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

Let the mean and the variance of $5$ observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first $4$ observation are $\frac{7}{2}$ and $a$ respectively, then $\left(4 a+x_{5}\right)$ is equal to

The mean of the numbers $a, b, 8,5,10$ is $6$ and their variance is $6.8$. If $M$ is the mean deviation of the numbers about the mean, then $25\; M$ is equal to

Let $y_1$ , $y_2$ , $y_3$ ,..... $y_n$ be $n$ observations. Let ${w_i} = l{y_i} + k\,\,\forall \,\,i = 1,2,3.....,n,$ where $l$ , $k$ are constants. If the mean of  $y_i's$ is  is $48$ and their standard deviation is $12$ , then mean of $w_i's$ is $55$ and standard deviation of $w_i's$  is $15$ , then values of $l$ and $k$ should be

The mean of $5$ observations is $4.4$ and their variance is $8.24$. If three observations are $1, 2$ and $6$, the other two observations are

The mean and the standard deviation (s.d.) of $10$ observations are $20$ and $2$ resepectively. Each of these $10$ observations is multiplied by $\mathrm{p}$ and then reduced by $\mathrm{q}$, where $\mathrm{p} \neq 0$ and $\mathrm{q} \neq 0 .$ If the new mean and new s.d. become half of their original values, then $q$ is equal to