Let $X = \{x \in N : 1 \leq x \leq 17\}$ and $Y = \{ax + b : x \in X \text{ and } a, b \in R, a > 0\}$. If the mean and variance of the elements of $Y$ are $17$ and $216$ respectively,then $a + b$ is equal to:

Let $x_1, x_2, x_3, \dots, x_n$ be $n$ observations,$\bar{x}$ be their arithmetic mean,and $\sigma^2$ be their variance.
Statement $-1$: The variance of observations $2x_1, 2x_2, 2x_3, \dots, 2x_n$ is $4\sigma^2$.
Statement $-2$: The arithmetic mean of $2x_1, 2x_2, 2x_3, \dots, 2x_n$ is $4\bar{x}$.

If $x_1, x_2, x_3, \ldots, x_n$ are $n$ observations such that $\sum(x_i+2)^2 = 28n$ and $\sum(x_i-2)^2 = 12n$,then the variance is:

Given that the total of $16$ values is $528$ and the sum of the squares of deviations from $33$ is $9158$. The variance is:

Let $x_1, x_2, \ldots, x_{10}$ be ten observations such that $\sum_{i=1}^{10}(x_i-2)=30$,$\sum_{i=1}^{10}(x_i-\beta)^2=98$,$\beta > 2$ and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2(x_1-1)+4\beta, 2(x_2-1)+4\beta, \ldots, 2(x_{10}-1)+4\beta$,then $\frac{\beta\mu}{\sigma^2}$ is equal to:

All students in a classroom performed poorly in mathematics. The teacher decided to increase the marks of each student by $10$. Which of the following statistical measures will not change even after adding the extra marks?