For the frequency distribution:

Variate $(x)$	$x_{1}$	$x_{2}$	$x_{3} \ldots x_{15}$
Frequency $(f)$	$f_{1}$	$f_{2}$	$f_{3} \ldots f_{15}$

where $0 < x_{1} < x_{2} < x_{3} < \ldots < x_{15} = 10$ and $\sum_{i=1}^{15} f_{i} > 0$,the standard deviation cannot be:

Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution:

$X_i$	$0$	$1$	$2$	$3$	$4$	$5$
$f_i$	$k+2$	$2k$	$k^2-1$	$k^2-1$	$k^2-1$	$k-3$

where $\sum f_i=62$. If $[x]$ denotes the greatest integer $\leq x$,then $[\mu^2+\sigma^2]$ is equal to:

For a data consisting of $15$ observations $x_i$,$i=1, 2, 3, \ldots, 15$,the following results are obtained: $\sum_{i=1}^{15} x_i = 170$ and $\sum_{i=1}^{15} x_i^2 = 2830$. If one of the observations,namely $20$,was found to be wrong and was replaced by its correct value $30$,then the corrected variance is:

The standard deviation of the numbers $31, 32, 33, \ldots, 46, 47$ is

The variance of $^{10}C_0, ^{10}C_1, ^{10}C_2, \dots, ^{10}C_{10}$ is:

Statement-$1$: The variance of the first $n$ even natural numbers is $\frac{n^2 - 1}{4}$.
Statement-$2$: The sum of the first $n$ natural numbers is $\frac{n(n + 1)}{2}$ and the sum of the squares of the first $n$ natural numbers is $\frac{n(n + 1)(2n + 1)}{6}$.