Consider the statistics of two sets of observations as follows :
Size | Mean | Variance | |
Observation $I$ | $10$ | $2$ | $2$ |
Observation $II$ | $n$ | $3$ | $1$ |
If the variance of the combined set of these two observations is $\frac{17}{9},$ then the value of $n$ is equal to ..... .
$8$
$10$
$5$
$15$
The mean and standard deviation of $10$ observations are $20$ and $84$ respectively. Later on, it was observed that one observation was recorded as $50$ instead of $40$. Then the correct variance is:
The variance of $\alpha$, $\beta$ and $\gamma$ is $9$, then variance of $5$$\alpha$, $5$$\beta$ and $5$$\gamma$ is
The variance of $20$ observation is $5$ . If each observation is multiplied by $2$ , then the new variance of the resulting observations, is
The varience of data $1001, 1003, 1006, 1007, 1009, 1010$ is -
The average marks of $10$ students in a class was $60$ with a standard deviation $4$ , while the average marks of other ten students was $40$ with a standard deviation $6$ . If all the $20$ students are taken together, their standard deviation will be