Let the mean and variance of $8$ numbers $x , y , 10$, $12,6,12,4,8$, be $9$ and $9.25$ respectively. If $x > y$, then $3 x-2 y$ is equal to $...........$.

The variance of $20$ observations is $5 .$ If each observation is multiplied by $2,$ find the new variance of the resulting observations.

The mean of two samples of size $200$ and $300$ were found to be $25, 10$ respectively their $S.D.$ is $3$ and $4$ respectively then variance of combined sample  of size $500$ is :-

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

Let $X _{1}, X _{2}, \ldots, X _{18}$ be eighteen observations such that $\sum_{ i =1}^{18}\left( X _{ i }-\alpha\right)=36 \quad$ and $\sum_{i=1}^{18}\left(X_{i}-\beta\right)^{2}=90,$ where $\alpha$ and $\beta$ are distinct real numbers. If the standard deviation of these observations is $1,$ then the value of $|\alpha-\beta|$ is ...... .

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.