$A$ student scored the following marks in five tests: $45, 54, 41, 57, 43$. His score is not known for the sixth test. If the mean score is $48$ in the six tests,then the standard deviation of the marks in the six tests is:

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 nine observations $x_1, x_2, \ldots, x_9$ is

If $x_1, x_2, x_3, \ldots, x_n$ are $n$ observations such that $\sum(x_i+2)^2 = 28n$ and $\sum(x_i-2)^2 = 12n$,then the variance is:

Suppose a population $A$ has $100$ observations $101, 102, . . ., 200$ and another population $B$ has $100$ observations $151, 152, . . ., 250$. If $V_A$ and $V_B$ represent the variances of the two populations,respectively,then $V_A / V_B$ is:

Consider $10$ observations $x_1, x_2, \ldots, x_{10}$ such that $\sum_{i=1}^{10}(x_i-\alpha)=2$ and $\sum_{i=1}^{10}(x_i-\beta)^2=40$,where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The value of $\frac{\beta}{\alpha}$ is equal to :

For $10$ observations $x_1, x_2, ..., x_{10}$, if $\sum_{i=1}^{10} (x_i + 2)^2 = 180$ and $\sum_{i=1}^{10} (x_i - 1)^2 = 90$, then their standard deviation is: