Let the mean and the variance of $6$ observations $a, b, 68, 44, 48, 60$ be $55$ and $194$,respectively. If $a > b$,then the value of $a + 3b$ is:

The mean of $100$ observations is $50$ and their standard deviation is $5$. Then,the sum of squares of all observations is

Statement-$1$: The variance of the first $n$ even natural numbers is $\frac{n^2 - 1}{3}$.
Statement-$2$: The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n(4n^2 - 1)}{3}$.

The raw data $x_1, x_2, \ldots, x_{n}$ is an $A$.$P$. with common difference $d$ and first term $0$. If $\bar{x}$ and $\sigma^2$ are the mean and variance of $x_{i}, i=1, 2, \ldots, n$,then $\sigma^2$ is:

$\text{Assertion (A):}$ Variance of $4x_1, 4x_2, \ldots, 4x_n$ is $16$ times the variance of $x_1, x_2, \ldots, x_n$. $\text{Reason (R):}$ If $y = ax + b$,then variance of $y$ is $a(\text{variance of } x) + b$. The correct option among the following is

If the median of the data $6, 7, x-2, x, 18$ and $21$ written in ascending order is $16$,then the variance of that data is