The mean and variance of $8$ observations are $10$ and $13.5,$ respectively. If $6$ of these observations are $5, 7, 10, 12, 14, 15,$ then the absolute difference of the remaining two observations is

$x_1, x_2, \ldots, x_n$ are $n$ observations with mean $\bar{x}$ and standard deviation $\sigma$. Match the items of List-$I$ with those of List-$II$:

List-$I$	List-$II$
$(a) \sum_{i=1}^n(x_i-\bar{x})$	$(i) \text{ Median}$
$(b) \text{ Variance } (\sigma^2)$	$(ii) \text{ Coefficient of variation}$
$(c) \text{ Mean deviation}$	$(iii) \text{ Zero}$
$(d) \text{ Measure used to find the homogeneity of given two series}$	$(iv) \text{ Mean of the absolute deviations from any measure of central tendency}$
	$(v) \text{ Mean of the squares of the deviations from mean}$

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 of $n$ observations is $\bar{x}$. If three observations $n+1, n-1, 2n-1$ are added such that the mean remains the same,then the value of $n$ is

Let $a, b, c \in N$ and $a < b < c$. Let the mean,the mean deviation about the mean,and the variance of the $5$ observations $9, 25, a, b, c$ be $18, 4$,and $\frac{136}{5}$,respectively. Then $2a + b - c$ is equal to:

If the sum of the deviations of $50$ observations from $30$ is $50$,then the mean of these observations is: