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

For a statistical data $x_1, x_2, \ldots, x_{10}$ of $10$ values,a student obtained the mean as $5.5$ and $\sum_{i=1}^{10} x_i^2 = 371$. He later found that he had noted two values in the data incorrectly as $4$ and $5$,instead of the correct values $6$ and $8$,respectively. The variance of the corrected data is

Let $X = \{x \in N : 1 \leq x \leq 17\}$ and $Y = \{ax + b : x \in X \text{ and } a, b \in R, a > 0\}$. If the mean and variance of the elements of $Y$ are $17$ and $216$ respectively,then $a + b$ is equal to:

The mean and the standard deviation $(s.d.)$ of five observations are $9$ and $0,$ respectively. If one of the observations is changed such that the mean of the new set of five observations becomes $10,$ then their $s.d.$ is?

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

Let $x_1, x_2, \ldots, x_{10}$ be ten observations such that $\sum_{i=1}^{10}(x_i-2)=30$,$\sum_{i=1}^{10}(x_i-\beta)^2=98$,$\beta > 2$ and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2(x_1-1)+4\beta, 2(x_2-1)+4\beta, \ldots, 2(x_{10}-1)+4\beta$,then $\frac{\beta\mu}{\sigma^2}$ is equal to: