If the variance of the first $n$ natural numbers is $10$ and the variance of the first $m$ even natural numbers is $16$,then $m + n$ is equal to

If the mean of the data $7, 8, 9, 7, 8, 7, \lambda, 8$ is $8$,then the variance of the data is:

The outcome of each of $30$ items was observed; $10$ items gave an outcome $\frac{1}{2} - d$ each,$10$ items gave an outcome $\frac{1}{2}$ each,and the remaining $10$ items gave an outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac{4}{3}$,then $|d|$ equals:

Statement-$1$: The variance of the first $n$ even natural numbers is $\frac{n^2 - 1}{4}$.
Statement-$2$: The sum of the first $n$ natural numbers is $\frac{n(n + 1)}{2}$ and the sum of the squares of the first $n$ natural numbers is $\frac{n(n + 1)(2n + 1)}{6}$.

The mean and standard deviation of $10$ observations are $20$ and $2$ respectively. Later on,it was observed that one observation was recorded as $50$ instead of $40$. Then the correct variance is:

If $A$ and $B$ are the variances of the first $n$ even numbers and the first $n$ odd numbers respectively,then: