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

There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:

$\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Marks } & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Frequency } & x-2 & x & x^{2} & (x+1)^{2} & 2 x & x+1 \\ \hline \end{array}$

where $x$ is a positive integer. Determine the mean and standard deviation of the marks.

Let $X=\{11,12,13, \ldots ., 40,41\}$ and $Y=\{61,62$, $63, \ldots ., 90,91\}$ be the two sets of observations. If $\bar{x}$ and $\bar{y}$ are their respective means and $\sigma^2$ is the variance of all the observations in $X \cup Y$, then $\left|\overline{ x }+\overline{ y }-\sigma^2\right|$ is equal to $.................$.

The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking, it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

If wrong item is omitted.

If the mean and variance of the data $65,68,58,44$, $48,45,60, \alpha, \beta, 60$ where $\alpha>\beta$ are $56$ and $66.2$ respectively, then $\alpha^2+\beta^2$ is equal to

Consider a set of $3 n$ numbers having variance $4.$ In this set, the mean of first $2 n$ numbers is $6$ and the mean of the remaining $n$ numbers is $3.$ A new set is constructed by adding $1$ into each of first $2 n$ numbers, and subtracting $1$ from each of the remaining $n$ numbers. If the variance of the new set is $k$, then $9 k$ is equal to .... .