The mean and standard deviation of $40$ observations are $30$ and $5$ respectively. It was noticed that two of these observations $12$ and $10$ were wrongly recorded. If $\sigma$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38 \sigma^{2}$ 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 it is replaced by $12$

For the frequency distribution :

where $0< x _{1}< x _{2}< x _{3}<\ldots .< x _{15}=10$ and

$\sum \limits_{i=1}^{15} f_{i}>0,$ the standard deviation cannot be 

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 ...... .

Let $n \geq 3$. A list of numbers $0 < x_1 < x_2 < \ldots < x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers is made as follows: $y_1=0, y_2=x_2, \ldots, x_{n-1}$ $=x_n-1, y_n=x_1+x_n$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?

Given that $\bar{x}$ is the mean and $\sigma^{2}$ is the variance of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ Prove that the mean and variance of the observations $a x_{1}, a x_{2}, a x_{3}, \ldots ., a x_{n}$ are $a \bar{x}$ and $a^{2} \sigma^{2},$ respectively, $(a \neq 0)$