If $\sum_{i = 1}^9 (x_i - 5) = 9$ and $\sum_{i = 1}^9 (x_i - 5)^2 = 45$,then the standard deviation of the $9$ items $x_1, x_2, ..., x_9$ is:

The standard deviation of $n$ observations $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ is $\sigma$. Then, the standard deviation of the observations $\lambda a_{1}, \lambda a_{2}, \ldots, \lambda a_{n}$ is

From the prices of shares $X$ and $Y$ below,find out which is more stable in value:

$X$	$35$	$54$	$52$	$53$	$56$	$58$	$52$	$50$	$51$	$49$
$Y$	$108$	$107$	$105$	$105$	$106$	$107$	$104$	$103$	$104$	$101$

The mean of the numbers $a, b, 8, 5$ and $10$ is $6$ and the variance is $6.80$. Then the possible values of $a$ and $b$ are:

If the variance of four numbers $w, x, y,$ and $z$ is $9$,then the variance of $5w, 5x, 5y,$ and $5z$ is:

Find the mean and variance of the frequency distribution given below:
$\begin{array}{|l|l|l|l|l|} \hline x & 1 \leq x < 3 & 3 \leq x < 5 & 5 \leq x < 7 & 7 \leq x < 10 \\ \hline f & 6 & 4 & 5 & 1 \\ \hline \end{array}$