Mean of $5$ observations is $7.$ If four of these observations are $6, 7, 8, 10$ and one is missing then the variance of all the five observations is

The mean and standard deviation of $20$ observations were calculated as $10$ and $2.5$ respectively. It was found that by mistake one data value was taken as $25$ instead of $35 .$ If $\alpha$ and $\sqrt{\beta}$ are the mean and standard deviation respectively for correct data, then $(\alpha, \beta)$ 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 

A data consists of $n$ observations

${x_1},{x_2},......,{x_n}.$ If $\sum\limits_{i - 1}^n {{{({x_i} + 1)}^2}}  = 9n$ and $\sum\limits_{i - 1}^n {{{({x_i} - 1)}^2}}  = 5n,$ then the standard deviation of this data is

If the mean deviation about the mean of the numbers $1,2,3, \ldots ., n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to

The first of the two samples in a group has $100$ items with mean $15$ and standard deviation $3 .$ If the whole group has $250$ items with mean $15.6$ and standard deviation $\sqrt{13.44}$, then the standard deviation of the second sample is: