The $S.D$ of $15$ items is $6$ and if each item is decreased or increased by $1$, then standard deviation will be

If the mean and variance of the following data:

$6,10,7,13, a, 12, b, 12$ are 9 and $\frac{37}{4}$ respectively, then $(a-b)^{2}$ is equal to:

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$.........$

Let the observations $\mathrm{x}_{\mathrm{i}}(1 \leq \mathrm{i} \leq 10)$ satisfy the equations, $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)=10$ and $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)^{2}=40$ If $\mu$ and $\lambda$ are the mean and the variance of the observations, $\mathrm{x}_{1}-3, \mathrm{x}_{2}-3, \ldots ., \mathrm{x}_{10}-3,$ then the ordered pair $(\mu, \lambda)$ is equal to :

The $S.D.$ of a variate $x$ is $\sigma$. The $S.D.$ of the variate $\frac{{ax + b}}{c}$ where $a, b, c$ are constant, is

If $v$ is the variance and $\sigma$ is the standard deviation, then