If for a distribution $\Sigma(x-5)=3, \Sigma(x-5)^{2}=43$ and the total number of item is $18,$ find the mean and standard deviation.
Given, $n=18, \Sigma(x-5)=3$ and $\Sigma(x-5)^{2}=43$
$\therefore \quad \operatorname{Mean}=A+\frac{\Sigma(x-5)}{18}=5+\frac{3}{18}=5+0.1666=5.1666=5.17$
and $\quad SD =\sqrt{\frac{\Sigma(x-5)^{2}}{n}-\left(\frac{\Sigma(x-5)}{n}\right)^{2}}=\sqrt{\frac{43}{18}-\left(\frac{3}{18}\right)^{2}}$
$=\sqrt{2.3889-(0.166)^{2}}=\sqrt{2.3889-0.0277}=1.53$
If the mean and variance of eight numbers $3,7,9,12,13,20, x$ and $y$ be $10$ and $25$ respectively, then $\mathrm{x} \cdot \mathrm{y}$ is equal to
Determine the mean and standard deviation for the following distribution:
$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline \text { Marks } & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline \text { Frequency } & 1 & 6 & 6 & 8 & 8 & 2 & 2 & 3 & 0 & 2 & 1 & 0 & 0 & 0 & 1 \\ \hline \end{array}$
Let $x_1, x_2,........,x_n$ be $n$ observations such that $\sum {{x_i}^2 = 300} $ and $\sum {{x_i} = 60} $ on value of $n$ among the following is
The mean and standard deviation of a group of $100$ observations were found to be $20$ and $3,$ respectively. Later on it was found that three observations were incorrect, which were recorded as $21,21$ and $18 .$ Find the mean and standard deviation if the incorrect observations are omitted.
The variance $\sigma^2$ of the data is $ . . . . . .$
$x_i$ | $0$ | $1$ | $5$ | $6$ | $10$ | $12$ | $17$ |
$f_i$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |