The standard deviations of two sets of observations $X=\left\{x_i\right\}$ and $Y=\left\{y_i\right\}(i=1,2, \ldots, 100)$ are respectively 5 and 6 . If $\bar{x}, \bar{y}$ are their means and $\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=600$, then the standard deviation of $Z=\left\{z_i / z_i=x_i-y_i\right)$ is

$$ \text { The variance of the following frequency distribution is } $$

$$ \begin{array}{ccccccc} \hline \text { Classes } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\ \hline \text { Frequency } & 11 & 29 & 18 & 4 & 5 & 3 \\ \hline \end{array} $$

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