x_2, \ldots, 4 x_n$ is 16 times the variance of $x_1, x_2, x_3, \ldots, x_n$

Reason (R) If $y=a x+b$, then variance of $y$ is a $($ variance of $x)+b$

The correct option among the following is

If $\alpha, \beta$ are respectively the mean deviation about the mean and variance of the first five prime numbers, then the ordered pair ( $\alpha, \beta$ )

For the following frequency distribution, the variance is approximately equal to

$$ \begin{array}{cccccc} \hline \begin{array}{c} \text { Class } \\ \text { Interval } \end{array} & 0-5 & 5-10 & 10-15 & 15-20 & 20-25 \\ \hline \text { Frequency } & 4 & 1 & 10 & 3 & 2 \\ \hline \end{array} $$