Determine mean and standard deviation of first n terms of an $A.P.$ whose first term is a and common difference is d.
$\begin{array}{|c|c|c|} \hline x_{i} & x_{i}-a & \left(x_{i}-a\right)^{2} \\ \hline a & 0 & 0 \\ \hline a+d & d & d^{2} \\ \hline a+2 d & 2 d & 4 d^{2} \\ \hline \end{array}$
$\begin{array}{|c|c|c|} \hline \ldots & \ldots & 9 d^{2} \\ \hline \ldots & \ldots & \ldots \\ \hline \ldots & \ldots & \ldots \\ \hline a+(n-1) d & (n-1) d & (n-1)^{2} d^{2} \\ \hline \Sigma x_{i}=\frac{n}{2}[2 a+(n-1) d ] & & \\ \hline \end{array}$
$\text { Mean }=\frac{\Sigma x_{i}}{n}=\frac{1}{n}\left[\frac{n}{2}(2 a+(n-1) d]=a+\frac{(n-1)}{2} d\right.$
$\Sigma\left(x_{i}-a\right)=d[1+2+3+\ldots+(n-1) d]=d \frac{(n-1) n}{2}$
and $\quad \Sigma\left(x_{i}-a\right)^{2}=d^{2} \cdot\left[1^{2}+2^{2}+3^{2}+\ldots+(n-1)^{2}\right]=\frac{d^{2} n(n-1)(2 n-1)}{6}$
$\sigma=\sqrt{\frac{\left(x_{i}-a\right)^{2}}{n}-\left(\frac{x_{i}-a}{n}\right)^{2}}$
$=\sqrt{\frac{d^{2} n(n-1)(2 n-1)}{6 n}-\left[\frac{d(n-1) n}{2 n}\right]^{2}}=\sqrt{\frac{d^{2}(n-1)(2 n-1)}{6}-\frac{d^{2}(n-1)^{2}}{4}}$
$=d \sqrt{\frac{(n-1)}{2}\left(\frac{2 n-1}{3}-\frac{n-1}{2}\right)=d \sqrt{\frac{(n-1)}{2}\left[\frac{4 n-2-3 n+3}{6}\right]}}$
${2 \sqrt{\frac{(n-1)(n+1)}{12}}=d \sqrt{\frac{n^{2}-1}{12}}}$
If the mean of the frequency distribution
Class: | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ |
Frequency | $2$ | $3$ | $x$ | $5$ | $4$ |
is $28$ , then its variance is $........$.
The mean and variance of $20$ observations are found to be $10$ and $4,$ respectively. On rechecking, it was found that an observation $9$ was incorrect and the correct observation was $11$. Then the correct variance is
If the mean and variance of the frequency distribution
$x_i$ | $2$ | $4$ | $6$ | $8$ | $10$ | $12$ | $14$ | $16$ |
$f_i$ | $4$ | $4$ | $\alpha$ | $15$ | $8$ | $\beta$ | $4$ | $5$ |
are $9$ and $15.08$ respectively, then the value of $\alpha^2+\beta^2-\alpha \beta$ is $............$.
Find the variance and standard deviation for the following data:
${x_i}$ | $4$ | $8$ | $11$ | $17$ | $20$ | $24$ | $32$ |
${f_i}$ | $3$ | $5$ | $9$ | $5$ | $4$ | $3$ | $1$ |
Let the mean and variance of four numbers $3,7, x$ and $y(x>y)$ be $5$ and $10$ respectively. Then the mean of four numbers $3+2 \mathrm{x}, 7+2 \mathrm{y}, \mathrm{x}+\mathrm{y}$ and $x-y$ is ..... .