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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 $1$.

$x_i$	$1$	$2$	$3$	$4$	$5$
$f_i$	$5$	$4$	$f$	$2$	$3$

$x_i$	$2$	$4$	$6$	$8$	$10$	$12$	$14$	$16$
$f_i$	$4$	$4$	$\alpha$	$15$	$8$	$\beta$	$4$	$5$

Daily Expenditure (	$0-50$	$50-100$	$100-150$	$150-200$	$200-250$
Number of Families $(f)$	$24$	$33$	$37$	$b$	$25$

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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