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(A) Let the $2n$ distinct observations be $x_1 < x_2 < ... < x_{2n}$.Since there are $2n$ observations,the median is the average of the $n^{th}$ and $

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increases by $1$

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(A) Let the $2n$ distinct observations be $x_1 Since there are $2n$ observations,the median is the average of the $n^{th}$ and $(n+1)^{th}$ observations.There are $n$ observations below the median (i.e.,$x_1, ..., x_n$) and $n$ observations above the median (i.e.,$x_{n+1}, ..., x_{2n}$).The sum of the original observations is $S = \sum_{i=1}^{2n} x_i$.The new sum $S'$ is obtained by adding $5$ to each of the first $n$ observations and subtracting $3$ from each of the remaining $n$ observations:$S' = \sum_{i=1}^{n} (x_i + 5) + \sum_{i=n+1}^{2n} (x_i - 3)$$S' = \sum_{i=1}^{n} x_i + 5n + \sum_{i=n+1}^{2n} x_i - 3n$$S' = \sum_{i=1}^{2n} x_i + 2n = S + 2n$.The new mean is $M' = \frac{S'}{2n} = \frac{S + 2n}{2n} = \frac{S}{2n} + 1$.Since the original mean is $M = \frac{S}{2n}$,the new mean is $M' = M + 1$.Thus,the mean increases by $1$.

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

In a set of $2n$ distinct observations,each of the observations below the median is increased by $5$ and each of the remaining observations is decreased by $3$. Then the mean of the new set of observations

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