The variance $\sigma^2$ of the data is $ . . . . . .$
$x_i$ | $0$ | $1$ | $5$ | $6$ | $10$ | $12$ | $17$ |
$f_i$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
The variance of the data $2, 4, 6, 8, 10$ is
The outcome of each of $30$ items was observed; $10$ items gave an outcome $\frac{1}{2} - d$ each, $10$ items gave outcome $\frac {1}{2}$ each and the remaining $10$ items gave outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac {4}{3}$ then $\left| d \right|$ equals
The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking it was found that, in the observations, $20$ was misread as $5$ . Then, the correct variance is equal to......
The $S.D.$ of a variate $x$ is $\sigma$. The $S.D.$ of the variate $\frac{{ax + b}}{c}$ where $a, b, c$ are constant, is