The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & 2 & 3 & 4 & 5 & 6 & 7 \\ f & 4 & 9 & 16 & 14 & 11 & 6 \\ \hline \end{array}$
Find the standard deviation.
$\begin{array}{|c|c|c|c|c|} \hline x_{i} & f_{i} & d_{i}=x_{i}-4 & f_{i} d_{i} & f_{i} d_{i}^{2} \\ \hline 2 & 4 & -2 & -8 & 16 \\ \hline 3 & 9 & -1 & -9 & 9 \\ \hline 4 & 16 & 0 & 0 & 0 \\ \hline 5 & 14 & 1 & 14 & 14 \\ \hline 6 & 11 & 2 & 22 & 44 \\ \hline 7 & 6 & 3 & 18 & 54 \\ \hline \text { Total } & n=60 & & \Sigma f_{i}=37 & \Sigma f_{i} d_{i}^{2}=137 \\ \hline \end{array}$
$ \therefore \quad SD =\sqrt{\frac{\Sigma f_{i} d_{i}^{2}}{n}-\left(\frac{\Sigma f_{i} d_{i}}{n}\right)^{2}}=\sqrt{\frac{137}{60}-\left(\frac{37}{60}\right)^{2}}=\sqrt{2.2833-(0.616)^{2}} $
$=\sqrt{2.2833-0.3794}=\sqrt{1.9037}=1.38 $
If the variance of observations ${x_1},\,{x_2},\,......{x_n}$ is ${\sigma ^2}$, then the variance of $a{x_1},\,a{x_2}.......,\,a{x_n}$, $\alpha \ne 0$ is
The mean and standard deviation of six observations are $8$ and $4,$ respectively. If each observation is multiplied by $3,$ find the new mean and new standard deviation of the resulting observations.
Let $r$ be the range and ${S^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} $ be the $S.D.$ of a set of observations ${x_1},\,{x_2},\,.....{x_n}$, then
The mean and $S.D.$ of $1, 2, 3, 4, 5, 6$ is
The mean and variance of $7$ observations are $8$ and $16$ respectively. If one observation $14$ is omitted a and $b$ are respectively mean and variance of remaining $6$ observation, then $a+3 b-5$ is equal to $..........$.