The mean and standard deviation of $20$ observations were calculated as $10$ and $2.5$ respectively. It was found that by mistake one data value was taken as $25$ instead of $35 .$ If $\alpha$ and $\sqrt{\beta}$ are the mean and standard deviation respectively for correct data, then $(\alpha, \beta)$ is :
$(11,26)$
$(10.5,25)$
$(11,25)$
$(10.5,26)$
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
In any discrete series (when all values are not same) the relationship between $M.D.$ about mean and $S.D.$ is
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.
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 the mean and variance of the frequency distribution
$\mathrm{x}$ | $\mathrm{x}_{1}=2$ | $\mathrm{x}_{2}=6$ | $\mathrm{x}_{3}=8$ | $\mathrm{x}_{4}=9$ |
$\mathrm{f}$ | $4$ | $4$ | $\alpha$ | $\beta$ |
be $6$ and $6.8$ respectively. If $x_{3}$ is changed from $8$ to $7 ,$ then the mean for the new data will be: