The weighted mean of the first $n$ natural numbers,where the weights are equal to the squares of the corresponding numbers,is:

The mean and standard deviation of $20$ observations are found to be $10$ and $2$,respectively. It was later discovered that one observation was mistakenly recorded as $8$ instead of $12$. The correct standard deviation is:

The mean and standard deviation of $50$ observations are $15$ and $2$ respectively. It was found that one incorrect observation was taken such that the sum of the correct and incorrect observations is $70$. If the correct mean is $16$,then the correct variance is equal to

If both the mean and the standard deviation of $50$ observations $x_1, x_2, \ldots, x_{50}$ are equal to $16$,then the mean of $(x_1-5)^2, (x_2-5)^2, \ldots, (x_{50}-5)^2$ is

If the coefficients of variation of two distributions are $60$ and $70$ and their standard deviations are $21$ and $16$ respectively,then their arithmetic means are respectively:

The mean of $5$ observations is $4.4$ and their variance is $8.24$. If three observations are $1, 2$ and $6$,the other two observations are